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For every integer n prove that the fraction $\frac{21n+4}{14n+3}$ cannot be reduced any further. | The desired result $(14n + 3, 21n + 4) = 1$ follows from $3(14n + 3) − 2(21n + 4) = 1.$ |
Let x be an angle and let the real numbers $a$, $b$, $c$, $\cos{x}$ satisfy the following equation: $a \cos^2{x} + b \cos{x} + c = 0$. Write the analogous quadratic equation for $a$, $b$, $c$, $\cos{2x}$. Compare the given and the obtained equality for a = 4, b = 2, c = −1. | Multiplying the equality by $4(a \cos^{2x} − b \cos{x}+c)$, we obtain $4a^2 \cos^4{x} + 2(4ac − 2b^2)\cos^2{x} + 4c^2 = 0$. Plugging in $2 \cos{2x} = 1 + \cos{2x}$ we obtain (after quite a bit of manipulation): $^2 \cos^2{2x} + (2a^2 + 4ac − 2b^2)\cos{2x} + (a2 + 4ac − 2b^2 + 4c^2)=0$. For a = 4, b = 2, and c = −1 we get $4\cos{2x} + 2\cos{x} − 1 = 0$ and $16\cos^2{2x} + 8\cos{2x} − 4=0 \implies 4 \cos^2{2x} + 2\cos{2x} − 1 = 0$. |
Construct a right-angled triangle whose hypotenuse c is given if it is known that the median from the right angle equals the geometric mean of the remaining two sides of the triangle. | Analysis. Let a and b be the other two sides of the triangle. From the conditions of the problem we have $c^2 = a^2 +b^2$ and $c/2 = \sqrt{ab} \iff 3/2c^2 = a^2+b^2+2ab = (a+b)^2 \iff \sqrt{3/2}c = a+b$. Given a desired \triangle ABC let D be a point on (AC such that CD = CB. In that case, $AD = a + b = \sqrt{3/2}c$, and also, since BC = CD, it follows that \angle ADB = 45. Construction. From a segment of length c we elementarily construct a segment AD of length \sqrt{3/2}c. We then construct a ray (DX such that \angle ADX = 45 and a circle k(A, c) that intersects the ray at point B. Finally, we construct the perpendicular from B to AD; point C is the foot of that perpendicular. Proof. It holds that AB = c, and, since CB = CD, it also holds that $AC+CB = AC + CD = AD = \sqrt{3/2}c$. From this it follows that $\sqrt{AC * CB} = c/2$. Since BC is perpendicular to AD, it follows that $\angle BCA = 90$. Thus \triangle ABC is the desired triangle. Discussion. Since$ AB \sqrt{2} = \sqrt{2}c > \sqrt{3/2} c = AD > AB$, the circle k intersects the ray DX in exactly two points, which correspond to two symmetric solutions |
Determine all prime numbers p and all positive integers x and y satisfying $x^3+y^3 = p(xy + p)$. | Up to a swap of the first two entries, the only solutions are (x, y, p) = (1, 8, 19), (x, y, p) = (2, 7, 13) and (x, y, p) = (4, 5, 7). The verification is routine. Set $s = x+y$. Rewrite the equation in the form $s(s^2 − 3xy) = p(p + xy)$, and express $xy = \frac{s^3 − p^2}{3s + p}$ . (∗) In particular, $s^2 \ge 4xy = \frac{4(s^3 − p^2)}{3s + p}$, or $(s − 2p)(s^2 + sp + 2p^2 ) \le p^2 − p^3 < 0$, so $s < 2p$. If $p \nmid s$, then $s = p$ and $xy = p(p − 1)/4$ which is impossible for $x + y = p$ (the equation $t^2 − pt + p(p − 1)/4 = 0$ has no integer solutions). If $p \nmid s$, rewrite (∗) in the form $27xy = (9s^2 − 3sp + p^2) − \frac{p^2 (p + 27)}{3s + p}$. Since $p \nmid s$, this could be integer only if $3s + p \mid p + 27$, and hence $3s + p \mid 27 − s$. If $s \ne 9$, then $|3s − 27| \ge 3s + p$, so $27 − 3s ≥ 3s + p$, or $27 − p \ge 6s$, whence $s \le 4$. These cases are ruled out by hand. If $s = x + y = 9$, then (∗) yields $xy = 27 − p$. Up to a swap of x and y, all such triples (x, y, p) are (1, 8, 19), (2, 7, 13), and (4, 5, 7). |
Determine all prime numbers p and all positive integers x and y satisfying $x^3+y^3 = p(xy + p)$. | Set again $s = x + y$. It is readily checked that $s \le 8$ provides no solutions, so assume $s \ge 9$. Notice that $x^3+y^3 = s(x^2−xy+y^2 ) \ge \frac{1}{4}s^3 and xy ≤ \frac{1}{4} s^2$. The condition in the statement then implies $s^2 (s − p) \le 4p^2$, so $s < p + 4$. Notice that p divides one of s and $x^2 − xy + y^2$. The case $p | s$ is easily ruled out by the condition $s < p + 4$: The latter forces $s = p$, so $x^2 − xy + y^2 = xy + p$, i. e., $(x − y)^2 = p$, which is impossible. Hence $p | x^2 − xy + y^2$, so $x^2 − xy + y^2 = kp$ and $xy + p = ks$ for some positive integer k, implying $s^2 + 3p = k(3s + p)$. (∗∗) Recall that $p \nmid s$ to infer that $3k \equiv s (mod p)$. We now present two approaches. 1st Approach. Write $3k = s + mp$ for some integer m and plug $k = \frac{1}{3}(s + mp) into (∗∗) to get $s = \frac{(9 − mp)}{(3m + 1)}$. The condition $s \ge 9$ then forces m = 0, so s = 9, in which case, up to a swap of the first two entries, the solutions turn out to be (x, y, p) = (1, 8, 19), (x, y, p) = (2, 7, 13) and (x, y, p) = (4, 5, 7). 2nd Approach. Notice that $k = \frac{s^2+3p}{3s+p} = 3 + \frac{s(s−9)}{3s+p} \le 3+1/3 (s−9) = 1/3 s \le 1/3 (p+3)$, since $s < p+4$. Hence $3k \le p + 3$, and the congruence $3k \equiv s (mod p)$ then forces either $3k = s − p$ or $3k = s$. The case $3k = s − p$ is easily ruled out: Otherwise, (∗∗) boils down to $2s + p + 9 = 0$, which is clearly impossible. Finally, if $3k = s$, then (∗∗) reduces to $s = 9$. In this case, up to a swap of the first two entries, the only solutions are (x, y, p) = (1, 8, 19), (x, y, p) = (2, 7, 13) and (x, y, p) = (4, 5, 7). |
Let $P(x), Q(x), R(x)$ and $S(x)$ be non-constant polynomials with real coefficients such that $P(Q(x))=R(S(x))$. Suppose that the degree of $P(x)$ is divisible by the degree of $R(x)$. Prove that there is a polynomial $T(x)$ with real coefficients such that $P(x)=R(T(x))$. | Solution 1. Degree comparison of $P(Q(x))$ and $R(S(x))$ implies that $q=\operatorname{deg} Q \mid \operatorname{deg} S=s$. We will show that $S(x)=T(Q(x))$ for some polynomial $T$. Then $P(Q(x))=R(S(x))=R(T(Q(x)))$, so the polynomial $P(t)-R(T(t))$ vanishes upon substitution $t=S(x)$; it therefore vanishes identically, as desired. Choose the polynomials $T(x)$ and $M(x)$ such that $$S(x)=T(Q(x))+M(x)$$ where $\operatorname{deg} M$ is minimised; if $M=0$, then we get the desired result. For the sake of contradiction, suppose $M \neq 0$. Then $q \nmid m=\operatorname{deg} M$; otherwise, $M(x)=\beta Q(x)^{m / q}+M_{1}(x)$, where $\beta$ is some number and $\operatorname{deg} M_{1}<\operatorname{deg} M$, contradicting the choice of $M$. In particular, $0<m<s$ and hence $\operatorname{deg} T(Q(x))=s$. Substitute now (*) into $R(S(x))-P(Q(x))=$ 0 ; let $\alpha$ be the leading coefficient of $R(x)$ and let $r=\operatorname{deg} R(x)$. Expand the brackets to get a sum of powers of $Q(x)$ and other terms including powers of $M(x)$ as well. Amongst the latter, the unique term of highest degree is $\operatorname{ar} M(x) T(Q(x))^{r-1}$. So, for some polynomial $N(x), \quad N(Q(x))=\operatorname{\alpha rM}(x) T(Q(x))^{r-1}+$ a polynomial of lower degree. This is impossible, since $q$ divides the degree of the left-hand member, but not that of the righthand member. |
Let $P(x), Q(x), R(x)$ and $S(x)$ be non-constant polynomials with real coefficients such that $P(Q(x))=R(S(x))$. Suppose that the degree of $P(x)$ is divisible by the degree of $R(x)$. Prove that there is a polynomial $T(x)$ with real coefficients such that $P(x)=R(T(x))$. | Solution 2. All polynomials in the solution have real coefficients. As usual, the degree of a polynomial $f(x)$ is denoted $\operatorname{deg} f(x)$. Of all pairs of polynomials $P(x), R(x)$, satisfying the conditions in the statement, choose one, say, $P_{0}(x), R_{0}(x)$, so that $P_{0}(Q(x))=R_{0}(S(x))$ has a minimal (positive) degree. We will show that $\operatorname{deg} R_{0}(x)=1$, say, $R_{0}(x)=\alpha x+\beta$ for some real numbers $\alpha \neq 0$ and $\beta$, so $P_{0}(Q(x))=\alpha S(x)+\beta$. Hence $S(x)=T(Q(x))$ for some polynomial $T(x)$. Now, if $P(x)$ and $R(x)$ are polynomials satisfying $P(Q(x))=R(S(x))$, then $P(Q(x))=$ $R(T(Q(x)))$. Since $Q(x)$ is not constant, it takes infinitely many values, so $P(x)$ and $R(T(x))$ agree at infinitely many points, implying that $P(x)=$ $R(T(x))$, as required. It is therefore sufficient to solve the problem in the particular case where $F(x)=P(Q(x))=$ $R(S(x))$ has a minimal degree. Let $d=$ $\operatorname{gcd}(\operatorname{deg} Q(x), \operatorname{deg} S(x))$ to write $\operatorname{deg} Q(x)=a d$ and $\operatorname{deg} S(x)=b d$, where $\operatorname{gcd}(a, b)=1$. Then $\operatorname{deg} P(x)=b c, \operatorname{deg} R(x)=a c$ and $\operatorname{deg} F(x)=a b c d$ for some positive integer $c$. We will show that minimality of $\operatorname{deg} F(x)$ forces $c=1$, so $\operatorname{deg} P(x)=b$, $\operatorname{deg} R(x)=a$ and $\operatorname{deg} F(x)=a b d$. The conditions $a=\operatorname{deg} R(x) \mid \operatorname{deg} P(x)=b$ and $\operatorname{gcd}(a, b)=1$ then force $a=1$, as stated above. Consequently, the only thing we are left with is the proof of the fact that $c=1$. For convenience, we may and will assume that $P(x), Q(x), R(x), S(x)$ are all monic; hence so is $F(x)$. The argument hinges on the lemma below. Lemma. If $f(x)$ is a monic polynomial of degree $m n$, then there exists a degree $n$ monic polynomial $g(x)$ such that $\operatorname{deg}\left(f(x)-g(x)^{m}\right)<(m-1) n$. (If $m=0$ or 1 , or $n=0$, the conclusion is still consistent with the usual convention that the identically zero polynomial has degree $-\infty$.) Proof. Write $f(x)=\sum_{k=0}^{m n} \alpha_{k} x^{k}, \alpha_{m n}=1$, and seek $g(x)=\sum_{k=0}^{n} \beta_{k} x^{k}, \beta_{n}=1$, so as to fit the bill. To this end, notice that, for each positive integer $k \leq n$, the coefficient of $x^{m n-k}$ in the expansion of $g(x)^{m}$ is of the form $m \beta_{n-k}+\varphi_{k}\left(\beta_{n}, \ldots, \beta_{n-k+1}\right)$, where $\varphi_{k}\left(\beta_{n}, \ldots, \beta_{n-k+1}\right)$ is an algebraic expression in $\beta_{n}, \ldots, \beta_{n-k+1}$. Recall that $\beta_{n}=1$ to determine the $\beta_{n-k}$ recursively by requiring $\beta_{n-k}=$ $\frac{1}{m}\left(a_{m n-k}-\varphi_{k}\left(\beta_{n}, \ldots, \beta_{n-k+1}\right)\right), k=1, \ldots, n$. The outcome is then the desired polynomial $g(x)$. We are now in a position to prove that $c=$ 1. Suppose, if possible, that $c>1$. By the lemma, there exist monic polynomials $U(x)$ and $V(x)$ of degree $b$ and $a$, respectively, such that $\operatorname{deg}\left(P(x)-U(x)^{c}\right)<(c-1) b$ and $\operatorname{deg}(R(x)-$ $\left.V(x)^{c}\right)<(c-1) a$. Then $\operatorname{deg}\left(F(x)-U(Q(x))^{c}\right)=$ $\operatorname{deg}\left(P(Q(x))-U(Q(x))^{c}\right)<(c-1) a b d, \operatorname{deg}(F(x)-$ $\left.V(S(x))^{c}\right)=\operatorname{deg}\left(R(S(x))-V(S(x))^{c}\right)<(c-1) a b d$, so $\operatorname{deg}\left(U(Q(x))^{c}-V(S(x))^{c}\right)=\operatorname{deg}((F(x)-$ $\left.\left.V(S(x))^{c}\right)-\left(F(x)-U(Q(x))^{c}\right)\right)<(c-1) a b d$. On the other hand, $U(Q(x))^{c}-V(S(x))^{c}=$ $(U(Q(x))-V(S(x)))\left(U(Q(x))^{c-1}+\cdots+\right.$ $\left.V(S(x))^{c-1}\right)$. By the preceding, the degree of the left-hand member is (strictly) less than $(c-1) a b d$ which is precisely the degree of the second factor in the righthand member. This forces $U(Q(x))=V(S(x))$, so $U(Q(x))=V(S(x))$ has degree $a b d<a b c d=$ $\operatorname{deg} F(x)$ - a contradiction. Consequently, $c=1$. This completes the argument and concludes the proof. |
Problem 6. Let $r, g, b$ be non-negative integers. Let $\Gamma$ be a connected graph on $r+g+b+1$ vertices. The edges of $\Gamma$ are each coloured red, green or blue. It turns out that $\Gamma$ has - a spanning tree in which exactly $r$ of the edges are red, - a spanning tree in which exactly $g$ of the edges are green and - a spanning tree in which exactly $b$ of the edges are blue. Prove that $\Gamma$ has a spanning tree in which exactly $r$ of the edges are red, exactly $g$ of the edges are green and exactly $b$ of the edges are blue. | Solution 1. Induct on $n=r+g+b$. The base case, $n=1$, is clear. Let now $n>1$. Let $V$ denote the vertex set of $\Gamma$, and let $T_{r}, T_{g}$, and $T_{b}$ be the trees with exactly $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. Consider two cases. Case 1: There exists a partition $V=A \sqcup B$ of the vertex set into two non-empty parts such that the edges joining the parts all bear the same colour, say, blue. Since $\Gamma$ is connected, it has a (necessarily blue) edge connecting $A$ and $B$. Let $e$ be one such. Assume that $T$, one of the three trees, does not contain $e$. Then the graph $T \cup\{e\}$ has a cycle $C$ through $e$. The cycle $C$ should contain another edge $e^{\prime}$ connecting $A$ and $B$; the edge $e^{\prime}$ is also blue. Replace $e^{\prime}$ by $e$ in $T$ to get another tree $T^{\prime}$ with the same number of edges of each colour as in $T$, but containing $e$. Performing such an operation to all three trees, we arrive at the situation where the three trees $T_{r}^{\prime}$, $T_{g}^{\prime}$, and $T_{b}^{\prime}$ all contain $e$. Now shrink $e$ by identifying its endpoints to obtain a graph $\Gamma^{*}$, and set $r^{*}=r, g^{*}=g$, and $b^{*}=b-1$. The new graph satisfies the conditions in the statement for those new values - indeed, under the shrinking, each of the trees $T_{r}^{\prime}, T_{g}^{\prime}$, and $T_{b}^{\prime}$ loses a blue edge. So $\Gamma^{*}$ has a spanning tree with exactly $r$ red, exactly $g$ green, and exactly $b-1$ blue edges. Finally, pass back to $\Gamma$ by restoring $e$, to obtain the a desired spanning tree in $\Gamma$.Case 2: There is no such a partition. Consider all possible collections $(R, G, B)$, where $R, G$ and $B$ are acyclic sets consisting of $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. By the problem assumptions, there is at least one such collection. Amongst all such collections, consider one such that the graph on $V$ with edge set $R \cup G \cup B$ has the smallest number $k$ of components. If $k=1$, then the collection provides the edges of a desired tree (the number of edges is one less than the number of vertices). Assume now that $k \geq 2$; then in the resulting graph some component $K$ contains a cycle $C$. Since $R, G$, and $B$ are acyclic, $C$ contains edges of at least two colours, say, red and green. By assumption, the edges joining $V(K)$ to $V \backslash V(K)$ bear at least two colours; so one of these edges is either red or green. Without loss of generality, consider a red such edge $e$. Let $e^{\prime}$ be a red edge in $C$ and set $R^{\prime}=R \backslash\left\{e^{\prime}\right\} \cup$ $\{e\}$. Then $\left(R^{\prime}, G, B\right)$ is a valid collection providing a smaller number of components. This contradicts minimality of the choice above and concludes the proof. |
Problem 6. Let $r, g, b$ be non-negative integers. Let $\Gamma$ be a connected graph on $r+g+b+1$ vertices. The edges of $\Gamma$ are each coloured red, green or blue. It turns out that $\Gamma$ has - a spanning tree in which exactly $r$ of the edges are red, - a spanning tree in which exactly $g$ of the edges are green and - a spanning tree in which exactly $b$ of the edges are blue. Prove that $\Gamma$ has a spanning tree in which exactly $r$ of the edges are red, exactly $g$ of the edges are green and exactly $b$ of the edges are blue. | Solution 2. For a spanning tree $T$ in $\Gamma$, denote by $r(T), g(T)$, and $b(T)$ the number of red, green, and blue edges in $T$, respectively. Assume that $\mathcal{C}$ is some collection of spanning trees in $\Gamma$. Write $$\begin{array}{rlrl} r(\mathcal{C}) & =\min _{T \in \mathcal{C}} r(T), & g(\mathcal{C})=\min _{T \in \mathcal{C}} g(T), \ b(\mathcal{C}) & =\min _{T \in \mathcal{C}} b(T), & R(\mathcal{C}) & =\max _{T \in \mathcal{C}}(T), \ G(\mathcal{C}) & =\max _{T \in \mathcal{C}} g(T), & B(\mathcal{C}) & =\max _{T \in \mathcal{C}} b(T) . \end{array}$$ Say that a collection $\mathcal{C}$ is good if $r \in[r(\mathcal{C}, R(\mathcal{C})]$, $g \in[g(\mathcal{C}, G(\mathcal{C})]$, and $b \in[b(\mathcal{C}, B(\mathcal{C})]$. By the problem conditions, the collection of all spanning trees in $\Gamma$ is good. For a good collection $\mathcal{C}$, say that an edge $e$ of $\Gamma$ is suspicious if $e$ belongs to some tree in $\mathcal{C}$ but not to all trees in $\mathcal{C}$. Choose now a good collection $\mathcal{C}$ minimizing the number of suspicious edges. If $\mathcal{C}$ contains a desired tree, we are done. Otherwise, without loss of generality, $r(\mathcal{C})<r$ and $G(\mathcal{C})>g$. We now distinguish two cases. Case 1: $B(\mathcal{C})=b$. Let $T^{0}$ be a tree in $\mathcal{C}$ with $g\left(T^{0}\right)=g(\mathcal{C}) \leq g$. Since $G(\mathcal{C})>g$, there exists a green edge $e$ contained in some tree in $\mathcal{C}$ but not in $T^{0}$; clearly, $e$ is suspicious. Fix one such green edge $e$. Now, for every $T$ in $\mathcal{C}$, define a spanning tree $T_{1}$ of $\Gamma$ as follows. If $T$ does not contain $e$, then $T_{1}=T$; in particular, $\left(T^{0}\right)_{1}=T^{0}$. Otherwise, the graph $T \backslash\{e\}$ falls into two components. The tree $T_{0}$ contains some edge $e^{\prime}$ joining those components; this edge is necessarily suspicious. Choose one such edge and define $T_{1}=T \backslash\{e\} \cup\left\{e^{\prime}\right\}$. Let $\mathcal{C}_{1}=\left\{T_{1}: T \in \mathcal{C}\right\}$. All edges suspicious for $\mathcal{C}_{1}$ are also suspicious for $\mathcal{C}$, but no tree in $\mathcal{C}_{1}$ con- tains $e$. So the number of suspicious edges for $\mathcal{C}_{1}$ is strictly smaller than that for $\mathcal{C}$. We now show that $\mathcal{C}_{1}$ is good, reaching thereby a contradiction with the choice of $\mathcal{C}$. For every $T$ in $\mathcal{C}$, the tree $T_{1}$ either coincides with $T$ or is obtained from it by removing a green edge and adding an edge of some colour. This already shows that $g\left(\mathcal{C}_{1}\right) \leq g(\mathcal{C}) \leq g, G\left(\mathcal{C}_{1}\right) \geq G(\mathcal{C})-1 \geq g$, $R\left(\mathcal{C}_{1}\right) \geq R(\mathcal{C}) \geq r, r\left(\mathcal{C}_{1}\right) \leq r(\mathcal{C})+1 \leq r$, and $B\left(\mathcal{C}_{1}\right) \geq B(\mathcal{C}) \geq b$. Finally, we get $b\left(T^{0}\right) \leq$ $B(\mathcal{C})=b$; since $\mathcal{C}_{1}$ contains $T^{0}$, it follows that $b\left(\mathcal{C}_{1}\right) \leq b\left(T^{0}\right) \leq b$, which concludes the proof. Case 2: $B(\mathcal{C})>b$. Consider a tree $T^{0}$ in $\mathcal{C}$ satisfying $r\left(T^{0}\right)=$ $R(\mathcal{C}) \geq r$. Since $r(\mathcal{C})<r$, the tree $T^{0}$ contains a suspicious red edge. Fix one such edge $e$. Now, for every $T$ in $\mathcal{C}$, define a spanning tree $T_{2}$ of $\Gamma$ as follows. If $T$ contains $e$, then $T_{2}=T$; in particular, $\left(T^{0}\right)_{2}=T^{0}$. Otherwise, the graph $T \cup\{e\}$ contains a cycle $C$ through $e$. This cycle contains an edge $e^{\prime}$ absent from $T^{0}$ (otherwise $T^{0}$ would contain the cycle $C$ ), so $e^{\prime}$ is suspicious. Choose one such edge and define $T_{2}=T \backslash\left\{e^{\prime}\right\} \cup\{e\}$. Let $\mathcal{C}_{2}=\left\{T_{2}: T \in \mathcal{C}\right\}$. All edges suspicious for $\mathcal{C}_{2}$ are also suspicious for $\mathcal{C}$, but all trees in $\mathcal{C}_{2}$ contain $e$. So the number of suspicious edges for $\mathcal{C}_{2}$ is strictly smaller than that for $\mathcal{C}$. We now show that $\mathcal{C}_{2}$ is good, reaching again a contradiction. For every $T$ in $\mathcal{C}$, the tree $T_{2}$ either coincides with $T$ or is obtained from it by removing some edge and adding a red edge. This shows that $r\left(\mathcal{C}_{2}\right) \leq r(\mathcal{C})+1 \leq r, R\left(\mathcal{C}_{2}\right) \geq R(\mathcal{C}) \geq r$, $G\left(\mathcal{C}_{2}\right) \geq G(\mathcal{C})-1 \geq g, g\left(\mathcal{C}_{2}\right) \leq g(\mathcal{C}) \leq g$, $b\left(\mathcal{C}_{1}\right) \leq b(\mathcal{C}) \leq b$ and $B\left(\mathcal{C}_{2}\right) \geq B(\mathcal{C})-1 \geq b$. This concludes the proof. |
Find all pairs of positive integers $(a,b)$ such that, \[ab^2+b+7|a^2b+a+b\]\n \n | We have the following divisibility relations, \[ab^2+b+7|a^2b+a+b|b(a^2b+a+b)=a^2b^2+ab+b^2\]\n\[ab^2+b+7|a(ab^2+b+7)=a^2b^2+ab+7a\]\nSubtracting, \[ab^2+b+7| \;|b^2-7a|\]\n If $b=1$, we may have \[a+8|7a-1\] \nOtherwise, we would have \[|b^2-7a|<ab^2+b+7\]\nIn the first case, \[a+8|7a+56\]\nThis gives, \[a+8|57\]\nSince, $a+8>3$ and $57=3\cdot19$, we must have $a+8=19$ or $57$ which yields the solutions $(a,b)=(11,1),(49,1)$.\n In the second case, $b^2-7a=0$ or, \[b^2=7a\implies 7|b\]\nSay, $b=7k$. Then $a=7k^2$. Checking we find that it is indeed a solution.\nThus, all solutions are \[(a,b)=(11,1),(49,1),(7k^2,7k).\]\n |
(Mongolia)\nLet $a_1, a_2, \dotsc, a_n$ be positive real numbers such that $a_1 + a_2 + \dotsb + a_n < 1$. Prove that\n\[\frac{a_1 a_2 \dotsm a_n \bigl[ 1- (a_1 + a_2 + \dotsb + a_n) \bigr]}{ (a_1 + a_2 + \dotsb + a_n)(1-a_1)(1-a_2) \dotsm (1-a_n)} \le \frac{1}{n^{n+1}} .\]\n | Let $a_{n+1}$ be the positive real number such that $a_{n+1} = 1 - \sum_{k=1}^n a_k$. Then\n\[\frac{a_1 a_2 \dotsm a_n \bigl[ 1- (a_1 + a_2 + \dotsb + a_n) \bigr]}{ (a_1 + a_2 + \dotsb + a_n)(1-a_1)(1-a_2) \dotsm (1-a_n)} = \frac{\prod_{k=1}^{n+1} a_k} {\prod_{k=1}^{n+1} \sum_{j\neq k} a_j} .\]\nBut by AM-GM, $\sum_{j\neq k} a_j \ge n \prod_{j\neq k} a_j^{1/n}$. Hence\n\[\frac{\prod_{k=1}^{n+1} a_k}{\prod_{k=1}^{n+1} \sum_{j\neq k} a_j} \le \frac{\prod_{k=1}^{n+1} a_k}{\prod_{k=1}^{n+1} n \prod_{j\neq k} a_j^{1/n}} = \frac{\prod_{k=1}^{n+1} a_k}{n^{n+1} \prod_{k=1}^{n+1} a_j} = \frac{1}{n^{n+1}},\]\nas desired. Equality holds when all the $a_k$ are equal to $1/(n+1)$. $\blacksquare$\n |
(Australia)\nLet $r_1, r_2, \dotsc, r_n$ be real numbers greater than or equal to 1. Prove that\n\[\frac{1}{r_1 + 1} + \frac{1}{r_2 + 1} + \dotsb + \frac{1}{r_n + 1} \ge \frac{n}{\sqrt[n]{r_1 r_2 \dotsm r_n} + 1} .\]\n | Let $f$ denote the function $x \mapsto 1/(e^x + 1)$.\n Lemma 1. For $x \ge 1$, the function $g : x \mapsto x/(x +1)^2$ is decreasing.\n Proof. Note that $g(x) = \frac{1}{x + 2 + 1/x}$. Since $x + 1/x$ is increasing for $x\ge 1$, the lemma follows. $\blacksquare$\n Lemma 2. For positive $x$, $f(x)$ is convex.\n Proof. Note that the derivative of $f$ is\n\[\frac{df}{dx} = - \frac{e^{x}}{(e^x+1)^2} .\]\nBy Lemma 1, $df/dx$ is increasing when $e^x \ge 1$, i.e., when $x\ge 0$. Therefore $f$ is convex for nonnegative $x$. $\blacksquare$\n For all integers $1 \le k \le n$, $r_k \ge 1$, so $\ln r_k \ge 0$. Since $f(x)$ is convex for nonnegative $x$, it follows from Jensen's Inequality that\n\[\sum_{k=1}^n \frac{1}{r_k+1} = \sum_{k=1}^n f(\ln r_k) \ge n \cdot f \left( \sum_{k=1}^n \ln r_k/n \right) = \frac{1}{ \prod_{k=1}^n r_k^{1/n} +1},\]\nas desired. $\blacksquare$\n |
An $m \times n$ array of real numbers has the sum of each row and column integral. Show that each non-integral element $x$ can be changed to either $\left\lfloor x \right\rfloor$ or $\left\lfloor x \right\rfloor + 1$ so that the row and column sums are unchanged.\n \n | Coming soon...\n |
Let $T$ denote the set of all ordered triples $(p,q,r)$ of nonnegative integers. Find all functions $f:T \rightarrow \mathbb{R}$ such that\n $f(p,q,r) = \begin{cases} 0 & \text{if} \; pqr = 0, \\ 1 + \tfrac{1}{6}\{f(p + 1,q - 1,r) + f(p - 1,q + 1,r) & \\ + f(p - 1,q,r + 1) + f(p + 1,q,r - 1) & \\ + f(p,q + 1,r - 1) + f(p,q - 1,r + 1)\} & \text{otherwise.} \end{cases}$ | We can see that $h(p,q,r)=\frac{3pqr}{p+q+r}$ for $pqr\neq0$ and $h(p,q,r)=0$ for $pqr=0$ satisfies the equation. Suppose there exists another solution $f(p,q,r)$. Let $g(p,q,r)=f(p,q,r)-h(p,q,r)$. Plugging in $f=g+h,$ we see that $g$ satisfies the relationship $g(p,q,r)=\begin{cases} \tfrac{1}{6}\{g(p + 1,q - 1,r) + g(p - 1,q + 1,r) & \\ + g(p - 1,q,r + 1) + g(p + 1,q,r - 1) & \\ + g(p,q + 1,r - 1) + g(p,q - 1,r + 1)\}\end{cases}$, so that each value of $g$ is equal to 6 points around it with an equal sum $p+q+r$. This implies that for fixed $p+q+r$, $g(p,q,r)$ is constant. Furthermore, some values of $g$ are always zero; for example, $f(p,2,0)=0$ by the problem statement, and similarly, $h(p,2,0)=0$, so $g(p,2,0)=0-0=0$. Thus, $g$ must be identically zero, so $h$ is the only function satisfying this equation.\n |
Let $a_0, a_1, a_2, \ldots$ be an arbitrary infinite sequence of positive numbers. Show that the inequality $1 + a_n > a_{n - 1} \sqrt [n]{2}$ holds for infinitely many positive integers $n$.\n | We proceed with a proof by contradiction. Suppose the statement were false. Then, there exists a sequence $a_0, a_1, \ldots$ of positive integers for which there are only finitely many $a_n$ with $1+a_n> a_{n-1}\sqrt[n]{2}$. Let the largest such $n$ be $N-1$, so that $1+a_n\le a_{n-1}\sqrt[n]{2}$ whenever $n\le N-1$. Then, it is clear that $1+a_{n+N}\le a_{n+N-1}\sqrt[n]{2}$ for all nonnegative $n$. Therefore, define $b_n=a_{n+N}$. If there does not exist a sequence $b_0, b_1, \ldots$ of positive integers for which $1+b_n\le b_{n-1}\sqrt[n]{2}$, it is clear that there will not exist any sequence $a_0, a_1, \ldots$ for which that property is eventually true.\n Thus, I claim there does not exist a sequence of positive integers $b_0, b_1, \ldots$ for which $1+b_n\le b_{n-1}\sqrt[n]{2}$. Again, suppose there does exist such a sequence. Then, define $x_0=b_0$ and $x_n=x_{n-1}\sqrt[n]{2} -1$. It is clear that $x_n\ge b_n$ for all $n$. I claim that this sequence will always become eventually negative. Note that\n$x_n=s_nx_{n-1}-1=s_n(s_{n-1}x_{n-2}-1)-1=\ldots=x_0s_n\cdot s_{n-1}\cdot\ldots\cdot s_0-\sum_{i=1}^{n-1}(\prod_{j=i}^n s_j)-1$,\nwhich becomes negative if and only if $\frac{x_n}{\prod_{i=1}^ns_i}=x_0-1-\frac{1}{s_1}-\frac{1}{s_1}{s_2}-\ldots$ does. In other words, $x_n$\nbecomes zero if $t_k=\sum_{i=1}^{k}\frac{1}{\prod_{j=1}^i s_j}$ is unbounded. However, $\sqrt[n]{2}$ is eventually less than $\frac{n+1}{n}$, so this sum is indeed unbounded and the proof is complete.\n |
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$, satisfying\n $f(xy)(f(x) - f(y)) = (x - y)f(x)f(y)$ for all $x,y$.\n | Assume $f(1) = 0$. Take $y = 1$. We get $f^2(x) = 0,\ \forall x$, so $f(x) = 0,\ \forall x$. This is a solution, so we can take it out of the way: assume $f(1)\ne 0$. \n $y = 1\Rightarrow f(x)[f(x) - f(1)] = (x - 1)f(x)f(1)$. We either have $f(x) = 0$ or $f(x) = f(1)x$, so for every $x$, $f(x)\in\{0,f(1)x\}$. In particular, $f(0) = 0$.\n Assume $f(y) = 0$. We get $f(x)f(xy) = 0,\ \forall x$. This means that $f(a),f(b)\ne 0\Rightarrow f\left(\frac ab\right)\ne 0\ (*)$ ($\frac ab$ is defined because $f(b)\ne 0\Rightarrow b\ne 0$). Assume now that $x\ne y$ and $f(x),f(y)\ne 0$. We get $f(x) = f(1)x,\ f(y) = f(1)y$, and after replacing everything we get $f(xy) = f(1)xy\ne 0$, so $x\ne y,\ f(x),f(y)\ne 0\Rightarrow f(xy)\ne 0\ (**)$. Assume now $f(x)\ne 0$. From $(*)$ we get $f\left(\frac 1x\right)\ne 0$, and after applying $(*)$ again to $a = x,b = \frac 1x$ we get $f(x^2)\ne 0\ (***)$. We can now see that $(**),(***)$ combine to $f(x),f(y)\ne 0\Rightarrow f(xy)\ne 0\ (\#)$.\n Let $G = \{x\in\mathbb R|f(x)\ne 0\}$. $(*)$ and $(\#)$ simply say that $(G,\ \cdot)$ is a subgroup of $(\mathbb R^{*},\ \cdot)$.\n Conversely, let $G$ be a subgroup of the multiplicative group $\mathbb R^*$. Take $f(x) = \left\{\begin{array}{c}f(1)x,\ x\in G \\ 0,\ x\not \in G\end{array}\right\}$. It's easy to check the condition $f(xy)[f(x) - f(y)] = (x - y)f(x)f(y)$.\n |
Find all positive integers $a_1, a_2, \ldots, a_n$ such that\n $\frac {99}{100} = \frac {a_0}{a_1} + \frac {a_1}{a_2} + \cdots + \frac {a_{n - 1}}{a_n},$ where $a_0 = 1$ and $(a_{k + 1} - 1)a_{k - 1} \geq a_k^2(a_k - 1)$ for $k = 1,2,\ldots,n - 1$.\n | We claim that there is only one such sequence: $a_1=2, a_2=5, a_3=56, a_4=56\times 1400$. This works because\n\t\t\t\[\frac{1}{2}+\frac{2}{5}+\frac{5}{56}+\frac{56}{56\times 1400}\]\n\[=\frac{700}{1400}+\frac{560}{1400}+\frac{125}{1400}+\frac{1}{1400}=\frac{1386}{1400}=\frac{99}{100}.\]\n \t\t\tIt is also easy to check that $(a_{k+1}-1)a_{k-1} \geq a_k^2(a_k - 1)$ for $k=1,2,3$.\n \t\t\tNow we prove such a sequence is unique. We first claim that a sequence of positive integers $a_0, a_1,\dots, a_n$ satisfying $(a_{k+1}-1)a_{k-1} \geq a_k^2(a_k - 1)$ for $k = 1,2,\ldots,n-1$ has the property that \n\t\t\t\[\sum_{k=0}^{n-1}\frac{a_k}{a_{k+1}}<\frac{a_0}{a_1-1}.\]\n\t\t\tWe use induction on the length of the sequence.\n\t\t\tThe base case is trivial (the condition guarantees that $a_1>1$). For the inductive step, notice that by the inductive hypothesis,\n\t\t\t\[\sum_{k=1}^{n-1}\frac{a_k}{a_{k+1}}<\frac{a_1}{a_2-1},\]\n\t\t\tso it suffices to show\n\t\t\t\[\frac{a_0}{a_1}+\frac{a_1}{a_2-1}\le \frac{a_0}{a_1-1},\]\n\t\t\tbut this is equivalent to \n\t\t\t\[a_0(a_1-1)(a_2-1)+a_1^2(a_1-1)\le a_0a_1(a_2-1)\]\n\[a_1^2(a_1-1)\le a_0(a_2-1),\]\n\t\t\twhich we know to be true.\n \t\t\tNow suppose there were two sequences $a_1,\dots, a_n$ and $a'_1,\dots, a'_m$ that satisfied the conditions in the problem. Clearly one cannot be a subsequence of the other, or else the longer one will obviously have a larger sum. So there exists a smallest integer $c$ such that $a_c\neq a'_c$ (so for $k<c$, $a_k=a'_k$). Without loss of generality, let $a_c<a'_c$.\n \t\t\tBy our previous claim, we have that \[\sum_{k=c-1}^{n-1}\frac{a'_k}{a'_{k+1}}<\frac{a'_{c-1}}{a'_c-1}=\frac{a_{c-1}}{a'_c-1}\le\frac{a_{c-1}}{a_c},\]\n\t\t\tand as a corollary, since the first $c-2$ terms are equal,\n\t\t\t\[\sum_{k=1}^{n-1}\frac{a'_k}{a'_{k+1}}<\sum_{k=1}^{c-1}\frac{a_k}{a_{k+1}}\le \sum_{k=1}^{m-1}\frac{a_k}{a_{k+1}},\]\n\t\t\twhich contradicts the fact that they are equal. So there is a unique sequence satisfying the problem conditions.\n |
Prove that for all positive real numbers $a,b,c$,\n $\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \geq 1.$ Generalization The leader of the Bulgarian team had come up with the following generalization to the inequality:\n $\frac {a}{\sqrt {a^2 + kbc}} + \frac {b}{\sqrt {b^2 + kca}} + \frac {c}{\sqrt {c^2 + kab}} \geq \frac{3}{\sqrt{1+k}}.$ | We will use the Jenson's inequality. \n Now, normalize the inequality by assuming $a+b+c=1$\n Consider the function $f(x)=\frac{1}{\sqrt{x}}$. Note that this function is convex and monotonically decreasing which implies that if $a > b$, then $f(a) < f(b)$.\n Thus, we have\n $\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} = af(a^2+8bc)+bf(b^2+8ca)+cf(c^2+8ab) \geq f(a^3+b^3+c^3+24abc)$\n Thus, we only need to show that $a^3+b^3+c^3+24abc \leq 1$ i.e.\n \[a^3+b^3+c^3+24abc \leq (a+b+c)^3=a^3+b^3+c^3+3(a+b+c)(ab+bc+ca)-3abc\]\n \[i.e. (a+b+c)(ab+bc+ca) \geq 9abc\]\n Which is true since\n \[(a+b+c)(ab+bc+ca) \geq (3\sqrt[3]{abc})(3\sqrt[3]{a^{2}b^{2}c^{2}}) = 9abc\]\nThe last part follows by the AM-GM inequality.\n Equality holds if $a=b=c$\n Alternative Solution By Carlson's Inequality, we can know that \[\Bigg(\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}}\Bigg)\Bigg(\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}}\Bigg)\Big((a^3+8abc)+(b^3+8abc)+(c^3+8abc)\Big) \ge (a+b+c)^3\]\n Then, \[\Bigg(\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}}\Bigg)^2 \ge \frac{(a+b+c)^3}{a^3+b^3+c^3+24abc}\]\n On the other hand, \[3a^2b+3b^2c+3c^2a \ge 9abc\] and \[3ab^2+3bc^2+3ca^2 \ge 9abc\]\n Then, \[(a+b+c)^3 = a^3+b^3+c^3+3(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+6abc \ge a^3+b^3+c^3+24abc\]\n Therefore, \[\Bigg(\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}}\Bigg)^2 \ge \frac{(a+b+c)^3}{a^3+b^3+c^3+24abc} \ge 1\]\n Thus, \[\frac {a}{\sqrt {a^2 + 8bc}} + \frac {b}{\sqrt {b^2 + 8ca}} + \frac {c}{\sqrt {c^2 + 8ab}} \ge 1\]\n -- Haozhe Yang\n |
Let $A = (a_1, a_2, \ldots, a_{2001})$ be a sequence of positive integers. Let $m$ be the number of 3-element subsequences $(a_i, a_j, a_k)$ with $1 \le i < j < k \le 2001$ such that $a_j = a_i + 1$ and $a_k = a_j + 1$. Considering all such sequences $A$ find the greatest value of $m$.\n | Solution 1 Consider what happens if $A$ is ordered from least to greatest. Then, all the original subsequences will still be subsequences because, since $a_k > a_j > a_i$, the order they appear in is $a_i, a_j, a_k$, so it is still a subsequence. So, if A attains the maximal value, so does it unstrictly increasing version. Therefore, we can look at an A such that it is unstrictly increasing and attains the maximal number of such subsequences. Let's divide it into n blocks, such two elements of the sequence have the same value if and only if they belong to the same block. We can do this because it is unstrictly increasing. For example, if the sequence consists of 1000 ones, then 500 twos, then 501 threes, the blocks would be the 1000 ones, the 500 twos, and the 501 threes. Let the number of elements in the ith block be $b_i$ and there be n blocks. Cases:\n $n < 3$: Then there are no subsequences, because there are only two possible values for each element, and the subsequence must have three elements with distinct values, but by pigeonhole, two of them will have the same value.\n $n > 3$: Note that the number of subsequences is $b_1b_2b_3 + b_2b_3b_4 + \ldots + b_{n-2}b_{n-1}b_n$. WLOG, $b_1b_2 \ge b_{n-2}b_{n-1}$. Let us say that all the elements in the first block have value m. If we change all of the elements in the n th block to elements with value m, then the first block would have $b_1 + b_n$ elements, and there would only be n - 1 blocks. Note that the number of 3-element subsequences $(a_i, a_j, a_k)$ with $1 \le i < j < k \le 2001$ such that $a_j = a_i + 1$ and $a_k = a_j + 1$ would increase by $b_n(b_2b_3 - b_{n-2}b_{n-3}) \ge 0$. We repeatedly do this until there are 3 blocks, and at that time there will be at least as many subsequences as there were originally. \n $n = 3$: So we have three nonegative integers x, y, z such that $x + y + z = 2001$ and we want to maximize xyz. By AM-GM, xyz is maximal when x = y = z = 667, so m = $667^3$.\n QED\n Solution 2 We claim that $m=667^3$ is the maximum. First, we show that $667^3$ is achievable. Consider the sequence where $a_i=1$ for $1\le i\le 667$, $a_i=2$ for $668\le i\le 1334$, and $a_i=3$ for $1335\le i\le 2001$. Any subsequence satisfying the condition must be equal to $(1,2,3)$; there are 667 ways to select each element to fill that, and every possible selection has indexes in increasing order, resulting in $667^3$ subsequences.\n |
Let $n$ be an odd integer greater than 1 and let $c_1, c_2, \ldots, c_n$ be integers. For each permutation $a = (a_1, a_2, \ldots, a_n)$ of $\{1,2,\ldots,n\}$, define $S(a) = \sum_{i = 1}^n c_i a_i$. Prove that there exist permutations $a \neq b$ of $\{1,2,\ldots,n\}$ such that $n!$ is a divisor of $S(a) - S(b)$.\n | We shall prove this by contradiction. Assume that for some $n$-tuple of $c_i$ there does not exist two permutations $a$ and $b$ of $s_n=\{ 1, 2, \ldots,n\}$ such that $n!|S(a)-S(b)$. Note that there are $n!$ distinct permutations of $s_n$, which means there are $n!$ distinct sums $S(a)$. Because no two of them are congruent modulo $n!$, we have that for some $n$-tuple of $c_i$ there exists a unique permutation $x$ of $s_n$ such that $S(x)\equiv y\bmod{n}$ for all integers $0\leq y\leq n-1$. This means that the sum of every $S(a)$ is congruent to $\sum_{i=0}^{n!-1}=\frac{(n!-1)n!}{2}$. However, that same sum is congruent to\n \[\sum_{i=1}^{n}\left(c_i\sum_{j=1}^{n} (n-1)!*a_j\right)=\sum_{i=1}^{n} c_i\cdot \frac{n(n+1)(n-1)!}{2}=\frac{c_i(n+1)!}{2}\bmod{n!}\]\n Note that $2|n+1$, so $n!|\frac{(n+1)!}{2}$, so $n!$ divides the sum. However, the sum is also congruent to $\frac{(n!-1)n!}{2}$ modulo $n!$, and $n!-1$ is odd, so $n!$ couldn't possibly divide the sum. This leads to a contradiction, so our previous assumption must be false. This proves the problem statement.\n |
Define a $k$-clique to be a set of $k$ people such that every pair of them are acquainted with each other. At a certain party, every pair of 3-cliques has at least one person in common, and there are no 5-cliques. Prove that there are two or fewer people at the party whose departure leaves no 3-clique remaining.\n | Solution 1 If there exists only one 3-clique, remove anyone in that clique. (If there are no 3-cliques, we are done!) Otherwise, consider the following cases:\n Case 1: There exist a pair of 3-cliques that share 2 people.\n Let these 3-cliques be $\{A,C,D\}$ and $\{B,C,D\}$. If every other 3-clique contained either $C$ or $D$, then removing $C$ and $D$ will leave no 3-clique remaining. \n Otherwise, if there was a 3-clique that did not contain $C$ or $D$, then it would have to contain $A$ and $B$ to satisfy the condition. Call this 3-clique $\{A,B,E\}$. First notice that $\{A,B,C\}$ and $\{A,B,D\}$ are also 3-cliques now since $A$ and $B$ are now acquainted. We claim that removing $A$ and $B$ will now suffice; suppose there was another 3-clique not containing $A$ or $B$. This 3-clique must contain $C$, $D$, and $E$ (to share with $\{A,B,C\}$, $\{A,B,D\}$, and $\{A,B,E\}$). However, this means $\{A,B,C,D,E\}$ is a 5-clique; contradiction.\n Case 2: Every pair of 3-cliques has at most one person in common.\n There exist two 3-cliques that share a person whom we will call $A$: name the 3-cliques $\{A,B,C\}$ and $\{A,D,E\}$. We now claim that every 3-clique in this case must contain $A$.\n Suppose there existed a 3-clique not containing $A$. It must share one person with each of $\{A,B,C\}$ and $\{A,D,E\}$; without loss of generality, let this 3-clique contain $B$ and $D$. But this means $B$ and $D$ are acquainted, and due to their mutual acquaintance with $A$, $\{A,B,D\}$ is a 3-clique, which shares 2 people with $\{A,B,C\}$, contradicting the premise of this case.\n Thus, every 3-clique contains $A$, and so removing $A$ will remove every 3-clique.\n $\blacksquare$\n |
A set of three nonnegative integers $\{x,y,z\}$ with $x < y < z$ is called historic if $\{z - y,y - x\} = \{1776,2001\}$. Show that the set of all nonnegative integers can be written as the union of pairwise disjoint historic sets.\n | We describe a greedy algorithm to cover all integers. If an integer has been included in a set, we call it 'colored'. We also say that a number x is 'in column A' (A being x, y, or z) if x was in the x, y, or z position in its historic set (e.g. in $\{1,2,4\}$, 4 is in column z, 2 in column y, and 1 in column x). \n Steps of Algorithm:\n Take the smallest integer k which is not colored\nIf k+a is likewise not colored, then the new historic set is $\{k,k+a,k+a+b\}$\nIf k+a is colored, then the new historic set is $\{k,k+b,k+a+b\}$ Proof for why k+a+b is never colored in 2) or 3):\nAssume otherwise. If k+a+b is in column z, then k would already be colored. If k+a+b is in column y, then k+a or k+b would be in column x, but this will never happen by the algorithm, as it says to take the smallest integer k which is not colored, so if k+a or k+b were chosen, k would already have been colored. Similarly, k+a+b can't be in column x. \n We see that if this works, all sets will be pairwise disjoint, as at every stage, we only include integers which are not colored. Also, this covers (and colors) all nonnegative integers, as if there is a set of integers which have not been colored, the algorithm just takes the smallest one and continues until there are none left. \n Now, we prove 2) and 3):\n For 2), there is nothing left to prove as we have shown that k+a+b will always be available and not colored. \n For 3), the main thing to prove is that if k+a is colored already, k+b cannot have been colored. Again, assume otherwise. \n If k+b has been colored, it can't be in column x, due to size constraints as above.\nIf in column y, the only option is that k+b-a is the first value in its historic set (as otherwise k would already have been colored). However, $k + b - a > k$, as b > a.\nThe last remaining option is that k+b is in column z, meaning k-a must be in column x. However, the algorithm states in 2) that the default is to color k-a+a = k, so in this case k would already be colored. And we are done. |
Find all finite sequences $(x_0, x_1, \ldots,x_n)$ such that for every $j$, $0 \leq j \leq n$, $x_j$ equals the number of times $j$ appears in the sequence.\n | See Problem 23, Chapter 4, in 102 Combinatorial Problems by Andreescu and Feng.\n |
For a positive integer $n$ define a sequence of zeros and ones to be balanced if it contains $n$ zeros and $n$ ones. Two balanced sequences $a$ and $b$ are neighbors if you can move one of the $2n$ symbols of $a$ to another position to form $b$. For instance, when $n = 4$, the balanced sequences $01101001$ and $00110101$ are neighbors because the third (or fourth) zero in the first sequence can be moved to the first or second position to form the second sequence. Prove that there is a set $S$ of at most $\frac {1}{n + 1} \binom{2n}{n}$ balanced sequences such that every balanced sequence is equal to or is a neighbor of at least one sequence in $S$.\n | This problem needs a solution. If you have a solution for it, please help us out by adding it.\n |
A pile of $n$ pebbles is placed in a vertical column. This configuration is modified according to the following rules. A pebble can be moved if it is at the top of a column which contains at least two more pebbles than the column immediately to its right. (If there are no pebbles to the right, think of this as a column with 0 pebbles.) At each stage, choose a pebble from among those that can be moved (if there are any) and place it at the top of the column to its right. If no pebbles can be moved, the configuration is called a final configuration. For each $n$, show that, no matter what choices are made at each stage, the final configuration obtained is unique. Describe that configuration in terms of $n$.\n | Note that if a column moves a pebble to the right, either the first column is still bigger or they have the same size. This means that at every point in the process, the amount of pebbles in every column decreases from left to right.\n At no point during this process can there be three or more columns with the same number of pebbles. We will show this by contradiction. These columns would have to be consecutive since the amount of pebbles in each column is decreasing. Letting there be $a$ pebbles in these columns, the configuration is $(..., a, a, a, ...)$ Immediately before the first time at least three of these columns are present, The configuration must be either $(... a - 1, a, a, ...), (..., a + 1, a - 1, a, ...), (..., a, a + 1, a - 1),$ or $(..., a, a, a + 1, ...)$, none of which are legal.\n Additionally, at no point during this process can there be two columns with $a$ pebbles and two with $a + 1$ pebbles for any natural number $a$. We will show this by contradiction. If so, then the configuration is $(..., a + 1, a + 1, a, a, ...)$. Immediately before, we have $(..., a, a + 1, a, a, ...), (..., a + 2, a, a, a, ...), (..., a + 1, a + 2, a - 1, a, ...), (..., a + 1, a + 1, a + 1, a - 1, ...),$ or $(..., a + 1, a + 1, a, a + 1, ...)$, none of which are legal.\n Finally, we show that we can never have two columns with $a$ pebbles, two with $b$ columns, and one of every size between. We can do this by inducting on the number of pebbles on $a - b$.\n Base case: $a - b = 1$. This is covered above.\n Inductive step: $a - b = k$ implies $a - b = k + 1$. Immediately before arriving at $(..., a, a, a - 1, ..., b + 1, b, b, ...),$ we must be at the position $(..., a + 1, a - 1, a - 1, ..., b + 1, b, b, ...)$ or $(..., a, a, a - 1, ..., b + 1, b + 1, b - 1, ...)$ which is impossible by our inductive hypothesis.\n Thus, the final configuration must be a sequence of decreasing numbers with at most one appearing twice. There is exactly one way to represent any number in this form (start with the largest triangular number below it, add a column with the size of their difference), so there is exactly one possible final configuration.\n |
Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that each contestant solved at most six problems, and for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy. Show that there is a problem that was solved by at least three girls and at least three boys.\n | This problem needs a solution. If you have a solution for it, please help us out by adding it.\n |
Let $A_1$ be the center of the square inscribed in acute triangle $ABC$ with two vertices of the square on side $BC$. Thus one of the two remaining vertices of the square is on side $AB$ and the other is on $AC$. Points $B_1,\ C_1$ are defined in a similar way for inscribed squares with two vertices on sides $AC$ and $AB$, respectively. Prove that lines $AA_1,\ BB_1,\ CC_1$ are concurrent.\n | Let $BC=a$, $CA=b$, $AB=c$, $\angle BAC=\alpha$, $\angle CBA=\beta$, and $\angle ACB=\gamma$. Let $A_2$ be the point on the other side of $BC$ than $A$ such that $BA_2C$ is an isosceles right triangle. Define $B_2$ and $C_2$ similarly. Let $E$ and $F$ be the points on $AB$ and $CA$ that are the vertices of the square centered at $A_1$. We then have that $EA_1F$ is also an isosceles right triangle. It's clear that $EF$ is parallel to $BC$, so $\triangle AEF\sim \triangle ABC$ and $\triangle A_1EF\sim \triangle A_2BC$. The ratio of similarity of both relations is $\frac{EF}{BC}$, which implies that quadrilaterals $AEA_1F$ and $ABA_2C$ are similar. Therefore $\angle BAA_2=\angle EAA_1$ and $\angle CAA_2=EAA_1$. It then follows that $A$, $A_1$, and $A_2$ are collinear. Similarly $B$, $B_1$, and $B_2$ are collinear, as are $C$, $C_1$, and $C_2$. It therefore suffices to show that $AA_2$, $BB_2$, and $CC_2$ are concurrent.\n Let $AC_2=s$ and $CC_2=d$, for positive reals $s$ and $d$. Also let $\angle ACC_2=\theta$ and $\angle BCC_2=\phi$. Note that $\angle C_2AC=\angle C_2AB+\angle BAC=45^{\circ}+\alpha$, and likewise $\angle C_2BC = 45^{\circ}+\beta$. It then follows from the Law of Sines on triangles $ACC_2$ and $BCC_2$ that\n \[\frac{d}{\sin{(\alpha+45^{\circ})}}=\frac{s}{\sin{\theta}}\]\n and\n \[\frac{d}{\sin{(\beta+45^{\circ})}}=\frac{s}{\sin{\phi}}\]\n Solving for $\sin{\theta}$ and $\sin{\phi}$ gives that\n \[\sin{\theta}=\frac{s\sin{(\alpha+45^{\circ})}}{d}\]\n and\n \[\sin{\phi}=\frac{s\sin{(\beta+45^{\circ})}}{d}\]\n Therefore\n \[\frac{\sin{\theta}}{\sin{\phi}}=\frac{\sin\angle ACC_2}{\sin\angle BCC_2}=\frac{\sin{(\alpha+45^{\circ})}}{\sin{(\beta+45^{\circ})}}\]\n Similar lines of reasoning show that\n \[\frac{\sin\angle BAA_2}{\sin\angle CAA_2}=\frac{\sin{(\beta+45^{\circ})}}{\sin{(\gamma+45^{\circ})}}\]\n and\n \[\frac{\sin\angle CBB_2}{\sin\angle ABB_2}=\frac{\sin{(\gamma+45^{\circ})}}{\sin{(\alpha+45^{\circ})}}\]\n An application of the trigonometric version of Ceva's Theorem shows that $AA_2$, $BB_2$, and $CC_2$ are concurrent, which shows that $AA_1$, $BB_1$, and $CC_1$ are concurrent.\n |
Consider an acute-angled triangle $ABC$. Let $P$ be the foot of the altitude of triangle $ABC$ issuing from the vertex $A$, and let $O$ be the circumcenter of triangle $ABC$. Assume that $\angle C \geq \angle B + 30^{\circ}$. Prove that $\angle A + \angle COP < 90^{\circ}$.\n | See 2001 IMO 1 page.\nhttps://artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/Problem_1\n |
Let $ABC$ be a triangle with centroid $G$. Determine, with proof, the position of the point $P$ in the plane of $ABC$ such that $AP{\cdot}AG + BP{\cdot}BG + CP{\cdot}CG$ is a minimum, and express this minimum value in terms of the side lengths of $ABC$.\n | We claim that the expression is minimized at $P=G$, resulting it having a value of $(a^2+b^2+c^2)/3$ ($a,b,c$ being the side lengths of $ABC$).\n \t\t\tWe will use vectors, with $G=\vec{0}$ (meaning that $\vec A+\vec B+\vec C=\vec 0$). Note that by Cauchy-Schwarz, \n\t\t\t\[\|\vec A\| \|\vec A-\vec P \|+\|\vec B\|\|\vec B-\vec P \|+\|\vec C\| \|\vec C-\vec P \|\] \[\ge \vec A\cdot (\vec A-\vec P) +\vec B\cdot (\vec B-\vec P)+\vec C\cdot (\vec C-\vec P)\]\n\[=\|\vec A\|^2+\|\vec B\|^2+\|\vec C\|^2-\vec P\cdot (\vec A+\vec B+\vec C)=\|\vec A\|^2+\|\vec B\|^2+\|\vec C\|^2,\]\n\t\t\tand this bound is clearly reached by $\vec P=\vec 0$. Furthermore, equality is only reached when $\vec A$, $\vec B$, $\vec C$ are scalar multiples of $\vec A-\vec P$, $\vec B-\vec P$, $\vec C-\vec P$, respectively. This means that $\vec P$ is a scalar multiple of $\vec A$, $\vec B$, and $\vec C$, so $\vec P=\vec 0$. (Note that $\vec A$ and $\vec B$ are linearly independent, since the centroid is not on $AB$.)\n \t\t\tNow all that remains is to calculate $\|\vec A\|^2+\|\vec B\|^2+\|\vec C\|^2=AG^2+BG^2+CG^2$. To calculate $AG$, first let $D$ be the midpoint of $BC$. Then by Stewart's theorem,\n\t\t\t\[AD^2\cdot a+\frac{a^3}{4}=\frac{b^2a}{2}+\frac{c^2a}{2}\]\n\[AD^2=\frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4}=\frac{2b^2+2c^2-a^2}{4}.\]\n \t\t\tFurthermore, $AG=2/3\cdot AD$, so \n\t\t\t\[AG^2=\frac{2b^2+2c^2-a^2}{9}.\]\n\t\t\tBy similar reasoning, we can calculate $BG^2=(2c^2+2a^2-b^2)/9$ and $CG^2=(2a^2+2b^2-c^2)/9$, so\n\t\t\t\[AG^2+BG^2+CG^2=\frac{a^2+b^2+c^2}{3}.\]\n $\blacksquare$\n |
Let $M$ be a point in the interior of triangle $ABC$. Let $A'$ lie on $BC$ with $MA'$ perpendicular to $BC$. Define $B'$ on $CA$ and $C'$ on $AB$ similarly. Define\n $p(M) = \frac {MA' \cdot MB' \cdot MC'}{MA \cdot MB \cdot MC}.$ Determine, with proof, the location of $M$ such that $p(M)$ is maximal. Let $\mu(ABC)$ denote this maximum value. For which triangles $ABC$ is the value of $\mu(ABC)$ maximal?\n | This problem needs a solution. If you have a solution for it, please help us out by adding it.\n |
Let $ABC$ be an acute triangle. Let $DAC,EAB$, and $FBC$ be isosceles triangles exterior to $ABC$, with $DA = DC, EA = EB$, and $FB = FC$, such that\n $\angle ADC = 2\angle BAC, \quad \angle BEA = 2 \angle ABC, \quad \angle CFB = 2 \angle ACB.$ Let $D'$ be the intersection of lines $DB$ and $EF$, let $E'$ be the intersection of $EC$ and $DF$, and let $F'$ be the intersection of $FA$ and $DE$. Find, with proof, the value of the sum\n $\frac {DB}{DD'} + \frac {EC}{EE'} + \frac {FA}{FF'}.$ | This problem needs a solution. If you have a solution for it, please help us out by adding it.\n |
Let $ABC$ be a triangle and $P$ an exterior point in the plane of the triangle. Suppose the lines $AP$, $BP$, $CP$ meet the sides $BC$, $CA$, $AB$ (or extensions thereof) in $D$, $E$, $F$, respectively. Suppose further that the areas of triangles $PBD$, $PCE$, $PAF$ are all equal. Prove that each of these areas is equal to the area of triangle $ABC$ itself.\n | Solution 1 by Mewto55555:\n We use barycentric coordinates.\n So $A$ is $(1,0,0)$, $B$ is $(0,1,0)$, $C$ is $(0,0,1)$, and $P$ is $(p,q,r)$, with $p+q+r=1$.\n Now, the equation of line $AP$ is just the line $qz=ry$, $BP$ is just $pz=rx$, and $CP$ is $qx=py$.\n Also, $AB$ is just $z=0$, $BC$ is $x=0$, and $AC$ is $y=0$.\n Thus, the coordinates of $D$ is $\left(0,\frac{q}{q+r},\frac{r}{q+r}\right)$. Similarly, $E$ is at $\left(\frac{p}{p+r},0,\frac{r}{p+r}\right)$ and $F$ is at $\left(0,\frac{q}{q+r},\frac{r}{q+r}\right)$\n Now, the ratio $[PBD]$ to $[ABC]$ is just\n $\begin{vmatrix} p & 0 & 0 \\ q & 1 & \frac{q}{q+r}\\ r & 0 & \frac{r}{q+r} \end{vmatrix}= \frac{pr}{q+r}$\n The other ratios are similarly $\frac{pq}{p+r}$ and $\frac{qr}{p+q}$\n Since $p+q+r=1$, we have $\frac{qr}{1-r}=\frac{pq}{1-q}=\frac{pr}{1-p}=K$ and we want to show that $|K|=1$.\n Thus, we have $\frac{pqr}{p(1-r)}=\frac{pqr}{r(1-q)}=\frac{pqr}{q(1-p)}$.\n Since none of $p,q,r=0$ (else $P$ would be on one of the sides of $ABC$):\n $p(1-r)=r(1-q)=q(1-p)$.\n We know $r=1-p-q$. Substuting:\n $p^2+pq=1-p-2q+pq+q^2=q-pq$.\n From the first and third, we get that $q(1-2p)=p^2 \implies q=\frac{p^2}{1-2p}$\n Now consider first and second;\n $p^2+p-1=q^2-2q$\n Subbing back in $q$:\n $(p^2+p-1)(1-2p)^2=p^4-2p^2(1-2p)$\n which rearranges to\n $0=3p^4-4p^3-5p^2+5p-1=(3p-1)(p^3-p^2-2p+1)=0$\n If $p=\frac{1}{3}$, then $q=r=\frac{1}{3}$, so $P$ is in the triangle (as all of $p,q,r>0$) contradiction.\n Thus, we have $p^3-p^2-2p+1=0$\n So, $1-2p=p^2(1-p) \implies q=\frac{p^2}{1-2p}=\frac{1}{1-p}$\n Thus, $K=\frac{pq}{1-q}=\frac{\frac{p}{1-p}}{1-\frac{1}{1-p}}=\frac{p}{1-p-1}=-1$\n Therefore, if $[PBD]=[PCE]=[PAF]$, necessarily $[PBD]=[PCE]=[PAF]=[ABC]$.\n |
Let $O$ be an interior point of acute triangle $ABC$. Let $A_1$ lie on $BC$ with $OA_1$ perpendicular to $BC$. Define $B_1$ on $CA$ and $C_1$ on $AB$ similarly. Prove that $O$ is the circumcenter of $ABC$ if and only if the perimeter of $A_1B_1C_1$ is not less than any one of the perimeters of $AB_1C_1, BC_1A_1$, and $CA_1B_1$.\n | This problem needs a solution. If you have a solution for it, please help us out by adding it.\n |
Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect $\angle ABC$, with $P$ on $BC$ and $Q$ on $AC$. If $AB + BP = AQ + QB$, what are the angles of the triangle?\n | |
Prove that there is no positive integer $n$ such that, for $k = 1,2,\ldots,9$, the leftmost digit (in decimal notation) of $(n + k)!$ equals $k$.\n | Suppose that there is such a number $n$. Let $a$ be the number of digits of $(n + 8)!$, and let $b$ be the number of digits of $n + 9$. Then we have:\n $8 \cdot 10^{a-1} \leqslant (n + 8)! < 9 \cdot 10^{a-1}$\n and\n $10^{b-1} \leqslant n + 9 < 10^b$\n Combining these inequalities via multiplication we get:\n $8 \cdot 10^{a+b-2} \leqslant (n+9)! < 9 \cdot 10^{a+b-1}$\n From this inequality, it can be seen that $(n+9)!$ has at least $a+b-1$ and at most $a+b$ digits. However, if it has $a+b$ digits then the first digit is less than $9$, which contradicts with how $n$ is defined. Thus, $(n+9)!$ must have $a+b-1$ digits. Expressing this as an inequality, we get:\n $9 \cdot 10^{a+b-2} \leqslant (n+9)! < 10^{a+b-1}$\n Combining these values with the earlier bounds for $(n+8)!$ via division, we get:\n $10^{b-1} < n+9 < \frac{5}4 10^{b - 1}$\n Suppose that $n+1 < 10^{b-1}$. Then $10^{b-1}$ is between $n+1$ and $n+9$. Thus, one of the values $n+2,n+2,\ldots n+8$ must be $10^{b-1}$. Call that value $n+l (1 < l < 9)$. This means that the digits of $(n+l)!$ are just the digits of $(n+l-1)!$, followed by $b-1$ zeroes. Thus the first digit of the two numbers are equal, which contradicts with how $n$ is defined.\nThis means that:\n $10^{b-1} \leqslant n+1 < n+2 < n+3 < n+4 < n+9 < \frac{5}4 10^{b - 1}$\n Let $c$ be the number of digits of $(n+1)!$. Then we have:\n $10^{c-1} \leqslant (n+1)! < 2 \cdot 10^{c-1}$\n Combining these with the bounds for $n+2$, $n+3$ and $n+4$ via multiplication we get:\n $10^{c+3b-4} \leqslant (n+4)! < \frac{125}{32}10^{c+3b-4} < 4 \cdot 10^{c+3b-4}$\n So $(n + 4)!$ has $c+3b-3$ digits, but is smaller than the smallest $c+3b-3$ digit number with the first digit 4, which means that its first digit can't be 4. This contradicts with how $n$ is defined, thus no such value of $n$ exists.\n ~Circling\n |
Consider the system $x + y = z + u,$ $2xy = zu.$ Find the greatest value of the real constant $m$ such that $m \leq x/y$ for any positive integer solution $(x,y,z,u)$ of the system, with $x \geq y$.\n | First consider the real solutions to the system. We have by AM-GM that $\frac{z+u}{2}\ge\sqrt{zu}$ and substituting we get $\frac{x+y}{2}\ge\sqrt{2xy}$. Squaring and simplifying and dividing by $y^2$, we get the inequality $r^2-6r+1\ge0$, where $r=\frac{x}{y}$. Then $r^2-6r+9\ge8$, so $r\ge3+2\sqrt2$ or $r\le3-2\sqrt2$. Since $r\ge1$, we discard the second inequality and have that $3+2\sqrt2$ is a lower bound for $r$\n This bound is also attainable for real values when $z=u$. Since $\mathbb{Q+}$ is dense, it is always possible to assign rational values to $x, y, z,$ and $w$ so that $r$ approaches $3+2\sqrt2$, though equality is never reached. From any rational solution, it is possible to create an integer solution by multiplying by the least common multiple of the denominators and keep the same value of $r$. Thus, $\boxed{m=3+2\sqrt2}$.\n |
Let $a_1 = 11^{11}, \, a_2 = 12^{12}, \, a_3 = 13^{13}$, and $a_n = |a_{n - 1} - a_{n - 2}| + |a_{n - 2} - a_{n - 3}|, n \geq 4.$ Determine $a_{14^{14}}$.\n | This problem needs a solution. If you have a solution for it, please help us out by adding it.\n |
Let $a > b > c > d$ be positive integers and suppose that\n $ac + bd = (b + d + a - c)(b + d - a + c).$ Prove that $ab + cd$ is not prime.\n | Equality is equivalent to \n$a^2 - ac + c^2 = b^2 + bd + d^2 (1)$.\n Let $ABCD$ be the quadrilateral with $AB = a$, $BC = d$, $CD = b$, $AD = c$, $\angle BAD = 60^\circ$, and $\angle BCD = 120^\circ$. Such a quadrilateral exists by $(1)$ and the Law of Cosines.\n By Strong Form of Ptolemy's Theorem, we find that;\n $BD^2 = \frac{(ab+cd)(ad+bc)}{ac+bd}$\n and by rearrangement inequality;\n $ab+cd > ac+bd > ad+bc$.\n Assume $ab+cd = p$ is a prime, since $a^2 - ac + c^2 = BD^2$ is an integer $p \times \frac{ad+bc}{ac+bd}$ must be an integer but this is false since $(p,ac+bd) = 1$ and $ac+bd > ad+bc$. Thus $ab+cd$ can not be a prime.\n |
Is it possible to find 100 positive integers not exceeding 25,000, such that all pairwise sums of them are different?\n | The answer is yes.\n First Solution For example, let's say that the integers are, $k, k+a_1, k+a_2, ..., k+a_{99}$. Now this turns into a problem of solving for the $99$ integers $a_i$. This then each ai takes on the form, $j+b_1, j+b_2,..., j+b_{98}$. Then we must find the $98$ $b$ integers. By doing this process over and over again, we obtain the last $3$ numbers, $y, y+u_1, y+u_2$. Obviously these 3 integers can have different sums, and the number of different "parts" in every sequence (the number of terms that are different for $a_i, b_i, c_i$, etc.) is $99+98+\ldots+2$, not exceeding $25000$.\n Second Solution Lemma: We try to show that for a prime $p$, there are such $p$ integers less than or equal to $2p^2$. Then it suffices because $p=101>100$ and $2\cdot 101^2<25000$ is enough.\n Proof: $\qquad$Define $\bar a$ to be $a^2\equiv \bar a\bmod p$ and $\hat i=2pi+\bar i$. Then, we show that the integers $\hat i$ does the trick for $i=1$ to $p$. If $\hat a+\hat b\equiv \hat c+\hat d\pmod p$, then $2(a+b)p+\bar a+\bar b\equiv 2(c+d)p+\bar c+\bar d\pmod p$. Thus, we may say that \[a+b\equiv c+d\pmod p\qquad (\dag)\] and, \[\bar a+\bar b\equiv \bar c+\bar d\pmod p\qquad(*)\] since $\bar a<p$. From $(\dag)$, we have $a\equiv c+d-b\pmod p$. From $(*)$, we infer that \[a^2+b^2\equiv c^2+d^2\pmod p\] which gives us $(c+d-b)^2+b^2\equiv c^2+d^2\pmod p$. Factorizing, \[2(b^2-bc-bd+cd)\equiv0\pmod p\]\n\[\implies(b-c)(b-d)\equiv0\pmod p\]\n This yields $p|b-c$ or $p|b-d$. But $|b-c|<p\implies b=c$, and this implies that $a=d$. The other case is similar. So, we are done. $\blacksquare$\n Therefore, every sum must be distinct, as desired.\n |
\nWhat is the smallest positive integer $t$ such that there exist integers $x_1,x_2,\ldots,x_t$ with $x^3_1+x^3_2+\,\ldots\,+x^3_t=2002^{2002}$? | Observe that $2002^{2002}\equiv 4^{2002}\equiv 64^{667}\cdot 4\equiv 4\pmod{9}$. On the other hand, each cube is congruent to 0, 1, or -1 modulo 9. So a sum of at most three cubes modulo 9 must among $0,\pm 1,\pm 2,\pm 3$ none of which are congruent to 4. Therefore $t\geq 4$.\n \nTo show that 4 is the minimum value of $t$, note that\n$(10\cdot 2002^{667})^3+(10\cdot 2002^{667})^3+(2002^{667})^3+(2002^{667})^3=2002^{2002}$\n |
Circles $\displaystyle S_1$ and $\displaystyle S_2$ intersect at points $\displaystyle P$ and $\displaystyle Q$. Distinct points $\displaystyle A_1$ and $\displaystyle B_1$ (not at $\displaystyle P$ or $\displaystyle Q$) are selected on $\displaystyle S_1$. The lines $\displaystyle A_1P$ and $\displaystyle B_1P$ meet $\displaystyle S_2$ again at $\displaystyle A_2$ and $\displaystyle B_2$ respectively, and the lines $\displaystyle A_1B_1$ and $\displaystyle A_2B_2$ meet at $\displaystyle{} C$. Prove that, as $\displaystyle A_1$ and $\displaystyle B_1$ vary, the circumcenters of triangles $\displaystyle A_1A_2C$ all lie on one fixed circle.\n | We will use directed angles mod $\displaystyle \pi$.\n Since $\displaystyle A_1, B_1, C$ are collinear, $\angle CA_1Q = \angle B_1A_1Q$. Since $\displaystyle A_1, B_1, P, Q$ all lie on $\displaystyle S_1$, $\angle B_1A_1Q = \angle B_1PQ$. Hence, $\angle CA_1Q = \angle B_1PQ$. Similarly, $\angle CA_2Q = \angle B_2PQ$. But since $\displaystyle B_1, B_2, P$ are collinear, $\angle B_1PQ = \angle B_2PQ$. This means that $\angle CA_1Q = \angle CA_2Q$, so $\displaystyle C, A_1, A_2, Q$ are concyclic. This means that, regardless of the location of $\displaystyle B_1$, the circumcenter of $\displaystyle A_1A_2C$ is the circumcenter of $\displaystyle A_1A_2Q$.\n Note that as $\displaystyle A_1$ varies, the values of $\angle QA_1A_2 = \angle QA_1P$ and $\angle QA_2A_1 = \angle QA_2P$ stay fixed, at half the measure of arc $\displaystyle QP$ on circles $\displaystyle S_1$ and $\displaystyle S_2$, respectively. Therefore all triangles $\displaystyle QA_1A_2$, are similar. If $\displaystyle X$ denotes the circumcenter of triangle $\displaystyle QA_1A_2$, then we must also have all triangles $\displaystyle QA_1X$ are similar. Since $\displaystyle Q$ is fixed, this means that there exists a spiral similarity that maps every point $\displaystyle A_1$ to its corresponding point $\displaystyle X$. This means that the locus of $\displaystyle X$ must be the image of the locus of $\displaystyle A_1$ under the spiral similarity. But the locus of $\displaystyle A_1$ is a circle, and the image of a circle under a spiral similarity is another circle. Q.E.D.\n |
Let $\displaystyle n$ be a positive integer. A sequence of $\displaystyle n$ positive integers (not necessarily distinct) is called full if it satisfies the following conditions: for each positive integers $\displaystyle k \ge 2$, if the number $\displaystyle k$ appears in the sequence then so does the number $\displaystyle k-1$, and moreover the first occurrance of $\displaystyle k-1$ comes before the last occurrance of $\displaystyle k$. For each $\displaystyle n$, how many full sequences are there?\n | We claim that there is a bijection between the permutations of the numbers $\displaystyle 1, \ldots , n$ and the full sequences of length $\displaystyle n$.\n To obtain a full sequence $\displaystyle a_1, \ldots a_n$ from a permutation $\displaystyle \sigma$ of the first $\displaystyle n$ positive integers, begin by letting $\displaystyle a_{\sigma(1)}$ be 1. Then, if $\displaystyle a_{\sigma(j)}$ is $\displaystyle i$ and $\displaystyle \sigma(j+1) < \sigma(j)$, let $\displaystyle a_{\sigma(j+1)}$ be $\displaystyle i$, but if $\displaystyle \sigma(j+1) > \sigma(j)$, let $\displaystyle a_{\sigma(j+1)}$ be $\displaystyle i+1$. This ensures that if $\displaystyle k > 2$ appears in the sequence, then the first appearance of $\displaystyle k-1$ occurs before the last appearance of $\displaystyle k$, as desired.\n To obtain a permutation from a full sequence, list the labels of the terms of the sequence that are 1 in decreasing order, followed by the labels of the terms of the sequence that are 2 in decreasing order, etc. This produces a permutation which will map back to our full sequence as described above. Thus the bijection is established; therefore there are $\displaystyle n!$ full sequences of length $\displaystyle n$, Q.E.D.\n Note: There is also a recursive solution, but it is very similar to the bijective solution given here.\n |
(Bulgaria)\nLet $\displaystyle T$ be the set of ordered triples $\displaystyle (x,y,z)$, where $\displaystyle x$, $\displaystyle y$, $\displaystyle z$ are integers with $0 \le x,y,z \le 9$. Players $\displaystyle A$ and $\displaystyle B$ play the following game. Player $\displaystyle A$ chooses a triple $\displaystyle (x,y,z)$ in $\displaystyle T$, and Player $\displaystyle B$ has to discover $\displaystyle A$'s triple in as few moves as possible. A move consists of the following: $\displaystyle B$ gives $\displaystyle A$ a triple $\displaystyle (a,b,c)$ in $\displaystyle T$, and $\displaystyle A$ replies by giving $\displaystyle B$ the number $\displaystyle |x+y -a-b| + |y+z -b-c| + |z+x -c-a|$. find the minimum number of moves that $\displaystyle B$ needs to be sure of determining $\displaystyle A$'s triple.\n | In mod 2, we see that\n \n$\displaystyle |x+y -a-b| + |y+z -b-c| + |z+x -c-a| \equiv 2(x+y+z -a-b-c)$,\n\n so the outcome of $\displaystyle B$'s move must always be even. Furthermore, the outcome must be no greater than 54 and no less than 0, so there are at most 28 different possible outcomes per move. Since there are $\displaystyle 10^3$ possible triples $\displaystyle (x,y,z)$ and at most $\displaystyle 28^2 < 10^3$ possible outcomes after two moves, at least three moves are required.\n We will now show how to determine $\displaystyle (x,y,z)$ in three moves. A first move of $\displaystyle (0,0,0)$ will give us $\displaystyle 2(x+y+z)$. We shall denote $\displaystyle x+y+z$ as $\displaystyle s$.\n If $s \le 9$, then the moves $\displaystyle (9,0,0)$ and $\displaystyle (0,9,0)$ will give us $\displaystyle 18 -2x$ and $\displaystyle 18 - 2y$, respectively, enabling us to determine $\displaystyle (x,y,z)$. Similarly, if $s \ge 18$, the moves $\displaystyle (0,9,9)$ and $\displaystyle (9,0,9)$ will give us $\displaystyle 2x$ and $\displaystyle 2y$.\n If $\displaystyle 9 < s < 18$, then our second move is $\displaystyle (9, s-9, 0)$. Let us call the result $\displaystyle 2k$. We have two cases. In Case I, $\displaystyle y > s-9$, which gives us $\displaystyle x = 9-k$, $\displaystyle z<k$. In Case II, $y \le s-9$, so $\displaystyle x \ge 9-k$ and $\displaystyle z=k$. In either case, we have $\displaystyle z \le k \le y+z$ (the right-hand side comes from $\displaystyle y+z = s-9 +k$ or $z=k , y \ge 0$) and $x+z \le 9 + k$.\n Now, if $s-k \le 9$, then our third move is $\displaystyle (s-k, 0, k)$. This gives us\n \n$\displaystyle |x+y - s+k| + |y+z -k| + |z+x - s| = k-z + y+z-k + y = 2y$,\n\n which gives us $\displaystyle y$ and tells us whether Case I or Case II holds, letting us determine $\displaystyle (x,y,z)$.\n On the other hand, if $\displaystyle s-k > 9$, our third move is $\displaystyle (9, s-k-9, k)$. This gives us\n \n$\displaystyle |x+y -s+k| + |y+z -s+9| + |z+x -k-9| = |k-z| + |9-x| + |k+9 -z-x| = 2(k+9-s+y)$,\n\n which again gives us $\displaystyle y$, telling us which of Cases I and II hold, letting us determine the triple $\displaystyle (x,y,z)$.\n |
(Hojoo Lee) Let $n \geq 3$ be an integer. Let $t_1, t_2, \dots , t_n$ be positive real numbers such that \n \[n^2 + 1 > \left( t_1 + t_2 + ... + t_n \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_n} \right).\]\n Show that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with $1 \leq i < j < k \leq n$.\n | For $n=3$, suppose (for sake of contradiction) that $t_3 = t_2 + t_1 + k$ for $k \ge 0$; then (by Cauchy-Schwarz Inequality)\n \begin{align*}10 &> [2(t_1 + t_2) + k]\left(\frac {1}{t_1} + \frac {1}{t_2} + \frac 1{t_1 + t_2 + k}\right) = 2(t_1+t_2)\left(\frac 1{t_1} + \frac{1}{t_2}\right) + \left(\frac{k}{t_1} + \frac k{t_2}\right) + \frac{2t_1 + 2t_2 + k}{t_1 + t_2 + k}\\ &\ge 8 + \left(\frac{k}{t_1} + \frac k{t_2}\right) + 2 - \frac{k}{t_1 + t_2 + k}\\ &= 10 + k\left(\frac{1}{t_1} + \frac {1}{t_2} - \frac{1}{t_1 + t_2+k}\right) \ge 10\end{align*}\n so it is true for $n=3$. We now claim the result by induction; for $n \ge 4$, we have \n \begin{align*}f(n) &:= \left( t_1 + t_2 + ... + t_n \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_n} \right) \\ &= \left( t_1 + t_2 + ... + t_{n-1} \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_{n-1}} \right) + t_n \sum_{i=1}^{n-1} \frac 1{t_i} + \frac{1}{t_n} \sum_{i=1}^{n-1} t_i + 1\end{align*}\n By AM-GM, $\frac{t_n}{t_i} + \frac{t_i}{t_n} \ge 2$, so $f(n) \ge f(n-1) + 2(n-1) + 1 = f(n-1) + 2n - 1$. Then the problem is reduced to proving the statement true for $n-1$ numbers, as desired.$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$\n See also <url>viewtopic.php?p=99756#99756 AoPS/MathLinks discussion</url> |
(Mihai Bălună, Romania)\nAn infinite sequence $a_0, a_1, a_2, \ldots$ of real numbers satisfies the condition\n \n$a_n = | a_{n+1} - a_{n+2} | \qquad$ for every $n \ge 0$,\n\n with $a_0$ and $a_1$ positive and distinct. Can this sequence be bounded?\n This was also Problem 4 of the 2005 German Pre-TST and Problem 1 of the first 2005 black MOP test. It was Problem 3, Day 2 of the 2005 Romanian TST in the following form:\n A sequence of real numbers $\{a_n\}_n$ is called a bs sequence if $a_n = |a_{n+1} - a_{n+2}|$, for all $n\geq 0$. Prove that a bs sequence is bounded if and only if the function $f$ given by $f(n,k)=a_na_k(a_n-a_k)$, for all $n,k\geq 0$ is the null function.\n Contents\n\n1 Problem\n2 Solution\n\n2.1 Solution 1\n2.2 Solution 2\n\n\n3 See also\n\n | Solution 1 We note that each of the $a_i$ must be nonnegative.\n Lemma 1. If the two initial terms of the sequence are nonzero and distinct, then every term of the sequence is nonzero and no two consecutive terms are equal.\n Proof. We proceed by induction; we are given a base case. If ${}a_k \neq a_{k+1}$, then $| a_{k+1} - a_{k+2} | \neq a_{k+1}$, so $a_{k+2} \neq 0$. Furthermore, since $|a_{k+1} - a_{k+2}| = a_k \neq 0$, $a_{k+1} \neq a_{k+2}$.\n Consider a sequence $\{b_i\}_{i=0}^{\infty}$ of positive reals which obeys the recursive relation\n $b_{n+2} = |b_{n+1} - b_n |$. Lemma 2. $\min (b_i, b_{i+1}, b_{i+2}) \ge \min (b_{i+3}, b_{i+4}, b_{i+5})$.\n Proof. We note that if $b_k \le b_{k+1}$, then $b_{k+2} = b_{k+1} - b_k$ and $b_{k+3} = b_k$. Thus if $b_{i+3}, b_{i+4}, b_{i+5} > \min(b_i, b_{i+1}, b_{i+2})$, we must have $b_i, b_{i+1} > b_{i+2} > b_{i+3}$, a contradiction.\n Lemma 3. Let $r = \left\lfloor \frac{b_{k-1}}{b_k} \right\rfloor$, $j = \left\lceil \frac{3}{2}r \right\rceil$. Then $\min (b_{k+j},b_{k+j+1},b_{k+j+2}) \le \frac{1}{2}b_{k+1}$.\n Proof. For $2i+2 \le r$, easy induction yields $b_{k+3i} = b_k$, $b_{k+3i+1} = b_{k-1} - (2i+1)b_k$, $b_{k+3i+2} = b_{k-1} - (2i+2)b_k$.\n Now, if $r$ is even, we have $b_{k+3r/2 - 1} = b_{k-1} - b_kr$, which is less than $b_k$ by the definition of $r$, and $b_{k+3r/2} = b_k$, $b_{k+3r/2 +1} = b_{3r/2} - b_{k+3r/2-1}$, $b_{k+3r/2+2} = b_{k+3r/2}$. Since $b_{k+3r/2+1}$ and $b_{k+3r/2+2}$ add to $b_{3r/2} = b_k$, one of them must be at most $\frac{b_k}{2}$, and the lemma follows.\n On the other hand, if $r$ is odd, then set $s = r-1$ and we have $b_{k+3s/2+1} = b_{k-1} - b_kr < b_k$, $b_{k+3s/2 + 2} = b_k$, and $b_{k+3s/2+3} = b_{k+3s/2 + 2} - b_{k+3s/2+1}$, so as before, at least one of $b_{k+3s/2+1}$, $b_{k+3s/2+3}$ must be at most $\frac{b_k}{2}$. In fact, we have ${} b_{k+3s/2+4}= b_{k+3s/2+1}$, so the minimum of $b_{k+3s/2+3}, b_{k+3s/2+4}$ is at most $\frac{b_k}{2}$, upholding the lemma, since in this case $\left\lceil \frac{3}{2}r \right\rceil = \frac{3}{2}s+2$.\n \nNow, suppose that $\{ a_i \}_{i=0}^{\infty}$ is bounded, i.e., there exists some real $N$ greater than each $a_i$. By setting $b_1, b_0$ equal to $a_{n-1}, a_n$, we generate the first $n$ of the $a_i$ in reverse, and from Lemma 2, we can see that $m = \min (a_1,a_2,a_3)$ is a lower bound of the $a_i$. But by making $n$ greater than $\lceil \log_2 (N/m) +1 \rceil \left\lceil \frac{3}{2} \left\lfloor \frac{N}{m} \right\rfloor \right\rceil$, by applying Lemma 3, we obtain the result $m \le \frac{1}{2} m$, which is a contradiction when $a_0, a_1$ are distinct and greater than 0, by Lemma 1.\n Now, if $a_na_k(a_n-a_k) = 0$ for all natural $n$, all the nonzero $a_i$ must be equal and the sequence is bounded. On the other hand, if $a_{n+1}a_n(a_{n+1}-a_n) = 0$ always holds, then either $a_{n+1} = a_n$ and $a_{n+2} = 0$, or one of $a_{n+1}, a_{n}$ is equal to zero and $a_{n+2}$ is equal to the other one; hence by induction, all the nonzero terms of the sequence are equal, and $a_na_k(a_n-a_k)$ is always equal to zero. Hence if for some $n, k$, ${} a_na_k(a_n-a_k) \neq 0$, then there exist two distinct positive consecutive terms of sequence, which is then unbounded as proven above.\n Solution 2 The given recursive condition is equivalent to $a_{n+2} = a_{n+1} + a_n$ if $a_{n+1} < a_n$, and $a_{n+2} = a_{n+1} \pm a_n$ otherwise. Note that if $a_{n+1} > a_n$, then $a_{n+1} = a_n + a_{n-1}$. \n Lemma 1: Let $m > 0$ be the minimal element of $\{a_n\}$. Then $m = \text{min}\,(a_0,a_1,a_2)$.\n Proof: Suppose $k \ge 3$ is the minimal value of $k$ such that $m = a_k$. Then $a_k = a_{k-1} - a_{k-2} \Longrightarrow a_{k-1} > a_{k-2}$, so $a_{k-1} = a_{k-2} + a_{k-3}$, and $a_k = a_{k-3}$. This contradicts our assumption of minimality. \n Let $T_n = a_n + a_{n+1} + a_{n+2}$; we claim that $T_{n+2} > T_n + m$, which would imply that $T_n$ is unbounded, in turn implying that $a_n$ is unbounded.\n |
(Canada)\nDoes there exist a function $s : \mathbf{Q} \rightarrow \{ -1,1 \}$ such that if $\displaystyle x$ and $\displaystyle y$ are distinct rational numbers satisfying $\displaystyle {} xy=1$ or $x+y \in \{ 0,1 \}$, then $\displaystyle s(x)s(y) = -1$? Justify your answer.\n | For a number $\displaystyle x$, we define the function $\displaystyle t(x)$ to be 0, if $\displaystyle x$ is an integer, or $\frac{1}{x-\lfloor x \rfloor}$ if $\displaystyle x$ is not an integer. We use the notation $\displaystyle t^n$ to denote $\displaystyle t$ composed $\displaystyle n$ times. We note that if $\displaystyle x = 0$, then $\displaystyle t^n(x) = 0$, for all nonnegative integers $\displaystyle n$. We also note that if $\displaystyle x,y$ differ by an integer, then $\displaystyle t(x) = t(y)$.\n We claim that for any nonnegative rational $\displaystyle x$, there exists a nonnegative number $\displaystyle n$ such that $\displaystyle t^n(x) = 0$. Let $\displaystyle p_n, q_n$ be the relatively prime positive integers such that $t^n(x) = \frac{p_n}{q_n}$. It is enough to show that if $\displaystyle t^n(x)$ is not an integer, then $\displaystyle q_{n+1} < q_n$, for then we must eventually have $\displaystyle q_n = 1$ and $\displaystyle t^{n+1}(x) = 0$. But this follows from $t^{n+1}(x) = \frac{1}{t^n(x) - \lfloor t^n(x) \rfloor} = \frac{q_n}{p_n - q_n \lfloor p_n/q_n \rfloor}$, since by definition, $p_n - q_n \lfloor p_n/q_n \rfloor < q_n$. Thus for nonnegative rational $\displaystyle x$, there exists some nonnegative integer $\displaystyle n$ such that $\displaystyle t^n(x) =0$. Then for nonnegative rational $\displaystyle x$, let $\displaystyle f(x) : \mathbb{Q} \rightarrow \mathbb{Z}/2\mathbb{Z}$ denote the parity class of the least such integer $\displaystyle n$. For negative rational $\displaystyle x$, let $\displaystyle f(x) = f(-x)+1$. We claim that $\displaystyle s(x) = (-1)^{f(x)}$ satisfies the desired conditions.\n First, we prove that if $\displaystyle x,y$ are distinct rationals such that $\displaystyle x+y =0$, then $\displaystyle s(x)s(y) = -1$. But this follows from $f(-x) \equiv f(x) + 1$.\n Next, we prove that if $\displaystyle x,y$ are distinct rationals such that $\displaystyle x+y = 1$, then $\displaystyle s(x)s(y) = -1$. Without loss of generality, let $\displaystyle x> y$. If $\displaystyle x$ and $\displaystyle y$ are integers, then $\displaystyle x$ must be at least 1, and $\displaystyle y$ must be at most 0, so $\displaystyle s(x) = -1$ and $\displaystyle s(y)=1$. If $\displaystyle x$ is a non-integer greater than 1, then $\displaystyle t(x) = t(x-1) = t(-y)$, so $\displaystyle s(x) = s(-y) = -s(y)$, as desired.\n On the other hand, if $\displaystyle x$ is a positive rational less than one, then $\displaystyle y$ must be positive, and there exist relatively prime positive integers $\displaystyle p, q$ such that $\displaystyle 2p < q$ and $\displaystyle y = \frac{p}{q} , x= \frac{q-p}{q}$. We note that $\displaystyle t(x) = \frac{q}{q-p}$; and since $\displaystyle \frac{q}{q-p} = 1 + \frac{p}{q-p} < 1+ \frac{p}{2p-p} = 2$, $\displaystyle t^2(x) = \frac{q-p}{p}$. On the other hand, $\displaystyle t(y) = \frac{q}{p} = \frac{q-p}{p} + 1 = t^2(x) +1$, so $\displaystyle t^2(y) = t^3(x)$. It follows that $\displaystyle {} f(x) = f(y) +1$, so $\displaystyle s(x)s(y) = -1$, as desired.\n Finally, we prove that for nonzero rational $\displaystyle x$, $\displaystyle s(x)s(1/x) = -1$, for $\displaystyle x \neq 1/x$. We may clearly assume $\displaystyle |x| > 1$ without loss of generality. We may also assume $\displaystyle x , 1/x > 0$, since in general, $\displaystyle s(x)s(y) = [-s(-x)][-s(-y)] = s(-x)s(-y)$. But we now have $\displaystyle t(x) = 1/x$, so $\displaystyle f(x) = f(1/x) +1$, and $\displaystyle s(x)s(1/x) = -1$. Q.E.D.\n \nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.\n Comment. The solution's construction may seem arbitrary, but it follows from the observation that if such a function $\displaystyle s$ exists, then for nonzero $\displaystyle x$, $\displaystyle s(x) = -s(-x)$, and indeed we must have $\displaystyle s(-x) = -s(x+1)$, so $\displaystyle s(x) = s(x+1)$. Hence for positive $\displaystyle x$, $\displaystyle s(x)$ must be distinct from $\frac{1}{x - k}$, where $\displaystyle k$ is any integer less than $\displaystyle x$: in particular, for non-integers $\displaystyle x$, when $k = \lfloor x \rfloor$. The functions $\displaystyle t,f$ encode this condition; the solution constitutes a proof that this set of constraints is consistent with itself. In fact, from our observations, we can see that our function $\displaystyle s$ is unique up to sign.\n |
Find all polynomials $f$ with real coefficients such that for all reals $a,b,c$ such that $ab + bc + ca = 0$ we have the following relations\n \[f(a - b) + f(b - c) + f(c - a) = 2f(a + b + c).\]\n Contents\n\n1 Problem\n2 Solution\n\n2.1 Solution 1\n2.2 Solution 2\n\n\n3 See also\n\n | Solution 1 From $b=c=0$, we have $f(a) + f(-a) = 2f(a) \Longrightarrow f(a) = f(-a)$, so $f$ is even, and all the degrees all of its terms are even. Let $\text{deg}\, f(x) = n$\n Let $(a,b,c) = (6x, 3x, -2x)$*; then we have $f(3x) + f(5x) + f(8x) = 2f(7x)$. Comparing lead coefficients, we have $3^n + 5^n + 8^n = 2 \cdot 7^n$, which cannot be true for $n \ge 5$. Hence, we have $f(x) = c_1 x^4 + c_2 x^2$. We can easily verify by expanding that all such polynomials work. \n The substitution arises from writing $a = \frac{-bc}{b+c}$. Solution 2 Let $a = (1 - \sqrt {3})x$, $b = x$, and $c = (1 + \sqrt {3})x$. Then it is easy to check that $ab + bc + ca = 0$, so\n \[P( - \sqrt {3}x) + P( - \sqrt {3}x) + P(2 \sqrt {3} x) = 2P(3x)\]\n for all $x$. Hence, for the coefficient of $x^n$ to be nonzero, we must have $2( - \sqrt {3})^n + (2 \sqrt {3})^n = 2 \cdot 3^n$.\n This does not hold for $n = 1$, and if $n$ is odd and $n \ge 3$, then the LHS is irrational and the RHS is a positive integer, so $n$ must be even.\n Let $n = 2m$. Then $2 \cdot 3^m + 12^m = 2 \cdot 3^{2m}$, so $2 + 4^m = 2 \cdot 3^m$. This holds for $m = 1$ and $m = 2$, and $(4/3)^3 = 64/27 > 2$, so $2 + 4^m > 4^m > 2 \cdot 3^m$ for $m \ge 3$. Therefore, $P(x)$ must be of the form $a_2 x^2 + a_4 x^4$.\n See also <url>viewtopic.php?p=99448#99448 AoPS/MathLinks discussion</url> |
(Puerto Rico)\nThere are 10001 students at an university (sic). Somer students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form societies (a club may belong to different societies. There are a total of $\displaystyle k$ societies. Suppose that the following conditions hold:\n(i) Each pair of students are in exactly one club. \n(ii) For each student and each society, the student is in exactly one club of the society. \n(iii) Each club has an odd number of students. In addition, a club with $\displaystyle 2m+1$ students ($\displaystyle m$ a positive integer) is in exactly $\displaystyle m$ societies.\n Find all possible values of $\displaystyle k$.\n | Solution 1 Let us replace 10001 with an arbitrary odd integer $\displaystyle n$. Suppose that some special student is a member of clubs $C_1, \ldots, C_j$, each with $2m_1 + 1, \ldots, 2m_j + 1$ members (including our special student). We note that for $1 \le i \le j$, no two $\displaystyle C_i$ may contain a common student other than our special student, or this common student and the special students would together be members of more than one society. Furthermore, no two of the $\displaystyle C_i$ may belong to a common society, or our special student would be a member of two clubs of the same society. Since our special student is a member of some club of every society, it follows that there $k= \sum_{i=1}^j m_i$ societies. But each non-special student must be in exactly one of the $\displaystyle C_i$, so $\sum_{i=1}^j 2m_i = n-1$. It follows that the only possible value for $\displaystyle k$ is $\frac{n-1}{2}$. To show that this is indeed a possible value of $\displaystyle k$ we present the trivial case of every student belonging to the same club, which is itself the only member of $\frac{n-1}{2}$ separate societies.\n \n Solution 2 Replacing the number 10001 with the variable $\displaystyle n$, we will count the number of ordered triples $\displaystyle (a,C,S)$, where $\displaystyle a$ is a student belonging to a club $\displaystyle {} C$, which belongs to a society $\displaystyle S$. We will denote such triples acceptable.\n Now, for any student $\displaystyle a$ and any society $\displaystyle S$, there is exactly one club which will form an acceptable triple. Thus the number of triples is $\displaystyle nk$.\n Consider any club $\displaystyle {} C$ with $\displaystyle |C|$ members. It is in $\frac{|C| -1}{2}$ societies, so $\displaystyle {} C$ can form $\frac{|C|(|C|-1}{2}$ acceptable triples. If $\mathcal{C}$ denotes the set of all clubs, then this implies that\n \n$nk = \sum_{C \in \mathcal{C}} \frac{|C|(|C|-1)}{2} = \sum_{C \in \mathcal{C}} {|C| \choose 2}$.\n\n But since any pair of students belong to exactly one club, it follows that ${n \choose 2} = \sum_{C \in \mathcal{C}} {|C| \choose 2}$, or $\frac{n(n-1)}{2} = nk$. Therefore $k = \frac{n-1}{2}$.\n \nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.\n |
(Germany)\nLet $\displaystyle n$ and $\displaystyle k$ be positive integers. There are given $\displaystyle n$ circles in the plane. Every two of them intersect at two distinct points, and all points of intersection they determine are pairwise distinct. Each intersection point must be colored with one of $\displaystyle n$ distinct colors so that each color is used at least once and exactly $\displaystyle k$ distinct colors occur on each circle. Find all values of $\displaystyle n \ge 2$ and $\displaystyle k$ for which such a coloring is possible.\n This was also Problem 6 of the 2005 German Pre-TST.\n | The answer is $\displaystyle k=2$ and $n \le 3$, or $3 \le k \le n$. We must have $\displaystyle k \le n$, since there are $\displaystyle n$ colors.\n If $\displaystyle k=1$, then all points on any circle must be the same color, so all points must be the same color and $\displaystyle n=1$, a contradiction.\n For $\displaystyle {} k=n=2$, we have two circles, and we color their intersection points distinct colors. For $\displaystyle k=2, n=3$, we have three circles. We associate with each distinct pair of circles a distinct color, which we use for both points common to the pair of circles.\n We will prove that the problem's situation is impossible for $\displaystyle k=2, n>3$. Suppose, on the contrary, that we have such a coloring. Then with each circle there are associated two distinct colors, and any two circles have at least one color in common. We claim that there is one color common to all circles. Suppose the contrary. Then there is a circle with colors 1,2. Since 1 is not common to all circles, there must be some circle with colors 2,3. But 2 is not common to all circles, so there must be some circle without the color 2, which must nonetheless have a color in common with both of the first two circles, so this third circle must have colors 1,3. But any more circles must have at least one color in common with each of our initial circles, i.e., they must have at least two of the colors 1,2,3. But since a circle only has two distinct colors, there are only 3 distinct colors in all the circles, so $\displaystyle n=3$, a contradiction. Thus there is a color 0 common to all circles. This means that no circle can hold more than 1 of the other colors. But since each color must occur on at least two circles, there are at most $\displaystyle n/2$ colors other than 0, a contradiction.\n We will now prove that there exist colorings as described in the problem for all $\displaystyle 3\le k\le n$. We proceed by induction on $\displaystyle k$.\n For the base case $\displaystyle k=3$, let our circles be $\displaystyle C_1, \ldots, C_n$. For each $1\le i \le n-1$, we will color one of the intersection points between $\displaystyle C_i$ and $\displaystyle C_{i+1}$ the color $\displaystyle i$. Furthermore, we will color one intersection point between $\displaystyle C_1, C_n$ the color 1, and the other the color $\displaystyle n-1$. Thus far, we have used $\displaystyle n-1$ colors, each circle has two distinct colors, and each circle has at least one uncolored intersection points. Therefore if we color all remaining points color $\displaystyle n$ we have a desired coloring.\n Now, suppose that there is such a coloring for $k-1 \ge 3$, $n-1 \ge k-1$, with circles $C_1, \ldots, C_{n-1}$ and colors $1, \ldots, n-1$. We claim that there exist distinct circles $R_1, \ldots, R_{k-1} \in \{ C_{i} \}_{i=1}^{n-1}$ such that circle $\displaystyle R_i$ has at least one point of color $\displaystyle f(i)$ such that all the $\displaystyle f(i)$ are distinct. Suppose the contrary. Consider the maximal $\displaystyle m$ for which there exist such circles $R_1, \ldots R_m$. This means that no circle other than the $\displaystyle R_i$ has any color other than $f(1), \ldots, f(m)$. But since $\displaystyle m < k-1 \le n-1$ by assumption, this means that there are some circles with less than $\displaystyle k-1$ colors on them, a contradiction. Now, we add a circle $\displaystyle C_n$. For each $\displaystyle i \le k-1$, we color exactly one intersection point of $\displaystyle R_i$ and $\displaystyle C_n$ the color $\displaystyle f(i)$. We color all other new intersection points a new color $\displaystyle n$. Now there are $\displaystyle n$ distinct colors used, and each circle uses $\displaystyle k$ distinct colors.\n |
(Australia)\nThe following operation is allowed on a finite graph: Choose an arbitrary cycle of length 4 (if there is any), choose an arbitrary edge in that cycle, and delete it from the graph. For a fixed integer $n \ge 4$, find the least number of edges of a graph that can be obtained by repeated applications of this operation from a complete graph on $\displaystyle n$ vertices (where each pair of vertices are joined by an edge).\n This was also Problem 2 of the 2005 Colombia TST.\n The proposer's formulation is:\n In a certain country there are $\displaystyle n$ towns, where $n \ge 4$. A road may be built between towns $\displaystyle A$ and $\displaystyle B$ if there exist two other towns $\displaystyle X$ and $\displaystyle Y$ such that there is no road between towns $\displaystyle A$ and $\displaystyle X$; there is no road between towns $\displaystyle X$ and $\displaystyle Y$; there is no road between towns $\displaystyle Y$ and $\displaystyle B$. What is the maximum number of roads that can be built?\n | Surprisingly, the answer is $\displaystyle n$.\n We first claim that our graph must always stay connected. Indeed, if an edge $\displaystyle AB$ is removed, then it must have been part of a 4-cycle, so there must remain some other way of travelling from $\displaystyle A$ to $\displaystyle B$. Hence the graph must always stay connected.\n We next claim that there is always some cycle of odd length. This is certainly true initially, for there are triangles initially. Now, suppose that we have a cycle $A_1\cdots A_n$ of odd length, and suppose that $A_1\cdots A_i B_1 \cdots B_{4-i}$ ($2\le i \le 3$) is a 4-cycle. Since $\displaystyle i$ and $\displaystyle 4-i$ have the same parity, $A_1 B_{4-i} \cdots B_1 A_i \cdots A_n$ is a cycle of odd length. But we cannot disrupt both this cycle and the original cycle by removing only one edge from our 4-cycle. Thus after each step, there must remain at least one cycle of odd length.\n The minimal number of edges in a connected graph with $\displaystyle n$ vertices is $\displaystyle n-1$, as each vertex apart from some special initial vertex needs at least one edge linking it indirectly to the initial vertex. Such minimal connected graphs must always be trees, for if there is a cycle, then we can remove one edge while keeping the graph connected. Since our graph must always be connected and must always have at least one cycle, it must always have at least $\displaystyle n$ edges. We will now describe recursively a sequence of moves that will lead to a graph with $\displaystyle n$ edges.\n Suppose our graph has vertices $A_1, \ldots, A_n$. For $\displaystyle n=4$, we remove the edge $\displaystyle A_4A_1$ from the cycle $\displaystyle A_4A_1A_2A_3$, and then remove the edge $\displaystyle A_4A_2$ from the cycle $\displaystyle A_4A_2A_1A_3$. Since we have only 4 edges left, we cannot go further. Now, for $\displaystyle n>4$ vertices, for each positive integer $\displaystyle k< n-1$, we remove the edge $\displaystyle A_nA_k$ from the 4-cycle $\displaystyle A_nA_kA_jA_{n-1}$, where $\displaystyle A_j$ is some vertex other than $\displaystyle A_n, A_k, A_{n-1}$. After this, $\displaystyle A_n$ will only be connected to $\displaystyle A_{n-1}$. We then use our procedure for graphs of size $\displaystyle n-1$ on the subgraph $A_1 \cdots A_{n-1}$ to obtain a final graph with $\displaystyle n$ edges. Q.E.D.\n Alternative It is also possible to prove that our final graph cannot be a tree by noting that trees are bipartite. But if we reconstruct our original graph by completing 4-cycles, the graph will remain bipartite, a contradiction.\n |
(Kazakhstan)\nThe circle $\displaystyle \Gamma$ and the line $\ell$ do not intersect. Let $\displaystyle AB$ be the diameter of $\displaystyle\Gamma$ perpendicular to $\ell$, with $\displaystyle B$ closer to $\ell$ than $\displaystyle A$. An arbitrary point $C \neq A,B$ is chosen on $\displaystyle \Gamma$. The line $\displaystyle AC$ intersects $\ell$ at $\displaystyle D$. The line $\displaystyle DE$ is tangent to $\displaystyle \Gamma$ at $\displaystyle E$, with $\displaystyle B$ and $\displaystyle E$ on the same side of $\displaystyle AC$. Let $\displaystyle BE$ intersect $\ell$ at $\displaystyle F$, and let $\displaystyle AF$ intersect $\displaystyle \Gamma$ at $G \neq A$. Prove that the reflection of $\displaystyle G$ in $\displaystyle AB$ lies on the line $\displaystyle CF$.\n This was also a Problem 2 on the 2005 Greece TST, Problem 1 of Day 1 of the 2005 Moldova TST, and Problem 2 of the final exam of the 3rd 2005 Taiwan TST.\n | Solution 1 We use directed angles mod $\displaystyle \pi$.\n Let $\displaystyle CF$ meet $\displaystyle \Gamma$ at $\displaystyle H$. The problem is equivalent to showing that lines $\displaystyle GH$ and $\ell$ are parallel, which happens if and only if $\angle AGH \equiv \angle AFD$. But by cyclic quadrilaterals and vertical angles, $\angle AGH \equiv \angle ACH \equiv \angle DCF$. To prove $\angle DCF \equiv \angle DFA$, it suffices to show that triangles $\displaystyle DFC, DAF$ are similar. Since these triangles share a common angle, it then suffices to show $\frac{DC}{DF} = \frac{DF}{DA}$, or $DF^2 = DC\cdot DA$.\n By considering the power of the point $\displaystyle D$ with respect to $\displaystyle \Gamma$, we see $\displaystyle DE^2 = DC \cdot DA$. Hence it suffices to show that $\displaystyle DE \equiv DF$. Let $\displaystyle O$ be the center of $\displaystyle \Gamma$. Since $AO \perp DF, AE \perp EF, OE \perp ED$, it follows that there is a spiral similarity mapping $\displaystyle AOE$ to $\displaystyle FDE$, i.e., these triangles are similar. Since $\displaystyle OA = OE$, it follows that $\displaystyle DF = DE$. Q.E.D.\n Solution 2 We use directed angles mod $\displaystyle \pi$.\n Lemma. Let $\displaystyle \omega_1 , \omega_2$ be two circles with centers $\displaystyle O_1, O_2$, and common points $\displaystyle M, N$. Let $\displaystyle A$ be a point on $\displaystyle \omega_1$, and let $\displaystyle A'$ be the second intersection of line $\displaystyle AM$ and $\displaystyle \omega_2$. Then $\angle AO_1N \equiv \angle A'O_2N$.\n Proof. Since $\displaystyle A, M, A'$ are collinear, $\angle AO_1N \equiv 2\angle AMN \equiv 2\angle A'MN$. But since $\displaystyle A', M, N$ lie on a circle with center $\displaystyle O_2$, $2\angle A'MN \equiv \angle A'ON$, as desired.\n Let the center of $\displaystyle \Gamma$ be $\displaystyle O$. Let $\displaystyle G'$ be the reflection of $\displaystyle G$ across $\displaystyle AB$. It is sufficient to show that $\angle ACG' \equiv \angle DCF$, since $\displaystyle A,C,D$ are collinear.\n Since $\displaystyle \Gamma$ is symmetric about its diameter $\displaystyle AB$, $\displaystyle G'$ lies on $\displaystyle \Gamma$, and ${\rm m}\widehat{AG} = {\rm m}\widehat{AG'}$, so\n \n$\angle ACG' \equiv \angle GCA$.\n\n If we consider the line which passes through $\displaystyle B$ and is parallel to $\ell$, we see ${\rm m}\angle GFD = \frac{1}{2}{\rm m}(\widehat{AB} - \widehat{BG}) = \frac{1}{2}{\rm m}[\widehat{AB} - (\widehat{BA} - \widehat{GA})] = \frac{1}{2}{\rm m}\widehat{GA} = {\rm m}\angle GCA$, since $\widehat{AB} , \widehat{BA}$ are semicircles. Thus\n \n$\angle GCA \equiv \angle GFD$,\n\n or $\angle GCA \equiv \angle GFD \equiv \angle AFD$. Since we also have $\angle FAD \equiv -\angle CAG$, it follows that triangles $\displaystyle AFD, ACG$ are similar, with opposite orientation. In particular, $\frac{AF}{AC} = \frac{AD}{AG}$, or $\displaystyle AF \cdot AG = AC \cdot AD$, so $\displaystyle F,G, D,C$ are concyclic.\n At this point, we note that $OE \perp EP$, $AE \perp EF$ (since $\angle AEF$ is inscribed in a semicircle), and $AO \perp DF$. It follows that there is a spiral similarity centered at $\displaystyle E$ with rotation of ${} \frac{\pi}{2}$ mapping triangle $\displaystyle AOE$ to triangle $\displaystyle FDE$ and $\displaystyle \Gamma$ to a circle $\displaystyle \Gamma'$ centered at $\displaystyle D$ with radius $\displaystyle DE = DF$ . Let $\displaystyle G''$ be the second intersection of $\displaystyle \Gamma$ and $\displaystyle \Gamma'$. We note that $\displaystyle AG''$ must intersect $\ell$ at some point $\displaystyle F'$ on the same side of $\displaystyle D$ as $\displaystyle F$, since $\displaystyle G', A$ must be on the same side of $\displaystyle DO$ as $\displaystyle F$. By the lemma, $\angle F'DE \equiv \angle AOE \equiv \angle FDE$, and since $\displaystyle F'$ is on $\displaystyle \Gamma'$, $\displaystyle DF' = DE = DF$. It follows that $\displaystyle F = F'$. Since $\displaystyle G''$ is the intersection point of $\displaystyle AF' = AF$ and $\displaystyle \Gamma$, $\displaystyle G'' = G$, and $\displaystyle G$ lies on $\displaystyle \Gamma'$. In particular, $\displaystyle DG = DF$, and\n \n$\angle GFD \equiv \angle DGF$.\n\n Now, since $\displaystyle F,G, D, C$ are concyclic, as we noted above,\n \n$\angle DGF \equiv \angle DCF$.\n\n To summarize,\n \n$\angle ACG' \equiv \angle GCA \equiv \angle GFD \equiv \angle DGF \equiv \angle DCF$,\n\n as desired. Q.E.D.\n \n Solution 3 We use projective geometry. Let $\displaystyle G'$ be the reflection of $\displaystyle G$ over $\displaystyle AB$. Let $\displaystyle \Phi$ be the intersection of two distinct parallels of $\ell$; let $\displaystyle Y$ be the intersection of $\displaystyle AE$ and $\displaystyle BG$; let $\displaystyle S$ be the intersection of $\displaystyle AB$ and $\displaystyle G'G$; and let $\displaystyle F'$ be the intersection of $\displaystyle BE$ and $\displaystyle G'C$. It is sufficient to show that $\displaystyle F'$ lies on $\ell$.\n We apply Pascal's Theorem for cyclic hexagons several times. By applying it to the degenerate hexagon $\displaystyle AAEBBG$, we see that $\displaystyle \Phi, Y, F$ are collinear, i.e., $\displaystyle Y$ lies on $\ell$. By applying the theorem to the hexagon $\displaystyle ABGG'EA$, we see that $\displaystyle S, Y, \Phi$ are collinear, i.e., $\displaystyle S$ also lies on $\ell$. Finally, by appling the theorem to $\displaystyle G'EEBAC$, we see that $\displaystyle S, D, F'$ are collinear, so $\displaystyle F'$ lies on $\ell$, as desired.\n |
(South Korea)\nLet $\displaystyle O$ be the circumcenter of an acute-angled triangle $\displaystyle ABC$ with $\angle B < \angle C$. The line $\displaystyle AO$ meets the side $\displaystyle BC$ at $\displaystyle D$. The circumcenters of the triangles $\displaystyle ABD$ and $\displaystyle ACD$ are $\displaystyle E$ and $\displaystyle F$, respectively. Extend the sides $\displaystyle BA$ and $\displaystyle CA$ beyond $\displaystyle A$, and choose, on the respective extensions points $\displaystyle G$ and $\displaystyle H$ such that $\displaystyle AG= AC$ and $\displaystyle AH = AB$. Prove that the quadrilateral $\displaystyle EFGH$ is a rectangle if and only if $\angle ACB - \angle ABC = 60^{\circ}$.\n (This was also Problem 2 of the 2005 3rd German TST; Problem 2, Day 3 of the 2005 Moldova TST; and Problem 5 of the 2005 Taiwan 2nd TST final exam.)\n | Lemma. In any triangle $\displaystyle ABC$ with circumcenter $\displaystyle O$, the altitude from $\displaystyle A$ is the reflection of $\displaystyle AO$ over the angle bisector of $\displaystyle A$.\n Proof. This is well-known, but we prove it anyway. Let $\displaystyle AO, BO, CO$ meet sides $\displaystyle a,b,c$ at $\displaystyle A', B', C'$, and let $\displaystyle H_a$ be the foot of the altitude from $\displaystyle A$. Let us denote $\angle BOA' = \angle B'OA = x$, $\angle COB' = \angle C'OB = y$, $\angle AOC' = \angle A'OC = z$, and let us use the notation $\displaystyle \alpha, \beta, \gamma$ for the angles of triangle $\displaystyle ABC$. By virtue of inscribed arcs in the circumcircle of $\displaystyle ABC$, we know $\displaystyle x+y = 2\beta$, $\displaystyle y+z = 2\gamma$, $\displaystyle z+x = 2\alpha$, so $\displaystyle x = \beta + \alpha - \gamma = \pi - 2\gamma$, and again by inscribed arcs, $\angle BAO = \frac{1}{2}x = \frac{\pi}{2}-\gamma = \angle H_aAC$. The lemma follows. ∎\n |
A cyclic quadrilateral $ABCD$ is given. The lines $AD$ and $BC$ intersect at $E$, with $C$ between $B$ and $E$; the diagonals $AC$ and $BD$ intersect at $F$. Let $M$ be the midpoint of the side $CD$, and let $N \neq M$ be a point on the circumcircle of $\triangle ABM$ such that $\frac{AN}{BN} = \frac{AM}{BM}$. Prove that $E, F, N$ are collinear.\n | Let $P = CD \cap EF$. Let $Q = AB \cap CD$. Let $R = AB \cap EF$. Let$(ABM)$ denote the circumcircle of $\triangle ABM$. Let $N' = EF \cap (ABM)$. Note that $N' = PR \cap (ABM)$\n \nClaim: $P$ is on $(ABM)$. Proof: $(C, D; P, Q) = -1$ as complete quadrilaterals induce harmonic bundles. $QP \cdot QM = QC \cdot QD$ by Lemma 9.17 on Euclidean Geometry in Maths Olympiad. By power of a point theorem, $QP \cdot QM = QA \cdot QB$ and this is equivalent to our original claim. \n \n$(A, B; R, Q) = -1$ as complete quadrilaterals induce harmonic bundles. By a projection through $P$ from $AQ$ onto $(ABM)$, $(A, B; N', M) = -1$. Since $\frac{AN}{BN} = \frac{AM}{BM}$, $M$ and $N$ are on the intersections of $(ABM)$ and an Appollonian circle centered on AB, so N and M are on the opposite sides of AB. Therefore, $(A, B; N, M) = -\frac{AN}{BN}\div\frac{AM}{BM} = -1$. By uniqueness of harmonic conjugate, $N = N'$\n |
(Russia)\nThe function $\displaystyle \psi$ from the set $\mathbf{N}$ of positive integers to itself is defined by the equality\n \n$\psi(n) = \sum_{k=1}^{n}(k,n), \qquad n \in \mathbf{N}$,\n\n where $\displaystyle (k,n)$ denotes the greatest common divisor of $\displaystyle k$ and $\displaystyle n$.\n a) Prove that $\displaystyle \psi(mn) = \psi(m)\psi(n)$ for every two relatively prime $m,n \in \mathbf{N}$. \nb) Prove that for each $a \in \mathbf{N}$ the equation $\displaystyle \psi(x) = ax$ has a solution. \nc) Find all $a \in \mathbf{N}$ such that the equation $\displaystyle \psi(x) = ax$ has a unique solution.\n This was also \n | Let $d$ be a divisor of $n$. We note that for $k \leq n/d$, $(dk,n) = d$ if and only if $k$ is relatively prime to $n/d$. It follows that each divisor $d$ of $n$ is found in the sum $\sum_{i=1}^{n}(k,n)$ exactly $\phi(n/d )$ times, where $\phi(m)$ is defined as the number of natural numbers less than or equal to $m$ and relatively prime to $m$ (the Euler Totient Function). It follows that\n \n$\psi (n) = \sum_{k=1}^{n}(k,n) = \sum_{d \mid n} d \phi(n/d)$.\n\n In other words, $\psi$ is the convolution of the identity function $f: n \mapsto n$ and the $\phi$ function. Since these are both multiplicative functions, $\psi$ is also multiplicative, which is part (a) of the problem. For the next parts, we note that the equation $\psi(x) = ax$ is equivalent to the equation $\frac{\psi(x)}{x} = a$.\n \nWe now note that for a prime $p$ and a non-negative integer $e$, $\psi(p^k) = \sum_{i=0}^{e}p^i \cdot \phi(p^{e-i}) = p^e + \sum_{i=0}^{e-1}p^i(p^{e-i} - p^{e-i-1}) = p^e + e(p^e - p^{e-1})$. Since $\psi$ is multiplicative, if $\prod_{i=1}^k p_i^{e_i}$ is the prime factorization of $x$, then\n \n$\psi(x) = \prod_{i=1}^k [(e_i+1)p_i^{e_i} - e_i\cdot p_i^{e_i}]$,\n\n and\n \n$\frac{\psi(x)}{x} = \frac{\prod [(e_i+1)p_i^{e_i} - e_i \cdot p_i^{e_i}]}{\prod p_i^{e_i}} = \prod \left(e_i \frac{p_i-1}{p_i} + 1 \right)$.\n\n From this we can see that $x = 2^{2(a-1)}$ is always a solution to the equation $\frac{\psi(x)}{x} = a$, which solves part (b). Let this solution to the equation be the canonical solution. However, we also note that if $2r +1$ is an odd postive integer, then $x = 3^{3r}$ is another solution to the equation. In particular, if $a$ has an odd prime factor $2r+1 > 1$, then $x = 2^{2[a/(2r+1) -1]} \cdot 3^{3r}$ is a solution distinct from the canonical solution. Thus if the only solution to the equation $\frac{\psi(x)}{x} = a$ is canonical, then $a$ cannot have any odd divisors greater than 1, i.e., $a$ must be a power of 2.\n We now claim that for $a = 2^k$, the equation has no non-canonical solutions.\n Lemma. Let $p_1, \ldots, p_k$ be odd primes, $e_1, \ldots, e_k$ be positive integers. If\n $\prod_{i=1}^{k}\left(e_i \frac{p-1}{p} +1\right) = \frac{p}{q}$ and $p, q$ are relatively prime positive integers, then $p$ is divisible by an odd prime.\n Proof. We first note that $e_i \frac{p_i-1}{p_i} + 1 \ge \frac{p_i-1}{p_i} + 1 > 1$, so $\frac{p}{q} = \prod \left( e_i \frac{p_i-1}{p_i} +1 \right) > 1$, or $p>q\ge 1$. Thus $p$ must be divisible by some prime. But since $\frac{p}{q} = \prod \left( e_i \frac{p_i-1}{p_i} + 1 \right) = \frac{ \prod [e_i(p_i-1) + p_i] }{\prod p_i}$, $p$ divides a product of odd numbers, so it cannot be divisible by 2. Hence $p$ is divisible by an odd prime. ∎\n |
Find all monic polynomials $\displaystyle p(x)$ of degree two for which there exists an integer polynomial $\displaystyle q(x)$ such that $\displaystyle p(x)q(x)$ is a polynomial having all coefficients $\pm 1$.\n \nThis was also the last problem of the final round of the 2006 Polish Mathematics Olympiad.\n | Since the constant term of $\displaystyle p(x)q(x)$ is $\pm 1$, and $\displaystyle p(x)$ and $\displaystyle q(x)$ both have integral constant terms, the constant term of $\displaystyle p(x)$ must be $\pm 1$.\n We note that for $\displaystyle |z| \ge 2$, $\displaystyle n \ge 2$ ($n \in \mathbb{N}$), we have\n \n$|z|^n > \frac{|z|^n -1}{|z|-1} = \sum_{i=0}^{n-1}|\pm z|^i \ge \left| \sum_{i=0}^{n-1} \pm z^i \right|$\n\n Since we must have $\displaystyle |z^n| = | p(x)q(x) - z^n |$ when $\displaystyle n$ is the degree of $\displaystyle p(x)q(x)$ and $\displaystyle z$ is a root thereof, this means that $\displaystyle p(x)q(x)$ cannot have any roots of magnitude greater than or equal to 2.\n Now, if $\displaystyle p(x) = x^2 + kx + 1$, then we cannot have $\displaystyle |k| \ge 3$, for then one of the roots would have magnitude $\frac{|k| + \sqrt{k^2 - 4}}{2} \ge \frac{3 + \sqrt{3^2 - 4}}{2} > 2$, and similarly, if $\displaystyle p(x) = x^2 +kx - 1$, then we cannot have $\displaystyle |k| \ge 2$, for then one of the roots would have magnitude $\frac{|n| + \sqrt{n^2+4}}{2} \ge \frac{2 + \sqrt{2^2 + 4}}{2} > 2$.\n This leaves us only the possibilities $\displaystyle p(x) = x^2 \pm 1,\; x^2 \pm x \pm 1,\; x^2 + 2x + 1,\; x^2 - 2x + 1$. For these we have respective solutions $\displaystyle q(x) = x+1,\; 1,\; x^2 - 1,\; x^2 + 1$. These are therefore the only solutions, Q.E.D.\n |
(Australia)\nA house has an even number of lamps distributed among its rooms in such a way that there are at least three lamps in every room. Each lamp shares a switch with exactly one other lamp, not necessarily from the same room. Each change in the switch shared by two lamps changes their states simultaneously. Prove that for each initial state of the lamps there exists a sequence of changes in some of the switches at the end of which each room contains lamps which are on as well as lamps which are off.\n This problem was not used on the contest, as it appeared in the 1990 Australian Mathematical Olympiad.\n The problem was submitted to the Problem Selection Committee in the following form:\n A school has an even number of students, each of whom attends exactly one of its (finitely many) classes. Each class has at least three students, and each student has exactly one "best friend" in the same school such that, whenever $\displaystyle B$ is $\displaystyle A$'s "best friend", then $\displaystyle A$ is $\displaystyle B$'s "best friend". Furthermore, each student prefers apple juice over orange juice or orange juice over apple juice, but students change their preferences from time to time. "Best friends", however, will change their preferences (whic may or may not be the same) always together, at the same moment.\n Whatever preference each student may initially have, prove that there is always a sequence of changes of preferences which will lead to a situation in which no class will have students all of whom have the same preference.\n | It is sufficient to prove that we can always reduce the number of monochromatic rooms by one after finitely many steps.\n Suppose there is a monochromatic room. We start in that room, and perform a sequence of moves according to the following rules:\n If we are in a monochromatic room, we toggle a switch connected to a light in our room which we have not toggled before. We then move to the room containing the other light which the switch changed.\nIf we are in a dichromatic room or a room in which we have been before, we stop. Eventually, we must either reach a dichromatic room, in which case we have reduced the number of monochromatic rooms by one, or we reach a room that we had been in before. But when we entered it for the first time, it was monochromatic. Since it was monochromatic, we have toggled two distinct lights in that room (one leaving and one coming back), and since there is at least one additional light in the room that has not been toggled since we were in it last, the room is now dichromatic. Thus we eventually encounter a dichromatic room, decreasing the number of monochromatic rooms by one, so we can eventually make all rooms dichromatic. Q.E.D.\n \nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.\n |
(Iran)\nLet $\displaystyle k$ be a fixed positive integer. A company has a special method to sell sombreros. Each customer can convince two persons to buy a sombrero after he/she buys one; convincing someone already convinced does not count. Each of these new customers can convince two others and so on. If each one of the two customers convinced by someone makes at least $\displaystyle k$ persons buy sombreros (directly or indirectly), then that someone wins a free instructional video. Prove that if $\displaystyle n$ persons bought sombreros, then at most $\displaystyle n/(k+2)$ of them got videos.\n This was also Problem 3, Day 2, of the 2006 Australia TST.\n | Suppose $m$ persons receive videos. We wish to prove $n \ge m(k+2)$. Since this is clear for $m=0$, let us WLOG assume that $m$ is positive. Under this assumption, we will now prove the stronger bound\n $n \ge (m+1)(k+1) + m$ by induction on $m$.\n We say a person $A$ is a direct successor of $B$ if $B$ directly convinced $A$ to buy a sombrero. We say $A$ is an indirect successor of $B$ if $B$ caused $A$ to buy a sombrero, directly or indirectly. We will call a person who receives a video a blossom. We will call direct successor of a blossom who is not a blossom him- or her-self a bud.\n For the base case of our induction, when there is only one blossom, that blossom must have at least two buds, each of which must have at least $k$ indirect successors. Hence $n \ge 2k+3 = (m+1)(k+1)+m$.\n Now, suppose there are $m$ blossoms. Since there are only finitely many blossoms, there exists at least one blossom which has no blossoms as indirect successors. We remove this blossom, one of its buds, and all indirect successors of that bud; we then make the disconnected bud a direct successor of whatever person was a direct successor of the blossom we removed, if there was such a person. All other blossoms stay blossoms; all other buds stay buds. We have thus removed at least $2+k$ people, and we have removed one blossom. If there are $n'$ people remaining, then the inductive hypothesis now tells us\n $n - (2+k) \ge n' \ge (m)(k+1) + (m-1)$, or $n \ge n' + k+2 \ge (m+1)(k+1) + m$. Therefore for all $m$, we have\n \n$n \ge (m+1)(k+1) + m = m(k+2) + k+1 > m(k+2)$, as desired.\n We note that the bound $n \ge (m+1)(k+1) +m$ is sharp, for it is possible to have blossoms $A_1, \ldots, A_m$, with $A_{i+1}$ a direct successor of $A_i$, and all buds having exactly $k$ indirect successors.\n |
(Iran)\nIn an $m \times n$ rectangular board of $\displaystyle mn$ unit squares, adjacent squares are ones with a common edge, and a path is a sequence of squares in which any two consecutive squares are adjacent. Each square of the board can be coloured black or white. Let $\displaystyle N$ denote the number of colourings of the board such that there exists at least one black path from the left edge of the board to the right edge of the board, and let $\displaystyle M$ denote the number of colourings in which there exist at least two non-intersecting black paths from the left edge to the right edge. Prove that $N^2 \ge M \cdot 2^{mn}$.\n This was also Problem 6 of the 2006 German Pre-TST.\n | We will call black paths which connect the left edge of the board to the right edge good paths.\n Lemma. Consider a board as described in the problem, and let columns run from top to bottom, and rows run from left to right, with $\displaystyle m$ columns and $\displaystyle n$ rows. We say that one path $\displaystyle M_1$ lies below another path $\displaystyle M_2$ if in each column, the highest square of $\displaystyle M_1$ in that column lies below the highest square of $\displaystyle M_2$ in that column. Then there exists a lowest good path, which we define as a good path that lies below all other good paths.\n Proof. Let $M_1, \ldots, M_k$ be all the good paths, with loops removed (i.e., each square appears in a path at most once—this can be done by replacing all sequences of squares of the form $ABC \cdot A$ by $\displaystyle A$). For each $1 \le j \le m-1$, let $\displaystyle Q_j$ be the lowest square in column $\displaystyle j$ that is in one of the loopless good paths, say $\displaystyle M_i$. Since $\displaystyle M_i$ contains no loops, $\displaystyle Q_j$ must be at the bottom of a subcolumn $\displaystyle T$ of black squares such that the square at the top of $\displaystyle T$ is adjacent to a square in column $\displaystyle j+1$. It follows that there exists a shortest subcolumn $\displaystyle T'$ of black squares (possibly of length 1) such that the square at the top of $\displaystyle T'$ is adjacent to some square in column $\displaystyle i+1$. Then the squares in the union of all such subcolumns constitute a good path which lies below all others. ∎\n |
(Ukraine)\nLet $\displaystyle ABCD$ be a parallelogram. A variable line $\ell$ passing through the point $\displaystyle A$ intersects the rays $\displaystyle BC$ and $\displaystyle DC$ at points $\displaystyle X$ and $\displaystyle Y$, respectively. Let $\displaystyle K$ and $\displaystyle L$ be the centres of the excircles of triangles $\displaystyle ABX$ and $\displaystyle ADY$, touching the sides $\displaystyle BX$ and $\displaystyle DY$, respectively. Prove that the size of angle $\displaystyle KCL$ does not depend on the choice of $\ell$.\n This was also Problem 3 of the 2006 3rd German TST, and a problem at the 2006 India IMO Training Camp. It also appeared in modified form as Problem 3, Day 3 of the 2006 Iran TST.\n | Let $\ell_1, \ell_2$ be the interior angle bisectors of $\displaystyle ABX, YAD$. Let $\displaystyle m_1, m_2$ be the exterior angle bisectors of $\displaystyle ABC, CDA$. Then $\displaystyle K$ is the intersection of $\ell_1, m_1$ and $\displaystyle L$ is the intersection of $\ell_2, m_2$.\n ISL2005G3.png Let us denote $\displaystyle x,y$ as the measures of $\angle BAK, \angle LAD$, and denote $\displaystyle \alpha = x+y$. Then $\angle BAD \equiv \angle DCB \equiv 2 \alpha$. Furthermore, since $\displaystyle BK$ is the exterior angle bisector of $\displaystyle ABC$, we know that the exterior angle at $\displaystyle ABK$ is $\displaystyle \alpha = x+y$, so $\angle AKB \equiv y$. Similarly, $\angle ALD \equiv x$. It follows that triangles $\displaystyle ABK, LDA$ are similar. Then since $\displaystyle ABCD$ is a parallelogram,\n \n$\frac{BK}{DC} = \frac{BK}{AB} = \frac{DA}{LD} = \frac{CB}{LD}$.\n\n |
Let $a_1,a_2,\ldots$ be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer $n$ the numbers $a_1,a_2,\ldots,a_n$ leave $n$ different remainders upon division by $n$.\n Prove that every integer occurs exactly once in the sequence $a_1,a_2,\ldots$.\n | It is clear that $a_i=a_j$ if and only if $i=j$, or the sequence would not satisfy the specified property.\n If $|a_i-a_j|\geq \max(i,j)$, then $a_i$ and $a_j$ leave the same remainder when divided by $|a_i-a_j|$, which violates the given condition for the sequence when $n=|a_i-a_j|$. It then follows that $|a_i-a_j|<\max(i,j)$ for all positive integers $i$ and $j$. Now consider $\min(a_1,a_2,\ldots,a_n)$, and let this be $a_k$, with $k\in [1,n]$. It follows that $a_1,a_2,\ldots a_{n}$ are all in the closed interval $[a_k,a_k+n-1]$, and hence $a_1,a_2,\ldots a_n$ is a permutation of $n$ consecutive numbers, for all $n$.\n Note that there are infinitely many positive and negative terms. Therefore for any arbitrarily large integer $\Delta$ there exists an $i$ such that $a_i\geq \Delta$ and a $j$ such that $a_j\leq -\Delta$. Since $a_1,a_2,\ldots, a_n$ is a permutation of $n$ consecutive integers, it follows that every integer in the range $[-\Delta,\Delta]$ is in the sequence, and consequently every integer occurs in the sequence.\n |
(Mongolia)\nLet $a$, $b$, $c$, $d$, $e$, and $f$ be positive integers. Suppose that the sum $S = a+b+c+d+e+f$ divides both $abc + def$ and $ab+bc+ca - de-ef-fd$. Prove that $S$ is composite.\n This was also Problem 1 of the 2nd 2006 German TST, and a problem at the 2006 Indian IMO Training Camp.\n | For all integers $\displaystyle x$ we have\n \n$(x+a)(x+b)(x+c) \equiv x^3 + (a+b+c)x^2 + (ab+bc+ca)x + abc \equiv x^3 - (d+e+f)x^2 + (de+ef+fd)x - def$ $\equiv (x-d)(x-e)(x-f) \pmod{S}$,\n\n since each coefficient of the first two polynomials is congruent to the corresponding coefficient of the second two polynomials, mod $\displaystyle S$. Now, suppose $\displaystyle S$ is prime. Since\n \n$(d+a)(d+b)(d+c) \equiv (d-d)(d-e)(d-f) \equiv 0 \pmod{S}$,\n\n one of $\displaystyle d+a, d+b, d+c$ is divisible by $\displaystyle S$, say $\displaystyle d+a$. Since $\displaystyle d,a > 0$, this means $d+a \ge S$. But since $a, \ldots, f$ are positive integers, we then have\n |
(Estonia)\nA sequence of real numbers $a_0, a_1, a_2, \dots$ is defined by the formula\n \n$a_{i+1} = \lfloor a_i \rfloor \cdot \langle a_i \rangle$\nfor $i \ge 0$;\n\n here $\displaystyle a_0$ is an arbitrary real number, $\lfloor a_i \rfloor$ denotes the greatest integer not exceeding $\displaystyle a_i$, and ${} \langle a_i \rangle = a_i - \lfloor a_i \rfloor$. Prove that $\displaystyle {} a_i = a_{i+2}$ for $\displaystyle i$ sufficiently large.\n | We first note that for all nonnegative integers $\displaystyle i$,\n \n$| a_{i+1} | = | \lfloor a_i \rfloor \cdot \langle a_i \rangle | < | \lfloor a_i \rfloor |$,\n\n so ${} \left| \lfloor a_i \rfloor \right|$ is a non-increasing function of $\displaystyle i$. We also note that if $\displaystyle a_i$ is not positive (resp. not negative), then $\displaystyle a_{i+1}$ is not positive (resp. not negative).\n When $\displaystyle a_i \in [0,1]$, $\displaystyle a_{i+1} = 0$. It follows that if $\displaystyle a_0$ is nonnegative, it is enough to show that for $\displaystyle a_i > 1$, then $0 \le \lfloor a_{i+1} \rfloor < \lfloor a_i \rfloor$, for then we must eventually have $\displaystyle a_i = a_{i+1} = \dotsb = 0$. But this comes from\n \n${} 0 \le \lfloor \lfloor a_i \rfloor \cdot \langle a_i \rangle \rfloor = \lfloor a_{i+1} \rfloor \le a_{i+1} = \lfloor a_i \rfloor \cdot \langle a_i \rangle < \lfloor a_i \rfloor$ .\n\n We prove the result for negative $\displaystyle a_0$ by induction on $\left| \lfloor a_0 \rfloor \right|$. For our base case, $\lfloor a_0 \rfloor = -1$, we either have $\displaystyle a_0 = -1$ and $\displaystyle a_j = 0$ for positive $\displaystyle j$. For $\displaystyle -1 < a_0 < 0$, we have $a_1 = 1 - a_0, a_2 = a_0, \dots$.\n Now, suppose that the statement holds for $\left| \lfloor a_0 \rfloor \right| < n$. If ever we have $|a_i| \le n-1$, then the problem reduces to a previous case. Therefore if we ever have $\langle a_i \rangle \le \frac{n-1}{n}$, then the problem reduces to a previous case. Therefore if $\displaystyle a_0$ generates a counterexample to the problem, then we must always have $a_i > \frac{n-1}{n} = 1 - \frac{1}{n}$.\n Then if $\displaystyle a_0$ is a counterexample with $\lfloor a_0 \rfloor = -n$, then for nonnegative $\displaystyle i$, we must have\n \n$\langle a_{i+1} \rangle = \left\langle -n \cdot \langle a_i \rangle \right\rangle = \left\langle -n + (n - n \cdot \langle a_i \rangle) \right\rangle = n - n\cdot \langle a_i \rangle$.\n\n It follows by induction that\n \n$\langle a_k \rangle = (-n)^k \langle a_0 \rangle - \sum_{i=1}^k (-n)^i$.\n\n In particular, for all even integers $\displaystyle k$, we have\n \n$(-n)^k \langle a_0 \rangle - \sum_{i=1}^k(-n)^i > 1 - \frac{1}{n}$ \n$\langle a_0 \rangle > \sum_{i=-1}^k (-n)^{i-k} = \sum_{i=0}^{k+1} (-1/n)^i$.\n\n Since $\sum_{i=0}^{k+1}(-1/n)^i$ becomes arbitrarily close to $\frac{n}{n+1}$ as $\displaystyle k$ becomes arbitrarily large, we thus have\n \n$\langle a_0 \rangle \ge \frac{n}{n+1}$.\n\n But for odd integers $\displaystyle k$, the inequality is reversed, and we have\n \n$\langle a_0 \rangle < \sum_{i=-1}^k (-n)^{i-k} = \sum_{i=0}^{k+1} (-1/n)^i$,\n\n and similarly,\n \n${} \langle a_0 \rangle \le \frac{n}{n+1}$.\n\n It follows that $\displaystyle a_0$ is of the form ${} -n + \frac{n}{n+1} = \frac{-n^2}{n+1}$, for some positive integer $\displaystyle n$. But this gives us a constant sequence, which is not a counterexample. Therefore by induction, there are no negative counterexamples. Since we have already proven that the problem statement holds for nonnegative reals, we are done. ∎\n |
(Poland)\nThe sequence of real number $a_0, a_1, a_2, \dots$ is defined recursively by\n \n$\displaystyle a_0 = -1$,\n$\sum_{k=0}^n \frac{a_{n-k}}{k+1} = 0$\nfor $n \ge 1$.\n\n Show that $\displaystyle {} a_n > 0$ for $n \ge 1$.\n This was also Problem 6 of the 2007 Poland Math Olympiad.\n | We proceed by induction on $\displaystyle n$. For the base case, we note that $\displaystyle a_1 = 1/2$. Suppose that $a_1, \dots, a_{n-1}$ are positive. We note that $\displaystyle a_n$ is positive if and only if $\sum_{k=1}^n \frac{a_{n-k}}{k+1}$ is negative. Now, since $a_1, \dots, a_{n-1}$ are all positive, we know\n \n${} - \frac{a_0}{n+1} = \frac{n}{n+1} \cdot \left( -\frac{a_0}{n} \right) = \frac{n}{n+1} \sum_{k=0}^{n-2}\frac{a_{n-1-k}}{k+1} > \sum_{k=0}^{n-2} \frac{k+1}{k+2} \cdot \frac{a_{n-1-k}}{k+1} = \sum_{k=0}^{n-2} \frac{a_{n-1-k}}{k+2}$\n.\n\n This means that\n \n$\sum_{k=1}^{n} = \frac{a_0}{n+1} + \sum_{k=1}^{n-1} \frac{a_{n-k}}{k+1} = \frac{a_0}{n+1} + \sum_{k=0}^{n-2}\frac{a_{n-1-k}}{k+2} < \frac{a_0}{n+1} - \frac{a_0}{n+1} = 0$,\n\n |
Prove the inequality\n\[\sum_{i<j} \frac{a_ia_j}{a_i+a_j} \le \frac{n}{2(a_1 + a_2 + \dotsb a_n)} \sum_{i<j} a_i a_j\]\nfor positive real numbers $a_1, \dotsc, a_n$.\n | Note that\n\[\sum_{i<j} \frac{a_ia_j}{a_i+a_j} = 1/2 \sum_{i\neq j} \frac{a_ia_j}{a_i + a_j} = 1/2 \sum_{j=1}^n \sum_{i\neq j} \frac{1}{1/a_i + 1/a_j} .\]\nSuppose that $1 \le k \neq \ell \le n$. Note that $1/(1/a_i + 1/a_j)$ is an increasing function of both $a_i$ and $a_j$. It follows that if $a_k \le a_\ell$, then\n\[\sum_{j \neq k} \frac{1}{1/a_k + 1/a_j} \le \sum_{j\neq \ell} \frac{1}{a_\ell + a_j},\]\ni.e., $\sum_{j\neq i} \frac{1}{1/a_i + 1/a_j}$ is an increasing function of $j$.\n Since $\sum_{i\neq j}(a_j + a_i) = (n-2)a_j + \sum_{i=1}^n a_i$ is also an increasing function of $j$, it follows from Chebyshev's Inequality that\n\[\frac{2n-2}{n} \sum_{j=1}^n a_j \cdot \sum_{j=1}^n \sum_{j\neq i} \frac{1}{1/a_i + 1/a_j} \le \sum_{j=1}^n \left[ \sum_{i\neq j} (a_i + a_j) \cdot \sum_{i\neq j} \frac{1}{1/a_i + 1/a_j} \right],\]\nor\n\[1/2 \sum_{j=1}^n a_j \cdot \sum_{j=1}^n \sum_{j\neq i} \frac{1}{1/a_i + 1/a_j} \le \frac{n}{4(n-1)} \sum_{j=1}^n \left[ \sum_{i\neq j} (a_i + a_j) \cdot \sum_{i\neq j} \frac{1}{1/a_i + 1/a_j} \right] .\]\nNow, for fixed $j$, both $a_i + a_j$ and $1/(1/a_i + 1/a_j)$ are increasing functions of $a_i$. It follows again from Chebyshev's Inequality that\n\[\frac{1}{n-1} \sum_{i\neq j} (a_i + a_j) \sum_{i \neq j} \frac{1}{1/a_i + 1/a_j} \le \sum_{i\neq j} \frac{a_i+a_j}{1/a_i + 1/a_j} = \sum_{i\neq j} a_i a_j,\]\nor\n\[\sum_{i\neq j} (a_i + a_j) \sum_{i \neq j} \frac{1}{1/a_i + 1/a_j} \le (n-1) \sum_{i\neq j} a_ia_j,\]\nwhich in sum becomes\n\[\frac{n}{4(n-1)} \sum_{j=1}^n \left[ \sum_{i\neq j}(a_i + a_j) \sum_{i \neq j} \frac{1}{1/a_i + 1/a_j} \right] \le \frac{n}{4} \sum_{j=1}^n \sum_{i \neq j} a_i a_j = \frac{n}{2} \sum_{i<j} a_i a_j .\]\nIf we denote $\sum_{i=1}^n a_i =S$, then in summary, we thus have\n\[\sum_{i < j} \frac{a_ia_j}{a_i + a_j} = (1/S) \cdot S/2 \cdot \sum_{j=1}^n \sum_{i\neq j} \frac{1}{1/a_i + 1/a_j} \le n/(2S) \cdot \sum_{i<j} a_i a_j,\]\nas desired. $\blacksquare$\n |
(Hojoo Lee, South Korea) Let $a,b,c$ be the sides of a triangle. Prove that\n\[\frac{\sqrt{b+c-a}}{\sqrt{b}+ \sqrt{c} - \sqrt{a}} + \frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a} - \sqrt{b}} + \frac{\sqrt{a+b-c}}{\sqrt{a} + \sqrt{b} - \sqrt{c}} \le 3.\]\n This problem also appeared on the 2007 IMO TSTs of Italy and Bangladesh.\n | Lemma. For any positive reals $x,y,z$,\n\[\sum_{\rm cyc} \left( \frac{y}{x} - 1 \right) \left( \frac{z}{x} -1 \right) \ge 0 .\]\n Proof 1. This is the Vornicu-Schur Inequality on the function $x \mapsto 1/x$. $\blacksquare$\n Proof 2. Without loss of generality, suppose that $x\ge y \ge z$. Evidently,\n\[\left( \frac{y}{x} - 1 \right) \left( \frac{z}{x} - 1\right) \ge 0 ,\]\nso it suffices to show that\n\[\left( \frac{x}{z} - 1 \right) \left( \frac{y}{z} - 1 \right) + \left( \frac{x}{y} - 1 \right) \left( \frac{z}{y} - 1 \right) \ge 0,\]\nor\n\[\left( \frac{x}{z} - 1 \right) \left( \frac{y}{z} - 1 \right) \ge \left( \frac{x}{y} - 1 \right) \left( 1 - \frac{z}{y} \right) .\]\nBy assumption,\n\[\frac{x}{z} - 1 \ge \frac{x}{y} - 1 ,\]\nand by AM-GM, $y/z + z/y \ge 2$, or\n\[\frac{y}{z} - 1 \ge 1 - \frac{z}{y} .\]\nSince all sides of both the previous two inequalities are positive, multiplication yields the desired bound, proving the lemma. $\blacksquare$\n We note that in the problem statement, each denominator is greater than zero, for\n\[\sqrt{b} + \sqrt{c} = \sqrt{b + c + 2 \sqrt{bc}} \ge \sqrt{a + 2\sqrt{bc}} > \sqrt{a} .\]\nWe now abbreviate $x = \sqrt{b} + \sqrt{c} - \sqrt{a}$, $y = \sqrt{c} + \sqrt{a} - \sqrt{b}$, $z= \sqrt{a} + \sqrt{b} - \sqrt{c}$. Then\n\begin{align*} \frac{\sqrt{b+c-a}}{\sqrt{b}+ \sqrt{c} - \sqrt{a}} + \frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a} - \sqrt{b}} + \frac{\sqrt{a+b-c}}{\sqrt{a} + \sqrt{b} - \sqrt{c}} &= \sum_{\rm cyc} \sqrt{ \frac{ \left( \frac{x+y}{2} \right)^2 + \left( \frac{x+z}{2} \right)^2 - \left( \frac{y+z}{2} \right)^2 }{x^2}} \\ &= \frac{1}{2} \sum_{\rm cyc} \sqrt{ \frac{2x^2 + 2xy + 2xz - 2yz}{x^2}} . \end{align*}\nBy the Power Mean Inequality,\n\[\frac{1}{2} \sum_{\rm cyc} \sqrt{ \frac{2x^2 + 2xy + 2xz - 2yz}{x^2}} \le \frac{\sqrt{3}}{2} \biggl( \sum_{\rm cyc} \frac{2x^2 + 2xy + 2xz - 2yz}{x^2} \biggr)^{1/2} .\]\nAlso, by the lemma,\n\[0 \le \sum_{\rm cyc} \left(\frac{y}{x}-1 \right) \left( \frac{z}{x} -1 \right) = \sum_{\rm cyc} \frac{yz - xz - xy}{x^2} + 1 = 6 - \sum_{\rm cyc} \frac{x^2 + xy + xz - yz}{x^2} ,\]\nso\n\[\frac{\sqrt{3}}{2} \biggl( \sum_{\rm cyc} \frac{2x^2 + 2xy + 2xz - 2yz}{x^2} \biggr)^{1/2} \le \frac{\sqrt{3}}{2} ( 2 \cdot 6 )^{1/2} = 3,\]\nproving the desired inequality. $\blacksquare$\n |
(France)\nWe have $n \ge 2$ lamps $L_1, \dots, L_n$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamps as follows:\n if the lamp $\displaystyle {} L_i$ and its neighbours (only one neighbour for $\displaystyle i=1$ or $\displaystyle {i=n}$, two neighbours for other $\displaystyle i$) are in the same state, then $\displaystyle {} L_i$ is switched off;\notherwise, $\displaystyle {} L_i$ is switched on. Initially, all the lamps are off except the leftmost one which is on.\n(a) Prove that there are infinitely many integers $\displaystyle n$ for which all the lamps will eventually be off. \n(b) Prove that there are infinitely many integers $\displaystyle n$ for which the lamps will never be all off.\n | We suppose that $\displaystyle L_1$ is the leftmost lamp. We note that for $k \le n$, in the $\displaystyle k$th second, $\displaystyle L_k$ is on, and is the rightmost lamp which is on. This follows from induction.\n Lemma. If $2^k \le n$, then in the $\displaystyle 2^k$th second, ${} L_1, \dots, L_{2^k}$ are exactly the lamps which are on.\n Proof. We know that no lamps to the left of $\displaystyle {} L_{2^k}$ are on, so it is enough to show that all the lamps ${} L_1, \dots, L_{2^{k}}$ are on. We show this by induction. For the base case, $\displaystyle k=0$, we note that initially $\displaystyle L_1$ is on. Now, suppose that in the $\displaystyle 2^{k-1}$th second, lamps ${L_1, \dots, L_{2^{k-1}}}$ are on. Then in the $\displaystyle 2^{k-1} + 1$th second, the only lights which are on are $\displaystyle L_{2^{k-1}}$ and $\displaystyle L_{2^{k-1} +1}$. At this point, we have symmetry between the lamps ${L_1, \dots, L_{2^{k-1}}}$ and the lamps $\displaystyle {} {L_{2^k}, \dots, L_{2^{k-1}+1}}$, which is preserved in every second until $\displaystyle L_{2^k +1}$ is turned on (if there is such a lamp). Thus from the $\displaystyle 2^{k-1}+1$th second to the $\displaystyle 2^k$th second, inclusive, the lamps $\displaystyle {} L_{2^{k-1}}, \dots, L_1$ and $\displaystyle {} L_{2^{k-1}+1},\dots, L_{2^k}$ behave as independent replicas of the original state. Then by the inductive hypothesis, after $\displaystyle 2^{k-1} + 2^{k-1} = 2^k$th second, the lamps $\displaystyle {} L_{2^{k-1}}, \dots, L_{1}$ as well as the lamps $\displaystyle {} L_{2^{k-1}+1}, \dots, L_{2^k}$ are all turned on. ∎\n Now, from the lemma, it is clear that when we have $\displaystyle 2^k$ lamps, for any integer $\displaystyle k$, in the $\displaystyle 2^k$th second, all lamps are turned on, so in the $\displaystyle 2^k +1$th second, they are all turned off. This proves part (a) of the problem. On the other hand, if we have $\displaystyle 2^k +1$ lamps, for any integer $k \ge 1$, then in the $\displaystyle 2^k + 1$th second, the lamps $\displaystyle {} L_{2^k}, L_{2^k+1}$ will be the only lamps on. This is the mirror image of the state of the lamps in the second second, so the lamps' state must be periodic and there will always be at least one lamp on. ∎\n |
(Argentina) An $(n,k)$-tournament is a contest with $n$ players held in $k$ rounds such that:\n (i) Each player plays in each round, and every two players meet at most once.\n(ii) If player $A$ meets player $B$ in round $i$, player $C$ meets player $D$ on round $i$, and player $A$ meets player $C$ in round $j$, then player $B$ meets player $D$ in round $j$. Determine all pairs $(n,k)$ for which there exists an $(n,k)$-tournament.\n This problem also appeared on the 2007 TST for Korea.\n | Let $t$ be the greatest integer such that $2^t$ divides $n$. Then there exists an $(n,k)$-tournament if and only if $k \le 2^t - 1$.\n We first prove that if $k \le 2^t - 1$, then there exists an $(n,k)$-tournament. Since we may partition our $n$ players into $n/2^t$ different groups of size $2^t$, it suffices to prove this for $n=2^t$.\n To this end, let us label the $2^t$ players with the elements of $(\mathbb{Z}/2\mathbb{Z})^t$, and label the different rounds with distinct nonzero elements of $(\mathbb{Z}/2\mathbb{Z})^t$. In the round with label $j$, let player $a$ meet player $a+j$. We then have a $(2^t,k)$ tournament, for if $a-b = c-d = i$, then $a-c = b-d$, for all $a,b,c,d \in (\mathbb{Z}/2\mathbb{Z})^t$.\n We next prove that if $k > 2^t-1$, then there is no $(n,k)$-tournament. For this, we first prove an intermediate result.\n Lemma. The number of players in any minimal subtournament of an $(n,k)$-tournament is a power of 2.\n Proof. We induct on $k$. The case $k=0$ is trivial.\n Suppose now that all minimal subtournaments of any $(n,k-1)$-tournament have sizes that are powers of 2. Then in any $(n,k)$-tournament, the minimal subtournaments, ignoring the last round, are of powers of 2. Let $S$ be the set of players for such a minimal tournament, and for any player $a$ in the $(n,k)$-tournament, let $K(a)$ be the player $a$ meets in round $k$. Then either $K(S) = S$, or $K(S)$ and $S$ are disjoint; furthermore, $K$ induces an isomorphism of tournaments from $S$ to $K(S)$, so $S$ and $K(S)$ have the same cardinality. It follows that the minimal subtournament of the $(n,k)$-tournament containing any element of $S$ has size either $|S|$ or $2|S|$, completing the inductve step and proving the lemma. $\blacksquare$\n Now, suppose that there exists an $(n,k)$-tournament for $k > 2^t-1$. Then every minimal sub-tournament of our tournament has a multiple of $2^{t+1}$ players, since each must have a size that is a power of 2, and no two players meet more than once. It follows that $2^{t+1}$ divides $n$, a contradiction. Therefore no $(n,k)$-tournament exists for $k > 2^t-1$, as desired. $\blacksquare$\n |
(Ukraine)\nLet $\displaystyle ABCD$ be a trapezoid with parallel sides ${} \displaystyle AB >CD$. Points $\displaystyle K$ and $\displaystyle L$ lie on the line segments $\displaystyle AB$ and $\displaystyle CD$, respectively, so that ${} \displaystyle AK/KB = DL/LC$. Suppose that there are points $\displaystyle P$ and $\displaystyle Q$ on the line segment $\displaystyle KL$ satisfying\n \n$\angle APB = \angle BCD$\nand\n$\angle CQD = \angle ABC$.\n\n Prove that the points $\displaystyle P$, $\displaystyle Q$, $\displaystyle B$, and ${} \displaystyle C$ are concyclic.\n | Since ${} \displaystyle A,B,K$ and $\displaystyle D,C,L$ are collinear, the condition ${} \displaystyle AK/KB = DL/LC$ is equivalent to the condition that lines $\displaystyle AD$, $\displaystyle KL$, and $\displaystyle BC$ are concurrent. Let $\displaystyle X$ be the point of concurrence.\n Let $\displaystyle \omega_1, \omega_2$ be the circumcircles of $\displaystyle APB, DQC$, respectively. Since $\angle XBA = \angle BCD = \angle APB$, the line $\displaystyle XB$ is tangent to $\displaystyle \omega_1$ at $\displaystyle B$. Similarly, $\displaystyle \omega_2$ is tangent to $\displaystyle XB$ at ${} \displaystyle C$. It follows there is a dilation $\displaystyle \gamma$ centered at $\displaystyle X$ which takes $\displaystyle \omega_2$ to $\displaystyle \omega_1$. Let $\displaystyle Q'$ denote the image of $\displaystyle Q$ under $\displaystyle \gamma$. Evidently, $\displaystyle A, B$ are the respective images of ${} \displaystyle D,C$ under $\displaystyle \gamma$.\n Now, since $\displaystyle XB$ is tangent to $\displaystyle \omega_1$ at $\displaystyle B$, it follows that\n \n$\angle XQ'B = \angle PQ'B = \angle XBP = \angle CBP$.\n\n But $\displaystyle Q'B$ is the image of ${} \displaystyle QC$ under the dilation $\displaystyle \gamma$, so these two lines are parallel. Hence\n \n$\angle PQC = \pi - \angle XQC = \pi - \angle XQ'B = \pi - \angle CBP$.\n\n Therefore $\displaystyle P, Q, B, C$ are concyclic, as desired. ∎\n |
(Canada)\nFor $x \in (0,1)$ let $y \in (0,1)$ be the number whose $\displaystyle n$th digit after the decimal point is the $\displaystyle (2^n)$th digit after the decimal point of $\displaystyle x$. Show that if $\displaystyle x$ is rational then so is $\displaystyle y$.\n | For any real ${ a \in (0,1) }$ and any natural number $\displaystyle n$, let $\displaystyle f_a(n)$ the $\displaystyle n$th digit after the decimal point of $\displaystyle a$. We note that $\displaystyle a$ is rational if and only if $\displaystyle f_a(n)$ is periodic for sufficiently large $\displaystyle n$, i.e., if $\displaystyle f_a(n)$ is determined by the residue of $\displaystyle n$ mod $\displaystyle m$, for some integer $\displaystyle m$.\n Suppose $\displaystyle x$ is rational, and let $\displaystyle m$ be an integer such that for sufficiently large $\displaystyle n$, $\displaystyle f_x(n)$ is determined by the residue of $\displaystyle n$ mod $\displaystyle m$. Let $\displaystyle m = 2^r \cdot k$, for some odd integer $\displaystyle k$ and some nonnegative integer $\displaystyle r$. We note that the residue of $\displaystyle n$ mod $\displaystyle m$ is uniquely determined by the residues of $\displaystyle n$ mod $\displaystyle 2^r$ and mod $\displaystyle k$. Then for sufficiently large $\displaystyle n$,\n \n$2^n \equiv 2^{n + \phi(k)} \equiv 0 \pmod{2^r}$,\n\n and\n \n$2^n \equiv 2^n \cdot 2^{\phi(k)} \equiv 2^{n+\phi(m)} \pmod{k}$,\n\n so\n \n$2^n \equiv 2^{n+\phi(k)} \pmod{m}$,\n\n and\n \n$\displaystyle f_y(n) = f_x(2^n) = f_x [2^{n+\phi(m)}] = f_y[n+\phi(m)]$.\n\n Hence $\displaystyle y$ is rational. ∎\n |
(New Zealand)\nYou are given a sequence $a_1,a_2,\dots ,a_n$ of numbers. For each $i$ ($1\leq i\leq n$) define\n $d_i=\max\{a_j:1\leq j\leq i\}-\min\{a_j:i\leq j\leq n\}$ and let\n $d=\max\{d_i:1\leq i\leq n\}$. (a) Prove that for arbitrary real numbers $x_1\leq x_2\leq \dots \leq x_n$,\n $\max\{|x_i-a_i|:1\leq i\leq n\}\geq \frac{d}{2}$. (b) Show that there exists a sequence $x_1\leq x_2\leq \dots \leq x_n$ of real numbers such that we have equality in (a).\n | This problem needs a solution. If you have a solution for it, please help us out by adding it.\n Resources |
(Bulgaria)\nConsider those functions $f:\mathbb{N}\to\mathbb{N}$ which satisfy the condition\n $f(m+n)\ge f(m)+f(f(n))-1$ for all $m, n\in\mathbb{N}$. Find all possible values of $f(2007).$\n ($\mathbb{N}$ denotes the set of all integers.)\n | |
(Thailand)\nFind all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that\n\[f(x+f(y)) = f(x+y) + f(y)\]\nfor all $x,y \in \mathbb{R}^+$. (Symbol $\mathbb{R}^+$ denotes the set of all positive real numbers [sic].)\n | We will show that $f(x) = 2x$ is the unique solution to this equation. To this end, let $g(x) = f(x) - x$. The given condition then translates to\n\[g(x+y+ g(y)) + x+y+g(y) = g(x+y) + x+y + g(y) + y ,\]\nor\n\[g(x+y+g(y)) = g(x+y) + y .\]\n Lemma 1. The function $g$ is injective.\n Proof. Suppose $g(a) = g(b)$. Then\n\[a = g(b+a+g(a)) - g(b+a) = g(a+b+g(b)) - g(a+b) = b,\]\nas desired. $\blacksquare$\n Lemma 2. If $a>b$, then $g(a+g(b)) = g(a) + b$.\n Proof. Set $y= b$, $x=a-b$. $\blacksquare$\n Lemma 3. For all $a,b$, $g(a+b) = g(a) + g(b)$.\n Proof. Pick an arbitrary positive real $c > \max(g(a)+g(b), g(a))$. Then by Lemma 2,\n\[g(c+g(a)+g(b)) = g(c+g(a))+b = g(c) + a+b = g(c+g(a+b)) .\]\nSince $g$ is injective, it follows that $c+g(a)+g(b) = c + g(a+b)$. The lemma then follows. $\blacksquare$\n Now, let $x$ be any positive real; pick some $y>x$. Then by Lemmata 3 and 2,\n\[g(x) + g(y) = g(x+y) = g(y) + x .\]\nHence $g(x)= x$ and $f(x) = g(x) + x = 2x$. Therefore the function $x \mapsto 2x$ is the only possible solution to the problem. Since this function evidently satisfies the problem's condition, it is the unique solution, as desired. $\blacksquare$\n |
Amy and Bob play the game. At the beginning, Amy writes down a positive integer on the board. Then the players take moves in turn, Bob moves first. On any move of his, Bob replaces the number $n$ on the blackboard with a number of the form $n-a^{2}$, where $a$ is a positive integer. On any move of hers, Amy replaces the number $n$ on the blackboard with a number of the form $n^{k}$, where $k$ is a positive integer. Bob wins if the number on the board becomes zero. Can Amy prevent Bob's win? | The answer is in the negative. For a positive integer $n$, we define its square-free part $S(n)$ to be the smallest positive integer $a$ such that $n / a$ is a square of an integer. In other words, $S(n)$ is the product of all primes having odd exponents in the prime expansion of $n$. We also agree that $S(0)=0$. Now we show that (i) on any move of hers, Amy does not increase the square-free part of the positive integer on the board; and (ii) on any move of his, Bob always can replace a positive integer $n$ with a non-negative integer $k$ with $S(k)<S(n)$. Thus, if the game starts by a positive integer $N$, Bob can win in at most $S(N)$ moves. Part (i) is trivial, as the definition of the square-part yields $S\left(n^{k}\right)=S(n)$ whenever $k$ is odd, and $S\left(n^{k}\right)=1 \leq S(n)$ whenever $k$ is even, for any positive integer $n$. Part (ii) is also easy: if, before Bob's move, the board contains a number $n=S(n) \cdot b^{2}$, then Bob may replace it with $n^{\prime}=n-b^{2}=(S(n)-1) b^{2}$, whence $S\left(n^{\prime}\right) \leq S(n)-1$. |
Given any positive real number $\varepsilon$, prove that, for all but finitely many positive integers $v$, any graph on $v$ vertices with at least $(1+\varepsilon) v$ edges has two distinct simple cycles of equal lengths. (Recall that the notion of a simple cycle does not allow repetition of vertices in a cycle.) | Fix a positive real number $\varepsilon$, and let $G$ be a graph on $v$ vertices with at least $(1+\varepsilon) v$ edges, all of whose simple cycles have pairwise distinct lengths.\n\nAssuming $\varepsilon^{2} v \geq 1$, we exhibit an upper bound linear in $v$ and a lower bound quadratic in $v$ for the total number of simple cycles in $G$, showing thereby that $v$ cannot be arbitrarily large, whence the conclusion.\n\nSince a simple cycle in $G$ has at most $v$ vertices, and each length class contains at most one such, $G$ has at most $v$ pairwise distinct simple cycles. This establishes the desired upper bound.\n\nFor the lower bound, consider a spanning tree for each component of $G$, and collect them all together to form a spanning forest $F$. Let $A$ be the set of edges of $F$, and let $B$ be the set of all other edges of $G$. Clearly, $|A| \leq v-1$, so $|B| \geq(1+\varepsilon) v-|A| \geq(1+\varepsilon) v-(v-1)=\varepsilon v+1>\varepsilon v$.\n\nFor each edge $b$ in $B$, adjoining $b$ to $F$ produces a unique simple cycle $C_{b}$ through $b$. Let $S_{b}$ be the set of edges in $A$ along $C_{b}$. Since the $C_{b}$ have pairwise distinct lengths, $\sum_{b \in B}\left|S_{b}\right| \geq$ $2+\cdots+(|B|+1)=|B|(|B|+3) / 2>|B|^{2} / 2>\varepsilon^{2} v^{2} / 2$.\n\nConsequently, some edge in $A$ lies in more than $\varepsilon^{2} v^{2} /(2 v)=\varepsilon^{2} v / 2$ of the $S_{b}$. Fix such an edge $a$ in $A$, and let $B^{\prime}$ be the set of all edges $b$ in $B$ whose corresponding $S_{b}$ contain $a$, so $\left|B^{\prime}\right|>\varepsilon^{2} v / 2$.\n\nFor each 2-edge subset $\left\{b_{1}, b_{2}\right\}$ of $B^{\prime}$, the union $C_{b_{1}} \cup C_{b_{2}}$ of the cycles $C_{b_{1}}$ and $C_{b_{2}}$ forms a $\theta$-graph, since their common part is a path in $F$ through $a$; and since neither of the $b_{i}$ lies along this path, $C_{b_{1}} \cup C_{b_{2}}$ contains a third simple cycle $C_{b_{1}, b_{2}}$ through both $b_{1}$ and $b_{2}$. Finally, since $B^{\prime} \cap C_{b_{1}, b_{2}}=\left\{b_{1}, b_{2}\right\}$, the assignment $\left\{b_{1}, b_{2}\right\} \mapsto C_{b_{1}, b_{2}}$ is injective, so the total number of simple cycles in $G$ is at least $\left(\begin{array}{c}\left|B^{\prime}\right| \ 2\end{array}\right)>\left(\begin{array}{c}\varepsilon^{2} v / 2 \ 2\end{array}\right)$. This establishes the desired lower bound and concludes the proof.\n\nRemarks. (1) The problem of finding two cycles of equal lengths in a graph on $v$ vertices with $2 v$ edges is known and much easier - simply consider all cycles of the form $C_{b}$.\n\nThe solution above shows that a graph on $v$ vertices with at least $v+\Theta\left(v^{3 / 4}\right)$ edges has two cycles of equal lengths. The constant $3 / 4$ is not sharp; a harder proof seems to show that $v+\Theta(\sqrt{v \log v})$ edges would suffice. On the other hand, there exist graphs on $v$ vertices with $v+\Theta(\sqrt{v})$ edges having no such cycles.\n\n(2) To avoid graph terminology, the statement of the problem may be rephrased as follows:\n\nGiven any positive real number $\varepsilon$, prove that, for all but finitely many positive integers $v$, any $v$-member company, within which there are at least $(1+\varepsilon) v$ friendship relations, satisfies the following condition: For some integer $u \geq 3$, there exist two distinct $u$ member cyclic arrangements in each of which any two neighbours are friends. (Two arrangements are distinct if they are not obtained from one another through rotation and/or symmetry; a member of the company may be included in neither arrangement, in one of them or in both.) |
Let $a, b, c, d$ be positive integers such that $a d \neq b c$ and $\operatorname{gcd}(a, b, c, d)=1$. Prove that, as $n$ runs through the positive integers, the values $\operatorname{gcd}(a n+b, c n+d)$ may achieve form the set of all positive divisors of some integer. | We extend the problem statement by allowing $a$ and $c$ take non-negative integer values, and allowing $b$ and $d$ to take arbitrary integer values. (As usual, the greatest common divisor of two integers is non-negative.) Without loss of generality, we assume $0 \leq a \leq c$. Let $S(a, b, c, d)=\left\{\operatorname{gcd}(a n+b, c n+d): n \in \mathbb{Z}_{>0}\right\}$.\n\nNow we induct on $a$. We first deal with the inductive step, leaving the base case $a=0$ to the end of the solution. So, assume that $a>0$; we intend to find a 4 -tuple $\left(a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime}\right)$ satisfying the requirements of the extended problem, such that $S\left(a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime}\right)=S(a, b, c, d)$ and $0 \leq a^{\prime}<a$, which will allow us to apply the induction hypothesis.\n\nThe construction of this 4-tuple is provided by the step of the Euclidean algorithm. Write $c=a q+r$, where $q$ and $r$ are both integers and $0 \leq r<a$. Then for every $n$ we have\n\n$$\n\operatorname{gcd}(a n+b, c n+d)=\operatorname{gcd}(a n+b, q(a n+b)+r n+d-q b)=\operatorname{gcd}(a n+b, r n+(d-q b)),\n$$\n\nso a natural intention is to define $a^{\prime}=r, b^{\prime}=d-q b, c^{\prime}=a$, and $d^{\prime}=b$ (which are already shown to satisfy $\left.S\left(a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime}\right)=S(a, b, c, d)\right)$. The check of the problem requirements is straightforward: indeed,\n\n$$\na^{\prime} d^{\prime}-b^{\prime} c^{\prime}=(c-q a) b-(d-q b) a=-(a d-b c) \neq 0\n$$\n\nand\n\n$$\n\operatorname{gcd}\left(a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime}\right)=\operatorname{gcd}(c-q a, b-q d, a, b)=\operatorname{gcd}(c, d, a, b)=1\n$$\n\nThus the step is verified.\n\nIt remains to deal with the base case $a=0$, i.e., to examine the set $S(0, b, c, d)$ with $b c \neq 0$ and $\operatorname{gcd}(b, c, d)=1$. Let $b^{\prime}$ be the integer obtained from $b$ by ignoring all primes $b$ and $c$ share (none of them divides $c n+d$ for any integer $n$, otherwise $\operatorname{gcd}(b, c, d)>1)$. We thus get $\operatorname{gcd}\left(b^{\prime}, c\right)=1$ and $S\left(0, b^{\prime}, c, d\right)=S(0, b, c, d)$.\n\nFinally, it is easily seen that $S\left(0, b^{\prime}, c, d\right)$ is the set of all positive divisors of $b^{\prime}$. Each member of $S\left(0, b^{\prime}, c, d\right)$ is clearly a divisor of $b^{\prime}$. Conversely, if $\delta$ is a positive divisor of $b^{\prime}$, then $c n+d \equiv \delta$ $\left(\bmod b^{\prime}\right)$ for some $n$, since $b^{\prime}$ and $c$ are coprime, so $\delta$ is indeed a member of $S\left(0, b^{\prime}, c, d\right)$.\n |
Let $n$ be a positive integer and fix $2 n$ distinct points on a circumference. Split these points into $n$ pairs and join the points in each pair by an arrow (i.e., an oriented line segment). The resulting configuration is good if no two arrows cross, and there are no arrows $\overrightarrow{A B}$ and $\overrightarrow{C D}$ such that $A B C D$ is a convex quadrangle oriented clockwise. Determine the number of good configurations. | The required number is $\left(\begin{array}{c}2 n \ n\end{array}\right)$. To prove this, trace the circumference counterclockwise to label the points $a_{1}, a_{2}, \ldots, a_{2 n}$.\n\nLet $\mathcal{C}$ be any good configuration and let $O(\mathcal{C})$ be the set of all points from which arrows emerge. We claim that every $n$-element subset $S$ of $\left\{a_{1}, \ldots, a_{2 n}\right\}$ is an $O$-image of a unique good configuration; clearly, this provides the answer.\n\nTo prove the claim induct on $n$. The base case $n=1$ is clear. For the induction step, consider any $n$-element subset $S$ of $\left\{a_{1}, \ldots, a_{2 n}\right\}$, and assume that $S=O(\mathcal{C})$ for some good configuration $\mathcal{C}$. Take any index $k$ such that $a_{k} \in S$ and $a_{k+1} \notin S$ (assume throughout that indices are cyclic modulo $2 n$, i.e., $a_{2 n+1}=a_{1}$ etc.).\n\nIf the arrow from $a_{k}$ points to some $a_{\ell}, k+1<\ell(<2 n+k)$, then the arrow pointing to $a_{k+1}$ emerges from some $a_{m}, m$ in the range $k+2$ through $\ell-1$, since these two arrows do not cross. Then the arrows $a_{k} \rightarrow a_{\ell}$ and $a_{m} \rightarrow a_{k+1}$ form a prohibited quadrangle. Hence, $\mathcal{C}$ contains an arrow $a_{k} \rightarrow a_{k+1}$.\n\nOn the other hand, if any configuration $\mathcal{C}$ contains the arrow $a_{k} \rightarrow a_{k+1}$, then this arrow cannot cross other arrows, neither can it occur in prohibited quadrangles.\n\nThus, removing the points $a_{k}, a_{k+1}$ from $\left\{a_{1}, \ldots, a_{2 n}\right\}$ and the point $a_{k}$ from $S$, we may apply the induction hypothesis to find a unique good configuration $\mathcal{C}^{\prime}$ on $2 n-2$ points compatible with the new set of sources (i.e., points from which arrows emerge). Adjunction of the arrow $a_{k} \rightarrow a_{k+1}$ to $\mathcal{C}^{\prime}$ yields a unique good configuration on $2 n$ points, as required.\n |
Let $n$ be a positive integer and fix $2 n$ distinct points on a circumference. Split these points into $n$ pairs and join the points in each pair by an arrow (i.e., an oriented line segment). The resulting configuration is good if no two arrows cross, and there are no arrows $\overrightarrow{A B}$ and $\overrightarrow{C D}$ such that $A B C D$ is a convex quadrangle oriented clockwise. Determine the number of good configurations. | \nUse the counterclockwise labelling $a_{1}, a_{2}, \ldots, a_{2 n}$ in the solution above.\n\nLetting $D_{n}$ be the number of good configurations on $2 n$ points, we establish a recurrence relation for the $D_{n}$. To this end, let $C_{n}=\frac{(2 n) !}{n !(n+1) !}$ the $n$th Catalan number; it is well-known that $C_{n}$ is the number of ways to connect $2 n$ given points on the circumference by $n$ pairwise disjoint chords.\n\nSince no two arrows cross, in any good configuration the vertex $a_{1}$ is connected to some $a_{2 k}$. Fix $k$ in the range 1 through $n$ and count the number of good configurations containing the arrow $a_{1} \rightarrow a_{2 k}$. Let $\mathcal{C}$ be any such configuration.\n\nIn $\mathcal{C}$, the vertices $a_{2}, \ldots, a_{2 k-1}$ are paired off with one other, each arrow pointing from the smaller to the larger index, for otherwise it would form a prohibited quadrangle with $a_{1} \rightarrow a_{2 k}$. Consequently, there are $C_{k-1}$ ways of drawing such arrows between $a_{2}, \ldots, a_{2 k-1}$.\n\nOn the other hand, the arrows between $a_{2 k+1}, \ldots, a_{2 n}$ also form a good configuration, which can be chosen in $D_{n-k}$ ways. Finally, it is easily seen that any configuration of the first kind and any configuration of the second kind combine together to yield an overall good configuration.\n\nThus the number of good configurations containing the arrow $a_{1} \rightarrow a_{2 k}$ is $C_{k-1} D_{n-k}$. Clearly, this is also the number of good configurations containing the arrow $a_{2(n-k+1)} \rightarrow a_{1}$, so\n\n$$\nD_{n}=2 \sum_{k=1}^{n} C_{k-1} D_{n-k}\n$$\n\nTo find an explicit formula for $D_{n}$, let $d(x)=\sum_{n=0}^{\infty} D_{n} x^{n}$ and let $c(x)=\sum_{n=0}^{\infty} C_{n} x^{n}=$ $\frac{1-\sqrt{1-4 x}}{2 x}$ be the generating functions of the $D_{n}$ and the $C_{n}$, respectively. Since $D_{0}=1$, relation (*) yields $d(x)=2 x c(x) d(x)+1$, so\n\n$$\n\begin{aligned}\nd(x)=\frac{1}{1-2 x c(x)}=(1-4 x)^{-1 / 2} & =\sum_{n \geq 0}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \ldots\left(-\frac{2 n-1}{2}\right) \frac{(-4 x)^{n}}{n !} \\n& =\sum_{n \geq 0} \frac{2^{n}(2 n-1) ! !}{n !} x^{n}=\sum_{n \geq 0}\left(\begin{array}{c}\n2 n \\nn\n\end{array}\right) x^{n} .\n\end{aligned}\n$$\n\nConsequently, $D_{n}=\left(\begin{array}{c}2 n \ n\end{array}\right)$.\n |
Let $n$ be a positive integer and fix $2 n$ distinct points on a circumference. Split these points into $n$ pairs and join the points in each pair by an arrow (i.e., an oriented line segment). The resulting configuration is good if no two arrows cross, and there are no arrows $\overrightarrow{A B}$ and $\overrightarrow{C D}$ such that $A B C D$ is a convex quadrangle oriented clockwise. Determine the number of good configurations. | Let $C_{n}=\frac{1}{n+1}\left(\begin{array}{c}2 n \ n\end{array}\right)$ denote the $n$th Catalan number and recall that there are exactly $C_{n}$ ways to join $2 n$ distinct points on a circumference by $n$ pairwise disjoint chords. Such a configuration of chords will be referred to as a Catalan n-configuration. An orientation of the chords in a Catalan configuration $\mathcal{C}$ making it into a good configuration (in the sense defined in the statement of the problem) will be referred to as a good orientation for $\mathcal{C}$.\n\nWe show by induction on $n$ that there are exactly $n+1$ good orientations for any Catalan $n$-configuration, so there are exactly $(n+1) C_{n}=\left(\begin{array}{c}2 n \ n\end{array}\right)$ good configurations on $2 n$ points. The base case $n=1$ is clear.\n\nFor the induction step, let $n>1$, let $\mathcal{C}$ be a Catalan $n$-configuration, and let $a b$ be a chord of minimal length in $\mathcal{C}$. By minimality, the endpoints of the other chords in $\mathcal{C}$ all lie on the major arc $a b$ of the circumference.\n\nLabel the $2 n$ endpoints $1,2, \ldots, 2 n$ counterclockwise so that $\{a, b\}=\{1,2\}$, and notice that the good orientations for $\mathcal{C}$ fall into two disjoint classes: Those containing the arrow $1 \rightarrow 2$, and those containing the opposite arrow.\n\nSince the arrow $1 \rightarrow 2$ cannot be involved in a prohibited quadrangle, the induction hypothesis applies to the Catalan $(n-1)$-configuration formed by the other chords to show that the first class contains exactly $n$ good orientations.\n\nFinally, the second class consists of a single orientation, namely, $2 \rightarrow 1$, every other arrow emerging from the smaller endpoint of the respective chord; a routine verification shows that this is indeed a good orientation. This completes the induction step and ends the proof.\n |
Problem 2. Determine whether there exist non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying\n\n$$\nP(x)^{10}+P(x)^{9}=Q(x)^{21}+Q(x)^{20} .\n$$ | \nThe answer is in the negative. Comparing the degrees of both sides in $(*)$ we get $\operatorname{deg} P=21 n$ and $\operatorname{deg} Q=10 n$ for some positive integer $n$. Take the derivative of $(*)$ to obtain\n\n$$\nP^{\prime} P^{8}(10 P+9)=Q^{\prime} Q^{19}(21 Q+20) \text {. }\n$$\n\nSince $\operatorname{gcd}(10 P+9, P)=\operatorname{gcd}(10 P+9, P+1)=1$, it follows that $\operatorname{gcd}\left(10 P+9, P^{9}(P+1)\right)=1$, so $\operatorname{gcd}(10 P+9, Q)=1$, by $(*)$. Thus $(* *)$ yields $10 P+9 \mid Q^{\prime}(21 Q+20)$, which is impossible since $0<\operatorname{deg}\left(Q^{\prime}(21 Q+20)\right)=20 n-1<21 n=\operatorname{deg}(10 P+9)$. A contradiction.\n |
Problem 2. Determine whether there exist non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying\n\n$$\nP(x)^{10}+P(x)^{9}=Q(x)^{21}+Q(x)^{20} .\n$$ | \nSolution 2. Letting $r$ and $s$ be integers such that $r \geq 2$ and $s \geq 2 r$, we show that if $P^{r}+P^{r-1}=$ $Q^{s}+Q^{s-1}$, then $Q$ is constant.\n\nLet $m=\operatorname{deg} P$ and $n=\operatorname{deg} Q$. A degree inspection in the given relation shows that $m \geq 2 n$.\n\nWe will prove that $P(P+1)$ has at least $m+1$ distinct complex roots. Assuming this for the moment, notice that $Q$ takes on one of the values 0 or -1 at each of those roots. Since $m+1 \geq 2 n+1$, it follows that $Q$ takes on one of the values 0 and -1 at more than $n$ distinct points, so $Q$ must be constant.\n\nFinally, we prove that $P(P+1)$ has at least $m+1$ distinct complex roots. This can be done either by referring to the Mason-Stothers theorem or directly, in terms of multiplicities of the roots in question.\n\nSince $P$ and $P+1$ are relatively prime, the Mason-Stothers theorem implies that the number of distinct roots of $P(P+1)$ is greater than $m$, hence at least $m+1$.\n\nFor a direct proof, let $z_{1}, \ldots, z_{t}$ be the distinct complex roots of $P(P+1)$, and let $z_{k}$ have multiplicity $\alpha_{k}, k=1, \ldots, t$. Since $P$ and $P+1$ have no roots in common, and $P^{\prime}=(P+1)^{\prime}$, it follows that $P^{\prime}$ has a root of multiplicity $\alpha_{k}-1$ at $z_{k}$. Consequently, $m-1=\operatorname{deg} P^{\prime} \geq$ $\sum_{k=1}^{t}\left(\alpha_{k}-1\right)=\sum_{k=1}^{t} \alpha_{k}-t=2 m-t$; that is, $t \geq m+1$. This completes the proof.\n |
Ann and Bob play a game on an infinite checkered plane making moves in turn; Ann makes the first move. A move consists in orienting any unit grid-segment that has not been oriented before. If at some stage some oriented segments form an oriented cycle, Bob wins. Does Bob have a strategy that guarantees him to win? | \nThe answer is in the negative: Ann has a strategy allowing her to prevent Bob's victory.\n\nWe say that two unit grid-segments form a low-left corner (or LL-corner) if they share an endpoint which is the lowest point of one and the leftmost point of the other. An up-right corner (or UR-corner) is defined similarly. The common endpoint of two unit grid-segments at a corner is the joint of that corner.\n\nFix a vertical line on the grid and call it the midline; the unit grid-segments along the midline are called middle segments. The unit grid-segments lying to the left/right of the midline are called left/right segments. Partition all left segments into LL-corners, and all right segments into UR-corners.\n\nWe now describe Ann's strategy. Her first move consists in orienting some middle segment arbitrarily. Assume that at some stage, Bob orients some segment $s$. If $s$ is a middle segment, Ann orients any free middle segment arbitrarily. Otherwise, $s$ forms a corner in the partition with some other segment $t$. Then Ann orients $t$ so that the joint of the corner is either the source of both arrows, or the target of both. Notice that after any move of Ann's, each corner in the partition is either completely oriented or completely not oriented. This means that Ann can always make a required move.\n\nAssume that Bob wins at some stage, i.e., an oriented cycle $C$ occurs. Let $X$ be the lowest of the leftmost points of $C$, and let $Y$ be the topmost of the rightmost points of $C$. If $X$ lies (strictly) to the left of the midline, then $X$ is the joint of some corner whose segments are both oriented. But, according to Ann's strategy, they are oriented so that they cannot occur in a cycle - a contradiction. Otherwise, $Y$ lies to the right of the midline, and a similar argument applies. Thus, Bob will never win, as desired.\n |
For a positive integer $a$, define a sequence of integers $x_{1}, x_{2}, \ldots$ by letting $x_{1}=a$ and $x_{n+1}=2 x_{n}+1$ for $n \geq 1$. Let $y_{n}=2^{x_{n}}-1$. Determine the largest possible $k$ such that, for some positive integer $a$, the numbers $y_{1}, \ldots, y_{k}$ are all prime. | \nThe largest such is $k=2$. Notice first that if $y_{i}$ is prime, then $x_{i}$ is prime as well. Actually, if $x_{i}=1$ then $y_{i}=1$ which is not prime, and if $x_{i}=m n$ for integer $m, n>1$ then $2^{m}-1 \mid 2^{x_{i}}-1=y_{i}$, so $y_{i}$ is composite. In particular, if $y_{1}, y_{2}, \ldots, y_{k}$ are primes for some $k \geq 1$ then $a=x_{1}$ is also prime.\n\nNow we claim that for every odd prime $a$ at least one of the numbers $y_{1}, y_{2}, y_{3}$ is composite (and thus $k<3$ ). Assume, to the contrary, that $y_{1}, y_{2}$, and $y_{3}$ are primes; then $x_{1}, x_{2}, x_{3}$ are primes as well. Since $x_{1} \geq 3$ is odd, we have $x_{2}>3$ and $x_{2} \equiv 3(\bmod 4)$; consequently, $x_{3} \equiv 7$ $(\bmod 8)$. This implies that 2 is a quadratic residue modulo $p=x_{3}$, so $2 \equiv s^{2}(\bmod p)$ for some integer $s$, and hence $2^{x_{2}}=2^{(p-1) / 2} \equiv s^{p-1} \equiv 1(\bmod p)$. This means that $p \mid y_{2}$, thus $2^{x_{2}}-1=x_{3}=2 x_{2}+1$. But it is easy to show that $2^{t}-1>2 t+1$ for all integer $t>3$. A contradiction.\n |
Does there exist a pair $(g, h)$ of functions $g, h: \mathbb{R} \rightarrow \mathbb{R}$ such that the only function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \in \mathbb{R}$ is the identity function $f(x) \equiv x$ ? | \nSuch a tester pair exists. We may biject $\mathbb{R}$ with the closed unit interval, so it suffices to find a tester pair for that instead. We give an explicit example: take some positive real numbers $\alpha, \beta$ (which we will specify further later). Take\n\n$$\ng(x)=\max (x-\alpha, 0) \quad \text { and } \quad h(x)=\min (x+\beta, 1) \text {. }\n$$\n\nSay a set $S \subseteq[0,1]$ is invariant if $f(S) \subseteq S$ for all functions $f$ commuting with both $g$ and $h$. Note that intersections and unions of invariant sets are invariant. Preimages of invariant sets under $g$ and $h$ are also invariant; indeed, if $S$ is invariant and, say, $T=g^{-1}(S)$, then $g(f(T))=$ $f(g(T)) \subseteq f(S) \subseteq S$, thus $f(T) \subseteq T$.\n\nWe claim that (if we choose $\alpha+\beta<1$ ) the intervals $[0, n \alpha-m \beta]$ are invariant where $n$ and $m$ are nonnegative integers with $0 \leq n \alpha-m \beta \leq 1$. We prove this by induction on $m+n$.\n\nThe set $\{0\}$ is invariant, as for any $f$ commuting with $g$ we have $g(f(0))=f(g(0))=f(0)$, so $f(0)$ is a fixed point of $g$. This gives that $f(0)=0$, thus the induction base is established.\n\nSuppose now we have some $m, n$ such that $\left[0, n^{\prime} \alpha-m^{\prime} \beta\right]$ is invariant whenever $m^{\prime}+n^{\prime}<$ $m+n$. At least one of the numbers $(n-1) \alpha-m \beta$ and $n \alpha-(m-1) \beta$ lies in $(0,1)$. Note however that in the first case $[0, n \alpha-m \beta]=g^{-1}([0,(n-1) \alpha-m \beta])$, so $[0, n \alpha-m \beta]$ is invariant. In the second case $[0, n \alpha-m \beta]=h^{-1}([0, n \alpha-(m-1) \beta])$, so again $[0, n \alpha-m \beta]$ is invariant. This completes the induction.\n\nWe claim that if we choose $\alpha+\beta<1$, where $0<\alpha \notin \mathbb{Q}$ and $\beta=1 / k$ for some integer $k>1$, then all intervals $[0, \delta]$ are invariant for $0 \leq \delta<1$. This occurs, as by the previous claim, for all nonnegative integers $n$ we have $[0,(n \alpha \bmod 1)]$ is invariant. The set of $n \alpha \bmod 1$ is dense in $[0,1]$, so in particular\n\n$$\n[0, \delta]=\bigcap_{(n \alpha \bmod 1)>\delta}[0,(n \alpha \bmod 1)]\n$$\n\nis invariant.\n\nA similar argument establishes that $[\delta, 1]$ is invariant, so by intersecting these $\{\delta\}$ is invariant for $0<\delta<1$. Yet we also have $\{0\},\{1\}$ both invariant, which proves $f$ to be the identity.\n |
Does there exist a pair $(g, h)$ of functions $g, h: \mathbb{R} \rightarrow \mathbb{R}$ such that the only function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \in \mathbb{R}$ is the identity function $f(x) \equiv x$ ? | \nLet us agree that a sequence $\mathbf{x}=\left(x_{n}\right)_{n=1,2, \ldots}$ is cofinally non-constant if for every index $m$ there exists an index $n>m$ such that $x_{m} \neq x_{n}$.\n\nBiject $\mathbb{R}$ with the set of cofinally non-constant sequences of 0's and 1's, and define $g$ and $h$ by\n\n$$\ng(\epsilon, \mathbf{x})=\left\{\begin{array}{ll}\n\epsilon, \mathbf{x} & \text { if } \epsilon=0 \\n\mathbf{x} & \text { else }\n\end{array} \quad \text { and } \quad h(\epsilon, \mathbf{x})= \begin{cases}\epsilon, \mathbf{x} & \text { if } \epsilon=1 \\n\mathbf{x} & \text { else }\end{cases}\right.\n$$\n\nwhere $\epsilon, \mathbf{x}$ denotes the sequence formed by appending $\mathbf{x}$ to the single-element sequence $\epsilon$. Note that $g$ fixes precisely those sequences beginning with 0 , and $h$ fixes precisely those beginning with 1.\n\nNow assume that $f$ commutes with both $f$ and $g$. To prove that $f(\mathbf{x})=\mathbf{x}$ for all $\mathbf{x}$ we show that $\mathbf{x}$ and $f(\mathbf{x})$ share the same first $n$ terms, by induction on $n$.\n\nThe base case $n=1$ is simple, as we have noticed above that the set of sequences beginning with a 0 is precisely the set of $g$-fixed points, so is preserved by $f$, and similarly for the set of sequences starting with 1.\n |
Does there exist a pair $(g, h)$ of functions $g, h: \mathbb{R} \rightarrow \mathbb{R}$ such that the only function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \in \mathbb{R}$ is the identity function $f(x) \equiv x$ ? | \nSuppose that $f(\mathbf{x})$ and $\mathbf{x}$ agree for the first $n$ terms, whatever $\mathbf{x}$. Consider any sequence, and write it as $\mathbf{x}=\epsilon, \mathbf{y}$. Without loss of generality, we may (and will) assume that $\epsilon=0$, so $f(\mathbf{x})=0, \mathbf{y}^{\prime}$ by the base case. Yet then $f(\mathbf{y})=f(h(\mathbf{x}))=h(f(\mathbf{x}))=h\left(0, \mathbf{y}^{\prime}\right)=\mathbf{y}^{\prime}$. Consequently, $f(\mathbf{x})=0, f(\mathbf{y})$, so $f(\mathbf{x})$ and $\mathbf{x}$ agree for the first $n+1$ terms by the inductive hypothesis.\n\nThus $f$ fixes all of cofinally non-constant sequences, and the conclusion follows.\n |
Does there exist a pair $(g, h)$ of functions $g, h: \mathbb{R} \rightarrow \mathbb{R}$ such that the only function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \in \mathbb{R}$ is the identity function $f(x) \equiv x$ ? | \nWe will show that there exists a tester pair of bijective functions $g$ and $h$.\n\nFirst of all, let us find out when a pair of functions is a tester pair. Let $g, h: \mathbb{R} \rightarrow \mathbb{R}$ be arbitrary functions. We construct a directed graph $G_{g, h}$ with $\mathbb{R}$ as the set of vertices, its edges being painted with two colors: for every vertex $x \in \mathbb{R}$, we introduce a red edge $x \rightarrow g(x)$ and a blue edge $x \rightarrow h(x)$.\n\nNow, assume that the function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \in \mathbb{R}$. This means exactly that if there exists an edge $x \rightarrow y$, then there also exists an edge $f(x) \rightarrow f(y)$ of the same color; that is $-f$ is an endomorphism of $G_{g, h}$.\n\nThus, a pair $(g, h)$ is a tester pair if and only if the graph $G_{g, h}$ admits no nontrivial endomorphisms. Notice that each endomorphism maps a component into a component. Thus, to construct a tester pair, it suffices to construct a continuum of components with no nontrivial endomorphisms and no homomorphisms from one to another. It can be done in many ways; below we present one of them.\n\nLet $g(x)=x+1$; the construction of $h$ is more involved. For every $x \in[0,1)$ we define the set $S_{x}=x+\mathbb{Z}$; the sets $S_{x}$ will be exactly the components of $G_{g, h}$. Now we will construct these components.\n\nLet us fix any $x \in[0,1)$; let $x=0 . x_{1} x_{2} \ldots$ be the binary representation of $x$. Define $h(x-n)=x-n+1$ for every $n>3$. Next, let $h(x-3)=x, h(x)=x-2, h(x-2)=x-1$, and $h(x-1)=x+1$ (that would be a "marker" which fixes a point in our component).\n\nNext, for every $i=1,2, \ldots$, we define\n\n(1) $h(x+3 i-2)=x+3 i-1, h(x+3 i-1)=x+3 i$, and $h(x+3 i)=x+3 i+1$, if $x_{i}=0$;\n\n(2) $h(x+3 i-2)=x+3 i, h(x+3 i)=3 i-1$, and $h(x+3 i-1)=x+3 i+1$, if $x_{i}=1$.\n\nClearly, $h$ is a bijection mapping each $S_{x}$ to itself. Now we claim that the graph $G_{g, h}$ satisfies the desired conditions.\n\nConsider any homomorphism $f_{x}: S_{x} \rightarrow S_{y}$ ( $x$ and $y$ may coincide). Since $g$ is a bijection, consideration of the red edges shows that $f_{x}(x+n)=x+n+k$ for a fixed real $k$. Next, there exists a blue edge $(x-3) \rightarrow x$, and the only blue edge of the form $(y+m-3) \rightarrow(y+m)$ is $(y-3) \rightarrow y$; thus $f_{x}(x)=y$, and $k=0$.\n\nNext, if $x_{i}=0$ then there exists a blue edge $(x+3 i-2) \rightarrow(x+3 i-1)$; then the edge $(y+3 i-2) \rightarrow(y+3 i-1)$ also should exist, so $y_{i}=0$. Analogously, if $x_{i}=1$ then there exists a blue edge $(x+3 i-2) \rightarrow(x+3 i)$; then the edge $(y+3 i-2) \rightarrow(y+3 i)$ also should exist, so $y_{i}=1$. We conclude that $x=y$, and $f_{x}$ is the identity mapping, as required.\n |
Let $A B C D$ be a quadrilateral inscribed in a circle $\omega$. The lines $A B$ and $C D$ meet at $P$, the lines $A D$ and $B C$ meet at $Q$, and the diagonals $A C$ and $B D$ meet at $R$. Let $M$ be the midpoint of the segment $P Q$, and let $K$ be the common point of the segment $M R$ and the circle $\omega$. Prove that the circumcircle of the triangle $K P Q$ and $\omega$ are tangent to one another. | \nLet $O$ be the centre of $\omega$. Notice that the points $P, Q$, and $R$ are the poles (with respect to $\omega$ ) of the lines $Q R, R P$, and $P Q$, respectively. Hence we have $O P \perp Q R, O Q \perp R P$, and $O R \perp P Q$, thus $R$ is the orthocentre of the triangle $O P Q$. Now, if $M R \perp P Q$, then the points $P$ and $Q$ are the reflections of one another in the line $M R=M O$, and the triangle $P Q K$ is symmetrical with respect to this line. In this case the statement of the problem is trivial.\n\nOtherwise, let $V$ be the foot of the perpendicular from $O$ to $M R$, and let $U$ be the common point of the lines $O V$ and $P Q$. Since $U$ lies on the polar line of $R$ and $O U \perp M R$, we obtain that $U$ is the pole of $M R$. Therefore, the line $U K$ is tangent to $\omega$. Hence it is enough to prove that $U K^{2}=U P \cdot U Q$, since this relation implies that $U K$ is also tangent to the circle $K P Q$.\n\nFrom the rectangular triangle $O K U$, we get $U K^{2}=U V \cdot U O$. Let $\Omega$ be the circumcircle of triangle $O P Q$, and let $R^{\prime}$ be the reflection of its orthocentre $R$ in the midpoint $M$ of the side $P Q$. It is well known that $R^{\prime}$ is the point of $\Omega$ opposite to $O$, hence $O R^{\prime}$ is the diameter of $\Omega$. Finally, since $\angle O V R^{\prime}=90^{\circ}$, the point $V$ also lies on $\Omega$, hence $U P \cdot U Q=U V \cdot U O=U K^{2}$, as required.\n |
Given an integer $k \geq 2$, set $a_{1}=1$ and, for every integer $n \geq 2$, let $a_{n}$ be the smallest $x>a_{n-1}$ such that:\n\n$$\nx=1+\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{x}{a_{i}}}\right\rfloor\n$$\n\nProve that every prime occurs in the sequence $a_{1}, a_{2}, \ldots$ | \nWe prove that the $a_{n}$ are precisely the $k$ th-power-free positive integers, that is, those divisible by the $k$ th power of no prime. The conclusion then follows.\n\nLet $B$ denote the set of all $k$ th-power-free positive integers. We first show that, given a positive integer $c$,\n\n$$\n\sum_{b \in B, b \leq c}\left\lfloor\sqrt[k]{\frac{c}{b}}\right\rfloor=c\n$$\n\nTo this end, notice that every positive integer has a unique representation as a product of an element in $B$ and a $k$ th power. Consequently, the set of all positive integers less than or equal to $c$ splits into\n\n$$\nC_{b}=\left\{x: x \in \mathbb{Z}_{>0}, x \leq c, \text { and } x / b \text { is a } k \text { th power }\right\}, \quad b \in B, b \leq c .\n$$\n\nClearly, $\left|C_{b}\right|=\lfloor\sqrt[k]{c / b}\rfloor$, whence the desired equality.\n\nFinally, enumerate $B$ according to the natural order: $1=b_{1}<b_{2}<\cdots<b_{n}<\cdots$. We prove by induction on $n$ that $a_{n}=b_{n}$. Clearly, $a_{1}=b_{1}=1$, so let $n \geq 2$ and assume $a_{m}=b_{m}$ for all indices $m<n$. Since $b_{n}>b_{n-1}=a_{n-1}$ and\n\n$$\nb_{n}=\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{b_{n}}{b_{i}}}\right\rfloor=\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{b_{n}}{b_{i}}}\right\rfloor+1=\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{b_{n}}{a_{i}}}\right\rfloor+1\n$$\n\nthe definition of $a_{n}$ forces $a_{n} \leq b_{n}$. Were $a_{n}<b_{n}$, a contradiction would follow:\n\n$$\na_{n}=\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{a_{n}}{b_{i}}}\right\rfloor=\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{a_{n}}{a_{i}}}\right\rfloor=a_{n}-1\n$$\n\nConsequently, $a_{n}=b_{n}$. This completes the proof.\n |
Given an integer $k \geq 2$, set $a_{1}=1$ and, for every integer $n \geq 2$, let $a_{n}$ be the smallest $x>a_{n-1}$ such that:\n\n$$\nx=1+\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{x}{a_{i}}}\right\rfloor\n$$\n\nProve that every prime occurs in the sequence $a_{1}, a_{2}, \ldots$ | \nFor every $n=1,2,3, \ldots$, introduce the function\n\n$$\nf_{n}(x)=x-1-\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{x}{a_{i}}}\right\rfloor\n$$\n\nDenote also by $g_{n}(x)$ the number of the indices $i \leq n$ such that $x / a_{i}$ is the $k$ th power of an integer. Then $f_{n}(x+1)-f_{n}(x)=1-g_{n}(x)$ for every integer $x \geq a_{n}$; hence $f_{n}(x)+1 \geq f_{n}(x+1)$. Moreover, $f_{n}\left(a_{n-1}\right)=-1$ (since $f_{n-1}\left(a_{n-1}\right)=0$ ). Now a straightforward induction shows that $f_{n}(x)<0$ for all integers $x \in\left[a_{n-1}, a_{n}\right)$.\n\nNext, if $g_{n}(x)>0$ then $f_{n}(x) \leq f_{n}(x-1)$; this means that such an $x$ cannot equal $a_{n}$. Thus $a_{j} / a_{i}$ is never the $k$ th power of an integer if $j>i$.\n\nNow we are prepared to prove by induction on $n$ that $a_{1}, a_{2}, \ldots, a_{n}$ are exactly all $k$ thpower-free integers in $\left[1, a_{n}\right]$. The base case $n=1$ is trivial.\n\nAssume that all the $k$ th-power-free integers on $\left[1, a_{n}\right]$ are exactly $a_{1}, \ldots, a_{n}$. Let $b$ be the least integer larger than $a_{n}$ such that $g_{n}(b)=0$. We claim that: (1) $b=a_{n+1}$; and (2) $b$ is the least $k$ th-power-free number greater than $a_{n}$.\n\nTo prove (1), notice first that all the numbers of the form $a_{j} / a_{i}$ with $1 \leq i<j \leq n$ are not $k$ th powers of rational numbers since $a_{i}$ and $a_{j}$ are $k$ th-power-free. This means that for every integer $x \in\left(a_{n}, b\right)$ there exists exactly one index $i \leq n$ such that $x / a_{i}$ is the $k$ th power of an integer (certainly, $x$ is not $k$ th-power-free). Hence $f_{n+1}(x)=f_{n+1}(x-1)$ for each such $x$, so $f_{n+1}(b-1)=f_{n+1}\left(a_{n}\right)=-1$. Next, since $b / a_{i}$ is not the $k$ th power of an integer, we have $f_{n+1}(b)=f_{n+1}(b-1)+1=0$, thus $b=a_{n+1}$. This establishes (1).\n\nFinally, since all integers in $\left(a_{n}, b\right)$ are not $k$ th-power-free, we are left to prove that $b$ is $k$ th-power-free to establish (2). Otherwise, let $y>1$ be the greatest integer such that $y^{k} \mid b$; then $b / y^{k}$ is $k$ th-power-free and hence $b / y^{k}=a_{i}$ for some $i \leq n$. So $b / a_{i}$ is the $k$ th power of an integer, which contradicts the definition of $b$.\n\nThus $a_{1}, a_{2}, \ldots$ are exactly all $k$ th-power-free positive integers; consequently all primes are contained in this sequence.\n |
\n$2 n$ distinct tokens are placed at the vertices of a regular $2 n$-gon, with one token placed at each vertex. A move consists of choosing an edge of the $2 n$-gon and interchanging the two tokens at the endpoints of that edge. Suppose that after a finite number of moves, every pair of tokens have been interchanged exactly once. Prove that some edge has never been chosen.\n | \nStep 1. Enumerate all the tokens in the initial arrangement in clockwise circular order; also enumerate the vertices of the $2 n$-gon accordingly. Consider any three tokens $i<j<k$. At each moment, their cyclic order may be either $i, j, k$ or $i, k, j$, counted clockwise. This order changes exactly when two of these three tokens have been switched. Hence the order has been reversed thrice, and in the final arrangement token $k$ stands on the arc passing clockwise from token $i$ to token $j$. Thus, at the end, token $i+1$ is a counter-clockwise neighbor of token $i$ for all $i=1,2, \ldots, 2 n-1$, so the tokens in the final arrangement are numbered successively in counter-clockwise circular order.\n\nThis means that the final arrangement of tokens can be obtained from the initial one by reflection in some line $\ell$.\n\nStep 2. Notice that each token was involved into $2 n-1$ switchings, so its initial and final vertices have different parity. Hence $\ell$ passes through the midpoints of two opposite sides of a $2 n$-gon; we may assume that these are the sides $a$ and $b$ connecting $2 n$ with 1 and $n$ with $n+1$, respectively.\n\nDuring the process, each token $x$ has crossed $\ell$ at least once; thus one of its switchings has been made at edge $a$ or at edge $b$. Assume that some two its switchings were performed at $a$ and at $b$; we may (and will) assume that the one at $a$ was earlier, and $x \leq n$. Then the total movement of token $x$ consisted at least of: (i) moving from vertex $x$ to $a$ and crossing $\ell$ along $a$; (ii) moving from $a$ to $b$ and crossing $\ell$ along $b$; (iii) coming to vertex $2 n+1-x$. This tales at least $x+n+(n-x)=2 n$ switchings, which is impossible.\n\nThus, each token had a switching at exactly one of the edges $a$ and $b$.\n\nStep 3. Finally, let us show that either each token has been switched at $a$, or each token has been switched at $b$ (then the other edge has never been used, as desired). To the contrary, assume that there were switchings at both $a$ and at $b$. Consider the first such switchings, and let $x$ and $y$ be the tokens which were moved clockwise during these switchings and crossed $\ell$ at $a$ and $b$, respectively. By Step 2, $x \neq y$. Then tokens $x$ and $y$ initially were on opposite sides of $\ell$.\n\nNow consider the switching of tokens $x$ and $y$; there was exactly one such switching, and we assume that it has been made on the same side of $\ell$ as vertex $y$. Then this switching has been made after token $x$ had traced $a$. From this point on, token $x$ is on the clockwise arc from token $y$ to $b$, and it has no way to leave out from this arc. But this is impossible, since token $y$ should trace $b$ after that moment. A contradiction.\n |
\nDoes there exist an infinite sequence of positive integers $a_{1}, a_{2}, a_{3}, \ldots$ such that $a_{m}$ and $a_{n}$ are coprime if and only if $|m-n|=1$ ?\n | \nThe answer is in the affirmative.\n\nThe idea is to consider a sequence of pairwise distinct primes $p_{1}, p_{2}, p_{3}, \ldots$, cover the positive integers by a sequence of finite non-empty sets $I_{n}$ such that $I_{m}$ and $I_{n}$ are disjoint if and only if $m$ and $n$ are one unit apart, and set $a_{n}=\prod_{i \in I_{n}} p_{i}, n=1,2,3, \ldots$\n\nOne possible way of finding such sets is the following. For all positive integers $n$, let\n\n$$\n\begin{aligned}\n& 2 n \in I_{k} \quad \text { for all } k=n, n+3, n+5, n+7, \ldots ; \quad \text { and } \\n& 2 n-1 \in I_{k} \quad \text { for all } k=n, n+2, n+4, n+6, \ldots\n\end{aligned}\n$$\n\nClearly, each $I_{k}$ is finite, since it contains none of the numbers greater than $2 k$. Next, the number $p_{2 n}$ ensures that $I_{n}$ has a common element with each $I_{n+2 i}$, while the number $p_{2 n-1}$ ensures that $I_{n}$ has a common element with each $I_{n+2 i+1}$ for $i=1,2, \ldots$ Finally, none of the indices appears in two consecutive sets.\n |
\nFor an integer $n \geq 5$, two players play the following game on a regular $n$-gon. Initially, three consecutive vertices are chosen, and one counter is placed on each. A move consists of one player sliding one counter along any number of edges to another vertex of the $n$-gon without jumping over another counter. A move is legal if the area of the triangle formed by the counters is strictly greater after the move than before. The players take turns to make legal moves, and if a player cannot make a legal move, that player loses. For which values of $n$ does the player making the first move have a winning strategy?\n | \nWe shall prove that the first player wins if and only the exponent of 2 in the prime decomposition of $n-3$ is odd.\n\nSince the game is identical for both players, has finitely many possible states and always terminates, we can label the possible states Wins od Losses according as whether a player faced with that position has a winning strategy or not. A state is a Win if and only if there is some legal move taking the state to a Loss, and a state is a Loss if and only if all moves take that state to a Win (including the case where there are no legal moves).\n\nLemma. Any configuration in which the triangle formed by the three counters is not isosceles is necessarily a Win.\n\nProof. Label the positions of the counters $X, Y, Z$ so that the arc $Y Z$ of the circumcircle is shortest and the arc $Z X$ is longest. Begin by moving the counter at $Z$ around the polygon on the arc $Y Z X$ until it forms an isosceles triangle $X Y Z^{\prime}$ with apex at $Y$ (note that the arc $X Y$ is less than half the circle, so that $Z$ does not jump over the counter at $X$ ). If this configuration is a Loss, we are done.\n\nIf instead this configuration is a Win, then the counters can be moved legally from triangle $X Y Z^{\prime}$ to reach a losing state. This cannot involve the counter at $Y$, so by symmetry a Loss state can be reached by moving the counter at $Z^{\prime}$ to a new location $Z^{\prime \prime}$. But then the counter at $Z$ could have been moved to $Z^{\prime \prime}$ in the first place, so the original configuration was a Win as well.\n\nFor every nonzero integer $x$, denote by $v_{2}(x)$ the exponent of 2 in the prime decomposition of $x$. Now, given a configuration in which the triangle formed by the three counters is isosceles, the arcs between the vertices having lengths $a, a, b$ respectively (in appropriate units so that $2 a+b=n)$, we show that the configuration is a Win if and only if $a \neq b$ and $v_{2}(a-b)$ is odd.\n\nWrite $b=a \pm|a-b|$ and notice that the only other isosceles triangle that can be reached from the original configuration is one with arc lengths $a, a \pm|a-b| / 2, a \pm|a-b| / 2$. If $|a-b|$ is odd, this is of course impossible, so the configuration is a Loss, since all non-isosceles configurations are Wins, by the lemma.\n\nIf instead $|a-b|$ is even, then all states that can be reached from the original configuration are Wins, except possibly the state with arc lengths $a, a \pm|a-b| / 2, a \pm|a-b| / 2$. Consequently, $(a, a, b)$ is a Win if and only if $(a, a \pm|a-b| / 2, a \pm|a-b| / 2)$ is a Loss. Since the side lengths of this new triangle differ by $|a-b| / 2$, the conclusion follows inductively once the exceptional and trivial case $a=b$ is dealt with.\n\nAs an immediate corollary, the configuration with arc lengths $1,1, n-2$ (the starting configuration of the question) is a Win if and only if $v_{2}(n-3)$ is odd.\n |
\nA finite list of rational numbers is written on a blackboard. In an operation, we choose any two numbers $a, b$, erase them, and write down one of the numbers\n\n$$\na+b, a-b, b-a, a \times b, a / b(\text { if } b \neq 0), b / a(\text { if } a \neq 0)\n$$\n\nProve that, for every integer $n>100$, there are only finitely many integers $k \geq 0$, such that, starting from the list\n\n$$\nk+1, k+2, \ldots, k+n\n$$\n\nit is possible to obtain, after $n-1$ operations, the value $n$ !.\n | \nWe prove the problem statement even for all positive integer $n$.\n\nThere are only finitely many ways of constructing a number from $n$ pairwise distinct numbers $x_{1}, \ldots, x_{n}$ only using the four elementary arithmetic operations, and each $x_{k}$ exactly once. Each such formula for $k>1$ is obtained by an elementary operation from two such formulas on two disjoint sets of the $x_{i}$.\n\nA straightforward induction on $n$ shows that the outcome of each such construction is a number of the form\n\n$$\n\frac{\sum_{\alpha_{1}, \ldots, \alpha_{n} \in\{0,1\}} a_{\alpha_{1}, \ldots, \alpha_{n}} x_{1}^{\alpha_{1}} \cdots x_{n}^{\alpha_{n}}}{\sum_{\alpha_{1}, \ldots, \alpha_{n} \in\{0,1\}} b_{\alpha_{1}, \ldots, \alpha_{n}} x_{1}^{\alpha_{1}} \cdots x_{n}^{\alpha_{n}}}\n$$\n\nwhere the $a_{\alpha_{1}, \ldots, \alpha_{n}}$ and $b_{\alpha_{1}, \ldots, \alpha_{n}}$ are all in the set $\{0, \pm 1\}$, not all zero of course, $a_{0, \ldots, 0}=b_{1, \ldots, 1}=0$, and also $a_{\alpha_{1}, \ldots, \alpha_{n}} \cdot b_{\alpha_{1}, \ldots, \alpha_{n}}=0$ for every set of indices.\n\nSince $\left|a_{\alpha_{1}, \ldots, \alpha_{n}}\right| \leq 1$, and $a_{0,0, \ldots, 0}=0$, the absolute value of the numerator does not exceed $\left(1+\left|x_{1}\right|\right) \cdots\left(1+\left|x_{n}\right|\right)-1$; in particular, if $c$ is an integer in the range $-n, \ldots,-1$, and $x_{k}=c+k$, $k=1, \ldots, n$, then the absolute value of the numerator is at most $(-c) !(n+c+1) !-1 \leq n !-1<n !$.\n\nConsider now the integral polynomials,\n\n$$\nP=\sum_{\alpha_{1}, \ldots, \alpha_{n} \in\{0,1\}} a_{\alpha_{1}, \ldots, \alpha_{n}}(X+1)^{\alpha_{1}} \cdots(X+n)^{\alpha_{n}}\n$$\n\nand\n\n$$\nQ=\sum_{\alpha_{1}, \ldots, \alpha_{n} \in\{0,1\}} b_{\alpha_{1}, \ldots, \alpha_{n}}(X+1)^{\alpha_{1}} \cdots(X+n)^{\alpha_{n}}\n$$\n\nwhere the $a_{\alpha_{1}, \ldots, \alpha_{n}}$ and $b_{\alpha_{1}, \ldots, \alpha_{n}}$ are all in the set $\{0, \pm 1\}$, not all zero, $a_{\alpha_{1}, \ldots, \alpha_{n}} b_{\alpha_{1}, \ldots, \alpha_{n}}=0$ for every set of indices, and $a_{0, \ldots, 0}=b_{1, \ldots, 1}=0$. By the preceding, $|P(c)|<n$ ! for every integer $c$ in the range $-n, \ldots,-1$; and since $b_{1, \ldots, 1}=0$, the degree of $Q$ is less than $n$.\n\nSince every non-zero polynomial has only finitely many roots, and the number of roots does not exceed the degree, to complete the proof it is sufficient to show that the polynomial $P-n ! Q$ does not vanish identically, provided that $Q$ does not (which is the case in the problem).\n\nSuppose, if possible, that $P=n ! Q$, where $Q \neq 0$. Since $\operatorname{deg} Q<n$, it follows that $\operatorname{deg} P<n$ as well, and since $P \neq 0$, the number of roots of $P$ does not exceed $\operatorname{deg} P<n$, so $P(c) \neq 0$ for some integer $c$ in the range $-n, \ldots,-1$. By the preceding, $|P(c)|$ is consequently a positive integer less than $n !$. On the other hand, $|P(c)|=n !|Q(c)|$ is an integral multiple of $n !$. A contradiction.\n |
\nLet $A B C$ be a triangle, let $D$ be the touchpoint of the side $B C$ and the incircle of the triangle $A B C$, and let $J_{b}$ and $J_{c}$ be the incentres of the triangles $A B D$ and $A C D$, respectively. Prove that the circumcentre of the triangle $A J_{b} J_{c}$ lies on the bisectrix of the angle $B A C$.\n | \nLet the incircle of the triangle $A B C$ meet $C A$ and $A B$ at points $E$ and $F$, respectively. Let the incircles of the triangles $A B D$ and $A C D$ meet $A D$ at points $X$ and $Y$, respectively. Then $2 D X=D A+D B-A B=D A+D B-B F-A F=D A-A F$; similarly, $2 D Y=D A-A E=2 D X$. Hence the points $X$ and $Y$ coincide, so $J_{b} J_{c} \perp A D$.\n\nNow let $O$ be the circumcentre of the triangle $A J_{b} J_{c}$. Then $\angle J_{b} A O=\pi / 2-\angle A O J_{b} / 2=$ $\pi / 2-\angle A J_{c} J_{b}=\angle X A J_{c}=\frac{1}{2} \angle D A C$. Therefore, $\angle B A O=\angle B A J_{b}+\angle J_{b} A O=\frac{1}{2} \angle B A D+$ $\frac{1}{2} \angle D A C=\frac{1}{2} \angle B A C$, and the conclusion follows.\n |
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