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G 2. Problem: Let $\triangle A B C$ be a right-angled triangle with $\angle B A C=90^{\circ}$, and let $E$ be the foot of the perpendicular from $A$ on $B C$. Let $Z \neq A$ be a point on the line $A B$ with $A B=B Z$. Let (c), $\left(c_{1}\right)$ be the circmucircles of the triangles $\triangle A E Z$ and $\triangle B E Z$, respectively. Let ( $c_{2}$ ) be an arbitrary circle passing through the points $A$ and $E$. Suppose $\left(c_{1}\right)$ meets the line $C Z$ again at the point $F$, and meets $\left(c_{2}\right)$ again at the point $N$. If $P$ is the other point of intesection of $\left(c_{2}\right)$ with $A F$, prove that the points $N, B, P$ are collinear.
|
Solution. Since the triangles $\triangle A E B$ and $\triangle C A B$ are similar, then
$$
\frac{A B}{E B}=\frac{C B}{A B}
$$
Since $A B=B Z$ we get
$$
\frac{B Z}{E B}=\frac{C B}{B Z}
$$
from which it follows that the triangles $\triangle Z B E$ and $\triangle C B Z$ are also similar. Since $F E B Z$ is cyclic,
then $\angle B E Z=\angle B F Z$. So by the similarity of triangles $\triangle Z B E$ and $\triangle C B Z$ we get
$$
\angle B F Z=\angle B E Z=\angle B Z C=\angle B Z F
$$
and therefore the triangle $\triangle B F Z$ is isosceles. Since $B F=B Z=A B$, then the triangle $\triangle A F Z$ is right-angled with $\angle A F Z=90^{\circ}$.
It now follows that the points $A, E, F, C$ are concyclic. Since $A, P, E, N$ are also concyclic, then
$$
\angle E N P=\angle E A P=\angle E A F=\angle E C F=\angle B C Z=\angle B Z E,
$$
where in the last equality we used again the similarity of the triangles $\triangle Z B E$ and $\triangle C B Z$. Since $N, B, E, Z$ are concyclic, then $\angle E N P=\angle B Z E=\angle E N B$, from which it follows that the points $N, B, P$ are collinear.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 115 |
G 3. Let $\triangle A B C$ be a right-angled triangle with $\angle B A C=90^{\circ}$ and let $E$ be the foot of the perpendicular from $A$ on $B C$. Let $Z \neq A$ be a point on the line $A B$ with $A B=B Z$. Let (c) be the circumcircle of the triangle $\triangle A E Z$. Let $D$ be the second point of intersection of $(c)$ with $Z C$ and let $F$ be the antidiametric point of $D$ with respect to $(c)$. Let $P$ be the point of intersection of the lines $F E$ and $C Z$. If the tangent to $(c)$ at $Z$ meets $P A$ at $T$, prove that the points $T, E, B, Z$ are concyclic.
|
Solution. We will first show that $P A$ is tangent to $(c)$ at $A$.
Since $E, D, Z, A$ are concyclic, then $\angle E D C=\angle E A Z=\angle E A B$. Since also the triangles $\triangle A B C$ and $\triangle E B A$ are similar, then $\angle E A B=\angle B C A$, therefore $\angle E D C=\angle B C A$.
Since $\angle F E D=90^{\circ}$, then $\angle P E D=90^{\circ}$ and so
$$
\angle E P D=90^{\circ}-\angle E D C=90^{\circ}-\angle B C A=\angle E A C
$$
Therefore the points $E, A, C, P$ are concyclic. It follows that $\angle C P A=90^{\circ}$ and therefore the triangle $\angle P A Z$ is right-angled. Since also $B$ is the midpoint of $A Z$, then $P B=A B=B Z$ and so $\angle Z P B=$ $\angle P Z B$.
Furthermore, $\angle E P D=\angle E A C=\angle C B A=\angle E B A$ from which it follows that the points $P, E, B, Z$ are also concyclic.
Now observe that
$$
\angle P A E=\angle P C E=\angle Z P B-\angle P B E=\angle P Z B-\angle P Z E=\angle E Z B
$$
Therefore $P A$ is tangent to $(c)$ at $A$ as claimed.
It now follows that $T A=T Z$. Therefore
$$
\begin{aligned}
\angle P T Z & =180^{\circ}-2(\angle T A B)=180^{\circ}-2(\angle P A E+\angle E A B)=180^{\circ}-2(\angle E C P+\angle A C B) \\
& =180^{\circ}-2\left(90^{\circ}-\angle P Z B\right)=2(\angle P Z B)=\angle P Z B+\angle B P Z=\angle P B A .
\end{aligned}
$$
Thus $T, P, B, Z$ are concyclic, and since $P, E, B, Z$ are also concyclic then $T, E, B, Z$ are concyclic as required.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 116 |
NT 5. The positive integer $k$ and the set $A$ of different integers from 1 to $3 k$ inclusive are such that there are no distinct $a, b, c$ in $A$ satisfying $2 b=a+c$. The numbers from $A$ in the interval $[1, k]$ will be called small; those in $[k+1,2 k]$ - medium and those in $[2 k+1,3 k]$ - large. Is it always true that there are no positive integers $x$ and $d$ such that if $x, x+d$ and $x+2 d$ are divided by $3 k$ then the remainders belong to $A$ and those of $x$ and $x+d$ are different and are:
a) small?
b) medium?
c) large?
(In this problem we assume that if a multiple of $3 k$ is divided by $3 k$ then the remainder is $3 k$ rather than 0. )
|
Solution. A counterexample for a) is $k=3, A=\{1,2,9\}, x=2$ and $d=8$. A counterexample for c) is $k=3, A=\{1,8,9\}, x=8$ and $d=1$.
We will prove that b) is true.
Suppose the contrary and let $x, d$ have the above properties. We can assume $03 k$, then since the remainder for $x+d$ is medium we have $4 k2 k$. Therefore $6 k=4 k+2 kk$ so $d=(x+d)-x<k$. Hence $0 \leq x+2 d=(x+d)+d<3 k$. Thus the remainders $x, x+d$ and $x+2 d$ are in $A$ and
$$
2(x+d)=(x+2 d)+x
$$
a contradiction.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 117 |
A 1. Let $x, y$ and $z$ be positive numbers. Prove that
$$
\frac{x}{\sqrt{\sqrt[4]{y}+\sqrt[4]{z}}}+\frac{y}{\sqrt{\sqrt[4]{z}+\sqrt[4]{x}}}+\frac{z}{\sqrt{\sqrt[4]{x}+\sqrt[4]{y}}} \geq \frac{\sqrt[4]{(\sqrt{x}+\sqrt{y}+\sqrt{z})^{7}}}{\sqrt{2 \sqrt{27}}}
$$
|
Solution. Replacing $x=a^{2}, y=b^{2}, z=c^{2}$, where $a, b, c$ are positive numbers, our inequality is equivalent to
$$
\frac{a^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}}+\frac{b^{2}}{\sqrt{\sqrt{c}+\sqrt{a}}}+\frac{c^{2}}{\sqrt{\sqrt{a}+\sqrt{b}}} \geq \frac{\sqrt[4]{(a+b+c)^{7}}}{\sqrt{2 \sqrt{27}}}
$$
Using the Cauchy-Schwarz inequality for the left hand side we get
$$
\frac{a^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}}+\frac{b^{2}}{\sqrt{\sqrt{c}+\sqrt{a}}}+\frac{c^{2}}{\sqrt{\sqrt{a}+\sqrt{b}}} \geq \frac{(a+b+c)^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}+\sqrt{\sqrt{c}+\sqrt{a}}+\sqrt{\sqrt{a}+\sqrt{b}}}
$$
Using Cauchy-Schwarz inequality for three positive numbers $\alpha . \beta . \uparrow$, we have
$$
\sqrt{\alpha}+\sqrt{\beta}+\sqrt{\gamma} \leq \sqrt{3(\alpha+\beta+\gamma)}
$$
Using this result twice, we have
$$
\begin{aligned}
\sqrt{\sqrt{b}+\sqrt{c}}+\sqrt{\sqrt{c}+\sqrt{a}}+\sqrt{\sqrt{a}+\sqrt{b}} & \leq \sqrt{6(\sqrt{a}+\sqrt{b}+\sqrt{c})} \\
& \leq \sqrt{6 \sqrt{3(a+b+c)}}
\end{aligned}
$$
Combining (1) and (2) we get the desired result.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 119 |
A 3. Let $a, b, c$ be positive real numbers. Prove that
$$
\frac{1}{a b(b+1)(c+1)}+\frac{1}{b c(c+1)(a+1)}+\frac{1}{c a(a+1)(b+1)} \geq \frac{3}{(1+a b c)^{2}}
$$
|
Solution. The required inequality is equivalent to
$$
\frac{c(a+1)+a(b+1)+b(c+1)}{a b c(a+1)(b+1)(c+1)} \geq \frac{3}{(1+a b c)^{2}}
$$
or equivalently to,
$$
(1+a b c)^{2}(a b+b c+c a+a+b+c) \geq 3 a b c(a b+b c+c a+a+b+c+a b c+1)
$$
Let $m=a+b+c, n=a b+b c+c a$ and $x^{3}=a b c$, then the above can be rewritten as
$$
(m+n)\left(1+x^{3}\right)^{2} \geq 3 x^{3}\left(x^{3}+m+n+1\right)
$$
or
$$
(m+n)\left(x^{6}-x^{3}+1\right) \geq 3 x^{3}\left(x^{3}+1\right)
$$
By the AM-GM inequality we have $m \geq 3 x$ and $n \geq 3 x^{2}$, hence $m+n \geq 3 x(x+1)$. It is sufficient to prove that
$$
\begin{aligned}
x(x+1)\left(x^{6}-x^{3}+1\right) & \geq x^{3}(x+1)\left(x^{2}-x+1\right) \\
3\left(x^{6}-x^{3}+1\right) & \geq x^{2}\left(x^{2}-x+1\right) \\
\left(x^{2}-1\right)^{2} & \geq 0
\end{aligned}
$$
which is true.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 120 |
G 1. Let $H$ be the orthocentre of an acute triangle $A B C$ with $B C>A C$, inscribed in a circle $\Gamma$. The circle with centre $C$ and radius $C B$ intersects $\Gamma$ at the point $D$, which is on the arc $A B$ not containing $C$. The circle with centre $C$ and radius $C A$ intersects the segment $C D$ at the point $K$. The line parallel to $B D$ through $K$, intersects $A B$ at point $L$. If $M$ is the midpoint of $A B$ and $N$ is the foot of the perpendicular from $H$ to $C L$, prove that the line $M N$ bisects the segment $C H$.
|
Solution. We use standard notation for the angles of triangle $A B C$. Let $P$ be the midpoint of $C H$ and $O$ the centre of $\Gamma$. As
$$
\alpha=\angle B A C=\angle B D C=\angle D K L
$$
the quadrilateral $A C K L$ is cyclic. From the relation $C B=C D$ we get $\angle B C D=180^{\circ}-2 \alpha$, so
$$
\angle A C K=\gamma+2 \alpha-180^{\circ}
$$
where $\gamma=\angle A C B$. From the relation $C K=C A$ we get
$$
\angle A L C=\angle A K C=180^{\circ}-\alpha-\frac{\gamma}{2}
$$
and thus from the triangle $A C L$ we obtain
$$
\angle A C L=180^{\circ}-\alpha-\angle A L C=\frac{\gamma}{2}
$$
which means that $C L$ is the angle bisector of $\angle A C B$, thus $\angle A C L=\angle B C L$. Moreover, from the fact that $C H \perp A B$ and the isosceles triangle $B O C$ has $\angle B O C=2 \alpha$, we get $\angle A C H=\angle B C O=90^{\circ}-\alpha$. It follows that,
$$
\angle N P H=2 \angle N C H=\angle O C H
$$
On the other hand, it is known that $2 C P=C H=2 O M$ and $C P \| O M$, so $C P M O$ is a parallelogram and
$$
\angle M P H=\angle O C H
$$
Now from (3) and (4) we obtain that
$$
\angle M P H=\angle N P H,
$$
which means that the points $M, N, P$ are collinear.
## Alternative formulation of the statement by PSC.
Let $H$ be the orthocentre of an acute triangle $A B C$ with $B C>A C$, inscribed in a circle $\Gamma$. A point $D$ on $\Gamma$, which is on the arc $A B$ not containing $C$, is chosen such that $C B=C D$. A point $K$ is chosen on the segment $C D$ such that $C A=C K$. The line parallel to $B D$ through $K$, intersects $A B$ at point $L$. If $M$ is the midpoint of $A B$ and $N$ is the foot of the perpendicular from $H$ to $C L$, prove that the line $M N$ bisects the segment $C H$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 123 |
G 2. Let $A B C$ be a right angled triangle with $\angle A=90^{\circ}$ and $A D$ its altitude. We draw parallel lines from $D$ to the vertical sides of the triangle and we call $E, Z$ their points of intersection with $A B$ and $A C$ respectively. The parallel line from $C$ to $E Z$ intersects the line $A B$ at the point $N$ Let $A^{\prime}$ be the symmetric of $A$ with respect to the line $E Z$ and $I, K$ the projections of $A^{\prime}$ onto $A B$ and $A C$ respectively. If $T$ is the point of intersection of the lines $I K$ and $D E$, prove that $\angle N A^{\prime} T=\angle A D T$.
|
Solution. Suppose that the line $A A^{\prime}$ intersects the lines $E Z, B C$ and $C N$ at the points $L, M$, $F$ respectively. The line $I K$ being diagonal of the rectangle $K A^{\prime} I A$ passes through $L$, which by construction of $A^{\prime}$, is the middle of the other diagonal $A A^{\prime}$. The triangles $Z A L, A L E$ are similar, so $\angle Z A L=\angle A E Z$. By the similarity of the triangles $A B C, D A B$, we get $\angle A C B=\angle B A D$. We have also that $\angle A E Z=\angle B A D$, therefore
$$
\angle Z A L=\angle C A M=\angle A C B=\angle A C M
$$
Since $A F \perp C N$, we have that the right triangles $A F C$ and $C D A$ are equal. Thus the altitudes from the vertices $F, D$ of the triangles $A F C, C D A$ respectively are equal. It follows that $F D \| A C$ and since $D E \| A C$ we get that the points $E, D, F$ are collinear.
In the triangle $L F T$ we have, $A^{\prime} I \| F T$ and $\angle L A^{\prime} I=\angle L I A^{\prime}$, so $\angle L F T=\angle L T F$. Therefore the points $F, A^{\prime}, I, T$ belong to the same circle. Also, $\angle A^{\prime} I N=\angle A^{\prime} F N=90^{\circ}$ so the quadrilateral $I A^{\prime} F N$ is cyclic. Thus, the points $F, A^{\prime}, I, T, N$ all lie on a circle. From the above, we infer that
$$
\angle N A^{\prime} T=\angle T F N=\angle A C F=\angle F E Z=\angle A D T \text {. }
$$
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 124 |
G 3. Let $A B C$ be an acute triangle, $A^{\prime}, B^{\prime}, C^{\prime}$ the reflexions of the vertices $A, B$ and $C$ with respect to $B C, C A$, and $A B$, respectively, and let the circumcircles of triangles $A B B^{\prime}$ and $A C C^{\prime}$ meet again at $A_{1}$. Points $B_{1}$ and $C_{1}$ are defined similarly. Prove that the lines $A A_{1}, B B_{1}$, and $C C_{1}$ have a common point.
|
Solution. Let $O_{1}, O_{2}$ and $O$ be the circumcenters of triangles $A B B^{\prime}, A C C^{\prime}$ and $A B C$ respectively. As $A B$ is the perpendicular bisector of the line segment $C C^{\prime}, O_{2}$ is the intersection of the perpendicular bisector of $A C$ with $A B$. Similarly, $O_{1}$ is the intersection of the perpendicular bisector of $A B$ with $A C$. It follows that $O$ is the orthocenter of triangle $A O_{1} O_{2}$. This means that $A O$ is perpendicular to $O_{1} O_{2}$. On the other hand, the segment $A A_{1}$ is the common chord of the two circles, thus it is perpendicular to $O_{1} O_{2}$. As a result, $A A_{1}$ passes through $O$. Similarly, $B B_{1}$ and $C C_{1}$ pass through $O$, so the three lines are concurrent at $O$.
Comment by PSC. We present here a different approach.
We first prove that $A_{1}, B$ and $C^{\prime}$ are collinear. Indeed, since $\angle B A B^{\prime}=\angle C A C^{\prime}=2 \angle B A C$, then from the circles $\left(A B B^{\prime}\right),\left(A C C^{\prime}\right)$ we get
$$
\angle A A_{1} B=90^{\circ}-\angle B A C=\angle A A_{1} C^{\prime}
$$
It follows that
$$
\angle A_{1} A C=\angle A_{1} C^{\prime} C=\angle B C^{\prime} C=90^{\circ}-\angle A B C
$$
On the other hand, if $O$ is the circumcenter of $A B C$, then
$$
\angle O A C=90^{\circ}-\angle A B C
$$
From (1) and (2) we conclude that $A_{1}, A$ and $O$ are collinear. Similarly, $B B_{1}$ and $C C_{1}$ pass through $O$, so the three lines are concurrent in $O$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 125 |
G 4. Let $A B C$ be a triangle with side-lengths $a, b, c$, inscribed in a circle with radius $R$ and let $I$ be it's incenter. Let $P_{1}, P_{2}$ and $P_{3}$ be the areas of the triangles $A B I, B C I$ and $C A I$, respectively. Prove that
$$
\frac{R^{4}}{P_{1}^{2}}+\frac{R^{4}}{P_{2}^{2}}+\frac{R^{4}}{P_{3}^{2}} \geq 16
$$
|
Solution. Let $r$ be the radius of the inscribed circle of the triangle $A B C$. We have that
$$
P_{1}=\frac{r c}{2}, \quad P_{2}=\frac{r a}{2}, \quad P_{3}=\frac{r b}{2}
$$
It follows that
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}}=\frac{4}{r^{2}}\left(\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)
$$
From Leibniz's relation we have that if $H$ is the orthocenter, then
$$
O H^{2}=9 R^{2}-a^{2}-b^{2}-c^{2}
$$
It follows that
$$
9 R^{2} \geq a^{2}+b^{2}+c^{2}
$$
Therefore, using the AM-HM inequality and then (1), we get
$$
\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}} \geq \frac{9}{a^{2}+b^{2}+c^{2}} \geq \frac{1}{R^{2}}
$$
Finally, using Euler's inequality, namely that $R \geq 2 r$, we get
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}} \geq \frac{4}{r^{2} R^{2}} \geq \frac{16}{R^{4}}
$$
Comment by PSC. We can avoid using Leibniz's relation as follows: as in the above solution we have that
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}}=\frac{4}{r^{2}}\left(\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)
$$
Let $a+b+c=2 \tau, E=(A B C)$ and using the inequality $x^{2}+y^{2}+z^{2} \geq x y+y z+z x$ we get
$$
\begin{aligned}
\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}} & \geq \frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}=\frac{2 \tau}{a b c} \\
& =\frac{\tau}{2 R E}=\frac{1}{2 R r}
\end{aligned}
$$
where we used the area formulas $E=\frac{a b c}{4 R}=\tau r$. Finally, using Euler's inequality, namely that $R \geq 2 r$, we get
$$
\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}} \geq \frac{2}{r^{3} R} \geq \frac{16}{R^{4}}
$$
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 126 |
G 5. Given a rectangle $A B C D$ such that $A B=b>2 a=B C$, let $E$ be the midpoint of $A D$. On a line parallel to $A B$ through point $E$, a point $G$ is chosen such that the area of $G C E$ is
$$
(G C E)=\frac{1}{2}\left(\frac{a^{3}}{b}+a b\right)
$$
Point $H$ is the foot of the perpendicular from $E$ to $G D$ and a point $I$ is taken on the diagonal $A C$ such that the triangles $A C E$ and $A E I$ are similar. The lines $B H$ and $I E$ intersect at $K$ and the lines $C A$ and $E H$ intersect at $J$. Prove that $K J \perp A B$.
|
Solution. Let $L$ be the foot of the perpendicular from $G$ to $E C$ and let $Q$ the point of intersection of the lines $E G$ and $B C$. Then,
$$
(G C E)=\frac{1}{2} E C \cdot G L=\frac{1}{2} \sqrt{a^{2}+b^{2}} \cdot G L
$$
So, $G L=\frac{a}{b} \sqrt{a^{2}+b^{2}}$.
Observing that the triangles $Q C E$ and $E L G$ are similar, we have $\frac{a}{b}=\frac{G L}{E L}$, which implies that $E L=\sqrt{a^{2}+b^{2}}$, or in other words $L \equiv C$.
Consider the circumcircle $\omega$ of the triangle $E B C$. Since
$$
\angle E B G=\angle E C G=\angle E H G=90^{\circ}
$$
the points $H$ and $G$ lie on $\omega$.
From the given similarity of the triangles $A C E$ and $A E I$, we have that
$$
\angle A I E=\angle A E C=90^{\circ}+\angle G E C=90^{\circ}+\angle G H C=\angle E H C
$$
therefore $E H C I$ is cyclic, thus $I$ lies on $\omega$.
Since $E B=E C$, we get that $\angle E I C=\angle E H B$, thus $\angle J I E=\angle E H K$. We conclude that $J I H K$ is cyclic, therefore
$$
\angle J K H=\angle H I C=\angle H B C
$$
It follows that $K J \| B C$, so $K J \perp A B$.
Comment. The proposer suggests a different way to finish the proof after proving that $I$ lies on $\omega$ : We apply Pascal's Theorem to the degenerated hexagon $E E H B C I$. Since $B C$ and $E E$ intersect at infinity, this implies that $K J$, which is the line through the intersections of the other two opposite pairs of sides of the hexagon, has to go through this point at infinity, thus it is parallel to $B C$, and so $K J \perp A B$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 127 |
NT 4. Show that there exist infinitely many positive integers $n$ such that
$$
\frac{4^{n}+2^{n}+1}{n^{2}+n+1}
$$
is an integer.
|
Solution. Let $f(n)=n^{2}+n+1$. Note that
$$
f\left(n^{2}\right)=n^{4}+n^{2}+1=\left(n^{2}+n+1\right)\left(n^{2}-n+1\right)
$$
This means that $f(n) \mid f\left(n^{2}\right)$ for every positive integer $n$. By induction on $k$, one can easily see that $f(n) \mid f\left(n^{2^{k}}\right)$ for every positive integers $n$ and $k$. Note that the required condition is equivalent to $f(n) \mid f\left(2^{n}\right)$. From the discussion above, if there exists a positive integer $n$ so that $2^{n}$ can be written as $n^{2^{k}}$, for some positive integer $k$, then $f(n) \mid f\left(2^{n}\right)$. If we choose $n=2^{2^{m}}$ and $k=2^{m}-m$ for some positive integer $m$, then $2^{n}=n^{2^{k}}$ and since there are infinitely many positive integers of the form $n=2^{2^{m}}$, we have the desired result.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 129 |
A4 Real numbers $x, y, z$ satisfy
$$
0<x, y, z<1
$$
and
$$
x y z=(1-x)(1-y)(1-z) .
$$
Show that
$$
\frac{1}{4} \leq \max \{(1-x) y,(1-y) z,(1-z) x\}
$$
|
Solution: It is clear that $a(1-a) \leq \frac{1}{4}$ for any real numbers $a$ (equivalent to $0\max \{(1-x) y,(1-y) x,(1-z) x\}
$$
Now
$$
(1-x) y\frac{1}{2}$.
Using same reasoning we conclude:
$$
z\frac{1}{2}
$$
Using these facts we derive:
$$
\frac{1}{8}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}>x y z=(1-x)(1-y)(1-z)>\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{8}
$$
Contradiction!
Remark: The exercise along with its proof generalizes for any given (finite) number of numbers, and you can consider this new form in place of the proposed one:
Exercise: If for the real numbers $x_{1}, x_{2}, \ldots, x_{n}, 0<x_{i}<1$, for all indices $i$, and
$$
x_{1} x_{2} \ldots x_{n}=\left(1-x_{1}\right)\left(1-x_{2}\right) \ldots\left(1-x_{n}\right)
$$
show that
$$
\frac{1}{4} \leq \max _{1 \leq i \leq n}\left(1-x_{i}\right) x_{i+1}
$$
(where $x_{n+1}=x_{1}$ ).
Or you can consider the following variation:
Exercise: If for the real numbers $x_{1}, x_{2}, \ldots, x_{2009}, 0<x_{i}<1$, for all indices $i$, and
$$
x_{1} x_{2} \ldots x_{2009}=\left(1-x_{1}\right)\left(1-x_{2}\right) \ldots\left(1-x_{2009}\right)
$$
show that
$$
\frac{1}{4} \leq \max _{1 \leq i \leq 2009}\left(1-x_{i}\right) x_{i+1}
$$
(where $x_{2010}=x_{1}$ ).
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 130 |
A5 Let $x, y, z$ be positive real numbers. Prove that:
$$
\left(x^{2}+y+1\right)\left(x^{2}+z+1\right)\left(y^{2}+z+1\right)\left(y^{2}+x+1\right)\left(z^{2}+x+1\right)\left(z^{2}+y+1\right) \geq(x+y+z)^{6}
$$
|
Solution I: Applying Cauchy-Schwarz's inequality:
$$
\left(x^{2}+y+1\right)\left(z^{2}+y+1\right)=\left(x^{2}+y+1\right)\left(1+y+z^{2}\right) \geq(x+y+z)^{2}
$$
Using the same reasoning we deduce:
$$
\left(x^{2}+z+1\right)\left(y^{2}+z+1\right) \geq(x+y+z)^{2}
$$
and
$$
\left(y^{2}+x+1\right)\left(z^{2}+x+1\right) \geq(x+y+z)^{2}
$$
Multiplying these three inequalities we get the desired result.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 131 |
G1 Let $A B C D$ be a parallelogram with $A C>B D$, and let $O$ be the point of intersection of $A C$ and $B D$. The circle with center at $O$ and radius $O A$ intersects the extensions of $A D$ and $A B$ at points $G$ and $L$, respectively. Let $Z$ be intersection point of lines $B D$ and $G L$. Prove that $\angle Z C A=90^{\circ}$.
|
## Solution:
From the point $L$ we draw a parallel line to $B D$ that intersects lines $A C$ and $A G$ at points $N$ and $R$ respectively. Since $D O=O B$, we have that $N R=N L$, and point $N$ is the midpoint of segment $L R$.
Let $K$ be the midpoint of $G L$. Now, $N K \| R G$, and
$$
\angle A G L=\angle N K L=\angle A C L
$$
Therefore, from the cyclic quadrilateral $N K C L$ we deduce:
$$
\angle K C N=\angle K L N
$$
Now, since $L R \| D Z$, we have
$$
\angle K L N=\angle K Z O
$$
It implies that quadrilateral $O K C Z$ is cyclic, and
$$
\angle O K Z=\angle O C Z
$$
Since $O K \perp G L$, we derive that $\angle Z C A=90^{\circ}$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 132 |
G3 A parallelogram $A B C D$ with obtuse angle $\angle A B C$ is given. After rotating the triangle $A C D$ around the vertex $C$, we get a triangle $C D^{\prime} A^{\prime}$, such that points $B, C$ and $D^{\prime}$ are collinear. The extension of the median of triangle $C D^{\prime} A^{\prime}$ that passes through $D^{\prime}$ intersects the straight line $B D$ at point $P$. Prove that $P C$ is the bisector of the angle $\angle B P D^{\prime}$.
|
Solution: Let $A C \cap B D=\{X\}$ and $P D^{\prime} \cap C A^{\prime}=\{Y\}$. Because $A X=C X$ and $C Y=Y A^{\prime}$, we deduce:
$$
\triangle A B C \cong \triangle C D A \cong \triangle C D^{\prime} A^{\prime} \Rightarrow \triangle A B X \cong \triangle C D^{\prime} Y, \triangle B C X \cong \triangle D^{\prime} A^{\prime} Y
$$
It follows that
$$
\angle A B X=\angle C D^{\prime} Y
$$
Let $M$ and $N$ be orthogonal projections of the point $C$ on the straight lines $P D^{\prime}$ and $B P$, respectively, and $Q$ is the orthogonal projection of the point $A$ on the straight line $B P$. Because $C D^{\prime}=A B$, we have that $\triangle A B Q \cong \triangle C D^{\prime} M$.
We conclude that $C M=A Q$. But, $A X=C X$ and $\triangle A Q X \cong \triangle C N X$. So, $C M=C N$ and $P C$ is the bisector of the angle $\angle B P D^{\prime}$.
Much shortened: $\triangle C D^{\prime} Y \equiv \triangle C D X$ means their altitudes from $C$ are also equal, i.e. $C M=C N$ and the conclusion.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 133 |
G4 Let $A B C D E$ be a convex pentagon such that $A B+C D=B C+D E$ and let $k$ be a semicircle with center on side $A E$ that touches the sides $A B, B C, C D$ and $D E$ of the pentagon, respectively, at points $P, Q, R$ and $S$ (different from the vertices of the pentagon). Prove that $P S \| A E$.
|
Solution: Let $O$ be center of $k$. We deduce that $B P=B Q, C Q=C R, D R=D S$, since those are tangents to the circle $k$. Using the condition $A B+C D=B C+D E$, we derive:
$$
A P+B P+C R+D R=B Q+C Q+D S+E S
$$
From here we have $A P=E S$.
Thus,
$$
\triangle A P O \cong \triangle E S O\left(A P=E S, \angle A P O=\angle E S O=90^{\circ}, P O=S O\right)
$$
This implies
$$
\angle O P S=\angle O S P
$$
Therefore,
$$
\angle A P S=\angle A P O+\angle O P S=90^{\circ}+\angle O P S=90^{\circ}+\angle O S P=\angle P S E
$$
Now, from quadrilateral $A P S E$ we deduce:
$$
2 \angle E A P+2 \angle A P S=\angle E A P+\angle A P S+\angle P S E+\angle S E A=360^{\circ}
$$
So,
$$
\angle E A P+\angle A P S=180^{\circ}
$$
and $A P S E$ is isosceles trapezoid. Therefore, $A E \| P S$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 134 |
G5 Let $A, B, C$ and $O$ be four points in the plane, such that $\angle A B C>90^{\circ}$ and $O A=$ $O B=O C$. Define the point $D \in A B$ and the line $\ell$ such that $D \in \ell, A C \perp D C$ and $\ell \perp A O$. Line $\ell$ cuts $A C$ at $E$ and the circumcircle of $\triangle A B C$ at $F$. Prove that the circumcircles of triangles $B E F$ and $C F D$ are tangent at $F$.
|
Solution: Let $\ell \cap A C=\{K\}$ and define $G$ to be the mirror image of the point $A$ with respect to $O$. Then $A G$ is a diameter of the circumcircle of the triangle $A B C$, therefore $A C \perp C G$. On the other hand we have $A C \perp D C$, and it implies that points $D, C, G$ are collinear.
Moreover, as $A E \perp D G$ and $D E \perp A G$, we obtain that $E$ is the orthocenter of triangle $A D G$ and $G E \perp A D$. As $A G$ is a diameter, we have $A B \perp B G$, and since $A D \perp G E$, the points $E, G$, and $B$ are collinear.
Notice that
$$
\angle C A G=90^{\circ}-\angle A G C=\angle K D C
$$
and
$$
\angle C A G=\angle G F C
$$
since both subtend the same arc.
Hence,
$$
\angle F D G=\angle G F C
$$
Therefore, $G F$ is tangent to the circumcircle of the triangle $C D F$ at point $F$.
We claim that line $G F$ is also tangent to the circumcircle of triangle $B E F$ at point $F$, which concludes the proof.
The claim is equivalent to $\angle G B F=\angle E F G$. Denote by $F^{\prime}$ the second intersection point - other than $F$ - of line $\ell$ with the circumcircle of triangle $A B C$. Observe that $\angle G B F=\angle G F^{\prime} F$, because both angles subtend the same arc, and $\angle F F^{\prime} G=\angle E F G$, since $A G$ is the perpendicular bisector of the chord $F F^{\prime}$, and we are done.
### 2.4 Number Theory
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 135 |
C1 Inside of a square whose side length is 1 there are a few circles such that the sum of their circumferences is equal to 10 . Show that there exists a line that meets alt least four of these circles.
|
Solution
Find projections of all given circles on one of the sides of the square. The projection of each circle is a segment whose length is equal to the length of a diameter of this circle. Since the sum of the lengths of all circles' diameters is equal to $10 / \pi$, it follows that the sum of the lengths of all mentioned projections is equal to $10 / \pi>3$. Because the side of the square is equal to 1 , we conclude that at least one point is covered with at least four of these projections. Hence, a perpendicular line to the projection side passing through this point meets at least four of the given circles, so this is a line with the desired property.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 136 |
A1. Let $a, b, c, d, e$ be real numbers such that $a+b+c+d+e=0$. Let, also $A=a b+b c+c d+d e+e a$ and $B=a c+c e+e b+b d+d a$.
Show that
$$
2005 A+B \leq 0 \text { or } \quad A+2005 B \leq 0
$$
|
## Solution
We have
$$
0=(a+b+c+d+e)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+2 A+2 B
$$
This implies that
$$
A+B \leq 0 \text { or } 2006(\dot{A}+B)=(2005 A+B)+(A+2005 B) \leq 0
$$
This implies the conclusion.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 137 |
G1. Let $A B C D$ be an isosceles trapezoid with $A B=A D=B C, A B / / D C, A B>D C$. Let $E$ be the point of intersection of the diagonals $A C$ and $B D$ and $N$ be the symmetric point of $\mathrm{B}$ with respect to the line $\mathrm{AC}$. Prove that quadrilateral $A N D E$ is cyclic.
|
## Solution
Let $\omega$ be a circle passing through the points $A, N, D$ and let $M$ the point where $\omega$ intersects $B D$ for the second time. The quadrilateral $A N D M$ is cyclic and it follows that
$$
\angle N D M+\angle N A M=\angle N D M+\angle B D C=180^{\circ}
$$
and
Figure 1
$$
\angle N A M=\angle B D C
$$
Now we have
$$
\angle B D C=\angle A C D=\angle N A C
$$
and
$$
\angle N A M=\angle N A C
$$
So the points $A, M, C$ are collinear and $M \equiv E$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 138 |
G2. Let $A B C$ be a triangle inscribed in a circle $K$. The tangent from $A$ to the circle meets the line $B C$ at point $P$. Let $M$ be the midpoint of the line segment $A P$ and let $R$ be the intersection point of the circle $K$ with the line $B M$. The line $P R$ meets again the circle $K$ at the point $S$. Prove that the lines $A P$ and $C S$ are parallel.
|
## Solution
Figure 2
Assume that point $C$ lies on the line segment $B P$. By the Power of Point theorem we have $M A^{2}=M R \cdot M B$ and so $M P^{2}=M R \cdot M B$. The last equality implies that the triangles $M R$ and $M P B$ are similar. Hence $\angle M P R=\angle M B P$ and since $\angle P S C=\angle M B P$, the claim is proved.
Slight changes are to be made if the point $B$ lies on the line segment $P C$.
Figure 3
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 139 |
G4. Let $\mathrm{ABC}$ be an isosceles triangle such that $A B=A C$ and $\angle \frac{A}{2}<\angle B$. On the extension of the altitude $\mathrm{AM}$ we get the points $\mathrm{D}$ and $\mathrm{Z}$ such that $\angle C B D=\angle A$ and $\angle Z B A=90^{\circ}$. $\mathrm{E}$ is the foot of the perpendicular from $\mathrm{M}$ to the altitude $\mathrm{BF}$ and $\mathrm{K}$ is the foot of the perpendicular from $\mathrm{Z}$ to $\mathrm{AE}$. Prove that $\angle K D Z=\angle K B D=\angle K Z B$.
|
## Solution
The points $A, B, K, Z$ and $C$ are co-cyclic.
Because ME//AC so we have
$$
\angle K E M=\angle E A C=\angle M B K
$$
Therefore the points $B, K, M$ and $E$ are co-cyclic. Now, we have
$$
\begin{aligned}
& \angle A B F=\angle A B C-\angle F B C \\
& =\angle A K C-\angle E K M=\angle M K C
\end{aligned}
$$
Also, we have
$$
\begin{aligned}
& \angle A B F=90^{\circ}-\angle B A F=90^{\circ}-\angle M B D \\
& =\angle B D M=\angle M D C
\end{aligned}
$$
From (1) and (2) we get $\angle M K C=\angle M D C$ and so the points $M, K, D$ and $C$ are co-cyclic.
Consequently,
$$
\angle K D M=\angle K C M=\angle B A K=\angle B Z K \text {, }
$$
and because the line $\mathrm{BD}$ is tangent to the circumcircle of triangle $A B C$, we have
$$
\angle K B D=\angle B A K
$$
Figure 5
Finally, we have
$$
\angle K D Z=\angle K B D=\angle K Z B
$$
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 140 |
G5. Let $A$ and $P$ are the points of intersection of the circles $k_{1}$ and $k_{2}$ with centers $O$ and $K$, respectively. Let also $B$ and $C$ be the symmetric points of $A$ with respect to $O$ and $K$, respectively. A line through $A$ intersects the circles $k_{1}$ and $k_{2}$ at the points $D$ and $E$, respectively. Prove that the centre of the circumcircle of the triangle $D E P$ lies on the circumcircle $O K P$.
|
## Solution
The points $B, P, C$ are collinear, and
$$
\angle A P C=\angle A P B=90^{\circ}
$$
Let $N$ be the midpoint of $D P$.
So we have:
$$
\begin{aligned}
& \angle N O P=\angle D A P \\
& =\angle E C P=\angle E C A+\angle A C P
\end{aligned}
$$
Since $O K / / B C$ and $O K$ is the bisector of $\angle A K P$ we get
Figure 6
$$
\angle A C P=O K P
$$
Also, since $A P \perp O K$ and $M K \perp P E$ we have that
$$
\angle A P E=\angle M K O
$$
The points $A, E, C, P$ are co-cyclic, and so $\angle E C A=\angle A P E$.
Therefore, from (1), (2) and (3) we have that $\angle N O P=\angle M K P$.
Thus $O, M, K$ and $P$ are co-cyclic.
## Comment
Points B and C may not be included in the statement of the problem
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 141 |
G7. Let $A B C D$ be a parallelogram, $\mathrm{P}$ a point on $C D$, and $Q$ a point on $A B$. Let also $M=A P \cap D Q, \quad N=B P \cap C Q, K=M N \cap A D$, and $L=M N \cap B C$. Show that $B L=D K$.
|
## Solution
Let $O$ be the intersection of the diagonals. Let $P_{1}$ be on $A B$ such that $P P_{1} / / A D$, and let $Q_{1}$ be on $C D$ such that $\mathrm{Q} Q_{1} / / A D$. Let $\sigma$ be the central symmetry with center $\mathrm{O}$. Let $\left.P^{\prime}=\sigma(P), Q^{\prime}=\sigma(Q), P_{1}^{\prime}=\sigma\left(P_{1}\right)\right)$ and, (figure 1).
Let $M_{1}=A Q_{1} \cap D P_{1}, N_{1}=B Q_{1} \cap C P_{1}, N^{\prime}=A Q^{\prime} \cap D P^{\prime}$ and $M^{\prime}=B Q^{\prime} \cap C P^{\prime}$.
Then: $M^{\prime}=\sigma(M), N^{\prime}=\sigma(N), M_{1}^{\prime}=\sigma\left(M_{1}\right)$ and $N_{1}^{\prime}=\sigma\left(N_{1}\right)$.
Since $A P$ and $D P_{1}$ are the diagonals of the parallelogram $A P_{1} P D, C P_{1}$ and $B P$ are the diagonals of the parallelogram $P_{1} B C P$, and $A Q_{1}$ and $D O$ are the diagonals of the parallelogram $A Q Q_{1} D$, it follows that the points $U, V, W$ (figure 2) are collinear and they lie on the line passing through the midpoints $R$ of $A D$ and $Z$ of $B C$. The diagonals AM and $D M_{1}$ the quadrilateral $A M_{1} M D$ intersect at $U$ and the diagonals $A M_{1}$ and - $D M$ intersect at $W$. Since the midpoint of $A D$ is on the line $U W$, it follows that the quadrilateral $A M_{1} M D$ is a trapezoid. Hence, $M M_{1}$ is parallel to $A D$ and the midpoint $S$ of $M M_{1}$ lies on the line $U W$, (figure 2).
Figure 11
Similarly $M^{\prime} M_{1}^{\prime}$ is parallel to $A D$ and its midpoint lies on $U W$. So $M_{1} M^{\prime} M_{1}^{\prime} M$ is a parallelogram whose diagonals intersect at $\mathrm{O}$.
Similarly, $N_{1}^{\prime} N N_{1} N^{\prime}$ is a parallelogram whose diagonals intersect at $O$.
All these imply that $M, N, M^{\prime}, N^{\prime}$ and $O$ are collinear, i.e. $O$ lies on the line $K L$. This implies that $K=\sigma(L)$, and since $D=\sigma(B)$, the conclusion follows.
Figure 12
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 142 |
NT3. Let $p$ be an odd prime. Prove that $p$ divides the integer
$$
\frac{2^{p!}-1}{2^{k}-1}
$$
for all integers $k=1,2, \ldots, p$.
|
## Solution
At first, note that $\frac{2^{p!}-1}{2^{k}-1}$ is indeed an integer.
We start with the case $\mathrm{k}=\mathrm{p}$. Since $p \mid 2^{p}-2$, then $p / 22^{p}-1$ and so it suffices to prove that $p \mid 2^{(p)!}-1$. This is obvious as $p \mid 2^{p-1}-1$ and $\left(2^{p-1}-1\right) \mid 2^{(p)!}-1$.
If $\mathrm{k}=1,2, \ldots, \mathrm{p}-1$, let $m=\frac{(p-1)!}{k} \in \mathbb{N}$ and observe that $p!=k m p$. Consider $a \in \mathbb{N}$ so that $p^{a} \mid 2^{k}-1$ and observe that it suffices to prove $p^{a+1} \mid 2^{p!}-1$. The case $a=0$ is solved as the case $k=p$. If else, write $2^{k}=1+p^{a} \cdot l, l \in \mathbb{N}$ and rising at the power mp gives
$$
2^{p!}=\left(1+p^{a} \cdot l\right)^{m p}=1+m p \cdot p^{a} \cdot l+M p^{2 a}
$$
where $M n$ stands for a multiply of $\mathrm{n}$. Now it is clear that $p^{a+1} \mid 2^{p!}-1$, as claimed.
Comment. The case $\mathrm{k}=\mathrm{p}$ can be included in the case $a=0$.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 143 |
NT5. Let $p$ be a prime number and let $a$ be an integer. Show that if $n^{2}-5$ is not divisible by $p$ for any integer $n$, there exist infinitely many integers $m$ so that $p$ divides $m^{5}+a$.
|
## Solution
We start with a simple fact:
Lemma: If $b$ is an integer not divisible by $p$ then there is an integer $s$ so that $s b$ has the remainder $l$ when divided by $p$.
For a proof, just note that numbers $b, 2 b, \ldots,(p-1) b$ have distinct non-zero remainders when divided by $p$, and hence one of them is equal to 1 .
We prove that if $x, y=0,1,2, \ldots, p-1$ and $\mathrm{p}$ divides $x^{5}-y^{5}$, then $x=y$.
Indeed, assume that $x \neq y$. If $x=0$, then $p \mid y^{5}$ and so $y=0$, a contradiction.
To this point we have $x, y \neq 0$. Since
$$
p \mid(x-y)\left(x^{4}+x^{3} y+x^{2} y^{2}+x y^{3}+y^{4}\right) \text { and } p /(x-y)
$$
we have
$$
\begin{aligned}
& p l\left(x^{2}+y^{2}\right)^{2}+x y\left(x^{2}+y^{2}\right)-x^{2} y^{2} \text {, and so } \\
& p \|\left(2\left(x^{2}+y^{2}\right)+x y\right)^{2}-5 x^{2} y^{2}
\end{aligned}
$$
As $p / x y$, from the lemma we find an integer $s$ so that $s x y=k p+1, k \in \mathbb{N}$. Then
$$
p \mid\left[s\left(2 x^{2}+2 y^{2}+x y\right)\right]^{2}-5\left(k^{2} p^{2}+2 k p+1\right)
$$
and so $p \mid z^{2}-5$, where $z=s\left(2 x^{2}+2 y^{2}+x y\right)$, a contradiction.
Consequeatly $r=y$.
Since we have proved that numbers $0^{5}, 1^{5}, \ldots,(p-1)^{5}$ have distinct remainders when divided by $p$, the same goes for the numbers $0^{5}+a, 1^{5}+a, \ldots,(p-1)^{5}+a$ and the conclusion can be reached easily.
## Comments
1. For beauty we may choose $a=-2$ or any other value.
2. Moreover, we may ask only for one value of $m$, instead of "infinitely many".
3. A simple version will be to ask for a proof that the numbers $0^{5}, 1^{5}, \ldots,(p-1)^{5}$ have distinct remainders when divided by $p$.
## Combinatorics
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 144 |
C1. A triangle with area 2003 is divided into non-overlapping small triangles. The number of all the vertices of all those triangles is 2005 . Show that at mest one of the smaller triangles has area less or equal to 1.
|
## Solution
Since all the vertices are 2005 , and the vertices of the big triangle are among them, it follows that the number of the small triangles is at least 2003. So, it follows that at least one of the small triangles has area at most 1
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 145 |
A1 Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that:
$\left(a^{5}+a^{4}+a^{3}+a^{2}+a+1\right)\left(b^{5}+b^{4}+b^{3}+b^{2}+b+1\right)\left(c^{5}+c^{4}+c^{3}+c^{2}+c+1\right) \geq 8\left(a^{2}+a+1\right)\left(b^{2}+b+1\right)\left(c^{2}+c+1\right)$.
|
## Solution
We have $x^{5}+x^{4}+x^{3}+x^{2}+x+1=\left(x^{3}+1\right)\left(x^{2}+x+1\right)$ for all $x \in \mathbb{R}_{+}$.
Take $S=\left(a^{2}+a+1\right)\left(b^{2}+b+1\right)\left(c^{2}+c+1\right)$.
The inequality becomes $S\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq 8 S$.
It remains to prove that $\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq 8$.
By $A M-G M$ we have $x^{3}+1 \geq 2 \sqrt{x^{3}}$ for all $x \in \mathbb{R}_{+}$.
So $\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq 2^{3} \cdot \sqrt{a^{3} b^{3} c^{3}}=8$ and we are done.
Equality holds when $a=b=c=1$.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 146 |
A2 Let $x, y, z$ be positive real numbers. Prove that:
$$
\frac{x+2 y}{z+2 x+3 y}+\frac{y+2 z}{x+2 y+3 z}+\frac{z+2 x}{y+2 z+3 x} \leq \frac{3}{2}
$$
|
## Solution 1
Notice that $\sum_{c y c} \frac{x+2 y}{z+2 x+3 y}=\sum_{c y c}\left(1-\frac{x+y+z}{z+2 x+3 y}\right)=3-(x+y+z) \sum_{c y c} \frac{1}{z+2 x+3 y}$.
We have to proof that $3-(x+y+z) \sum_{c y c} \frac{1}{z+2 x+3 y} \leq \frac{3}{2}$ or $\frac{3}{2(x+y+z)} \leq \sum_{c y c} \frac{1}{z+2 x+3 y}$.
By Cauchy-Schwarz we obtain $\sum_{\text {cyc }} \frac{1}{z+2 x+3 y} \geq \frac{(1+1+1)^{2}}{\sum_{\text {cyc }}(z+2 x+3 y)}=\frac{3}{2(x+y+z)}$.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 147 |
A3 Let $a, b$ be positive real numbers. Prove that $\sqrt{\frac{a^{2}+a b+b^{2}}{3}}+\sqrt{a b} \leq a+b$.
|
Solution 1
Applying $x+y \leq \sqrt{2\left(x^{2}+y^{2}\right)}$ for $x=\sqrt{\frac{a^{2}+a b+b^{2}}{3}}$ and $y=\sqrt{a b}$, we will obtain $\sqrt{\frac{a^{2}+a b+b^{2}}{3}}+\sqrt{a b} \leq \sqrt{\frac{2 a^{2}+2 a b+2 b^{2}+6 a b}{3}} \leq \sqrt{\frac{3\left(a^{2}+b^{2}+2 a b\right)}{3}}=a+b$.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 148 |
A9 Let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers satisfying $\sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)$.
Prove that $\sum_{k=2}^{n-1} x_{k} \geq 0$.
|
## Solution 1
Case I. If $\min \left(x_{1}, x_{n}\right)=x_{1}$, we know that $x_{k} \geq \min \left(x_{k} ; x_{k+1}\right)$ for all $k \in\{1,2,3, \ldots, n-1\}$. So $x_{1}+x_{2}+\ldots+x_{n-1} \geq \sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)=x_{1}$, hence $\sum_{k=2}^{n-1} x_{k} \geq 0$.
Case II. If $\min \left(x_{1}, x_{n}\right)=x_{n}$, we know that $x_{k} \geq \min \left(x_{k-1} ; x_{k}\right)$ for all $k \in\{2,3,4, \ldots, n\}$. So $x_{2}+x_{3}+\ldots+x_{n} \geq \sum_{k=1}^{n-1} \min \left(x_{k} ; x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)=x_{n}$, hence $\sum_{k=2}^{n-1} x_{k} \geq 0$.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 151 |
A7 Let $a, b$ and $c$ be a positive real numbers such that $a b c=1$. Prove the inequality
$$
\left(a b+b c+\frac{1}{c a}\right)\left(b c+c a+\frac{1}{a b}\right)\left(c a+a b+\frac{1}{b c}\right) \geq(1+2 a)(1+2 b)(1+2 c)
$$
|
## Solution 1
By Cauchy-Schwarz inequality and $a b c=1$ we get
$$
\begin{gathered}
\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(a b+b c+\frac{1}{c a}\right)}=\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(\frac{1}{c a}+a b+b c\right)} \geq \\
\left(\sqrt{a b} \cdot \sqrt{\frac{1}{a b}}+\sqrt{b c} \cdot \sqrt{b c}+\sqrt{\frac{1}{c a}} \cdot \sqrt{c a}\right)=(2+b c)=(2 a b c+b c)=b c(1+2 a)
\end{gathered}
$$
Analogously we get $\sqrt{\left(b c+c a+\frac{1}{a b}\right)\left(c a+a b+\frac{1}{b c}\right)} \geq c a(1+2 b)$ and
$\sqrt{\left(c a+a b+\frac{1}{b c}\right)\left(a b+b c+\frac{1}{c a}\right)} \geq a b(1+2 a)$.
Multiplying these three inequalities we get:
$$
\left(a b+b c+\frac{1}{c a}\right)\left(b c+c a+\frac{1}{a b}\right)\left(c a+a b+\frac{1}{b c}\right) \geq a^{2} b^{2} c^{2}(1+2 a)(1+2 b)(1+2 c)=
$$
$(1+2 a)(1+2 b)(1+2 c)$ because $a b c=1$.
Equality holds if and only if $a=b=c=1$.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 152 |
A8 Show that
$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq 4\left(\frac{x}{x y+1}+\frac{y}{y z+1}+\frac{z}{z x+1}\right)^{2}
$$
for any real positive numbers $x, y$ and $z$.
|
## Solution
The idea is to split the inequality in two, showing that
$$
\left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}\right)^{2}
$$
can be intercalated between the left-hand side and the right-hand side. Indeed, using the Cauchy-Schwarz inequality one has
$$
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq\left(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}\right)^{2}
$$
On the other hand, as
$$
\sqrt{\frac{x}{y}} \geq \frac{2 x}{x y+1} \Leftrightarrow(\sqrt{x y}-1)^{2} \geq 0
$$
by summation one has
$$
\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}} \geq \frac{2 x}{x y+1}+\frac{2 y}{y z+1}+\frac{2 z}{z x+1}
$$
The rest is obvious.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 153 |
C1 On a $5 \times 5$ board, $n$ white markers are positioned, each marker in a distinct $1 \times 1$ square. A smart child got an assignment to recolor in black as many markers as possible, in the following manner: a white marker is taken from the board; it is colored in black, and then put back on the board on an empty square such that none of the neighboring squares contains a white marker (two squares are called neighboring if they contain a common side). If it is possible for the child to succeed in coloring all the markers black, we say that the initial positioning of the markers was good.
a) Prove that if $n=20$, then a good initial positioning exists.
b) Prove that if $n=21$, then a good initial positioning does not exist.
|
Solution
a) Position 20 white markers on the board such that the left-most column is empty. This
positioning is good because the coloring can be realized column by column, starting with the second (from left), then the third, and so on, so that the white marker on position $(i, j)$ after the coloring is put on position $(i, j-1)$.
b) Suppose there exists a good positioning with 21 white markers on the board i.e. there exists a re-coloring of them all, one by one. In any moment when there are 21 markers on the board, there must be at least one column completely filled with markers, and there must be at least one row completely filled with markers. So, there exists a "cross" of markers on the board. At the initial position, each such cross is completely white, at the final position each such cross is completely black, and at every moment when there are 21 markers on the board, each such cross is monochromatic. But this cannot be, since every two crosses have at least two common squares and therefore it is not possible for a white cross to vanish and for a black cross to appear by re-coloring of only one marker. Contradiction!
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 154 |
C3 Integers $1,2, \ldots, 2 n$ are arbitrarily assigned to boxes labeled with numbers $1,2, \ldots, 2 n$. Now, we add the number assigned to the box to the number on the box label. Show that two such sums give the same remainder modulo $2 n$.
|
## Solution
Let us assume that all sums give different remainder modulo $2 n$, and let $S$ denote the value of their sum.
For our assumption,
$$
S \equiv 0+1+\ldots+2 n-1=\frac{(2 n-1) 2 n}{2}=(2 n-1) n \equiv n \quad(\bmod 2 n)
$$
But, if we sum, breaking all sums into its components, we derive
$$
S \equiv 2(1+\ldots+2 n)=2 \cdot \frac{2 n(2 n+1)}{2}=2 n(2 n+1) \equiv 0 \quad(\bmod 2 n)
$$
From the last two conclusions we derive $n \equiv 0(\bmod 2 n)$. Contradiction.
Therefore, there are two sums with the same remainder modulo $2 n$.
Remark: The result is no longer true if one replaces $2 n$ by $2 n+1$. Indeed, one could assign the number $k$ to the box labeled $k$, thus obtaining the sums $2 k, k=\overline{1,2 n+1}$. Two such numbers give different remainders when divided by $2 n+1$.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 155 |
G1 Two perpendicular chords of a circle, $A M, B N$, which intersect at point $K$, define on the circle four arcs with pairwise different length, with $A B$ being the smallest of them.
We draw the chords $A D, B C$ with $A D \| B C$ and $C, D$ different from $N, M$. If $L$ is the point of intersection of $D N, M C$ and $T$ the point of intersection of $D C, K L$, prove that $\angle K T C=\angle K N L$.
|
## Solution
First we prove that $N L \perp M C$. The arguments depend slightly on the position of $D$. The other cases are similar.
From the cyclic quadrilaterals $A D C M$ and $D N B C$ we have:
$$
\varangle D C L=\varangle D A M \text { and } \varangle C D L=\varangle C B N \text {. }
$$
So we obtain
$$
\varangle D C L+\varangle C D L=\varangle D A M+\varangle C B N .
$$
And because $A D \| B C$, if $Z$ the point of intersection of $A M, B C$ then $\varangle D A M=\varangle B Z A$, and we have
$$
\varangle D C L+\varangle C D L=\varangle B Z A+\varangle C B N=90^{\circ}
$$
Let $P$ the point of intersection of $K L, A C$, then $N P \perp A C$, because the line $K P L$ is a Simson line of the point $N$ with respect to the triangle $A C M$.
From the cyclic quadrilaterals $N P C L$ and $A N D C$ we obtain:
$$
\varangle C P L=\varangle C N L \text { and } \varangle C N L=\varangle C A D \text {, }
$$
so $\varangle C P L=\varangle C A D$, that is $K L\|A D\| B C$ therefore $\varangle K T C=\varangle A D C$ (1).
But $\varangle A D C=\varangle A N C=\varangle A N K+\varangle K N C=\varangle C N L+\varangle K N C$, so
$$
\varangle A D C=\varangle K N L
$$
From (1) and (2) we obtain the result.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 156 |
G2 For a fixed triangle $A B C$ we choose a point $M$ on the ray $C A$ (after $A$ ), a point $N$ on the ray $A B$ (after $B$ ) and a point $P$ on the ray $B C$ (after $C$ ) in a way such that $A M-B C=B N-A C=C P-A B$. Prove that the angles of triangle $M N P$ do not depend on the choice of $M, N, P$.
|
## Solution
Consider the points $M^{\prime}$ on the ray $B A$ (after $A$ ), $N^{\prime}$ on the ray $C B$ (after $B$ ) and $P^{\prime}$ on the ray $A C$ (after $C$ ), so that $A M=A M^{\prime}, B N=B N^{\prime}, C P=C P^{\prime}$. Since $A M-B C=B N-A C=B N^{\prime}-A C$, we get $C M=A C+A M=B C+B N^{\prime}=C N^{\prime}$. Thus triangle $M C N^{\prime}$ is isosceles, so the perpendicular bisector of $\left[M N^{\prime}\right]$ bisects angle $A C B$ and hence passes through the incenter $I$ of triangle $A B C$. Arguing similarly, we may conclude that $I$ lies also on the perpendicular bisectors of $\left[N P^{\prime}\right]$ and $\left[P M^{\prime}\right]$. On the other side, $I$ clearly lies on the perpendicular bisectors of $\left[M M^{\prime}\right],\left[N N^{\prime}\right]$ and $\left[P P^{\prime}\right]$. Thus the hexagon $M^{\prime} M N^{\prime} N P^{\prime} P$ is cyclic. Then angle $P M N$ equals angle $P N^{\prime} N$, which measures $90^{\circ}-\frac{\beta}{2}$ (the angles of triangle $A B C$ are $\alpha, \beta, \gamma$ ). In the same way angle $M N P$ measures $90^{\circ}-\frac{\gamma}{2}$ and angle $M P N$ measures $90^{\circ}-\frac{\alpha}{2}$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 157 |
G4 Let $A B C$ be a triangle, $(B C<A B)$. The line $\ell$ passing trough the vertices $C$ and orthogonal to the angle bisector $B E$ of $\angle B$, meets $B E$ and the median $B D$ of the side $A C$ at points $F$ and $G$, respectively. Prove that segment $D F$ bisect the segment $E G$.
|
## Solution
Let $C F \cap A B=\{K\}$ and $D F \cap B C=\{M\}$. Since $B F \perp K C$ and $B F$ is angle bisector of $\varangle K B C$, we have that $\triangle K B C$ is isosceles i.e. $B K=B C$, also $F$ is midpoint of $K C$. Hence $D F$ is midline for $\triangle A C K$ i.e. $D F \| A K$, from where it is clear that $M$ is a midpoint of $B C$.
We will prove that $G E \| B C$. It is sufficient to show $\frac{B G}{G D}=\frac{C E}{E D}$. From $D F \| A K$ and $D F=\frac{A K}{2}$ we have
$$
\frac{B G}{G D}=\frac{B K}{D F}=\frac{2 B K}{A K}
$$
Also
$$
\begin{gathered}
\frac{C E}{D E}=\frac{C D-D E}{D E}=\frac{C D}{D E}-1=\frac{A D}{D E}-1=\frac{A E-D E}{D E}-1=\frac{A E}{D E}-2= \\
=\frac{A B}{D F}-2=\frac{A K+B K}{\frac{A K}{2}}-2=2+2 \frac{B K}{A K}-2=\frac{2 B K}{A K}
\end{gathered}
$$
From (1) and (2) we have $\frac{B G}{G D}=\frac{C E}{E D}$, so $G E \| B C$, as $M$ is the midpoint of $B C$, it follows that the segment $D F$, bisects the segment $G E$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 158 |
G8 The side lengths of a parallelogram are $a, b$ and diagonals have lengths $x$ and $y$, Knowing that $a b=\frac{x y}{2}$, show that
$$
a=\frac{x}{\sqrt{2}}, b=\frac{y}{\sqrt{2}} \text { or } a=\frac{y}{\sqrt{2}}, b=\frac{x}{\sqrt{2}}
$$
|
## Solution 1.
Let us consider a parallelogram $A B C D$, with $A B=a, B C=b, A C=x, B D=y$, $\widehat{A O D}=\theta$.
For the area of $A B C D$ we know $(A B C D)=a b \sin A$.
But it is also true that $(A B C D)=4(A O D)=4 \cdot \frac{O A \cdot O D}{2} \sin \theta=2 O A \cdot O D \sin \theta=$ $=2 \cdot \frac{x}{2} \cdot \frac{y}{2} \sin \theta=\frac{x y}{2} \sin \theta$. So $a b \sin A=\frac{x y}{2} \sin \theta$ and since $a b=\frac{x y}{2}$ by hypothesis, we get
$$
\sin A=\sin \theta
$$
Thus
$$
\theta=\widehat{A} \text { or } \theta=180^{\circ}-\widehat{A}=\widehat{B}
$$
If $\theta=A$ then (see Figure below) $A_{2}+B_{1}=A_{1}+A_{2}$, so $B_{1}=A_{1}$ which implies that $A D$ is tangent to the circumcircle of triangle $O A B$. So
$$
D A^{2}=D O \cdot D B \Rightarrow b^{2}=\frac{y}{2} \cdot y \Rightarrow b=\frac{y}{\sqrt{2}}
$$
Then by $a b=\frac{x y}{2}$ we get $a=\frac{x}{\sqrt{2}}$.
If $\theta=B$ we similarly get $a=\frac{x}{\sqrt{2}}, b=\frac{y}{\sqrt{2}}$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 160 |
G9 Let $O$ be a point inside the parallelogram $A B C D$ such that
$$
\angle A O B+\angle C O D=\angle B O C+\angle C O D
$$
Prove that there exists a circle $k$ tangent to the circumscribed circles of the triangles $\triangle A O B, \triangle B O C, \triangle C O D$ and $\triangle D O A$.
|
## Solution
From given condition it is clear that $\varangle A O B+\varangle C O D=\varangle B O C+\varangle A O D=180^{\circ}$.
Let $E$ be a point such that $A E=D O$ and $B E=C E$. Clearly, $\triangle A E B \equiv \triangle D O C$ and from that $A E \| D O$ and $B E \| C O$. Also, $\varangle A E B=\varangle C O D$ so $\varangle A O B+\varangle A E B=$ $\varangle A O B+\varangle C O D=180^{\circ}$. Thus, the quadrilateral $A O B E$ is cyclic.
So $\triangle A O B$ and $\triangle A E B$ the same circumcircle, therefor the circumcircles of the triangles $\triangle A O B$ and $\triangle C O D$ have the same radius.
Also, $A E \| D O$ and $A E=D O$ gives $A E O D$ is parallelogram and $\triangle A O D \equiv \triangle O A E$. So $\triangle A O B, \triangle C O D$ and $\triangle D O A$ has the same radius of their circumcircle (the radius of the cyclic quadrilateral $A E B O)$. Analogously, triangles $\triangle A O B, \triangle B O C, \triangle C O D$ and $\triangle D O A$ has same radius $R$.
Obviously, the circle with center $O$ and radius $2 R$ is externally tangent to each of these circles, so this will be the circle $k$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 161 |
G10 Let $\Gamma$ be a circle of center $O$, and $\delta$ be a line in the plane of $\Gamma$, not intersecting it. Denote by $A$ the foot of the perpendicular from $O$ onto $\delta$, and let $M$ be a (variable) point on $\Gamma$. Denote by $\gamma$ the circle of diameter $A M$, by $X$ the (other than $M$ ) intersection point of $\gamma$ and $\Gamma$, and by $Y$ the (other than $A$ ) intersection point of $\gamma$ and $\delta$. Prove that the line $X Y$ passes through a fixed point.
|
## Solution
Consider the line $\rho$ tangent to $\gamma$ at $A$, and take the points $\{K\}=A M \cap X Y,\{L\}=$ $\rho \cap X M$, and $\{F\}=O A \cap X Y$.
(Remark: Moving $M$ into its reflection with respect to the line $O A$ will move $X Y$ into its reflection with respect to $O A$. These old and the new $X Y$ meet on $O A$, hence it should be clear that the fixed point mult be $F$.)
Since $\varangle L M A=\varangle F Y A$ and $\varangle Y A F=\varangle L A M=90^{\circ}$, it follows that triangles $F A Y$ and $L A M$ are similar, therefore $\varangle A F Y=\varangle A L M$, hence the quadrilateral $A L X F$ is cyclic. But then $\varangle A F L=\varangle A X L=90^{\circ}$, so $L F \perp A F$, hence $L F \| \delta$.
Now, $\rho$ is the radical axis of circles $\gamma$ and $A$ (consider $A$ as a circle of center $A$ and radius 0 ), while $X M$ is the radical axis of circles $\gamma$ and $\Gamma$, so $L$ is the radical center of the three circle, which means that $L$ lies on the radical axis of circles $\Gamma$ and $A$. From $L F \perp O A$, where $O A$ is the line of the centers of the circles $A$ and $\Gamma$, and $F \in X Y$, it follows that $F$ is (the) fixed point of $X Y$.
(The degenerate two cases when $M \in O A$, where $X \equiv M$ and $Y \equiv A$, also trivially satisfy the conclusion, as then $F \in A M)$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 162 |
G11 Consider $A B C$ an acute-angled triangle with $A B \neq A C$. Denote by $M$ the midpoint of $B C$, by $D, E$ the feet of the altitudes from $B, C$ respectively and let $P$ be the intersection point of the lines $D E$ and $B C$. The perpendicular from $M$ to $A C$ meets the perpendicular from $C$ to $B C$ at point $R$. Prove that lines $P R$ and $A M$ are perpendicular.
|
Solution
Let $F$ be the foot of the altitude from $A$ and let $S$ be the intersection point of $A M$ and $R C$. As $P C$ is an altitude of the triangle $P R S$, the claim is equivalent to $R M \perp P S$, since the latter implies that $M$ is the orthocenter of $P R S$. Due to $R M \perp A C$, we need to prove that $A C \| P S$, in other words
$$
\frac{M C}{M P}=\frac{M A}{M S}
$$
Notice that $A F \| C S$, so $\frac{M A}{M S}=\frac{M F}{M C}$. Now the claim is reduced to proving $M C^{2}=$ $M F \cdot M P$, a well-known result considering that $A F$ is the polar line of $P$ with respect to circle of radius $M C$ centered at $M$.
The "elementary proof" on the latter result may be obtained as follows: $\frac{P B}{P C}=\frac{F B}{F C}$, using, for instance, Menelaus and Ceva theorems with respect to $A B C$. Cross-multiplying one gets $(P M-x)(F M+x)=(x-F M)(P M+x)$
- $x$ stands for the length of $M C$ - and then $P M \cdot F M=x^{2}$.
Comment. The proof above holds for both cases $A BA C$; it is for the committee to decide if a contestant is supposed to (even) mention this.
### 2.4 Number Theory
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 163 |
NT2 Let $n \geq 2$ be a fixed positive integer. An integer will be called " $n$-free" if it is not a multiple of an $n$-th power of a prime. Let $M$ be an infinite set of rational numbers, such that the product of every $n$ elements of $M$ is an $n$-free integer. Prove that $M$ contains only integers.
|
## Solution
We first prove that $M$ can contain only a finite number of non-integers. Suppose that there are infinitely many of them: $\frac{p_{1}}{q_{1}}, \frac{p_{2}}{q_{2}}, \ldots, \frac{p_{k}}{q_{k}}, \ldots$, with $\left(p_{k}, q_{k}\right)=1$ and $q_{k}>1$ for each $k$. Let $\frac{p}{q}=\frac{p_{1} p_{2} \ldots p_{n-1}}{q_{1} q_{2} \ldots q_{n-1}}$, where $(p, q)=1$. For each $i \geq n$, the number $\frac{p}{q} \cdot \frac{p_{i}}{q_{i}}$ is an integer, so $q_{i}$ is a divisor of $p$ (as $q_{i}$ and $p_{i}$ are coprime). But $p$ has a finite set of divisors, so there are $n$ numbers of $M$ with equal denominators. Their product cannot be an integer, a contradiction.
Now suppose that $M$ contains a fraction $\frac{a}{b}$ in lowest terms with $b>1$. Take a prime divisor $p$ of $b$. If we take any $n-1$ integers from $M$, their product with $\frac{a}{b}$ is an integer, so some of them is a multiple of $p$. Therefore there are infinitely many multiples of $p$ in $M$, and the product of $n$ of them is not $n$-free, a contradiction.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 164 |
NT10 Prove that $2^{n}+3^{n}$ is not a perfect cube for any positive integer $n$.
|
## Solution
If $n=1$ then $2^{1}+3^{1}=5$ is not perfect cube.
Perfect cube gives residues $-1,0$ and 1 modulo 9 . If $2^{n}+3^{n}$ is a perfect cube, then $n$ must be divisible with 3 (congruence $2^{n}+3^{n}=x^{3}$ modulo 9 ).
If $n=3 k$ then $2^{3 k}+3^{2 k}>\left(3^{k}\right)^{3}$. Also, $\left(3^{k}+1\right)^{3}=3^{3 k}+3 \cdot 3^{2 k}+3 \cdot 3^{k}+1>3^{3 k}+3^{2 k}=$ $3^{3 k}+9^{k}>3^{3 k}+8^{k}=3^{3 k}+2^{3 k}$. But, $3^{k}$ and $3^{k}+1$ are two consecutive integers so $2^{3 k}+3^{3 k}$ is not a perfect cube.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 165 |
ALG 1. A number $A$ is written with $2 n$ digits, each of whish is 4 , and a number $B$ is written with $n$ digits, each of which is 8 . Prove that for each $n, A+2 B+4$ is a total square.
|
## Solution.
$$
\begin{aligned}
A & =\underbrace{44 \ldots 44}_{2 n}=\underbrace{44 \ldots 4}_{n} \underbrace{44 \ldots 4}_{n}=\underbrace{44 \ldots 4}_{n} \underbrace{400 \ldots 0}_{n}-\underbrace{44 \ldots 4}_{n}+\underbrace{88 \ldots 8}_{n}=\underbrace{44 \ldots 4}_{n} \cdot\left(10^{n}-1\right)+B \\
& =4 \cdot \underbrace{11 \ldots 1}_{n} \cdot \underbrace{99 \ldots 9}_{n}+B=2^{2} \cdot \underbrace{11 \ldots 1}_{n} \cdot 3^{2} \cdot \underbrace{11 \ldots 1}_{n}+B=\underbrace{66}_{n} \ldots 6 \\
& =[\frac{36}{4} \cdot \underbrace{88 \ldots 8}_{n}+B=[3 \cdot \underbrace{22 \ldots 2}_{n}]^{2}+B=\left(\frac{3}{4} B\right)^{2}+B .
\end{aligned}
$$
So,
$$
\begin{aligned}
A+2 B+4 & =\left(\frac{3}{4} B\right)^{2}+B+2 B+4=\left(\frac{3}{4} B\right)^{2}+2 \cdot \frac{3}{4} B \cdot 2+2^{2}=\left(\frac{3}{4} B+2\right)^{2}=(\frac{3}{4} \cdot \underbrace{88 \ldots 8}_{n}+2)^{2} \\
& =(3 \cdot \underbrace{22 \ldots 2}_{n}+2)^{2}=\underbrace{66 \ldots 68^{2}}_{n-1}
\end{aligned}
$$
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 166 |
ALG 2. Let $a, b, c$ be lengths of triangle sides, $p=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $q=\frac{a}{c}+\frac{c}{b}+\frac{b}{a}$.
Prove that $|p-q|<1$.
|
Solution: One has
$$
\begin{aligned}
a b c|p-q| & =a b c\left|\frac{c-b}{a}+\frac{a-c}{b}+\frac{b-a}{c}\right| \\
& =\left|b c^{2}-b^{2} c+a^{2} c-a c^{2}+a b^{2}-a^{2} b\right|= \\
& =\left|a b c-a c^{2}-a^{2} b+a^{2} c-b^{2} c+b c^{2}+a b^{2}-a b c\right|= \\
& =\left|(b-c)\left(a c-a^{2}-b c+a b\right)\right|= \\
& =|(b-c)(c-a)(a-b)| .
\end{aligned}
$$
Since $|b-c|<a,|c-a|<b$ and $|a-b|<c$ we infere
$$
|(b-c)(c-a)(a-b)|<a b c
$$
and
$$
|p-q|=\frac{|(b-c)(c-a)(a-b)|}{a b c}<1
$$
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 167 |
## ALG 4.
Let $a, b, c$ be rational numbers such that
$$
\frac{1}{a+b c}+\frac{1}{b+a c}=\frac{1}{a+b}
$$
Prove that $\sqrt{\frac{c-3}{c+1}}$ is also a rational number
|
Solution. By cancelling the denominators
$$
(a+b)^{2}(1+c)=a b+c\left(a^{2}+b^{2}\right)+a b c^{2}
$$
and
$$
a b(c-1)^{2}=(a+b)^{2}
$$
If $c=-1$, we obtrin the contradiction
$$
\frac{1}{a-b}+\frac{1}{b-a}=\frac{1}{a+b}
$$
Furtherrdore,
$$
\begin{aligned}
(c-3)(c+1) & =(c-1)^{2}-4=\frac{(a+b)^{2}}{a b}-4 \\
& =\frac{(a-b)^{2}}{a b}=\left(\frac{(a-b)(c-1)}{a+b}\right)^{2}
\end{aligned}
$$
Thus
$$
\sqrt{\frac{c-3}{c+1}}=\frac{\sqrt{(c-3)(c+1)}}{c+1}=\frac{|a-b||c-1|}{(c+1)|a+b|} \in \mathrm{Q}
$$
as needed.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 169 |
ALG 6'. Let $a, b, c$ be positive numbers such that $a b+b c+c a=3$. Prove that
$$
a+b+c \geq a b c+2
$$
|
Solution. Eliminating $c$ gives
$$
a+b+c-a b c=a+b+(1-a b) c=a+b+\frac{(1-a b)(3-a b)}{a+b}
$$
Put $x=\sqrt{a b}$. Then $a+b \geq 2 x$, and since $1<x^{2}<3, \frac{(1-a b)(3-a b)}{a+b} \geq \frac{\left(1-x^{2}\right)\left(3-x^{2}\right)}{2 x}$.
It then suffices to prove that
$$
2 x+\frac{\left(1-x^{2}\right)\left(3-x^{2}\right)}{2 x} \geq 2
$$
This iast inequality follows from the arithrnelic-geomeric means inequadily
$$
2 x+\frac{\left(1-x^{2}\right)\left(3-x^{2}\right)}{2 x}=\frac{3+x^{4}}{2 x}=\frac{1}{2 x}+\frac{1}{2 x}+\frac{1}{2 x}+\frac{x^{3}}{2} \geq 4\left(\frac{1}{-16}\right)^{\frac{1}{4}}=2
$$
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 172 |
ALG 7 .
Let $x, y, z$ be real numbers greater than -1 . Prove that
$$
\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \geq 2
$$
|
Solution. We have $y \leq \frac{1+y^{2}}{2}$, hence $\quad$
$$
\frac{1+x^{2}}{1+y+z^{2}} \geq \frac{1+x^{2}}{1+z^{2}+\frac{1+\dot{y}^{2}}{2}}
$$
and the similar inequalities.
Setting $a=1+x^{2}, b=1+y^{2}, c=1+z^{2}$, it sufices to prove that
$$
\frac{a}{2 c+b}+\frac{b}{2 a+c}+\frac{c}{2 b+a} \geq 1
$$
for all $a, b, c \geq 0$.
Put $A=2 c+b, B=2 a+c, C=2 b+a$. Then
$$
a=\frac{C+4 B-2 A}{9}, b=\frac{A+4 C-2 B}{9}, c=\frac{B+4 A-2 C}{9}
$$
and (1) rewrites as
$$
\frac{C+4 B-2 A}{A}+\frac{A+4 C-2 B}{B}+\frac{B+4 A-2 C}{C} \geq 9
$$
and consequently
$$
\frac{C}{A}+\frac{A}{B}+\frac{B}{C}+4\left(\frac{B}{A}+\frac{C}{B}+\frac{A}{C}\right) \geq 15
$$
As $A, B, C>0$, by $A M-G M$ inequality we have
$$
\frac{A}{B}+\frac{B}{C}+\frac{C}{A} \geq 3 \sqrt[3]{\frac{A}{B} \cdot \frac{B}{C} \cdot \frac{C}{A}}
$$
and
$$
\frac{B}{A}+\frac{C}{B}+\frac{A}{C} \geq 3
$$
and we are done.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 173 |
ALG 8. Prove that there exist two sets $A=\{x, y, z\}$ and $B=\{m, n, p\}$ of positive integers greater than 2003 such that the sets have no common elements and the equalities $x+y+z=m+n+p$ and $x^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}$ hold.
|
Solution. Let $A B C$ be a triangle with $B C=a, A C=b, A B=c$ and $ak+3=c
$$
a triangle with such length sides there exist. After the simple calculations we have
$$
\begin{gathered}
A=\left\{3(k+1)^{2}-2,3(k+2)^{2}+4,3(k+3)^{2}-2\right\} \\
B=\left\{3(k+1)^{2}, 3(k+2)^{2}, 3(k+3)^{2}\right\}
\end{gathered}
$$
It easy to prove that
$$
\begin{gathered}
x+y+z=m+n+p=3\left[(k+1)^{2}+(k+2)^{2}+(k+3)^{2}\right] \\
x^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}=9\left[(k+1)^{4}+(k+2)^{4}+(k+3)^{4}\right]
\end{gathered}
$$
$>$ From the inequality $3(k+1)^{2}-2>2003$ we obtain $k \geq 25$. For $k=25$ we have an example of two sets
$$
A=\{2026,2191,2350\}, \quad B=\{2028,2187,2352\}
$$
with desired properties.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 174 |
COM 3. Prove that amongst any 29 natural numbers there are 15 such that sum of them is divisible by 15 .
|
Solution: Amongst any 5 natural numbers there are 3 such that sum of them is divisible by 3 . Amongst any 29 natural numbers we can choose 9 groups with 3 numbers such that sum of numbers in every group is divisible by 3. In that way we get 9 natural numbers such that all of them are divisiblc by 3. It is easy to see that amongst any 9 natural numbers there are 5 such that sum of them is divisible by 5 . Since we have 9 numbers, all of them are divisible by 3 , there are 5 such that sum of them is divisible by 15 .
## $\operatorname{COM} 4$.
$n$ points are given in a plane, not three of them colinear. One observes that no matter how we label the points from 1 to $n$, the broken line joining the points $1,2,3, \ldots, n$ (in this order) do not intersect itself.
Find the maximal value of $n$.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 175 |
COM 5. If $m$ is a number from the set $\{1,2,3,4\}$ and each point of the plane is painted in red or blue, prove that in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$.
|
Solution. Suppose that in the plane there no exists an equilateral triangle with the vertices of the same colour and length side $m=1,2,3,4$.
First assertion: we shall prove that in the plane there no exists a segment with the length 2 such that the ends and the midpint of this segment have the same colour. Suppose that the segment $X Y$ with length 2 have the midpoint $T$ such that the points $X, Y, T$ have the same colour (for example, red). We construct the equilateral triangle. $X Y Z$. Hence, the point $Z$ is blue. Let $U$ and $V$ be the midpoints of the segments $X Z$ and $Y Z$ respectively. So, the points $U$ and $V$ are blue. We obtain a contradiction, because the equilateral triangle $U V Z$ have three blue vertices.
Second assertion: in the same way we prove that in the plane there no exists a segment with the length 4 such that the ends and the midpoint of this segment have the same colour.
Consider the equilateral triangle $A B C$ with length side 4 and divide it into 16 equilateral triangles with length sides 1. L $0: D$ be the midpoint of the segment $A B$. The vertices $A, B, C$ don't have the same colour. WLOG we suppose that $A$ and $B$ are red and $C$ is blue. So, the point $D$ is blue too. We shall investigate the following cases:
a) The midpoints $E$ and $F$ of the sides $A C$ and, respectively, $B C$ are red. From the first assertion it follows that the midpoints $M$ and $N$ of the segments $A E$ and, respectively, $B F$ are blue. Hence, the equilateral triangle $M N C$ have three blue vertices, a contradiction.
b) Let $E$ is red and $F$ is blue. The second one position of $E$ and $F$ is simmetrical. If $P, K, L$ are the midpoints of the segments $C F, A D, B D$ respectively, then by first assertion $P$ is red, $M$ is blue and $N$ is red. This imply that $K$ and $L$ are blue. So, the segment $K L$ with length 2 has the blue ends and blue midpoint, a contradiction.
c) If $E$ and $F$ are blue, then the equilateral triangle $E F C$ has three blue vertices, a contradiction.
Hence, in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$, where $m \in\{1,2,3,4\}$.
Comment: The formuation of the problem suggests that one has to find 4 triangles, one for each $m$ from the set $\{1,2,3,4\}$ whereas the solution is for one $m$. A better formulation is:
Each point of the plane is painted in red or blue. Prove that in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$, where $m$ is some number from the set $\{1,2,3,4\}$.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 176 |
## GEO 3.
Let $G$ be the centroid of the the triangle $A B C$. Reflect point $A$ across $C$ at $A^{\prime}$. Prove that $G, B, C, A^{\prime}$ are on the same circle if and only if $G A$ is perpendicular to $G C$.
|
Solution. Observe first that $G A \perp G C$ if and only if $5 A C^{2}=A B^{2}+B C^{2}$. Indeed,
$$
G A \perp G C \Leftrightarrow \frac{4}{9} m_{a}^{2}+\frac{4}{9} m_{c}^{2}=b^{2} \Leftrightarrow 5 b^{2}=a^{2}+c^{2}
$$
Moreover,
$$
G B^{2}=\frac{4}{9} m_{b}^{2}=\frac{2 a^{2}+2 c^{2}-b^{2}}{9}=\frac{9 b^{2}}{9}=b^{2}
$$
hence $G B=A C=C A^{\prime}$ (1). Let $C^{\prime}$ be the intersection point of the lines $G C$ and $A B$. Then $C C^{\prime}$ is the middle line of the triangle $A B A^{\prime}$, hence $G C \| B A^{\prime}$. Consequently, $G C A^{\prime} B$ is a trapezoid. From (1) we find that $G C A^{\prime} B$ is isosceles, thus cyclic, as needed.
Conversely, since $G C A^{\prime} B$ is a cyclic trapezoid, then it is also isosceles. Thus $C A^{\prime}=$ $G B$, which leads to (1).
Comment: An alternate proof is as follows:
Let $M$ be the midpoint of $A C$. Then the triangles $M C G$ and $M A^{\prime} B$ are similar. So $G C$ is parallel to $A^{\prime} B$.
$G A \perp G C$ if and only if $G M=M C$. By the above similarity, this happen if and only if $A^{\prime} C=G B$; if and only if the trapezoid is cyclic.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 177 |
GEO 5. Let three congruent circles intersect in one point $M$ and $A_{1}, A_{2}$ and $A_{3}$ be the other intersection points for those circles. Prove that $M$ is a.orthocenter for a triangle $A_{1} A_{2} A_{3}$.
|
Solution: The quadrilaterals $\mathrm{O}_{3} M O_{2} A_{1}, \mathrm{O}_{3} M O_{1} A_{2}$ and $O_{1} M O_{2} A_{3}$ are rombes. Therefore, $O_{2} A_{1} \| M O_{3}$ and $M O_{3} \| O_{1} A_{2}$, which imply $O_{2} A_{1} \| O_{1} A_{2}$. Because $O_{2} A_{1}=O_{3}{ }^{*} M=O_{1} A_{2}$ the quadrilateral $O_{2} A_{1} A_{2} O_{1}$ is parallelogram and then $A_{1} A_{2} \| O_{1} O_{2}$ and $A_{1} A_{2}=O_{1} O_{2}$. Similary, $A_{2} A_{3} \| O_{2} O_{3}$ and $A_{2} A_{3}=O_{2} O_{3} ; A_{3} A_{1} \| O_{3} O_{1}$ and $A_{3} A_{1}=O_{3} O_{1}$. The triangles $A_{1} A_{2} A_{3}$ and $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$ are congruent.
Since $A_{3} M \perp O_{1} O_{2}$ and $O_{1} O_{2} \| A_{1} A_{2}$ we infere $A_{3} M \perp A_{1} A_{2}$. Similary, $A_{2} M \perp A_{1} A_{3}$ and $A_{1} M \perp A_{2} A_{3}$. Thus, $M$ is the orthocenter for the triangle $A_{1} A_{2} A_{3}$.
## GEO.6.
Consider an isosceles triangle $A B C$ with $A B=A C$. A semicircle of diameter $E F$, lying on the side $B C$, is tangent to the lines $A B$ and $A C$ at $M$ and $N$, respectively. The line $A E$ intersects again the semicircle at point $P$.
Prove that the line PF passes through the midpoint of the chord $M N$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 178 |
GEO 7. Through a interior point of a triangle, three lines parallel to the sides of the triangle are constructed. In that way the triangle is divided on six figures, areas equal $a, b, c, \alpha, \beta, \gamma$ (see the picture).
Prove that
$$
\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma} \geqslant \frac{3}{2}
$$
|
Solution: We will prove the inequality in two steps. First one is the following
Lemma: Let $A B C$ be a triangle, $E$ arbitrary point on the side $A C$. Parallel lines to $A B$ and $B C$, drown through $E$ meet sides $B C$ and $A B$ in points $F$. and $D$ respectively. Then: $P_{B D E F}=2 \sqrt{P_{A D E} \cdot P_{E F C}}$ ( $P_{X}$ is area for the figure $X)$.
The triangles $A D E$ and $E F C$ are similar. Then:
$$
\frac{P_{B D E F}}{2 P_{A D E}}=\frac{P_{B D E}}{P_{A D E}}=\frac{B D}{A D}=\frac{E F}{A D}=\frac{\sqrt{P_{E F C}}}{\sqrt{P_{A D E}}}
$$
Hence, $P_{B D E F}=2 \sqrt{P_{A D E} \cdot P_{E F C}}$.
Using this lemma one has $\alpha=2 \sqrt{b c}, \beta=2 \sqrt{a c}, \gamma=2 \sqrt{a b}$. The GML-AM mean inequality provides
$$
\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma} \geqslant 3 \sqrt[3]{\frac{a b c}{\alpha \beta \gamma}}=3 \sqrt[3]{\frac{a b c}{2^{3} \sqrt{a^{2} b^{2} c^{2}}}}=\frac{3}{2}
$$
BULGARIA
| Leader: | Chavdar Lozanov |
| :--- | :--- |
| Deputy Leader: | Ivan Tonov |
| Contestants: | Asparuh Vladislavov Hriston |
| | Tzvetelina Kirilova Tzeneva |
| | Vladislav Vladilenon Petkov |
| | Alexander Sotirov Bikov |
| | Deyan Stanislavov Simeonov |
| | Anton Sotirov Bikov |
## CYPRUS
| Leader: | Efthyvoulos Liasides |
| :--- | :--- |
| Deputy Leader: | Andreas Savvides |
| Contestants: | Marina Kouyiali |
| | Yiannis loannides |
| | Anastasia Solea |
| | Nansia Drakou |
| | Michalis Rossides |
| | Domna Fanidou |
| Observer: | Myrianthi Savvidou |
FORMER YUGOSLAV
REPUBLIC of MACEDONIA
| Leader: | Slavica Grkovska |
| :--- | :--- |
| Deputy Leader: | Misko Mitkovski |
| Contestants: | Aleksandar lliovski |
| | Viktor Simjanovski |
| | Maja Tasevska |
| | Tanja Velkova |
| | Matej Dobrevski |
| | Oliver Metodijev |
## GREECE
Leader: Anargyros Felouris
Deputy Leader: Ageliki Vlachou
Contestants: Theodosios Douvropoulos
Marina lliopoulou
Faethontas Karagiannopoulos
Stefanos Kasselakis
Fragiskos Koufogiannis
Efrosyni Sarla
ROMANIA
| Leader: | Dan Branzei |
| :--- | :--- |
| Deputy Leader: | Dinu Serbanescu |
| Contestants: | Dragos Michnea |
| | Adrian Zahariuc |
| | Cristian Talau |
| | Beniämin Bogosel |
| | Sebastian Dumitrescu |
| | Lucian Turea |
## TURKEY
Leader:
Halil Ibrahim Karakaş
\&Deputy Leader: Duru Türkoğlu
Contestants: Sait Tunç
Anmet Kabakulak
Türkü Çobanoğlu
Burak Sağlam
Ibrahim Çimentepe
Hale Nur Kazaçeşme
## YUGOSLAVIA
(SERBIA and MONTENEGRO)
| Leader: | Branislav Popovic |
| :--- | :--- |
| Deputy Leader: | Marija Stanic |
| Contestants: | Radojevic Mladen |
| | Jevremovic Marko |
| | Djoric Milos |
| | Lukic Dragan |
| | Andric Jelena |
| | Pajovic Jelena |
## TURKEY-B
## Leader:
Deputy Leader: Contestants:
Ahmet Karahan
Deniz Ahçihoca ..... Havva Yeşildağl|
Çağıl Şentip
Buse Uslu
Ali Yilmaz
Demirhan Çetereisi
Yakup Yildirim
## REPUBLIC of MOLDOVA
| Leader: | Ion Goian |
| :--- | :--- |
| Deputy Leader: | Ana Costas |
| Contestants: | lurie Boreico |
| | Andrei Frimu |
| | Mihaela Rusu |
| | Vladimir Vanovschi |
| | Da Vier: |
| | Alexandru Zamorzaev |
1.Prove that $7^{n}-1$ is not divisible by $6^{n}-1$ for any positive integer $n$.
2. 2003 denars were divided in several bags and the bags were placed in several pockets. The number of bags is greater than the number of denars in each pocket. Is it true that the number of pockets is greater than the number of denars in one of the bags?
3. In the triangle $\mathrm{ABC}, R$ and $r$ are the radii of the circumcircle and the incircle, respectively; $a$ is the longest side and $h$ is the shortest altitude. Prove that $R / r>a / h$.
4. Prove that for all positive numbers $x, y, z$ such that $x+y+z=1$ the following inequality holds
$$
\frac{x^{2}}{1+y}+\frac{y^{2}}{1+z}+\frac{z^{2}}{1+x} \leq 1
$$
5.Is it possible to cover a $2003 \times 2003$ board with $1 \times 2$ dominoes placed horizontally and $1 \times 3$ threeminoes placed vertically?
## THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA Chişinău, March 9-12, 2003
7.1 Let $m>n$ be pozitive integers. For every positive integers $k$ we define the number $a_{k}=(\sqrt{5}+2)^{k}+$ $(\sqrt{5}-2)^{k}$. Show that $a_{m+n}+a_{m-n}=a_{m} \cdot a_{n}$.
T. Fild all five digits numbers $\overline{a b c d e}$, written in decimal system, if it is known that $\overline{a b} c d e-\overline{e b c d a}=69993$, $\overline{b c d}-\overline{d c b}=792, \overline{b c}-\overline{c b}=72$.
7.3 In the triangle $A B C$ with semiperemeter $p$ the points $M, N$ and $P$ lie on the sides $(B C),(C A)$ and - (AB) respectively. Show that $pb \geq$ 10. Prove that this equation has two irrational solutions. (The number $m$ is triangular, if $m=n(n-1) / 2$ for certain positive integer $n \geq 1$ ).
9.3 The distinct points $M$ and $N$ lie on the hypotenuse ( $A C)$ of the right isosceles triangle $A B C$ so that $M \in(A N)$ and $M N^{2}=A M^{2}+C N^{2}$. Prove that $m(\angle M B N)=45^{\circ}$.
9.4 Find all the functions $f: N^{*} \rightarrow N^{*}$ which verify the relation $f(2 x+3 y)=2 f(x)+3 f(y)+4$ for every positive integers $x, y \geq 1$.
9.5 The numbers $a_{1}, a_{2}, \ldots, a_{n}$ are the first $n$ positive integers with the property that the number $8 a_{k}+1$ is a perfect square for every $k=1,2, \ldots, n$. Find the sum $S_{n}=a_{1}+a_{2}+\ldots+a_{n}$.
9.6 Find all real solutions of the equation $x^{4}+7 x^{3}+6 x^{2}+5 \sqrt{2003} x-2003=0$.
9.7 The side lengths of the triangle $A B C$ satisfy the relations $a>b \geq 2 c$. Prove that the altitudes of the triangle $A B C$ can not be the sides of any triangle.
9.8 The base of a pyramid is a convex polygon with 9 sides. All the lateral edges of the pyramid and all the liagunads ui the base are coloured in a random way in red or blue. Pröve that there exist at least three vertices of the pyramid which belong to a triangle with the sides coloured in the same colour.
10.1 Find all prime numbers $a, b$ and $c$ for which the equality $(a-2)!+2 b!=22 c-1$ holds.
10.2 Solve the system $x+y+z+t=6, \sqrt{1-x^{2}}+\sqrt{4-y^{2}}+\sqrt{9-z^{2}}+\sqrt{16-t^{2}}=8$.
10.3 In the scalen triangle $A B C$ the points $A_{1}$ and $B_{1}$ are the bissectrices feets, drawing from the vertices $A$ and $B$ respectively. The straight line $A_{1} B_{1}$ intersect the line $A B$ at the point $D$. Prove that one of the angles $\angle A C D$ or $\angle B C D$ is obtuze and $m(\angle A C D)+m(\angle B C D)=180^{\circ}$.
10.4 Let $a>1$ be not integer number and $a \neq \sqrt[2]{q}$ for every positive integers $p \geq 2$ and $q \geq 1$, $k=\left[\log _{a} n\right] \geq 1$, where $[x]$ is the integral part of the real number $x$. Prove that for every positive integer $n \geq 1$ the equality
$$
\left[\log _{a} 2\right]+\left[\log _{a} 3\right]+\ldots+\left[\log _{a} n\right]+[a]+\left[a^{2}\right]+\ldots+\left[a^{k}\right]=n k
$$
holds.
10.5 The rational numbers $p, q, r$ satisfy the relation $p q+p r+q r=1$. Prove that the number $\left(1+p^{2}\right)\left(1+q^{3}\right)\left(1+r^{2}\right)$ is a square of any rational number.
10.6 Let $n \geq 1$ be a positive integer. For every $k=1,2, \ldots, n$ the functions $f_{k}: R \rightarrow R, f_{k}(x)=$ $a_{k} x^{2}+b_{k} x+c_{k}$ with $a_{k} \neq 0$ are given. Find the greatest possible number of parts of the rectangular plane $x O y$ which can be obtained by the intersection of the graphs of the functions $f_{k}(k=1,2, \ldots, n)$.
10.7 The circle with the center $O$ is tangent to the sides $[A B],[B C],[C D]$ and $[D A]$ of the convex quadrilateral $A B C D$ at the points $M, N, \mathcal{K}$ and $L$ respectively. The straight lines $M N$ and $A C$ are parallel and the straight line $M K$ intersect the line $L N$ at the point $P$. Prove that the points $A, M, P, O$ and $L$ are concyclic.
10.8 Find all integers $n$ for which the number $\log _{2 n-1}\left(n^{2}+2\right)$ is rational.
11.1 Let $a, b, c, d \geq 1$ be arbitrary positive numbers. Prove that the equations system $a x-y z=$ $c, \quad b x-y t=-d$. has at least a solution $(x, y, z, t)$ in positive integers.
11.2 The sequences $\left(a_{n}\right)_{n \geq 0}$ and $\left(b_{n}\right)_{n \geq 0}$ satisfy the conditions $(1+\sqrt{3})^{2 n+1}=a_{n}+b_{n} \sqrt{3}$ and $a_{n}, b_{n} \in Z$. Find the recurrent relation for each of the sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$.
11.3 The triangle $A B C$ is rightangled in $A, A C=b, A B=c$ and $B C=a$. The halfstraight line ( $A z$ is perpendicular to the plane $(A B C), M \in(A z$ so that $\alpha, \beta, \gamma$ are the mesures of the angles, formed by the edges $M B, M C$ and the plane ( $M B C$ ) with the plane ( $A B C$ ) respectively. In the set of the triangular pyramids MABC on consider the pyramids with the volumes $V_{1}$ and $V_{2}$ which satisfy the relations $\alpha+\beta+\gamma=\pi$ and $\alpha+\beta+\gamma=\pi / 2$ respectively. Prove the equality $\left(V_{1} / V_{2}\right)^{2}=(a+b+c)(1 / a+1 / b+1 / c)$.
11.4 Find all the functions $f:[0 ;+\infty) \rightarrow[0 ;+\infty)$ which satisfy the conditions: : $f(x f(y)) \cdot f(y)=$ $f(x+y)$ for every $x, y \in[0 ;+\infty) ; f(2)=0 ; f(x) \neq 0$ for every $x \in[0 ; 2)$.
11.5 Let $02 R \sin \alpha$.
12.4 The real numbers $\alpha, \beta, \gamma$ satisfy the relations $\sin \alpha+\sin \beta+\sin \gamma=0$ and $\cos \alpha+\cos \beta+\cos \gamma=0$. Find all positive integers $n \geq 0$ for-which $\sin (n \alpha+\pi / 4)+\sin (n \beta+\pi / 4)+\sin (n \gamma+\pi / 4)=0$.
12.5 For every positive integer $n \geq 1$ we define the polynomial $P(X)=X^{2 n}-X^{2 n-1}+\ldots-X+1$, Find the remainder of the division of the polynomial $P\left(X^{2 n+1}\right)$ by the polynomial $P(X)$.
12.6 Fie $n \in N$. Find all the primitives of the function
$$
f: R \rightarrow R, \quad f(x)=\frac{x^{3}-9 x^{2}+29 x-33}{\left(x^{2}-6 x+10\right)^{n}}
$$
12.7 In a rectangular system $x O y$ the graph of the function $f: R \rightarrow R, f(x)=x^{2}$ is drawn. The ordered triple $B, A, C$ has distinct points on the parabola, the point $D \in(B C)$ such that the straight line $A D$ is parallel to the axis $O y$ and the triangles $B A D$ and $C A D$ have the areas $s_{1}$ and $s_{2}$ respectively. Find the length of the segment $[A D]$.
12.8 Let $\left(F_{n}\right)_{n \in N^{*}}$ be the Fibonacci sequence so that: $F_{1}=1, F_{2}=1, F_{n+1}=F_{n}+F_{n-1}$ for every positive integer $n \geq 2$. Shown that $F_{n}<3^{n / 2}$ and calculate the limit $\lim _{n \rightarrow \infty}\left(F_{1} / 2+F_{2} / 2^{2}+\ldots+F_{n} / 2^{n}\right)$.
## The first selection test for IMO 2003 and BMO 2003, March 12, 2003
B1. Each side of the arbitrary triangle is divided into 2002 congruent segments. After that each interior division point of the side is joined with opposite vertex. Prove that the number of obtained regions of the triangle is divisible by 6 .
B2. The positive real numbers $x, y$ and $z$ satisfy the relation $x+y+z \geq 1$. Prove the inequality
$$
\frac{x \sqrt{x}}{y+z}+\frac{y \sqrt{y}}{x+z}+\frac{z \sqrt{z}}{x+y} \geq \frac{\sqrt{3}}{2}
$$
B3. The quadrilateral $A B C D$ is inscribed in the circle with center $O$, the points $M$ and $N$ are the middle points of the diagonals $[A C]$ and $[B D]$ respectively and $P$ is the intersection point of the diagonals. It is known that the points $O, M, N$ si $P$ are distinct. Prove that the points $O, M, B$ and $D$ are concyclic if and only if the points $O, N, A$ and $C$ are concyclic.
B4. Prove that the equation $1 / a+1 / b+1 / c+1 /(a b c) \doteq 12 /(a+b+c)$ has many solutions $(a, b, c)$ in strictly positive integers.
## The second selection test for IMO 2003, March 22, 2003
B5. Let $n \geq 1$ be positive integer. Find all polynomials of degree $2 n$ with real coefficients
$$
P(X)=X^{2 n}+(2 n-10) X^{2 n-1}+a_{2} X^{2 n-2}+\ldots+a_{2 n-2} X^{2}+(2 n-10) X+1
$$
-if it is known that they have positive real roots.
B6. The triangle $A B C$ has the semiperimeter $p$, the circumradius $R$, the inradius $r$ and $l_{a,}, l_{b}, l_{c}$ are the lengths of internal bissecticies, drawing from the vertices $A, B$ and $C$ respectively. Prove the inequality $l_{a} l_{b}+l_{b} l_{c}+l_{c} l_{a} \leq p \sqrt{3 r^{2}+12 R r}$.
B7. The points $M$ and $N$ are the tangent points of the sides $[A B]$ and $[A C]$ of the triangle $A B C$ to the incircle with the center $I$. The internal bissectrices, drawn from the vertices $B$ and $C$, intersect the straight line $M N$ at points $P$ and $Q$ respectively. If $F$ is the intersection point of the swtraight lines $C P$ and $B Q$, then prove that the straight lines $F I$ and $B C$ are perpendicular.
B8. Let $n \geq 4$ be the positive integer. On the checkmate table with dimensions $n \times n$ we put the coins. One consider the diagonal of the table each diagonal with at least two unit squares. What is the smallest number of coins put on the table so that on the each horizontal, each vertical and each diagonal there exists att least one coin. Prove the answer.
## The third selection test for IMO 2003, March 23, 2003
B9. Let $n \geq 1$ be positive integer. A permutation $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of the numbers $(1,2, \ldots, n)$ is called quadratique if among the numbers $a_{1}, a_{1}+a_{2}, \ldots, a_{1}+a_{2}+\ldots+a_{n}$ there exist at least a perfect square. Find the greatest number $n$, which is less than 2003 , such that every permutation of the numbers $(1,2, \ldots, n)$ will be quadratique.
B10. The real numbers $a_{1}, a_{2}, \ldots, a_{2003}$ satisfy simultaneousiy the relations: $a_{i} \geq 0$ for all $i=$ $1,2, \ldots, 2003 ; \quad a_{1}+a_{2}+\ldots+a_{2003}=2 ; \quad a_{1} a_{2}+a_{2} a_{3}+\ldots+a_{2003} a_{1}=1$. Find the smallest value of the sum $a_{1}^{2}+a_{2}^{2}+\ldots+a_{2003}^{2}$.
B11. The arbitrary point $M$ on the plane of the triangle $A B C$ does not belong on the straight lines $A B, B C$ and $A C$. If $S_{1}, S_{2}$ and $S_{3}$ are the areas of the triangles $A M B, B M C$ and $A M C$ respectively, find the geometrical locus of the points $M$ which satisfy the relation $\left(M A^{2}+M B^{2}+M C^{2}\right)^{2}=16\left(S_{1}^{2}+S_{2}^{2}+S_{3}^{2}\right)$.
812. Let $n \geq 1$ be a positive integer. A square table of dimensions $n \times n$ is full arbitrarly completed $\because$ the numb so, shat every number appear exactly conce the table. from cack fine one select the smallest number and the greatest of them is denote by $x$. From each column one select the greatest number and the smallest of them is denote by $y$. The table is called equilibrated if $x=y$. How match equilibrated tables there exist?
## The first selection test for JBMO 2003, April 12, 2003
JB1. Let $n \geq 2003$ be a positive integer such that the number $1+2003 n$ is a perfect square. Prove that the number $n+1$ is equal to the sum of 2003 positive perfect squares.
JB2. The positive real numbers $a, b, c$ satisfy the relation $a^{2}+b^{2}+c^{2}=3 a b c$. Prove the inequality
$$
\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \geq \frac{9}{a+b+c}
$$
JB3. The quadrilateral $A B C D$ with perpendicular diagonals is inscribed in the circle with center $O$, the points $M$ and $N$ are the middle points of the sides $[B C]$ and $[C D]$ respectively. Find the value of the ratio of areas of the figures $O M C N$ and $A B C D$.
JB4. Let $m$ and $n$ be the arbitrary digits of the decimal system and $a, b, c$ be the positive distinct integers of the form $2^{m} \cdot 5^{n}$. Find the number of the equations $a x^{2}-2 b x+c=0$, if it is known that each equation has a single real solution.
## The second selection test for JMBO 2003, April 13, 2003
JB5. Prove that each positive integer is equal to a difference of two positive integers with the same number of the prime divisors.
JB6. The real numbers $x$ and $y$ satisfy the equalities
$$
\sqrt{3 x}\left(1+\frac{1}{x+y}\right)=2, \quad \sqrt{7 y}\left(1-\frac{1}{x+y}\right)=4 \sqrt{2}
$$
Find the numerical value of the ratio $y / x$.
$J B 7$. The triangle $A B C$ is isosceles with $A B=B C$. The point $F$ on the side $[B C]$ and the point $D$ on the side $[A C]$ are the feets of the internal bissectrix drawn from $A$ and altitude drawn from $B$ respectively so that $A F=2 B D$. Find the measure of the angle $A B C$.
JB8. In the rectangular coordinate system every point with integer coordinates is called laticeal point. Let $P_{n}(n, n+5)$ be a laticeal point and denote by $f(n)$ the number of laticeal points on the open segment $\left(O P_{n}\right)$, where the point $O(0,0)$ is the coordinates system origine. Calculate the number $f(1)+f(2)+$ $f(3)+\ldots+f(2002)+f(2003)$.
7 th Junior Balkan Mathematical O-lympiad
$20-25$ Jun e, 20.03 I $\mathrm{m}$ i r $\quad$. $\quad$ u rke y
## English Version
1. Let $n$ be a positive integer. A number $A$ consists of $2 n$ digits, each of which is 4 ; and a number $B$ consists of $n$ digits, each of which is 8 . Prove that $A+2 B+4$ is a perfect square.
\&
2. Suppose there are $n$ points in a plane no three of which are collinear with the following property:
If we label these points as $A_{1}, A_{2}, \ldots, A_{n}$ in any way whatsoever, the broken line $A_{1} A_{2} \ldots A_{n}$ does not intersect itself.
Find the maximal value that $n$ can have.
3. Let $k$ be the circumcircle of the triangle $A B C$. Consider the arcs $\overparen{A B}, \widehat{B C}, \widetilde{C A}$ such that $C \notin \widetilde{A B}, A \notin \widetilde{B C}, B \notin \widetilde{C A}$. Let $D, E$ and $F$ be the midpoints of the arcs $\widehat{B C}, \overparen{C A}, \overparen{A B}$, respectively. Let $G$ and $H$ be the points of intersection of $D E$ with $C B$ and $C A$; let $I$ and $J$ be the points of intersection of $D F$ with $B C$ and $B A$, respectively. Denote the midpoints of $G H$ and $I J$ by $M$ and $N$, respectively.
a) Find the angles of the triangle $D M N$ in terms of the angles of the triangle $A B C$.
b) If $O$ is the circumcentre of the triangle $D M N$ and $P$ is the intersection point of $A D$ and $E F$, prove that $O, P, M$ and $N$ lie on the same circle.
4. Let $x, y, z$ be real numbers greater than -1 . Prove that
$$
\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \geq 2
$$
## Romanian Version
|
proof
|
Inequalities
|
proof
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olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 179 |
87.1. Nine journalists from different countries attend a press conference. None of these speaks more than three languages, and each pair of the journalists share a common language. Show that there are at least five journalists sharing a common language.
|
Solution. Assume the journalists are $J_{1}, J_{2}, \ldots, J_{9}$. Assume that no five of them have a common language. Assume the languages $J_{1}$ speaks are $L_{1}, L_{2}$, and $L_{3}$. Group $J_{2}, J_{3}$, $\ldots, J_{9}$ according to the language they speak with $J_{1}$. No group can have more than three members. So either there are three groups of three members each, or two groups with three members and one with two. Consider the first alternative. We may assume that $J_{1}$ speaks $L_{1}$ with $J_{2}, J_{3}$, and $J_{4}, L_{2}$ with $J_{5}, J_{6}$, and $J_{7}$, and $L_{3}$ with $J_{8}, J_{9}$, and $J_{2}$. Now $J_{2}$ speaks $L_{1}$ with $J_{1}, J_{3}$, and $J_{4}, L_{3}$ with $J_{1}, J_{8}$, and $J_{9}$. $J_{2}$ must speak a fourth language, $L_{4}$, with $J_{5}, J_{6}$, and $J_{7}$. But now $J_{5}$ speaks both $L_{2}$ and $L_{4}$ with $J_{2}, J_{6}$, and $J_{7}$. So $J_{5}$ has to use his third language with $J_{1}, J_{4}, J_{8}$, and $J_{9}$. This contradicts the assumption we made. So we now may assume that $J_{1}$ speaks $L_{3}$ only with $J_{8}$ and $J_{9}$. As $J_{1}$ is not special, we conclude that for each journalist $J_{k}$, the remaining eight are divided into three mutually exclusive language groups, one of which has only two members. Now $J_{2}$ uses $L_{1}$ with three others, and there has to be another language he also speaks with three others. If this were $L_{2}$ or $L_{3}$, a group of five would arise (including $J_{1}$ ). So $J_{2}$ speaks $L_{4}$ with three among $J_{5}, \ldots, J_{9}$. Either two of these three are among $J_{5}, J_{6}$, and $J_{7}$, or among $J_{8}, J_{9}$. Both alternatives lead to a contradiction to the already proved fact that no pair of journalists speaks two languages together. The proof is complete.
Figure 1.
|
proof
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Combinatorics
|
proof
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olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
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train
| 181 |
87.2. Let $A B C D$ be a parallelogram in the plane. We draw two circles of radius $R$, one through the points $A$ and $B$, the other through $B$ and $C$. Let $E$ be the other point of
intersection of the circles. We assume that $E$ is not a vertex of the parallelogram. Show that the circle passing through $A, D$, and $E$ also has radius $R$.
|
Solution. (See Figure 1.) Let $F$ and $G$ be the centers of the two circles of radius $R$ passing through $A$ and $B$; and $B$ and $C$, respectively. Let $O$ be the point for which the the rectangle $A B G O$ is a parallelogram. Then $\angle O A D=\angle G B C$, and the triangles $O A D$ and $G B C$ are congruent (sas). Since $G B=G C=R$, we have $O A=O D=R$. The quadrangle $E F B G$ is a rhombus. So $E F\|G B\| O A$. Moreover, $E F=O A=R$, which means that $A F E O$ is a parallelogram. But this implies $O E=A F=R$. So $A, D$, and $E$ all are on the circle of radius $R$ centered at $O$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 182 |
87.4. Let $a, b$, and $c$ be positive real numbers. Prove:
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \leq \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}
$$
|
Solution. The arithmetic-geometric inequality yields
$$
3=3 \sqrt[3]{\frac{a^{2}}{b^{2}} \cdot \frac{b^{2}}{c^{2}} \cdot \frac{c^{2}}{a^{2}}} \leq \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}
$$
or
$$
\sqrt{3} \leq \sqrt{\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}}
$$
On the other hand, the Cauchy-Schwarz inequality implies
$$
\begin{aligned}
\frac{a}{b}+\frac{b}{c}+ & \frac{c}{a} \leq \sqrt{1^{2}+1^{2}+1^{2}} \sqrt{\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}} \\
& =\sqrt{3} \sqrt{\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}}
\end{aligned}
$$
We arrive at the inequality of the problem by combining (1) and (2).
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 183 |
88.2. Let $a, b$, and $c$ be non-zero real numbers and let $a \geq b \geq c$. Prove the inequality
$$
\frac{a^{3}-c^{3}}{3} \geq a b c\left(\frac{a-b}{c}+\frac{b-c}{a}\right)
$$
When does equality hold?
|
Solution. Since $c-b \leq 0 \leq a-b$, we have $(a-b)^{3} \geq(c-b)^{3}$, or
$$
a^{3}-3 a^{2} b+3 a b^{2}-b^{3} \geq c^{3}-3 b c^{2}+3 b^{2} c-b^{3}
$$
On simplifying this, we immediately have
$$
\frac{1}{3}\left(a^{3}-c^{3}\right) \geq a^{2} b-a b^{2}+b^{2} c-b c^{2}=a b c\left(\frac{a-b}{c}+\frac{b-c}{a}\right)
$$
A sufficient condition for equality is $a=c$. If $a>c$, then $(a-b)^{3}>(c-b)^{3}$, which makes the proved inequality a strict one. So $a=c$ is a necessary condition for equality, too.
|
proof
|
Inequalities
|
proof
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olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
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train
| 184 |
90.1. Let $m, n$, and $p$ be odd positive integers. Prove that the number
$$
\sum_{k=1}^{(n-1)^{p}} k^{m}
$$
is divisible by $n$.
|
Solution. Since $n$ is odd, the sum has an even number of terms. So we can write it as
$$
\sum_{k=1}^{\frac{1}{2}(n-1)^{p}}\left(k^{m}+\left((n-1)^{p}-k+1\right)^{m}\right)
$$
Because $m$ is odd, each term in the sum has $k+(n-1)^{p}-k+1=(n-1)^{p}+1$ as a factor. As $p$ is odd, too, $(n-1)^{p}+1=(n-1)^{p}+1^{p}$ has $(n-1)+1=n$ as a factor. So each term in the sum (1) is divisible by $n$, and so is the sum.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 188 |
90.2. Let $a_{1}, a_{2}, \ldots, a_{n}$ be real numbers. Prove
$$
\sqrt[3]{a_{1}^{3}+a_{2}^{3}+\ldots+a_{n}^{3}} \leq \sqrt{a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}}
$$
When does equality hold in (1)?
|
Solution. If $0 \leq x \leq 1$, then $x^{3 / 2} \leq x$, and equality holds if and only if $x=0$ or $x=1$. - The inequality is true as an equality, if all the $a_{k}$ 's are zeroes. Assume that at least one of the numbers $a_{k}$ is non-zero. Set
$$
x_{k}=\frac{a_{k}^{2}}{\sum_{j=1}^{n} a_{j}^{2}}
$$
Then $0 \leq x_{k} \leq 1$, and by the remark above,
$$
\sum_{k=1}^{n}\left(\frac{a_{k}^{2}}{\sum_{j=1}^{n} a_{j}^{2}}\right)^{3 / 2} \leq \sum_{k=1}^{n} \frac{a_{k}^{2}}{\sum_{j=1}^{n} a_{j}^{2}}=1
$$
So
$$
\sum_{k=1}^{n} a_{k}^{3} \leq\left(\sum_{j_{1}}^{n} a_{j}^{2}\right)^{3 / 2}
$$
which is what was supposed to be proved. For equality, exactly on $x_{k}$ has to be one and the rest have to be zeroes, which is equivalent to having exactly one of the $a_{k}$ 's positive and the rest zeroes.
Figure 2.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 189 |
90.4. It is possible to perform three operations $f, g$, and $h$ for positive integers: $f(n)=$ $10 n, g(n)=10 n+4$, and $h(2 n)=n$; in other words, one may write 0 or 4 in the end of the number and one may divide an even number by 2. Prove: every positive integer can be constructed starting from 4 and performing a finite number of the operations $f, g$, and $h$ in some order.
|
Solution. All odd numbers $n$ are of the form $h(2 n)$. All we need is to show that every even number can be obtained fron 4 by using the operations $f, g$, and $h$. To this end, we show that a suitably chosen sequence of inverse operations $F=f^{-1}, G=g^{-1}$, and $H=h^{-1}$ produces a smaller even number or the number 4 from every positive even integer. The operation $F$ can be applied to numbers ending in a zero, the operation $G$ can be applied to numbers ending in 4 , and $H(n)=2 n$. We obtain
$$
\begin{gathered}
H(F(10 n))=2 n \\
G(H(10 n+2))=2 n, \quad n \geq 1 \\
H(2)=4 \\
H(G(10 n+4))=2 n \\
G(H(H(10 n+6)))=4 n+2 \\
G(H(H(H(10 n+8))))=8 n+6
\end{gathered}
$$
After a finite number of these steps, we arrive at 4 .
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 190 |
91.4. Let $f(x)$ be a polynomial with integer coefficients. We assume that there exists a positive integer $k$ and $k$ consecutive integers $n, n+1, \ldots, n+k-1$ so that none of the numbers $f(n), f(n+1), \ldots, f(n+k-1)$ is divisible by $k$. Show that the zeroes of $f(x)$ are not integers.
|
Solution. Let $f(x)=a_{0} x^{d}+a_{1} x^{d-1}+\cdots+a_{d}$. Assume that $f$ has a zero $m$ which is an integer. Then $f(x)=(x-m) g(x)$, where $g$ is a polynomial. If $g(x)=b_{0} x^{d-1}+b_{1} x^{d-2}+$ $\cdots+b_{d-1}$, then $a_{0}=b_{0}$, and $a_{k}=b_{k}-m b_{k-1}, 1 \leq k \leq d-1$. So $b_{0}$ is an integer, and by induction all $b_{k}$ 's are integers. Because $f(j)$ is not divisible by $k$ for $k$ consequtive values of $j$, no one of the $k$ consequtive integers $j-m, j=n, n+1, \ldots, n+k-1$, is divisible by $k$. But this is a contradiction, since exactly one of $k$ consequtive integers is divisible by $k$. So $f$ cannot have an integral zero.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 192 |
92.2. Let $n>1$ be an integer and let $a_{1}, a_{2}, \ldots, a_{n}$ be $n$ different integers. Show that the polynomial
$$
f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)-1
$$
is not divisible by any polynomial with integer coefficients and of degree greater than zero but less than $n$ and such that the highest power of $x$ has coefficient 1.
|
Solution. Suppose $g(x)$ is a polynomial of degree $m$, where $1 \leq m<n$, with integer coefficients and leading coefficient 1 , such that
$$
f(x)=g(x) h(x)
$$
whre $h(x)$ is a polynomial. Let
$$
\begin{aligned}
& g(x)=x^{m}+b_{m-1} x^{m-1}+\cdots+b_{1} x+b_{0} \\
& h(x)=x^{n-m}+c_{n-m-1} x^{n-m-1}+\cdots+c_{1} x+c_{0}
\end{aligned}
$$
We show that the coefficients of $h(x)$ are integers. If they are not, there is a greatest index $j=k$ such that $c_{k}$ is not an integer. But then the coefficient of $f$ multiplying $x^{k+m}-$ which is an integer - would be $c_{k}+b_{m-1} c_{k+1}+b_{m-2} c_{k+2}+\ldots b_{k-m}$. All terms except the first one in this sum are integers, so the sum cannot be an integer. A contradiction. So $h(x)$ is a polynomial with integral coefficients. Now
$$
f\left(a_{i}\right)=g\left(a_{i}\right) h\left(a_{i}\right)=-1
$$
for $i=1,2, \ldots, n$, and $g\left(a_{i}\right)$ and $h\left(a_{i}\right)$ are integers. This is only possible, if $g\left(a_{i}\right)=$ $-f\left(a_{i}\right)= \pm 1$ and $g\left(a_{i}\right)+h\left(a_{i}\right)=0$ for $i=1,2, \ldots, n$. So the polynomial $g(x)+h(x)$ has at least $n$ zeroes. But the degree of $g(x)+h(x)$ is less than $n$. So $g(x)=-h(x)$ for all $x$, and $f(x)=-g(x)^{2}$. This is impossible, however, because $f(x) \rightarrow+\infty$, as $x \rightarrow+\infty$. This contradiction proves the claim.
Figure 4 .
92.3 Prove that among all triangles with inradius 1, the equilateral one has the smallest perimeter.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 193 |
94.1. Let $O$ be an interior point in the equilateral triangle $A B C$, of side length $a$. The lines $A O, B O$, and $C O$ intersect the sides of the triangle in the points $A_{1}, B_{1}$, and $C_{1}$. Show that
$$
\left|O A_{1}\right|+\left|O B_{1}\right|+\left|O C_{1}\right|<a
$$
|
Solution. Let $H_{A}, H_{B}$, and $H_{C}$ be the orthogonal projections of $O$ on $B C, C A$, and $A B$, respectively. Because $60^{\circ}\left|O A_{1}\right| \frac{\sqrt{3}}{2}
$$
In the same way,
$$
\left|O H_{B}\right|>\left|O B_{1}\right| \frac{\sqrt{3}}{2} \quad \text { and } \quad\left|O H_{C}\right|>\left|O C_{1}\right| \frac{\sqrt{3}}{2}
$$
The area of $A B C$ is $a^{2} \frac{\sqrt{3}}{4}$ but also $\frac{a}{2}\left(O H_{A}+O H_{B}+O H_{C}\right)$ (as the sum of the areas of the three triangles with common vertex $O$ which together comprise $A B C$ ). So
$$
\left|O H_{A}\right|+\left|O H_{B}\right|+\left|O H_{C}\right|=a \frac{\sqrt{3}}{2}
$$
and the claim follows at once.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 195 |
94.3. A piece of paper is the square $A B C D$. We fold it by placing the vertex $D$ on the point $H$ of the side $B C$. We assume that $A D$ moves onto the segment $G H$ and that $H G$ intersects $A B$ at $E$. Prove that the perimeter of the triangle $E B H$ is one half of the perimeter of the square.
Figure 6 .
|
Solution. (See Figure 6.) The fold gives rise to an isosceles trapezium $A D H G$. Because of symmetry, the distance of the vertex $D$ from the side $G H$ equals the distance of the vertex
$H$ from side $A D$; the latter distance is the side length $a$ of the square. The line $G H$ thus is tangent to the circle with center $D$ and radius $a$. The lines $A B$ and $B C$ are tangent to the same circle. If the point common to $G H$ and the circle is $F$, then $A E=E F$ and $F H=H C$. This implies $A B+B C=A E+E B+B H+H C=E F+E B+B H+H F=E H+E B+B H$, which is equivalent to what we were asked to prove.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 196 |
95.1. Let $A B$ be a diameter of a circle with centre $O$. We choose a point $C$ on the circumference of the circle such that $O C$ and $A B$ are perpendicular to each other. Let $P$ be an arbitrary point on the (smaller) arc $B C$ and let the lines $C P$ and $A B$ meet at $Q$. We choose $R$ on $A P$ so that $R Q$ and $A B$ are perpendicular to each other. Show that $|B Q|=|Q R|$.
Figure 7 .
|
Solution 1. (See Figure 7.) Draw $P B$. By the Theorem of Thales, $\angle R P B=\angle A P B=$ $90^{\circ}$. So $P$ and $Q$ both lie on the circle with diameter $R B$. Because $\angle A O C=90^{\circ}$, $\angle R P Q=\angle C P A=45^{\circ}$. Then $\angle R B Q=45^{\circ}$, too, and $R B Q$ is an isosceles right triangle, or $|B Q|=|Q R|$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 197 |
95.3. Let $n \geq 2$ and let $x_{1}, x_{2}, \ldots x_{n}$ be real numbers satisfying $x_{1}+x_{2}+\ldots+x_{n} \geq 0$ and $x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}=1$. Let $M=\max \left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$. Show that
$$
M \geq \frac{1}{\sqrt{n(n-1)}}
$$
When does equality hold in (1)?
|
Solution. Denote by $I$ the set of indices $i$ for which $x_{i} \geq 0$, and by $J$ the set of indices $j$ for which $x_{j}<0$. Let us assume $M<\frac{1}{\sqrt{n(n-1)}}$. Then $I \neq\{1,2, \ldots, n\}$, since otherwise we would have $\left|x_{i}\right|=x_{i} \leq \frac{1}{\sqrt{n(n-1)}}$ for every $i$, and $\sum_{i=1}^{n} x_{i}^{2}<\frac{1}{n-1} \leq 1$. So $\sum_{i \in I} x_{i}^{2}<(n-1) \cdot \frac{1}{n(n-1)}=\frac{1}{n}$, and $\sum_{i \in I} x_{i}<(n-1) \frac{1}{\sqrt{n(n-1)}}=\sqrt{\frac{n-1}{n}}$. Because
$$
0 \leq \sum_{i=1}^{n} x_{i}=\sum_{i \in I} x_{i}-\sum_{i \in J}\left|x_{i}\right|
$$
we must have $\sum_{i \in J}\left|x_{i}\right| \leq \sum_{i \in I} x_{i}<\sqrt{\frac{n-1}{n}}$ and $\sum_{i \in J} x_{i}^{2} \leq\left(\sum_{i \in J}\left|x_{i}\right|\right)^{2}<\frac{n-1}{n}$. But then
$$
\sum_{i=1}^{n} x_{i}^{2}=\sum_{i \in I} x_{i}^{2}+\sum_{i \in J} x_{i}^{2}<\frac{1}{n}+\frac{n-1}{n}=1
$$
and we have a contradiction. - To see that equality $M=\frac{1}{\sqrt{n(n-1)}}$ is possible, we choose $x_{i}=\frac{1}{\sqrt{n(n-1)}}, i=1,2, \ldots, n-1$, and $x_{n}=-\sqrt{\frac{n-1}{n}}$. Now
$$
\sum_{i=1}^{n} x_{i}=(n-1) \frac{1}{\sqrt{n(n-1)}}-\sqrt{\frac{n-1}{n}}=0
$$
and
$$
\sum_{i=1}^{n} x_{i}^{2}=(n-1) \cdot \frac{1}{n(n-1)}+\frac{n-1}{n}=1
$$
We still have to show that equality can be obtained only in this case. Assume $x_{i}=$ $\frac{1}{\sqrt{n(n-1)}}$, for $i=1, \ldots, p, x_{i} \geq 0$, for $i \leq q$, and $x_{i}<0$, kun $q+1 \leq i \leq n$. As before we get
$$
\sum_{i=1}^{q} x_{i} \leq \frac{q}{\sqrt{n(n-1)}}, \quad \sum_{i=q+1}^{n}\left|x_{i}\right| \leq \frac{q}{\sqrt{n(n-1)}}
$$
and
$$
\sum_{i=q+1}^{n} x_{i}^{2} \leq \frac{q^{2}}{n(n-1)}
$$
so
$$
\sum_{i=1}^{n} x_{i}^{2} \leq \frac{q+q^{2}}{n^{2}-n}
$$
It is easy to see that $q^{2}+q<n^{2}+n$, for $n \geq 2$ and $q \leq n-2$, but $(n-1)^{2}+(n-1)=n^{2}-n$. Consequently, a necessary condition for $M=\frac{1}{\sqrt{n(n-1)}}$ is that the sequence only has one negative member. But if among the positive members there is at least one smaller than $M$ we have
$$
\sum_{i=1}^{n}<\frac{q+q^{2}}{n(n-1)}
$$
so the conditions of the problem are not satisfied. So there is equality if and only if $n-1$ of the numbers $x_{i}$ equal $\frac{1}{\sqrt{n(n-1)}}$, and the last one is $\frac{1-n}{\sqrt{n(n-1)}}$.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 198 |
95.4. Show that there exist infinitely many mutually non-congruent triangles $T$, satisfying
(i) The side lengths of $T$ are consecutive integers.
(ii) The area of $T$ is an integer.
|
Solution. Let $n \geq 3$, and let $n-1, n, n+1$ be the side lengths of the triangle. The semiperimeter of the triangle then equals on $\frac{3 n}{2}$. By Heron's formula, the area of the triangle is
$$
\begin{gathered}
T=\sqrt{\frac{3 n}{2} \cdot\left(\frac{3 n}{2}-n+1\right)\left(\frac{3 n}{2}-n\right)\left(\frac{3 n}{2}-n-1\right)} \\
=\frac{n}{2} \sqrt{\frac{3}{4}\left(n^{2}-4\right)} .
\end{gathered}
$$
If $n=4$, then $T=6$. So at least one triangle of the kind required exists. We prove that we always can form new triangles of the required kind from ones already known to exist. Let $n$ be even, $n \geq 4$, and let $\frac{3}{4}\left(n^{2}-4\right)$ be a square number. Set $m=n^{2}-2$. Then $m>n$, $m$ is even, and $m^{2}-4=(m+2)(m-2)=n^{2}\left(n^{2}-4\right)$. So $\frac{3}{4}\left(m^{2}-4\right)=n^{2} \cdot \frac{3}{4}\left(n^{2}-4\right)$ is a square number. Also, $T=\frac{m}{2} \sqrt{\frac{3}{4}\left(m^{2}-4\right)}$ is an integer. The argument is complete.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 199 |
96.1. Show that there exists an integer divisible by 1996 such that the sum of the its decimal digits is 1996 .
|
Solution. The sum of the digits of 1996 is 25 and the sum of the digits of $2 \cdot 1996=3992$ is 23 . Because $1996=78 \cdot 25+46$, the number obtained by writing 781996 's and two 3992 in succession satisfies the condition of the problem. - As $3 \cdot 1996=5998$, the sum of the digits of 5988 is 30 , and $1996=65 \cdot 30+46$, the number $39923992 \underbrace{5988 \ldots 5988}_{65 \text { times }}$ also can be be given as an answer, indeed a better one, as it is much smaller than the first suggestion.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 200 |
96.3. The circle whose diameter is the altitude dropped from the vertex $A$ of the triangle $A B C$ intersects the sides $A B$ and $A C$ at $D$ and $E$, respectively $(A \neq D, A \neq E)$. Show that the circumcentre of $A B C$ lies on the altitude dropped from the vertex $A$ of the triangle $A D E$, or on its extension.
Figure 8.
|
Solution. (See Figure 8.) Let $A F$ be the altitude of $A B C$. We may assume that $\angle A C B$ is sharp. From the right triangles $A C F$ and $A F E$ we obtain $\angle A F E=\angle A C F . \angle A D E$ and $\angle A F E$ subtend the same arc, so they are equal. Thus $\angle A C B=\angle A D E$, and the triangles $A B C$ and $A E D$ are similar. Denote by $P$ and $Q$ the circumcenters of $A B C$ and $A E D$, respectively. Then $\angle B A P=\angle E A Q$. If $A G$ is the altitude of $A E D$, then $\angle D A G=\angle C A F$. But this implies $\angle B A P=\angle D A G$, which means that $P$ is on the altitude $A G$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 201 |
97.2. Let $A B C D$ be a convex quadrilateral. We assume that there exists a point $P$ inside the quadrilateral such that the areas of the triangles $A B P, B C P, C D P$, and $D A P$ are equal. Show that at least one of the diagonals of the quadrilateral bisects the other diagonal.
Figure 9.
|
Solution. (See Figure 9.) We first assume that $P$ does not lie on the diagonal $A C$ and the line $B P$ meets the diagonal $A C$ at $M$. Let $S$ and $T$ be the feet of the perpendiculars from $A$ and $C$ on the line $B P$. The triangles $A P B$ and $C B P$ have equal area. Thus $A S=C T$. If $S \neq T$, then the right trianges $A S M$ and $C T M$ are congruent, and $A M=C M$. If, on the other hand, $S=T$, the $A C \perp P B$ and $S=M=T$, and again $A M=C M$. In both cases $M$ is the midpoint of the diagonal $A C$. We prove exactly in the same way that the line $D P$ meets $A C$ at the midpoint of $A C$, i.e. at $M$. So $B, M$, and $P$, and also $D, M$, and $P$ are collinear. So $M$ is on the line $D B$, which means that $B D$ divides the diagonal $A C$ in two equal parts.
We then assume that $P$ lies on the diagonal $A C$. Then $P$ is the midpoint of $A C$. If $P$ is not on the diagonal $B D$, we argue as before that $A C$ divides $B D$ in two equal parts. If $P$ lies also on the diagonal $B D$, it has to be the common midpoint of the diagonals.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 203 |
97.4. Let $f$ be a function defined in the set $\{0,1,2, \ldots\}$ of non-negative integers, satisfying $f(2 x)=2 f(x), f(4 x+1)=4 f(x)+3$, and $f(4 x-1)=2 f(2 x-1)-1$. Show that $f$ is an injection, i.e. if $f(x)=f(y)$, then $x=y$.
|
Solution. If $x$ is even, then $f(x)$ is even, and if $x$ is odd, then $f(x)$ is odd. Moreover, if $x \equiv 1 \bmod 4$, then $f(x) \equiv 3 \bmod 4$, and if $x \equiv 3 \bmod 4$, then $f(x) \equiv 1 \bmod 4$. Clearly $f(0)=0, f(1)=3, f(2)=6$, and $f(3)=5$. So at least $f$ restricted to the set $\{0,1,2,3\}$ ia an injection. We prove that $f(x)=f(y) \Longrightarrow x=y$, for $x, y<k$ implies $f(x)=f(y) \Longrightarrow x=y$, for $x, y<2 k$. So assume $x$ and $y$ are smaller than $2 k$ and $f(x)=f(y)$. If $f(x)$ is even, then $x=2 t, y=2 u$, and $2 f(t)=2 f(u)$. As $t$ and $u$ are less than $k$, we have $t=u$, and $x=y$. Assume $f(x) \equiv 1 \bmod 4$. Then $x \equiv 3 \bmod 4 ;$ $x=4 u-1$, and $f(x)=2 f(2 u-1)-1$. Also $y=4 t-1$ and $f(y)=2 f(2 t-1)-1$. Moreover, $2 u-1<\frac{1}{2}(4 u-1)<k$ and $2 t-1<k$, so $2 u-1=2 t-1, u=t$, and $x=y$. If, finally, $f(x) \equiv 3 \bmod 4$, then $x=4 u+1, y=4 t+1, u<k, t<k, 4 f(u)+3=4 f(t)+3, u=t$, and $x=y$. Since for all $x$ and $y$ there is an $n$ such that the larger one of the numbers $x$ and $y$ is $<2^{n} \cdot 3$, the induction argument above shows that $f(x)=f(y) \Rightarrow x=y$.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 204 |
98.2. Let $C_{1}$ and $C_{2}$ be two circles intersecting at $A$ and $B$. Let $S$ and $T$ be the centres of $C_{1}$ and $C_{2}$, respectively. Let $P$ be a point on the segment $A B$ such that $|A P| \neq|B P|$ and $P \neq A, P \neq B$. We draw a line perpendicular to $S P$ through $P$ and denote by $C$ and $D$ the points at which this line intersects $C_{1}$. We likewise draw a line perpendicular to TP through $P$ and denote by $E$ and $F$ the points at which this line intersects $C_{2}$. Show that $C, D, E$, and $F$ are the vertices of a rectangle.
|
Solution. (See Figure 10.) The power of the point $P$ with respect to the circles $C_{1}$ and $C_{2}$ is $P A \cdot P B=P C \cdot P D=P E \cdot P F$. Since $S P$ is perpendicular to the chord $C D, P$
Figure 10 .
has to be the midpoint of $C D$. So $P C=P D$. In a similar manner, we obtain $P E=P F$. Alltogether $P C=P D=P E=P F=\sqrt{P A \cdot P B}$. Consequently the points $C, D, E$, and $F$ all lie on a circle withe center $P$, and $C D$ and $E F$ as diameters. By Thales' theorem, the angles $\angle E C F, \angle C F D$ etc. are right angles. So $C D E F$ is a rectangle.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 205 |
99.4. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers and $n \geq 1$. Show that
$$
\begin{aligned}
& n\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right) \\
& \quad \geq\left(\frac{1}{1+a_{1}}+\cdots+\frac{1}{1+a_{n}}\right)\left(n+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)
\end{aligned}
$$
When does equality hold?
|
Solution. The inequality of the problem can be written as
$$
\frac{1}{1+a_{1}}+\cdots+\frac{1}{1+a_{n}} \leq \frac{n\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)}{n+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}}
$$
A small manipulation of the right hand side brings the inequality to the equivalent form
$$
\frac{1}{\frac{1}{a_{1}^{-1}}+1}+\cdots+\frac{1}{\frac{1}{a_{n}^{-1}}+1} \leq \frac{n}{\frac{1}{\frac{a_{1}^{-1}+\cdots+a_{n}^{-1}}{n}}+1}
$$
Consider the function
$$
f(x)=\frac{1}{\frac{1}{x}+1}=\frac{x}{1+x}
$$
We see that it is concave, i.e.
$$
t f(x)+(1-t) f(y)<f(t x+(1-t) y)
$$
for all $t \in(0,1)$. In fact, the inequality
$$
t \frac{x}{1+x}+(1-t) \frac{y}{1+y}<\frac{t x+(1-t) y}{1+t x+(1-t) y}
$$
can be written as
$$
t^{2}(x-y)^{2}<t(x-y)^{2}
$$
and because $0<t<1$, it is true. [Another standard way to see this is to compute
$$
f^{\prime}(x)=\frac{1}{(1+x)^{2}}, \quad f^{\prime \prime}(x)=-\frac{2}{(1+x)^{3}}<0
$$
A function with a positive second derivative is concave.] For any concave function $f$, the inequality
$$
\frac{1}{n}\left(f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{n}\right)\right) \leq f\left(\frac{x_{1}+\cdots+x_{n}}{n}\right)
$$
holds, with equality only for $x_{1}=x_{2}=\ldots=x_{n}$. So (1) is true, and equality holds only if all $a_{i}$ 's are equal.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 207 |
00.3. In the triangle $A B C$, the bisector of angle $B$ meets $A C$ at $D$ and the bisector of angle $C$ meets $A B$ at $E$. The bisectors meet each other at $O$. Furthermore, $O D=O E$. Prove that either $A B C$ is isosceles or $\angle B A C=60^{\circ}$.
|
Solution. (See Figure 11.) Consider the triangles $A O E$ and $A O D$. They have two equal pairs of sides and the angles facing one of these pairs are equal. Then either $A O E$ and $A O D$ are congruent or $\angle A E O=180^{\circ}-\angle A D O$. In the first case, $\angle B E O=\angle C D O$, and
Figure 11.
the triangles $E B O$ and $D C O$ are congruent. Then $A B=A C$, and $A B C$ is isosceles. In the second case, denote the angles of $A B C$ by $2 \alpha, 2 \beta$, and $2 \gamma$, and the angle $A E O$ by $\delta$. By the theorem on the adjacent angle of an angle of a triangle, $\angle B O E=\angle D O C=\beta+\gamma$, $\delta=2 \beta+\gamma$, and $180^{\circ}-\delta=\beta+2 \gamma$. Adding these equations yields $3(\beta+\gamma)=180^{\circ}$ eli $\beta+\gamma=60^{\circ}$. Combining this with $2(\alpha+\beta+\gamma)=180^{\circ}$, we obtain $2 \alpha=60^{\circ}$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 208 |
01.1. Let $A$ be a finite collection of squares in the coordinate plane such that the vertices of all squares that belong to $A$ are $(m, n),(m+1, n),(m, n+1)$, and $(m+1, n+1)$ for some integers $m$ and $n$. Show that there exists a subcollection $B$ of $A$ such that $B$ contains at least $25 \%$ of the squares in $A$, but no two of the squares in $B$ have a common vertex.
|
Solution. Divide the plane into two sets by painting the strips of squares parallel to the $y$ axis alternately red and green. Denote the sets of red and green squares by $R$ and $G$, respectively. Of the sets $A \cap R$ and $A \cap G$ at least one contains at least one half of the squares in $A$. Denote this set by $A_{1}$. Next partition the strips of squares which contain squares of $A_{1}$ into two sets $E$ and $F$ so that each set contains every second square of $A_{1}$ on each strip. Now neither of the dets $E$ and $F$ has a common point with a square in the same set. On the other hand, at least one of the sets $E \cap A_{1}, F \cap A_{1}$ contains at least one half of the squares in $A_{1}$ and thus at least one quarter of the sets in $A$. This set is good for the required set $B$.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 210 |
01.2. Let $f$ be a bounded real function defined for all real numbers and satisfying for all real numbers $x$ the condition
$$
f\left(x+\frac{1}{3}\right)+f\left(x+\frac{1}{2}\right)=f(x)+f\left(x+\frac{5}{6}\right)
$$
Show that $f$ is periodic. (A function $f$ is bounded, if there exists a number $L$ such that $|f(x)|<L$ for all real numbers $x$. A function $f$ is periodic, if there exists a positive number $k$ such that $f(x+k)=f(x)$ for all real numbers $x$.)
|
Solution. Let $g(6 x)=f(x)$. Then $g$ is bounded, and
$$
\begin{gathered}
g(t+2)=f\left(\frac{t}{6}+\frac{1}{3}\right), \quad g(t+3)=f\left(\frac{t}{6}+\frac{1}{2}\right) \\
g(t+5)=f\left(\frac{t}{6}+\frac{5}{6}\right), \quad g(t+2)+g(t+3)=g(t)+g(t+5) \\
g(t+5)-g(t+3)=g(t+2)-g(t)
\end{gathered}
$$
for all real numbers $t$. But then
$$
\begin{gathered}
g(t+12)-g(6) \\
=g(t+12)-g(t+10)+g(t+10)-g(t+8)+g(t+8)-g(t+6) \\
=g(t+9)-g(t+7)+g(t+7)-g(t+5)+g(t+5)-g(t+3) \\
=g(t+6)-g(t+4)+g(t+4)-g(t+2)+g(t+2)-g(t) \\
=g(t+6)-g(t)
\end{gathered}
$$
By induction, then $g(t+6 n)-g(t)=n(g(t+6)-g(0))$ for all positive integers $n$. Unless $g(t+6)-g(t)=0$ for all real $t, g$ cannot be bounded. So $g$ has to be periodic with 6 as a period, whence $f$ is periodic, with 1 as a period.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 211 |
01.4. Let $A B C D E F$ be a convex hexagon, in which each of the diagonals $A D, B E$, and $C F$ divides the hexagon in two quadrilaterals of equal area. Show that $A D, B E$, and $C F$ are concurrent.
Figure 12.
|
Solution. (See Figure 12.) Denote the area of a figure by $|\cdot|$. Let $A D$ and $B E$ intersect at $P, A D$ and $C F$ at $Q$, and $B E$ and $C F$ at $R$. Assume that $P, Q$, and $R$ are different. We may assume that $P$ lies between $B$ and $R$, and $Q$ lies between $C$ and $R$. Both $|A B P|$ and $|D E P|$ differ from $\frac{1}{2}|A B C D E F|$ by $|B C D P|$. Thus $A B P$ and $D E P$ have equal area. Since $\angle A P B=\angle D P E$, we have $A P \cdot B P=D P \cdot E P=(D Q+Q P)(E R+R P)$. Likewise $C Q \cdot D Q=(A P+P Q)(F R+R Q)$ and $E R \cdot F R=(C Q+Q R)(B P+P R)$. When we multiply the three previous equalities, we obtain $A P \cdot B P \cdot C Q \cdot D Q \cdot E R \cdot F R=$ $D Q \cdot E R \cdot A P \cdot F R \cdot C Q \cdot B P+$ positive terms containing $P Q, Q R$, and $P R$. This is a contradiction. So $P, Q$ and $R$ must coincide.
Figure 13.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 212 |
02.3. Let $a_{1}, a_{2}, \ldots, a_{n}$, and $b_{1}, b_{2}, \ldots, b_{n}$ be real numbers, and let $a_{1}, a_{2}, \ldots, a_{n}$ be all different.. Show that if all the products
$$
\left(a_{i}+b_{1}\right)\left(a_{i}+b_{2}\right) \cdots\left(a_{i}+b_{n}\right)
$$
$i=1,2, \ldots, n$, are equal, then the products
$$
\left(a_{1}+b_{j}\right)\left(a_{2}+b_{j}\right) \cdots\left(a_{n}+b_{j}\right)
$$
$j=1,2, \ldots, n$, are equal, too.
|
Solution. Let $P(x)=\left(x+b_{1}\right)\left(x+b_{2}\right) \cdots\left(x+b_{n}\right)$. Let $P\left(a_{1}\right)=P\left(a_{2}\right)=\ldots=P\left(a_{n}\right)=d$. Thus $a_{1}, a_{2}, \ldots, a_{n}$ are the roots of the $n$ :th degree polynomial equation $P(x)-d=0$. Then $P(x)-d=c\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)$. Clearly the $n$ :th degree terms of $P(x)$ and $P(x)-d$ are equal. So $c=1$. But $P\left(-b_{j}\right)=0$ for each $b_{j}$. Thus for every $j$,
$$
\begin{gathered}
-d=\left(-b_{j}-a_{1}\right)\left(-b_{j}-a_{2}\right) \cdots\left(-b_{j}-a_{n}\right) \\
=(-1)^{n}\left(a_{1}+b_{j}\right)\left(a_{2}+b_{j}\right) \cdots\left(a_{n}+b_{j}\right)
\end{gathered}
$$
and the claim follows.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 214 |
03.1. Stones are placed on the squares of a chessboard having 10 rows and 14 columns. There is an odd number of stones on each row and each column. The squares are coloured black and white in the usual fashion. Show that the number of stones on black squares is even. Note that there can be more than one stone on a square.
|
Solution. Changing the order of rows or columns does not influence the number of stones on a row, on a column or on black squares. Thus we can order the rows and columns in such a way that the $5 \times 7$ rectangles in the upper left and lower right corner are black and the other two $5 \times 7$ rectangles are white. If the number of stones on black squares would be odd, then one of the black rectangles would have an odd number of stones while the number of stones on the other would be even. Since the number of stones is even, one of the white rectangles would have an odd number of stones and the other an even number. But this would imply either a set of five rows or a set of seven columns with an even number of stones. But this is not possible, because every row and column has an odd number of stones. So the number of stones on black squares has to be even.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 215 |
03.3. The point $D$ inside the equilateral triangle $\triangle A B C$ satisfies $\angle A D C=150^{\circ}$. Prove that a triangle with side lengths $|A D|,|B D|,|C D|$ is necessarily a right-angled triangle.
Figure 14 .
|
Solution. (See Figure 14.) We rotate the figure counterclockwise $60^{\circ}$ around $C$. Because $A B C$ is an equilateral triangle, $\angle B A C=60^{\circ}$, so $A$ is mapped on $B$. Assume $D$ maps to $E$. The properties of rotation imply $A D=B E$ and $\angle B E C=150^{\circ}$. Because the triangle $D E C$ is equilateral, $D E=D C$ and $\angle D E C=60^{\circ}$. But then $\angle D E B=150^{\circ}-60^{\circ}=90^{\circ}$. So segments having the lengths as specified in the problem indeed are sides of a right triangle.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 216 |
04.2. Let $f_{1}=0, f_{2}=1$, and $f_{n+2}=f_{n+1}+f_{n}$, for $n=1$, 2, ..., be the Fibonacci sequence. Show that there exists a strictly increasing infinite arithmetic sequence none of whose numbers belongs to the Fibonacci sequence. [A sequence is arithmetic, if the difference of any of its consecutive terms is a constant.]
|
Solution. The Fibonacci sequence modulo any integer $n>1$ is periodic. (Pairs of residues are a finite set, so some pair appears twice in the sequence, and the sequence from the second appearance of the pair onwards is a copy of the sequence from the first pair onwards.) There are integers for which the Fibonacci residue sequence does not contain all possible residues. For instance modulo 11 the sequence is $0,1,1,2,3,5,8,2,10,1$, $0,1,1, \ldots$ Wee see that the number 4 is missing. It follows that no integer of the form $4+11 k$ appears in the Fibonacci sequence. But here we have an arithmetic sequence of the kind required.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 217 |
04.3. Let $x_{11}, x_{21}, \ldots, x_{n 1}, n>2$, be a sequence of integers. We assume that all of the numbers $x_{i 1}$ are not equal. Assuming that the numbers $x_{1 k}, x_{2 k}, \ldots, x_{n k}$ have been defined, we set
$$
\begin{aligned}
x_{i, k+1} & =\frac{1}{2}\left(x_{i k}+x_{i+1, k}\right), i=1,2, \ldots, n-1 \\
x_{n, k+1} & =\frac{1}{2}\left(x_{n k}+x_{1 k}\right)
\end{aligned}
$$
Show that for $n$ odd, $x_{j k}$ is not an integer for some $j, k$. Does the same conclusion hold for $n$ even?
|
Solution. We compute the first index modulo $n$, i.e. $x_{1 k}=x_{n+1, k}$. Let $M_{k}=\max _{j} x_{j k}$ and $m_{k}=\min _{j} x_{j k}$. Evidently $\left(M_{k}\right)$ is a non-increasing and $\left(m_{k}\right)$ a non-decreasing sequence, and $M_{k+1}=M_{k}$ is possible only if $x_{j k}=x_{j+1, k}=M_{k}$ for some $j$. If exactly $p$
consequtive numbers $x_{j k}$ equal $M_{k}$, then exactly $p-1$ consequtive numbers $x_{j, k+1}$ equal $M_{k+1}$ which is equal to $M_{k}$. So after a finite number of steps we arrive at the situation $M_{k+1}m_{k}$ for some $k$ 's. If all the numbers in all the sequences are integers, then all $m_{k}$ 's and $M_{k}$ 's are integers. So after a finite number of steps $m_{k}=M_{k}$, and all numbers $x_{j k}$ are equal. Then $x_{1, k-1}+x_{2, k-1}=x_{2, k-1}+x_{3, k-1}=\cdots=x_{n-1, k-1}+x_{n, k-1}=x_{n, k-1}+x_{1, k-1}$. If $n$ is odd, then $x_{1, k-1}=x_{3, k-1}=\cdots=x_{n, k-1}$ and $x_{1, k-1}=x_{n-1, k-1}=\cdots=x_{2, k-1}$. But then we could show in a similar way that all numbers $x_{j, k-2}$ are equal and finally that all numbers $x_{j, 1}$ are equal, contrary to the assumption. If $n$ is even, then all $x_{i, k}$ 's can be integers. Take, for instance, $x_{1,1}=x_{3,1}=\cdots=x_{n-1,1}=0, x_{2,1}=x_{4,1}=\cdots=x_{n, 1}=2$. Then every $x_{j, k}=1, k \geq 2$.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 218 |
04.4. Let $a, b$, and $c$ be the side lengths of a triangle and let $R$ be its circumradius. Show that
$$
\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a} \geq \frac{1}{R^{2}}
$$
|
Solution 1. By the well-known (Euler) theorem, the inradius $r$ and circumradius $R$ of any triangle satisfy $2 r \leq R$. (In fact, $R(R-2 r)=d^{2}$, where $d$ is the distance between the incenter and circumcenter.) The area $S$ of a triangle can be written as
$$
A=\frac{r}{2}(a+b+c)
$$
and, by the sine theorem, as
$$
A=\frac{1}{2} a b \sin \gamma=\frac{1}{4} \frac{a b c}{R}
$$
Combining these, we obtain
$$
\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}=\frac{a+b+c}{a b c}=\frac{2 A}{r} \cdot \frac{1}{4 R A}=\frac{1}{2 r R} \geq \frac{1}{R^{2}}
$$
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 219 |
05.2. Let $a, b$, and $c$ be positive real numbers. Prove that
$$
\frac{2 a^{2}}{b+c}+\frac{2 b^{2}}{c+a}+\frac{2 c^{2}}{a+b} \geq a+b+c
$$
|
Solution 1. Use brute force. Removing the denominators and brackets and combining simililar terms yields the equivalent inequality
$$
\begin{gathered}
0 \leq 2 a^{4}+2 b^{4}+2 c^{4}+a^{3} b+a^{3} c+a b^{3}+b^{3} c+a c^{3}+b c^{3} \\
-2 a^{2} b^{2}-2 b^{2} c^{2}-2 a^{2} c^{2}-2 a b c^{2}-2 a b^{2} c-2 a^{2} b c \\
=a^{4}+b^{4}-2 a^{2} b^{2}+b^{4}+c^{4}-2 b^{2} c^{2}+c^{4}+a^{4}-2 a^{2} c^{2} \\
+a b\left(a^{2}+b^{2}-2 c^{2}\right)+b c\left(b^{2}+c^{2}-2 a^{2}\right)+c a\left(c^{2}+a^{2}-2 b^{2}\right) \\
=\left(a^{2}-b^{2}\right)^{2}+\left(b^{2}-c^{2}\right)^{2}+\left(c^{2}-a^{2}\right)^{2} \\
+a b(a-b)^{2}+b c(b-c)^{2}+c a(c-a)^{2} \\
+a b\left(2 a b-2 c^{2}\right)+b c\left(2 b c-2 a^{2}\right)+c a\left(2 c a-2 b^{2}\right)
\end{gathered}
$$
The six first terms on the right hand side are non-negative and the last three can be written as
$$
\begin{gathered}
2 a^{2} b^{2}-2 a b c^{2}+2 b^{2} c^{2}-2 a^{2} b c+2 c^{2} a^{2}-2 a b^{2} c \\
=a^{2}\left(b^{2}+c^{2}-2 b c\right)+b^{2}\left(a^{2}+c^{2}-2 a c\right)+c^{2}\left(a^{2}+b^{2}-2 a b\right) \\
=a^{2}(b-c)^{2}+b^{2}(c-a)^{2}+c^{2}(a-b)^{2} \geq 0
\end{gathered}
$$
So the original inequality is true.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 220 |
05.4. The circle $\mathcal{C}_{1}$ is inside the circle $\mathcal{C}_{2}$, and the circles touch each other at $A$. A line through $A$ intersects $\mathcal{C}_{1}$ also at $B$ and $\mathcal{C}_{2}$ also at $C$. The tangent to $\mathcal{C}_{1}$ at $B$ intersects $\mathcal{C}_{2}$ at $D$ and $E$. The tangents of $\mathcal{C}_{1}$ passing through $C$ touch $\mathcal{C}_{1}$ at $F$ and $G$. Prove that $D$, $E, F$, and $G$ are concyclic.
Figure 15.
|
Solution. (See Figure 15.) Draw the tangent $\mathrm{CH}$ to $\mathcal{C}_{2}$ at $C$. By the theorem of the angle between a tangent and chord, the angles $A B H$ and $A C H$ both equal the angle at $A$ between $B A$ and the common tangent of the circles at $A$. But this means that the angles $A B H$ and $A C H$ are equal, and $C H \| B E$. So $C$ is the midpoint of the arc $D E$. This again implies the equality of the angles $C E B$ and $B A E$, as well as $C E=C D$. So the triangles $A E C, C E B$, having also a common angle $E C B$, are similar. So
$$
\frac{C B}{C E}=\frac{C E}{A C}
$$
and $C B \cdot A C=C E^{2}=C D^{2}$. But by the power of a point theorem, $C B \cdot C A=C G^{2}=C F^{2}$. We have in fact proved $C D=C E=C F=C G$, so the four points are indeed concyclic.
06.1 Let $B$ and $C$ be points on two fixed rays emanating from a point $A$ such that $A B+A C$ is constant. Prove that there exists a point $D \neq A$ such that the circumcircles of the triangels $A B C$ pass through $D$ for every choice of $B$ and $C$.
Figure 16 .
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 221 |
06.4. The squares of a $100 \times 100$ chessboard are painted with 100 different colours. Each square has only one colour and every colour is used exactly 100 times. Show that there exists a row or a column on the chessboard in which at least 10 colours are used.
|
Solution. Denote by $R_{i}$ the number of colours used to colour the squares of the $i$ 'th row and let $C_{j}$ be the number of colours used to colour the squares of the $j$ 'th column. Let $r_{k}$ be the number of rows on which colour $k$ appears and let $c_{k}$ be the number of columns on which colour $k$ appears. By the arithmetic-geometric inequality, $r_{k}+c_{k} \geq 2 \sqrt{r_{k} c_{k}}$. Since colour $k$ appears at most $c_{k}$ times on each of the $r_{k}$ columns on which it can be found, $c_{k} r_{k}$ must be at least the total number of occurences of colour $k$, which equals 100 . So $r_{k}+c_{k} \geq 20$. In the sum $\sum_{i=1}^{100} R_{i}$, each colour $k$ contributes $r_{k}$ times and in the sum $\sum_{j=1}^{100} C_{j}$ each colour $k$ contributes $c_{k}$ times. Hence
$$
\sum_{i=1}^{100} R_{i}+\sum_{j=1}^{100} C_{j}=\sum_{k=1}^{100} r_{k}+\sum_{k=1}^{100} c_{k}=\sum_{k=1}^{100}\left(r_{k}+c_{k}\right) \geq 2000
$$
But if the sum of 200 positive integers is at least 2000, at least one of the summands is at least 10. The claim has been proved.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 222 |
07.2. A triangle, a line and three rectangles, with one side parallel to the given line, are given in such a way that the rectangles completely cover the sides of the triangle. Prove that the rectangles must completely cover the interior of the triangle.
|
Solution. Take any point $P$ inside the triangle and draw through $P$ the line parallel to the given line as well as the line perpendicular to it. These lines meet the sides of the triangle in four points. Of these four, two must be in one of the three rectangles. Now if the two points are on the same line, then the whole segment between them, $P$ included, is in the same rectangle. If the two points, say $Q$ and $R$, are on perpendicular lines, the perpendicular segments $R P$ and $P Q$ are also in the same rectangle. So in any case, $P$ is in one of the rectangles.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 223 |
08.4. The difference between the cubes of two consecutive positive integers is a square $n^{2}$, where $n$ is a positive integer. Show that $n$ is the sum of two squares.
|
Solution. Assume that $(m+1)^{3}-m^{3}=n^{2}$. Rearranging, we get $3(2 m+1)^{2}=(2 n+$ $1)(2 n-1)$. Since $2 n+1$ and $2 n-1$ are relatively prime (if they had a common divisor, it would have divided the difference, which is 2 , but they are both odd), one of them is a square (of an odd integer, since it is odd) and the other divided by 3 is a square. An odd number squared minus 1 is divisible by 4 since $(2 t+1)^{2}-1=4\left(t^{2}+t\right)$. From the first equation we see that $n$ is odd, say $n=2 k+1$. Then $2 n+1=4 k+3$, so the square must be $2 n-1$, say $2 n-1=(2 t+1)^{2}$. Rearrangement yields $n=t^{2}+(t+1)^{2}$. (An example: $8^{3}-7^{3}=\left(2^{2}+3^{2}\right)^{2}$. $)$
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 226 |
09.4. There are 32 competitors in a tournament. No two of them are equal in playing strength, and in a one against one match the better one always wins. Show that the gold, silver, and bronze medal winners can be found in 39 matches.
|
Solution. To determine the gold medalist, we organize 16 pairs and matches, then 8 matches of the winners, 4 matches of the winners, 2 and finally one match, 31 matches altogether. Now the silver medal winner has at some point lost to number 1 ; as there were 5 rounds, there are 5 candidates. Let $C_{i}$ be the candidate who lost to the gold medalist in round $i$. Now let $C_{l}$ and $C_{2}$ play, the winner then play with $C_{3}$ etc. After 4 matches we know the silver medalist; assume she was $C_{k}$. Now the bronze medalist must have lost against the gold medalist or against $C_{k}$ or both. (If she lost to someone else, this someone else was below the second place). Now the silver medalist $C_{k}$ won $k-1$ times in the first rounds and the $5-k$ players $C_{k+1}, \ldots, C_{5}$, and if $\mathrm{k} i 1$ one player $C_{j}$ with $j<k$. So there are either $k-1+5-k=4$ or 5 candidates for the third place. At most 4 matches are again needed to determine the bronze winner.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 228 |
10.1. A function $f: \mathbb{Z} \rightarrow \mathbb{Z}_{+}$, where $\mathbb{Z}_{+}$is the set of positive integers, is non-decreasing and satisfies $f(m n)=f(m) f(n)$ for all relatively prime positive integers $m$ and $n$. Prove that $f(8) f(13) \geq(f(10))^{2}$.
|
Solution. Since $\mathrm{f}$ is non-decreasing, $f(91) \geq f(90)$, which (by factorization into relatively prime factors) implies $f(13) f(7) \geq f(9) f(10)$. Also $f(72) \geq f(70)$, and therefore $f(8) f(9) \geq f(7) f(10)$. Since all values of $\mathrm{f}$ are positive, we get $f(8) f(9) \cdot f(13) f(7) \geq$ $f(7) f(10) \cdot f(9) f(10)$, and dividing both sides by $f(7) f(9)>0, f(8) f(13) \geq f(l 0) f(10)=$ $(f(10))^{2}$.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 229 |
10.2. Three circles $\Gamma_{A}, \Gamma_{B}$ and $\Gamma_{C}$ share a common point of intersection $O$. The other common of $\Gamma_{A}$ and $\Gamma_{B}$ is $C$, that of $\Gamma_{A}$ and $\Gamma_{C}$ is $B$ and that of $\Gamma_{C}$ and $\Gamma_{B}$ is $A$. The
line $A O$ intersects the circle $\Gamma_{C}$ in the poin $X \neq O$. Similarly, the line $B O$ intersects the circle $\Gamma_{B}$ in the point $Y \neq O$, and the line $C O$ intersects the circle $\Gamma_{C}$ in the point $Z \neq O$. Show that
$$
\frac{|A Y||B Z||C X|}{|A Z||B X||C Y|}=1
$$
|
Solution 1. Let $\angle A O Y=\alpha, \angle A O Z=\beta$ and $\angle Z O B=\gamma$. So $\alpha+\beta+\gamma=180^{\circ}$. Also $\angle B O X=\alpha$ (vertical angles) and $\angle A C Y=\alpha=\angle B C X$ (angles subtending equal arcs); similarly $\angle C O X=\beta$, $\angle A B Z=\beta=\angle C B X ; \angle C O Y=\gamma ; \angle B A Z=\gamma=$ $\angle C A Y$. Each of the triangles $C Y A, C B X$ and $Z B A$ have two angles from the set $\{\alpha, \beta, \gamma\}$. All triangles are then similar.
Similarity implies
$$
\frac{A Y}{C Y}=\frac{A B}{B Z}, \quad \frac{C X}{B X}=\frac{A Z}{A B}
$$
Consequently
$$
\frac{A Y}{A Z} \cdot \frac{B Z}{B X} \cdot \frac{C X}{C Y}=\frac{A B}{B Z} \cdot \frac{A Z}{A B} \cdot \frac{B Z}{A Z}=1
$$
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 230 |
11.2. In a triangle $A B C$ assume $A B=A C$, and let $D$ and $E$ be points on the extension of segment $B A$ beyond $A$ and on the segment $B C$, respectively, such that the lines $C D$ and $A E$ are parallel. Prove that $C D \geq \frac{4 h}{B C} C E$, where $h$ is the height from $A$ in triangle ABC. When does equality hold?
|
Solution. Because $A E \| D C$, the triangles $A B E$ and $D B C$ are similar. So
$$
C D=\frac{B C}{B E} \cdot A E
$$
$\mathrm{ja}$
$$
C D=\frac{A E \cdot B C}{B E \cdot C E} \cdot C E
$$
Let $A F$ be an altitude of $A B C$. Then $A E \geq A F=h$, and equality holds if and only if $E=F$. Because $A B C$ is isosceles, $F$ is the midpoint of $B C$. The arithmetic-geometric mean inequality yields
$$
B E \cdot C E \leq\left(\frac{B E+E C}{2}\right)^{2}=\left(\frac{B C}{2}\right)^{2}
$$
and equality holds if and only if $E$ is the midpoint of $B C$ i.e. $E=F$. The conclusion folows when these estimates are inserted in (1); furthermore, equality is equivalent to $E=F$
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 231 |
## Problem 4
Let $A B C$ be an acute-angled triangle with circumscribed circle $k$ and centre of the circumscribed circle $O$. A line through $O$ intersects the sides $A B$ and $A C$ at $D$ and $E$. Denote by $B^{\prime}$ and $C^{\prime}$ the reflections of $B$ and $C$ over $O$, respectively. Prove that the circumscribed circles of $O D C^{\prime}$ and $O E B^{\prime}$ concur on $k$.
|
Solution. Let $P$ be the intersection of the circles $k$ and the circumscribed circle of triangle $A D E^{1}$. Let $C_{1}$ be the second intersection of the circumscribed circle of $\triangle D O P$ with $k$. We will prove that $C_{1}=C^{\prime}$, i.e. the reflection of $C$ over $O$. We know that $\left|O C_{1}\right|=|O P|$, and hence $\measuredangle C_{1} P O=\measuredangle O C_{1} P$, furthermore $\measuredangle O C_{1} P=\measuredangle O D P=$ $\measuredangle E D P$, since the quadrilateral $C_{1} P O D$ by assumption is inscribed and the points $O, D$ and $E$ are collinear. Now, since $P$ is the centre of spiral similarity sending $D E$ to $B C$ the triangles $P D E$ and $P B C$ are similar, and we have $\measuredangle E D P=\measuredangle C B P$, and finally, from the inscribed angle theorem we have
$$
\measuredangle O P C=90^{\circ}-\frac{\measuredangle C O P}{2}=90^{\circ}-\measuredangle C B P=90^{\circ}-\measuredangle C_{1} P O
$$
The conclusion follows, since $90^{\circ}=\measuredangle C_{1} P O+\measuredangle O P C$, and since $C_{1}$ is by assumption on $k$, it must be the antipodal point of $C$ with respect to $k$.
[^0]
[^0]: ${ }^{1}$ That is, the Miquel point of quadrilateral BCED.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 233 |
Problem 1 Let $n$ be a positive integer. Show that there exist positive integers $a$ and $b$ such that:
$$
\frac{a^{2}+a+1}{b^{2}+b+1}=n^{2}+n+1
$$
|
Solution 1 Let $P(x)=x^{2}+x+1$. We have $P(n) P(n+1)=\left(n^{2}+n+1\right)\left(n^{2}+3 n+3\right)=$ $n^{4}+4 n^{3}+7 n^{2}+6 n+3$. Also, $P\left((n+1)^{2}\right)=n^{4}+4 n^{3}+7 n^{2}+6 n+3$. By choosing $a=(n+1)^{2}$ and $b=n+1$ we get $P(a) / P(b)=P(n)$ as desired.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 234 |
Problem 2 Let $a, b, \alpha, \beta$ be real numbers such that $0 \leq a, b \leq 1$, and $0 \leq \alpha, \beta \leq \frac{\pi}{2}$. Show that if
$$
a b \cos (\alpha-\beta) \leq \sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}
$$
then
$$
a \cos \alpha+b \sin \beta \leq 1+a b \sin (\beta-\alpha)
$$
|
Solution 2 The condition can be rewritten as
$$
a b \cos (\alpha-\beta)=a b \cos \alpha \cos \beta+a b \sin \alpha \sin \beta \leq \sqrt{\left(1-a^{2}\right)\left(1-b^{2}\right)}
$$
Set $x=a \cos \alpha, y=b \sin \beta, z=b \cos \beta, t=a \sin \alpha$. We can now rewrite the condition as
$$
x z+y t \leq \sqrt{\left(1-x^{2}-t^{2}\right)\left(1-y^{2}-z^{2}\right)}
$$
whereas the inequality we need to prove now looks like
$$
x+y \leq 1+x y-z t
$$
Since $x, y, z, t \geq 0$, and $1+x y-z t=1+a b \sin (\beta-\alpha) \geq 0$, we can square both sides of both inequalities, and get equivalent ones. After a couple of cancelations the condition yields
$$
2 x y z t \leq 1-x^{2}-y^{2}-z^{2}-t^{2}+x^{2} y^{2}+z^{2} t^{2}
$$
so that
$$
x^{2}+y^{2}+z^{2}+t^{2} \leq(x y-z t)^{2}+1
$$
which is equivalent to
$$
x^{2}+y^{2}+z^{2}+t^{2}+2 x y-2 z t \leq(1+x y-z t)^{2}
$$
or
$$
(x+y)^{2}+(z-t)^{2} \leq(1+x y-z t)^{2}
$$
Since $(x+y)^{2} \leq(x+y)^{2}+(z-t)^{2}$, the desired inequality follows.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 235 |
Problem 3 Let $M$ and $N$ be the midpoints of the sides $A C$ and $A B$, respectively, of an acute triangle $A B C, A B \neq A C$. Let $\omega_{B}$ be the circle centered at $M$ passing through $B$, and let $\omega_{C}$ be the circle centered at $N$ passing through $C$. Let the point $D$ be such that $A B C D$ is an isosceles trapezoid with $A D$ parallel to $B C$. Assume that $\omega_{B}$ and $\omega_{C}$ intersect in two distinct points $P$ and $Q$. Show that $D$ lies on the line $P Q$.
|
Solution 3 Let $E$ be such that $A B E C$ is a parallelogram with $A B \| C E$ and $A C \| B E$, and let $\omega$ be the circumscribed circle of $\triangle A B C$ with centre $O$.
It is known that the radical axis of two circles is perpendicular to the line connecting the two centres. Since $B E \perp M O$ and $C E \perp N O$, this means that $B E$ and $C E$ are the radical axes of $\omega$ and $\omega_{B}$, and of $\omega$ and $\omega_{C}$, respectively, so $E$ is the radical centre of $\omega$, $\omega_{B}$, and $\omega_{C}$.
Now as $B E=A C=B D$ and $C E=A B=C D$ we find that $B C$ is the perpendicular bisector of $D E$. Most importantly we have $D E \perp B C$. Denote by $t$ the radical axis of $\omega_{B}$ and $\omega_{C}$, i.e. $t=P Q$. Then since $t \perp M N$ we find that $t$ and $D E$ are parallel. Therefore since $E$ lies on $t$ we get that $D$ also lies on $t$.
Alternative solution Reflect $B$ across $M$ to a point $B^{\prime}$ forming a parallelogram $A B C B^{\prime}$. Then $B^{\prime}$ lies on $\omega_{B}$ diagonally opposite $B$, and since $A B^{\prime} \| B C$ it lies on $A D$. Similarly reflect $C$ across $N$ to a point $C^{\prime}$, which satisfies analogous properties. Note that $C B^{\prime}=A B=C D$, so we find that triangle $C D B^{\prime}$ and similarly triangle $B D C^{\prime}$ are isosceles.
Let $B^{\prime \prime}$ and $C^{\prime \prime}$ be the orthogonal projections of $B$ and $C$ onto $A D$. Since $B B^{\prime}$ is a diameter of $\omega_{B}$ we get that $B^{\prime \prime}$ lies on $\omega_{B}$, and similarly $C^{\prime \prime}$ lies on $\omega_{C}$. Moreover $B B^{\prime \prime}$ is an altitude of the isosceles triangle $B D C^{\prime}$ with $B D=B C^{\prime}$, hence it coincides with the median from $B$, so $B^{\prime \prime}$ is in fact the midpoint of $D C^{\prime}$. Similarly $C^{\prime \prime}$ is the midpoint of $D B^{\prime}$. From this we get
$$
2=\frac{D C^{\prime}}{D B^{\prime \prime}}=\frac{D B^{\prime}}{D C^{\prime \prime}}
$$
which rearranges as $D C^{\prime} \cdot D C^{\prime \prime}=D B^{\prime} \cdot D B^{\prime \prime}$. This means that $D$ has same the power with respect to $\omega_{B}$ and $\omega_{C}$, hence it lies on their radical axis $P Q$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 236 |
Problem 2. Given a triangle $A B C$, let $P$ lie on the circumcircle of the triangle and be the midpoint of the arc $B C$ which does not contain $A$. Draw a straight line $l$ through $P$ so that $l$ is parallel to $A B$. Denote by $k$ the circle which passes through $B$, and is tangent to $l$ at the point $P$. Let $Q$ be the second point of intersection of $k$ and the line $A B$ (if there is no second point of intersection, choose $Q=B)$. Prove that $A Q=A C$.
|
Solution I. There are three possibilities: $Q$ between $A$ and $B, Q=B$, and $B$ between $A$ and $Q$. If $Q=B$ we have that $\angle A B P$ is right, and $A P$ is a diameter
of the circumcircle. The triangles $A B P$ and $A C P$ are then congruent (they have $A P$ in common, $P B=P C$, and both have a right angle opposite to $A P$ ). Hence ir follows that $A B=A C$.
The solutions in the other two cases are very similar. We present the one in the case when $Q$ lies between $A$ and $B$.
The segment $A P$ is the angle bisector of the angle at $A$, since $P$ is the midpoint of the arc $B C$ of the circumcircle which does not contain $A$. Also, $P C=P B$. Since the segment $Q B$ is parallel to the tangent to $k$ at $P$, it is orthogonal to the diameter of $k$ through $P$. Thus this diameter cuts $Q B$ in halves, to form two congruent right triangles, and it follows that $P Q=P B$. We have (in the usual notation) $\angle P C B=\angle P B C=\frac{\alpha}{2}$, and
$$
\angle A Q P=180^{\circ}-\angle B Q P=180^{\circ}-\angle Q B P=180^{\circ}-\beta-\frac{\alpha}{2}=\frac{\alpha}{2}+\gamma=\angle A C P
$$
Hence the triangles $A Q P$ and $A C P$ are congruent (two pairs of equal angles and one pair of equal corresponding sides), and it follows that $A C=A Q$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 239 |
## Problem 1.
Let $A B C$ be a triangle and $\Gamma$ the circle with diameter $A B$. The bisectors of $\angle B A C$ and $\angle A B C$ intersect $\Gamma$ (also) at $D$ and $E$, respectively. The incircle of $A B C$ meets $B C$ and $A C$ at $F$ and $G$, respectively. Prove that $D, E, F$ and $G$ are collinear.
|
Solution 1. Let the line $E D$ meet $A C$ at $G^{\prime}$ and $B C$ at $F^{\prime} . A D$ and $B E$ intersect at $I$, the incenter of $A B C$. As angles subtending the same arc $\widehat{B D}$, $\angle D A B=\angle D E B=\angle G^{\prime} E I$. But $\angle D A B=\angle C A D=$ $\angle G^{\prime} A I$. This means that $E, A, I$ and $G^{\prime}$ are concyclic, and $\angle A E I=\angle A G^{\prime} I$ as angles subtending the same chord $A I$. But $A B$ is a diameter of $\Gamma$, and so $\angle A E B=$ $\angle A E I$ is a right angle. So $I G^{\prime} \perp A C$, or $G^{\prime}$ is the foot of the perpendicular from $I$ to $A C$. This implies $G^{\prime}=G$. In a similar manner we prove that $F^{\prime}=F$, and the proof is complete.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 240 |
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