problem_id
stringlengths 32
32
| link
stringlengths 75
84
| problem
stringlengths 14
5.33k
| solution
stringlengths 15
6.63k
| letter
stringclasses 5
values | answer
stringclasses 957
values |
|---|---|---|---|---|---|
34022b0112a8cfaa6b22f6f8751b58e8 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_11 | sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000. | Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:
$a_{1}\equiv 1 \pmod {1000} \\ a_{2}\equiv 1 \pmod {1000} \\ a_{3}\equiv 1 \pmod {1000} \\ a_{4}\equiv 3 \pmod {1000} \\ a_{5}\equiv 5 \pmod {1000} \\ \cdots \\ a_{25} \equiv 793 \pmod {1000} \\ a_{26} \equiv 281 \pmod {1000} \\ a_{27} \equiv 233 \pmod {1000} \\ a_{28} \equiv 307 \pmod {1000}$
Adding all the residues shows the sum is congruent to $\boxed{834}$ mod $1000$ | null | 834 |
34022b0112a8cfaa6b22f6f8751b58e8 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_11 | sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000. | All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given $a_{28}, a_{29},$ and $a_{30}$ , so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some $p, q, r$ such that $\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}$ . From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that $(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})$ , at least for the first few terms. From this, we have that $\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}(\mod 1000)$ | null | 834 |
31d4d7dffd6ef44e954dec685b098f61 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_12 | Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime , and $q$ is not divisible by the square of any prime, find $p+q+r.$ | Notice that $\angle{E} = \angle{BGC} = 120^\circ$ because $\angle{A} = 60^\circ$ . Also, $\angle{GBC} = \angle{GAC} = \angle{FAE}$ because they both correspond to arc ${GC}$ . So $\Delta{GBC} \sim \Delta{EAF}$
\[[EAF] = \frac12 (AE)(EF)\sin \angle AEF = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4.\]
Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, $[GBC] = \frac {BC^2}{AF^2}\cdot[EAF] = \frac {12}{11^2 + 13^2 - 2\cdot11\cdot13\cdot\cos120^\circ}\cdot\frac {143\sqrt3}4 = \frac {429\sqrt3}{433}$ . Therefore, the answer is $429+433+3=\boxed{865}$ | null | 865 |
31d4d7dffd6ef44e954dec685b098f61 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_12 | Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime , and $q$ is not divisible by the square of any prime, find $p+q+r.$ | Solution by e_power_pi_times_i/edited by srisainandan6
Let the center of the circle be $O$ and the origin. Then, $A (0,2)$ $B (-\sqrt{3}, -1)$ $C (\sqrt{3}, -1)$ $D$ and $E$ can be calculated easily knowing $AD$ and $AE$ $D (-\dfrac{13}{2}, \dfrac{-13\sqrt{3}+4}{2})$ $E (\dfrac{11}{2}, \dfrac{-11\sqrt{3}+4}{2})$ . As $DF$ and $EF$ are parallel to $AE$ and $AD$ $F (-1, -12\sqrt{3}+2)$ $G$ and $A$ is the intersection between $AF$ and circle $O$ . Therefore $G (-\dfrac{48\sqrt{3}}{433}, -\dfrac{862}{433})$ . Using the Shoelace Theorem, $[CBG] = \dfrac{429\sqrt{3}}{433}$ , so the answer is $\boxed{865}$ . Note that although the solution may appear short, actually getting all the coordinates take a while as there is plenty of computation. | null | 865 |
31d4d7dffd6ef44e954dec685b098f61 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_12 | Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime , and $q$ is not divisible by the square of any prime, find $p+q+r.$ | Lines $l_1$ and $l_2$ are constructed such that $AEFD$ is a parallelogram, hence $DF = 13$ . Since $BAC$ is equilateral with angle of $60^{\circ}$ , angle $D$ is $120^{\circ}$ . Use law of cosines to find $AF = \sqrt{433}$ . Then use law of sines to find angle $BAG$ and $GAC$ . Next we use Ptolemy's Theorem on $ABGC$ to find that $CG + BG = AG$ . Next we use law of cosine on triangles $BAG$ and $GAC$ , solving for BG and CG respectively. Subtract the two equations and divide out a $BG + CG$ to find the value of $CG - BG$ . Next, $AG = 2\cdot R \cos{\theta}$ , where R is radius of circle $= 2$ and $\theta =$ angle $BAG$ . We already know sine of the angle so find cosine, hence we have found $AG$ . At this point it is system of equation yielding $CG = \frac{26\sqrt{3}}{\sqrt{433}}$ and $BG = \frac{22\sqrt{3}}{\sqrt{433}}$ . Given $[CBG] = \frac{BC \cdot CG \cdot BG}{4R}$ , and $BC = 2\sqrt{3}$ by $30-60-90$ triangle, we can evaluate to find $[CBG] = \frac{429\sqrt{3}}{433}$ , to give answer = $\boxed{865}$ | null | 865 |
31d4d7dffd6ef44e954dec685b098f61 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_12 | Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime , and $q$ is not divisible by the square of any prime, find $p+q+r.$ | Note that $AB=2\sqrt3$ $DF=11$ , and $EF=13$ . If we take a homothety of the parallelogram with respect to $A$ , such that $F$ maps to $G$ , we see that $\frac{[ABG]}{[ACG]}=\frac{11}{13}$ . Since $\angle AGB=\angle AGC=60^{\circ}$ , from the sine area formula we have $\frac{BG}{CG}=\frac{11}{13}$ . Let $BG=11k$ and $CG=13k$
By Law of Cosines on $\triangle BGC$ , we have \[12=k^2(11^2+11\cdot13+13^2)=433k^2\implies k^2=\frac{12}{433}\] Thus, $[CBG]=\frac12 (11k)(13k)\sin 120^{\circ} = \frac{\sqrt3}{4}\cdot 143\cdot \frac{12}{433}=\frac{429\sqrt3}{433}\implies\boxed{865}$ | null | 865 |
cb6c90421ee9299815a4bdba37c381a8 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_13 | How many integers $N$ less than $1000$ can be written as the sum of $j$ consecutive positive odd integers from exactly 5 values of $j\ge 1$ | Let the first odd integer be $2n+1$ $n\geq 0$ . Then the final odd integer is $2n+1 + 2(j-1) = 2(n+j) - 1$ . The odd integers form an arithmetic sequence with sum $N = j\left(\frac{(2n+1) + (2(n+j)-1)}{2}\right) = j(2n+j)$ . Thus, $j$ is a factor of $N$
Since $n\geq 0$ , it follows that $2n+j \geq j$ and $j\leq \sqrt{N}$
Since there are exactly $5$ values of $j$ that satisfy the equation, there must be either $9$ or $10$ factors of $N$ . This means $N=p_1^2p_2^2$ or $N=p_1p_2^4$ . Unfortunately, we cannot simply observe prime factorizations of $N$ because the factor $(2n+j)$ does not cover all integers for any given value of $j$
Instead we do some casework:
\[(3^2\cdot5^2),\ (3^2\cdot7^2),\ (3^4\cdot5),\ (3^4\cdot7),\ (3^4\cdot 11)\]
\[N = 4k(n+k) \Longrightarrow \frac{N}{4} = k(n+k)\]
\[\frac{N}{4} = (2^2\cdot3^2),(2^2\cdot5^2),(2^2\cdot7^2), (3^2\cdot5^2), (2^4\cdot3), (2^4\cdot5), (2^4\cdot7), (2^4\cdot11), (2^4\cdot13), (3^4\cdot2)\]
The total number of integers $N$ is $5 + 10 = \boxed{15}$ | null | 15 |
cb6c90421ee9299815a4bdba37c381a8 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_13 | How many integers $N$ less than $1000$ can be written as the sum of $j$ consecutive positive odd integers from exactly 5 values of $j\ge 1$ | Let the largest odd number below the sequence be the $q$ th positive odd number, and the largest odd number in the sequence be the $p$ th positive odd number. Therefore, the sum is $p^2-q^2=(p+q)(p-q)$ by sum of consecutive odd numbers. Note that $p+q$ and $p-q$ have the same parity, and $q$ can equal $0$ . We then perform casework based on the parity of $p-q$
If $p-q$ is odd, then $p^2-q^2$ must always be odd. Therefore, to have 5 pairs of odd factors, we must have either $9$ (in which case the number is a perfect square) or $10$ factors. Considering the upper bound, the only way this can happen is $p_1^4\cdot{p_2}$ or $p_1^2\cdot{p_2^2}$ . N must then be one of \[(3^2\cdot5^2),\ (3^2\cdot7^2),\ (3^4\cdot5),\ (3^4\cdot7),\ (3^4\cdot 11)\] So, there are $5$ solutions when $(p+q)(p-q)$ is odd.
If $p-q$ is even, then $(p+q)(p-q)$ must have at least two factors of $2$ , so we can rewrite the expression as $4(k)(k-q)$ where $k=\frac{p+q}{2}$ . We can disregard the $4$ by dividing by $4$ and restricting our upper bound to $250$ . Since $k$ and $q$ don't have to be the same parity, we can include all cases less than 250 that have 9 or 10 factors. We then have \[(2^2\cdot3^2),(2^2\cdot5^2),(2^2\cdot7^2), (3^2\cdot5^2), (2^4\cdot3), (2^4\cdot5), (2^4\cdot7), (2^4\cdot11), (2^4\cdot13), (3^4\cdot2)\] as the possibilities.
Therefore, there are $10+5=\boxed{015}$ possibilities for $p^2-q^2=N$ ~sigma | null | 015 |
4425e762fa659aad2b2d65c09ca6b225 | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_14 | Let $S_n$ be the sum of the reciprocals of the non-zero digits of the integers from $1$ to $10^n$ inclusive. Find the smallest positive integer $n$ for which $S_n$ is an integer. | Let $K = \sum_{i=1}^{9}{\frac{1}{i}}$ . Examining the terms in $S_1$ , we see that $S_1 = K + 1$ since each digit $n$ appears once and 1 appears an extra time. Now consider writing out $S_2$ . Each term of $K$ will appear 10 times in the units place and 10 times in the tens place (plus one extra 1 will appear), so $S_2 = 20K + 1$
In general, we will have that
because each digit will appear $10^{n - 1}$ times in each place in the numbers $1, 2, \ldots, 10^{n} - 1$ , and there are $n$ total places.
The denominator of $K$ is $D = 2^3\cdot 3^2\cdot 5\cdot 7$ . For $S_n$ to be an integer, $n10^{n-1}$ must be divisible by $D$ . Since $10^{n-1}$ only contains the factors $2$ and $5$ (but will contain enough of them when $n \geq 3$ ), we must choose $n$ to be divisible by $3^2\cdot 7$ . Since we're looking for the smallest such $n$ , the answer is $\boxed{063}$ | null | 063 |
967ddd9d775b096480707865af19621c | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_15 | Given that $x, y,$ and $z$ are real numbers that satisfy: \begin{align*} x &= \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}, \\ y &= \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}, \\ z &= \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}, \end{align*} and that $x+y+z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n.$ | Let $\triangle XYZ$ be a triangle with sides of length $x, y$ and $z$ , and suppose this triangle is acute (so all altitudes are in the interior of the triangle).
Let the altitude to the side of length $x$ be of length $h_x$ , and similarly for $y$ and $z$ . Then we have by two applications of the Pythagorean Theorem we that \[x = \sqrt{y^2 - h_x^2} + \sqrt{z^2 - h_x^2}\] As a function of $h_x$ , the RHS of this equation is strictly decreasing, so it takes each value in its range exactly once. Thus we must have that $h_x^2 = \frac1{16}$ and so $h_x = \frac{1}4$ and similarly $h_y = \frac15$ and $h_z = \frac16$
The area of the triangle must be the same no matter how we measure it; therefore $x\cdot h_x = y\cdot h_y = z \cdot h_z$ gives us $\frac x4 = \frac y5 = \frac z6 = 2A$ and $x = 8A, y = 10A$ and $z = 12A$
The semiperimeter of the triangle is $s = \frac{8A + 10A + 12A}{2} = 15A$ so by Heron's formula we have \[A = \sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = 15A^2\sqrt{7}\]
Thus, $A = \frac{1}{15\sqrt{7}}$ and $x + y + z = 30A = \frac2{\sqrt{7}}$ and the answer is $2 + 7 = \boxed{009}$ | null | 009 |
967ddd9d775b096480707865af19621c | https://artofproblemsolving.com/wiki/index.php/2006_AIME_II_Problems/Problem_15 | Given that $x, y,$ and $z$ are real numbers that satisfy: \begin{align*} x &= \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}, \\ y &= \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}, \\ z &= \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}, \end{align*} and that $x+y+z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n.$ | Note that none of $x,y,z$ can be zero.
Each of the equations is in the form \[a=\sqrt{b^2-d^2}+\sqrt{c^2-d^2}\]
Isolate a radical and square the equation to get \[b^2-d^2=a^2-2a\sqrt{c^2-d^2}+c^2-d^2\]
Now cancel, and again isolate the radical, and square the equation to get \[a^4+b^4+c^4+2a^2c^2-2a^2b^2-2b^2c^2=4a^2c^2-4a^2d^2\]
Rearranging gives \[a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2-4a^2d^2\]
Now note that everything is cyclic but the last term (i.e. $-4a^2d^2$ ), which implies \[-4x^2\cdot\frac1{16}=-4y^2\cdot\frac1{25}=-4z^2\cdot\frac1{36}\]
Or
\[x: y: z=4: 5: 6 \implies x=\frac{4y}5 \textrm{ and } z=\frac{6y}5\]
Plug these values into the middle equation to get \[\frac{256y^4+625y^4+1296y^4}{625}=\frac{800y^4}{625}+\frac{1800y^4}{625}+\frac{1152y^4}{625}-\frac{100y^2}{625}\]
Simplifying gives \[1575y^4=100y^2 \textrm{ but } y \neq 0 \implies y^2=\frac4{63} \textrm{ or } y=\frac2{3\sqrt7}\]
Substituting the value of $y$ for $x$ and $z$ gives \[x+y+z = \frac{4y+5y+6y}5 = 3y = 3 \cdot \frac{2}{3\sqrt7} = \frac{2}{\sqrt7}\]
And thus the answer is $\boxed{009}$ | null | 009 |
a63e8ea5c645e15d9015a5c77ad0f18a | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_1 | Six congruent circles form a ring with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle $C$ with radius 30. Let $K$ be the area of the region inside circle $C$ and outside of the six circles in the ring. Find $\lfloor K \rfloor$ (the floor function ).
2005 AIME I Problem 1.png | Define the radii of the six congruent circles as $r$ . If we draw all of the radii to the points of external tangency, we get a regular hexagon . If we connect the vertices of the hexagon to the center of the circle $C$ , we form several equilateral triangles . The length of each side of the triangle is $2r$ . Notice that the radius of circle $C$ is equal to the length of the side of the triangle plus $r$ . Thus, the radius of $C$ has a length of $3r = 30$ , and so $r = 10$ $K = 30^2\pi - 6(10^2\pi) = 300\pi$ , so $\lfloor 300\pi \rfloor = \boxed{942}$ | null | 942 |
252cd295f12a6ce715c8b9e18648bcae | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_2 | For each positive integer $k$ , let $S_k$ denote the increasing arithmetic sequence of integers whose first term is $1$ and whose common difference is $k$ . For example, $S_3$ is the sequence $1,4,7,10,\ldots.$ For how many values of $k$ does $S_k$ contain the term $2005$ | Suppose that the $n$ th term of the sequence $S_k$ is $2005$ . Then $1+(n-1)k=2005$ so $k(n-1)=2004=2^2\cdot 3\cdot 167$ . The ordered pairs $(k,n-1)$ of positive integers that satisfy the last equation are $(1,2004)$ $(2,1002)$ $(3,668)$ $(4,501)$ $(6,334)$ $(12,167)$ $(167,12)$ $(334,6)$ $(501,4)$ $(668,3)$ $(1002,2)$ and $(2004,1)$ , and each of these gives a possible value of $k$ . Thus the requested number of values is $12$ , and the answer is $\boxed{012}$ | null | 012 |
252cd295f12a6ce715c8b9e18648bcae | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_2 | For each positive integer $k$ , let $S_k$ denote the increasing arithmetic sequence of integers whose first term is $1$ and whose common difference is $k$ . For example, $S_3$ is the sequence $1,4,7,10,\ldots.$ For how many values of $k$ does $S_k$ contain the term $2005$ | Any term in the sequence $S_k$ can be written as 1+kx. If this is to equal 2005, then the remainder when 2005 is divided by k is 1.
Now all we have to do is find the numbers of factors of 2004. There are $(2 + 1)(1 + 1)(1 + 1) = \boxed{012}$ divisors of $2^2\cdot 3^1\cdot 167^1$ | null | 012 |
27ba515b2991c2f427345b2586fd706f | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_3 | How many positive integers have exactly three proper divisors (positive integral divisors excluding itself), each of which is less than 50? | Suppose $n$ is such an integer . Because $n$ has $3$ proper divisors, it must have $4$ divisors,, so $n$ must be in the form $n=p\cdot q$ or $n=p^3$ for distinct prime numbers $p$ and $q$
In the first case, the three proper divisors of $n$ are $1$ $p$ and $q$ . Thus, we need to pick two prime numbers less than $50$ . There are fifteen of these ( $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43$ and $47$ ) so there are ${15 \choose 2} =105$ ways to choose a pair of primes from the list and thus $105$ numbers of the first type.
In the second case, the three proper divisors of $n$ are 1, $p$ and $p^2$ . Thus we need to pick a prime number whose square is less than $50$ . There are four of these ( $2, 3, 5,$ and $7$ ) and so four numbers of the second type.
Thus there are $105+4=\boxed{109}$ integers that meet the given conditions. | null | 109 |
2884c87e18e379ff4e817ab8273f68e6 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_4 | The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have. | If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$ . If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$ . Thus, if the number of columns is $n$ , the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$ . In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$ , so this number works and no larger number can. Thus, the answer is $\boxed{294}$ | null | 294 |
2884c87e18e379ff4e817ab8273f68e6 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_4 | The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have. | Let there be $m$ members and $n$ members for the square and $c$ for the number of columns of the other formation. We have $n^2 +5 = c(c+7) \implies n^2+5 = \left(c+\frac{7}{2}\right)^2 -\frac{49}{4} \implies n^2 - \left(c+\frac{7}{2}\right)^2 = -\frac{69}{4} \implies \left(n-c-\frac{7}{2}\right)\left(n + c +\frac{7}{2}\right) \implies (2n-2c-7)(2n+2n+7) = -69.$
To maximize this we let $2n+2c+7 = 68$ and $2n-2c-7 = 1.$ Solving we find $n = 17$ so the desired number of members is $17^2 + 5 = \boxed{294}.$ | null | 294 |
2884c87e18e379ff4e817ab8273f68e6 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_4 | The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have. | Think of the process of moving people from the last column to new rows. Since there are less columns than rows, for each column removed, there are people discarded to the "extra" pile to be placed at the end. To maximize the number of "extra" people to fill in the last few rows. We remove 3 columns and add 4 rows. For the first new row, one more person will be discarded. For the second, 1 extra person are added since there is one more row now, and there are 2 less columns. Thus, there are 3 extra people discarded. Similarly, 5 extra people are discarded for the third column. Now there are $5+1+3+5=14$ people in the extra pile to put as the last row, so there are $14(14+7)=\boxed{294}$ people. | null | 294 |
2884c87e18e379ff4e817ab8273f68e6 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_4 | The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have. | Note: Only do this if you have a LOT of time (and you've memorized all your perfect squares up to 1000).
We can see that the number of members in the band must be of the form $n(n + 7)$ for some positive integer $n$ . When $n = 28$ , this product is $980$ , and since AIME answers are nonnegative integers less than $1000$ , we don't have to check any higher $n$ . Also, we know that this product must be 5 more than a perfect square, so we can make a table as shown and bash: \[\begin{tabular}{|c|c|c|} n & n(n+7) & 5 more than a perfect square?\\ \hline 1 & 8 & no\\ \hline 2 & 18 & no\\ \hline 3 & 30 & yes\\ \hline 4 & 44 & no\\ \hline 5 & 60 & no\\ \hline 6 & 78 & no\\ \hline 7 & 98 & no\\ \hline 8 & 120 & no\\ \hline 9 & 144 & no\\ \hline 10 & 170 & no\\ \hline 11 & 198 & no\\ \hline 12 & 228 & no\\ \hline 13 & 260 & no\\ \hline 14 & 294 & yes\\ \hline 15 & 330 & no\\ \hline 16 & 368 & no\\ \hline 17 & 408 & no\\ \hline 18 & 450 & no\\ \hline 19 & 494 & no\\ \hline 20 & 540 & no\\ \hline 21 & 588 & no\\ \hline 22 & 638 & no\\ \hline 23 & 690 & no\\ \hline 24 & 744 & no\\ \hline 25 & 800 & no\\ \hline 26 & 858 & no\\ \hline 27 & 918 & no\\ \hline 28 & 980 & no\\ \hline \end{tabular}\] Thus, we can see that our largest valid $n(n+7)$ is $\boxed{294}$ | null | 294 |
f670714c5d675133bf1ef53ec2cff7a0 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_5 | Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distinguishable arrangements of the 8 coins. | There are two separate parts to this problem: one is the color (gold vs silver), and the other is the orientation.
There are ${8\choose4} = 70$ ways to position the gold coins in the stack of 8 coins, which determines the positions of the silver coins.
Create a string of letters H and T to denote the orientation of the top of the coin. To avoid making two faces touch, we cannot have the arrangement HT. Thus, all possible configurations must be a string of tails followed by a string of heads, since after the first H no more tails can appear. The first H can occur in a maximum of eight times different positions, and then there is also the possibility that it doesn’t occur at all, for $9$ total configurations. Thus, the answer is $70 \cdot 9 = \boxed{630}$ | null | 630 |
f670714c5d675133bf1ef53ec2cff7a0 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_5 | Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distinguishable arrangements of the 8 coins. | First, we can break this problem up into two parts: the amount of ways to order the coins based on color, and which side is facing up for each coin (assuming they are indistinguishable). In the end, we multiply these values together.
First, there are obviously $\binom{8}{4}$ ways to order the coins based on color.
Next, we set up a recurrence. Let $a_n$ be the number of ways $n$ indistinguishable coins to be stacked such that no heads are facing each other.
Consider the top coin. If it is facing up, then there are $a_{n-1}$ ways for the rest of the coins below it to be arranged. If it is facing down however, there will only be one way to arrange the coins. We can the recurrence: \[a_n=a_{n-1}+1.\] Note $a_1=2$ , so $a_8 = 9$
Finally, to get the result, we do $70\times 9$ to get $\boxed{630}$ | null | 630 |
6bae14bef77ff58bb21121f30061ef92 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_6 | Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$ | The left-hand side of that equation is nearly equal to $(x - 1)^4$ . Thus, we add 1 to each side in order to complete the fourth power and get $(x - 1)^4 = 2006$
Let $r = \sqrt[4]{2006}$ be the positive real fourth root of 2006. Then the roots of the above equation are $x = 1 + i^n r$ for $n = 0, 1, 2, 3$ . The two non-real members of this set are $1 + ir$ and $1 - ir$ . Their product is $P = 1 + r^2 = 1 + \sqrt{2006}$ $44^2 = 1936 < 2006 < 2025 = 45^2$ so $\lfloor P \rfloor = 1 + 44 = \boxed{045}$ | null | 045 |
6bae14bef77ff58bb21121f30061ef92 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_6 | Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$ | Starting like before, $(x-1)^4= 2006$ This time we apply differences of squares. $(x-1)^4-2006=0$ so $((x-1)^2+\sqrt{2006})((x-1)^2 -\sqrt{2006})=0$ If you think of each part of the product as a quadratic, then $((x-1)^2+\sqrt{2006})$ is bound to hold the two non-real roots since the other definitely crosses the x-axis twice since it is just $x^2$ translated down and right.
Therefore $P$ is the product of the roots of $((x-1)^2+\sqrt{2006})$ or $P=1+\sqrt{2006}$ so
$\lfloor P \rfloor = 1 + 44 = \boxed{045}$ | null | 045 |
6bae14bef77ff58bb21121f30061ef92 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_6 | Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$ | If we don't see the fourth power, we can always factor the LHS to try to create a quadratic substitution. Checking, we find that $x=0$ and $x=2$ are both roots. Synthetic division gives $(x^2-2x)(x^2-2x+2)=2005$ . We now have our quadratic substitution of $y=x^2-2x+1=(x-1)^2$ , giving us $(y-1)(y+1)=2005$ . From here we proceed as in Solution 1 to get $\boxed{045}$ | null | 045 |
6bae14bef77ff58bb21121f30061ef92 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_6 | Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$ | Realizing that if we add 1 to both sides we get $x^4-4x^3+6x^2-4x+1=2006$ which can be factored as $(x-1)^4=2006$ . Then we can substitute $(x-1)$ with $y$ which leaves us with $y^4=2006$ . Now subtracting 2006 from both sides we get some difference of squares $y^4-2006=0 \rightarrow (y-\sqrt[4]{2006})(y+\sqrt[4]{2006})(y^2+\sqrt{2006})=0$ . The question asks for the product of the complex roots so we only care about the last factor which is equal to zero. From there we can solve $y^2+\sqrt{2006}=0$ , we can substitute $(x-1)$ for $y$ giving us $(x-1)^2+\sqrt{2006}=0$ , expanding this we get $x^2-2x+1+\sqrt{2006}=0$ . We know that the product of a quadratics roots is $\frac{c}{a}$ which leaves us with $\frac{1+\sqrt{2006}}{1}=1+\sqrt{2006}\approx\boxed{045}$ | null | 045 |
6bae14bef77ff58bb21121f30061ef92 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_6 | Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$ | As in solution 1, we find that $(x-1)^4 = 2006$ . Now $x-1=\pm \sqrt[4]{2006}$ so $x_1 = 1+\sqrt[4]{2006}$ and $x_2 = 1-\sqrt[4]{2006}$ are the real roots of the equation. Multiplying, we get $x_1 x_2 = 1 - \sqrt{2006}$ . Now transforming the original function and using Vieta's formula, $x^4-4x^3+6x^2-4x-2005=0$ so $x_1 x_2 x_3 x_4 = \frac{-2005}{1} = -2005$ . We find that the product of the nonreal roots is $x_3 x_4 = \frac{-2005}{1-\sqrt{2006}} \approx 45.8$ and we get $\boxed{045}$ | null | 045 |
6bae14bef77ff58bb21121f30061ef92 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_6 | Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$ | As all the other solutions, we find that $(x-1)^4 = 2006$ . Thus $x=\sqrt[4]{2006}+1$ . Thus $x= \sqrt[4]{2006}(\cos(\frac{2\pi(k)}{4}+i\sin(\frac{2\pi(k)}{4}))+1$ when $k=0,1,2,3$ . The complex values of $x$ are the ones where $i\sin(\frac{2\pi(k)}{4})$ does not equal 0. These complex roots are $1+\sqrt[4]{2006}(i)$ and $1-\sqrt[4]{2006}(i)$ . The product of these two nonreal roots is ( $1+\sqrt[4]{2006}(i)$ )( $1-\sqrt[4]{2006}(i)$ ) which is equal to $1+\sqrt {2006}$ . The floor of that value is $\boxed{045}$ | null | 045 |
57e84508f17acef466f806c770639ba6 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_7 | In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers , find $p+q.$ | [asy]draw((0,0)--(20.87,0)--(15.87,8.66)--(5,8.66)--cycle); draw((5,8.66)--(5,0)); draw((15.87,8.66)--(15.87,0)); draw((5,8.66)--(16.87,6.928)); label("$A$",(0,0),SW); label("$B$",(20.87,0),SE); label("$E$",(15.87,8.66),NE); label("$D$",(5,8.66),NW); label("$P$",(5,0),S); label("$Q$",(15.87,0),S); label("$C$",(16.87,7),E); label("$12$",(10.935,7.794),S); label("$10$",(2.5,4.5),W); label("$10$",(18.37,4.5),E); [/asy]
Draw line segment $DE$ such that line $DE$ is concurrent with line $BC$ . Then, $ABED$ is an isosceles trapezoid so $AD=BE=10$ , and $BC=8$ and $EC=2$ . We are given that $DC=12$ . Since $\angle CED = 120^{\circ}$ , using Law of Cosines on $\bigtriangleup CED$ gives \[12^2=DE^2+4-2(2)(DE)(\cos 120^{\circ})\] which gives \[144-4=DE^2+2DE\] . Adding $1$ to both sides gives $141=(DE+1)^2$ , so $DE=\sqrt{141}-1$ $\bigtriangleup DAP$ and $\bigtriangleup EBQ$ are both $30-60-90$ , so $AP=5$ and $BQ=5$ $PQ=DE$ , and therefore $AB=AP+PQ+BQ=5+\sqrt{141}-1+5=9+\sqrt{141} \rightarrow (p,q)=(9,141) \rightarrow \boxed{150}$ | null | 150 |
57e84508f17acef466f806c770639ba6 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_7 | In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers , find $p+q.$ | Draw the perpendiculars from $C$ and $D$ to $AB$ , labeling the intersection points as $E$ and $F$ . This forms 2 $30-60-90$ right triangles , so $AE = 5$ and $BF = 4$ . Also, if we draw the horizontal line extending from $C$ to a point $G$ on the line $DE$ , we find another right triangle $\triangle DGC$ $DG = DE - CF = 5\sqrt{3} - 4\sqrt{3} = \sqrt{3}$ . The Pythagorean Theorem yields that $GC^2 = 12^2 - \sqrt{3}^2 = 141$ , so $EF = GC = \sqrt{141}$ . Therefore, $AB = 5 + 4 + \sqrt{141} = 9 + \sqrt{141}$ , and $p + q = \boxed{150}$ | null | 150 |
1f657e592c823173aace70b7b582beaf | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_8 | The equation $2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1$ has three real roots . Given that their sum is $m/n$ where $m$ and $n$ are relatively prime positive integers , find $m+n.$ | Let $y = 2^{111x}$ . Then our equation reads $\frac{1}{4}y^3 + 4y = 2y^2 + 1$ or $y^3 - 8y^2 + 16y - 4 = 0$ . Thus, if this equation has roots $r_1, r_2$ and $r_3$ , by Vieta's formulas we have $r_1\cdot r_2\cdot r_3 = 4$ . Let the corresponding values of $x$ be $x_1, x_2$ and $x_3$ . Then the previous statement says that $2^{111\cdot(x_1 + x_2 + x_3)} = 4$ so taking a logarithm of that gives $111(x_1 + x_2 + x_3) = 2$ and $x_1 + x_2 + x_3 = \frac{2}{111}$ . Thus the answer is $111 + 2 = \boxed{113}$ | null | 113 |
73dea59b04c53bf855a86558788abdae | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_9 | Twenty seven unit cubes are painted orange on a set of four faces so that two non-painted faces share an edge . The 27 cubes are randomly arranged to form a $3\times 3 \times 3$ cube. Given the probability of the entire surface area of the larger cube is orange is $\frac{p^a}{q^br^c},$ where $p,q,$ and $r$ are distinct primes and $a,b,$ and $c$ are positive integers , find $a+b+c+p+q+r.$ | 2005 I AIME-9.png
We can consider the orientation of each of the individual cubes independently.
The unit cube at the center of our large cube has no exterior faces, so all of its orientations work.
For the six unit cubes and the centers of the faces of the large cube, we need that they show an orange face. This happens in $\frac{4}{6} = \frac{2}{3}$ of all orientations, so from these cubes we gain a factor of $\left(\frac{2}{3}\right)^6$
The twelve unit cubes along the edges of the large cube have two faces showing, and these two faces are joined along an edge. Thus, we need to know the number of such pairs that are both painted orange. We have a pair for each edge, and 7 edges border one of the unpainted faces while only 5 border two painted faces. Thus, the probability that two orange faces show for one of these cubes is $\frac{5}{12}$ , so from all of these cubes we gain a factor of $\left(\frac{5}{12}\right)^{12} = \frac{5^{12}}{2^{24}3^{12}}$
Finally, we need to orient the eight corner cubes. Each such cube has 3 faces showing, and these three faces share a common vertex. Thus, we need to know the number of vertices for which all three adjacent faces are painted orange. There are six vertices which are a vertex of one of the unpainted faces and two vertices which have our desired property, so each corner cube contributes a probability of $\frac{2}{8} = \frac{1}{4}$ and all the corner cubes together contribute a probability of $\left(\frac{1}{4}\right)^8 = \frac{1}{2^{16}}$
Since these probabilities are independent, the overall probability is just their product, $\frac{2^6}{3^6} \cdot \frac{5^{12}}{2^{24}3^{12}} \cdot \frac{1}{2^{16}} = \frac{5^{12}}{2^{34}\cdot 3^{18}}$ and so the answer is $2 + 3 + 5 + 12 + 34 + 18 = \boxed{074}$ | null | 074 |
1b958c722ad85c33afd59df0e9f4f49c | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_10 | Triangle $ABC$ lies in the cartesian plane and has an area of $70$ . The coordinates of $B$ and $C$ are $(12,19)$ and $(23,20),$ respectively, and the coordinates of $A$ are $(p,q).$ The line containing the median to side $BC$ has slope $-5.$ Find the largest possible value of $p+q.$ | The midpoint $M$ of line segment $\overline{BC}$ is $\left(\frac{35}{2}, \frac{39}{2}\right)$ . The equation of the median can be found by $-5 = \frac{q - \frac{39}{2}}{p - \frac{35}{2}}$ . Cross multiply and simplify to yield that $-5p + \frac{35 \cdot 5}{2} = q - \frac{39}{2}$ , so $q = -5p + 107$
Use determinants to find that the area of $\triangle ABC$ is $\frac{1}{2} \begin{vmatrix}p & 12 & 23 \\ q & 19 & 20 \\ 1 & 1 & 1\end{vmatrix} = 70$ (note that there is a missing absolute value ; we will assume that the other solution for the triangle will give a smaller value of $p+q$ , which is provable by following these steps over again) (alternatively, we could use the Shoelace Theorem ). We can calculate this determinant to become $140 = \begin{vmatrix} 12 & 23 \\ 19 & 20 \end{vmatrix} - \begin{vmatrix} p & q \\ 23 & 20 \end{vmatrix} + \begin{vmatrix} p & q \\ 12 & 19 \end{vmatrix}$ $\Longrightarrow 140 = 240 - 437 - 20p + 23q + 19p - 12q$ $= -197 - p + 11q$ . Thus, $q = \frac{1}{11}p + \frac{337}{11}$
Setting this equation equal to the equation of the median, we get that $\frac{1}{11}p + \frac{337}{11} = -5p + 107$ , so $\frac{56}{11}p = \frac{107 \cdot 11 - 337}{11}$ . Solving produces that $p = 15$ Substituting backwards yields that $q = 32$ ; the solution is $p + q = \boxed{047}$ | null | 047 |
9670bf579d3edea8e04021c216678582 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_11 | semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$ | We can just look at a quarter circle inscribed in a $45-45-90$ right triangle. We can then extend a radius, $r$ to one of the sides creating an $r,r, r\sqrt{2}$ right triangle. This means that we have $r + r\sqrt{2} = 8\sqrt{2}$ so $r = \frac{8\sqrt{2}}{1+\sqrt{2}} = 16 - 8\sqrt{2}$ . Then the diameter is $32 - \sqrt{512}$ giving us $32 + 512 = \boxed{544}$ | null | 544 |
9670bf579d3edea8e04021c216678582 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_11 | semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$ | We proceed by finding the area of the square in 2 different ways. The square is obviously 8*8=64, but we can also find the area in terms of d. From the center of the circle, draw radii that hit the points where the square is tangent to the semicircle. Then the square's area is the area of the small square +2* the area of the trapezoids on the corners+ the area of an isoceles triangle. Adding these all up gives $64=\frac{d^2}{4}+\frac{d^2}{4}+(8-\frac{d\sqrt{2}}{2}+\frac{d}{2})(8-\frac{d}{2})$ Simplifying gives $d-16\sqrt{2}+d\sqrt{2}=0$ . Solving gives $d=32-16\sqrt{2}=32-\sqrt{512}$ so the answer is $32+512= \boxed{544}$ | null | 544 |
9670bf579d3edea8e04021c216678582 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_11 | semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$ | It is easy after getting the image, after drawing labeling the lengths of those segments, assume the radius is $x$ , we can see $x=\sqrt{2}(8-x)$ and we get $2x=32-\sqrt{512}$ and we have the answer $\boxed{544}$ ~bluesoul | null | 544 |
d2535f24ee8ab583985b8105b7fd341b | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_12 | For positive integers $n,$ let $\tau (n)$ denote the number of positive integer divisors of $n,$ including 1 and $n.$ For example, $\tau (1)=1$ and $\tau(6) =4.$ Define $S(n)$ by $S(n)=\tau(1)+ \tau(2) + \cdots + \tau(n).$ Let $a$ denote the number of positive integers $n \leq 2005$ with $S(n)$ odd , and let $b$ denote the number of positive integers $n \leq 2005$ with $S(n)$ even . Find $|a-b|.$ | Notice that the difference between consecutive squares are consecutively increasing odd numbers. Thus, there are $3$ numbers between $1$ (inclusive) and $4$ (exclusive), $5$ numbers between $4$ and $9$ , and so on. The number of numbers from $n^2$ to $(n + 1)^2$ is $(n + 1 - n)(n + 1 + n) = 2n + 1$ . Whenever the lowest square beneath a number is odd, the parity will be odd, and the same for even. Thus, $a = [2(1) + 1] + [2(3) + 1] \ldots [2(43) + 1] = 3 + 7 + 11 \ldots 87$ $b = [2(2) + 1] + [2(4) + 1] \ldots [2(42) + 1] + 70 = 5 + 9 \ldots 85 + 70$ , the $70$ accounting for the difference between $2005$ and $44^2 = 1936$ , inclusive. Notice that if we align the two and subtract, we get that each difference is equal to $2$ . Thus, the solution is $|a - b| = |b - a| = |2 \cdot 21 + 70 - 87| = \boxed{025}$ | null | 025 |
d2535f24ee8ab583985b8105b7fd341b | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_12 | For positive integers $n,$ let $\tau (n)$ denote the number of positive integer divisors of $n,$ including 1 and $n.$ For example, $\tau (1)=1$ and $\tau(6) =4.$ Define $S(n)$ by $S(n)=\tau(1)+ \tau(2) + \cdots + \tau(n).$ Let $a$ denote the number of positive integers $n \leq 2005$ with $S(n)$ odd , and let $b$ denote the number of positive integers $n \leq 2005$ with $S(n)$ even . Find $|a-b|.$ | Similarly, $b = (3^2 - 2^2) + (5^2 - 4^2) + \ldots + (45^2 - 44^2) - 19$ , where the $-19$ accounts for those numbers between $2005$ and $2024$
Thus $a = (2^2 - 1^2) + (4^2 - 3^2) + \ldots + (44^2 - 43^2)$
Then, $|a - b| = |2(2^2 + 4^2 + \ldots + 44^2) - 2(1^2 + 3^2 + 5^2 + \ldots 43^2) + 1^2 - 45^2 + 19|$ .
We can apply the formula $1^2 + 2^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$ . From this formula, it follows that $2^2 + 4^2 + \ldots + (2n)^2 = \frac{2n(n + 1)(2n + 1)}{3}$ and so that
$|a - b| = \left| 2\cdot \frac{44\cdot23\cdot45}{3} - 2\cdot \frac{22 \cdot 43 \cdot 45}{3} - 45^2 + 20\right| = |-25| =\boxed{025}$ | null | 025 |
d2535f24ee8ab583985b8105b7fd341b | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_12 | For positive integers $n,$ let $\tau (n)$ denote the number of positive integer divisors of $n,$ including 1 and $n.$ For example, $\tau (1)=1$ and $\tau(6) =4.$ Define $S(n)$ by $S(n)=\tau(1)+ \tau(2) + \cdots + \tau(n).$ Let $a$ denote the number of positive integers $n \leq 2005$ with $S(n)$ odd , and let $b$ denote the number of positive integers $n \leq 2005$ with $S(n)$ even . Find $|a-b|.$ | Let $\Delta n$ denote the sum $1+2+3+ \dots +n-1+n$ . We can easily see from the fact "It is well-known that $\tau(n)$ is odd if and only if $n$ is a perfect square.", that
$a = (2^2-1^2) + (4^2-3^2) \dots (44^2 - 43^2) = (2+1)(2-1)+(4+3)(4-3) \dots (44+43)(44-43) = 1+2+3...44 = \Delta 44$
$b = 3^2-2^2+5^2-4^2...2006-44^2 = (3+2)(3-2) \dots (43+42)(43-42) + 1 + (2005 - 44^2) = \Delta 43 + 69$
$a-b = \Delta 44-\Delta 43-69 = 44-69 = -25$ . They ask for $|a-b|$ , so our answer is $|-25| = \boxed{025}$ | null | 025 |
a2f17726f85ff19d52ab94b2ca2670f2 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_13 | A particle moves in the Cartesian plane according to the following rules:
How many different paths can the particle take from $(0,0)$ to $(5,5)$ | The length of the path (the number of times the particle moves) can range from $l = 5$ to $9$ ; notice that $d = 10-l$ gives the number of diagonals. Let $R$ represent a move to the right, $U$ represent a move upwards, and $D$ to be a move that is diagonal. Casework upon the number of diagonal moves:
Together, these add up to $2 + 18 + 42 + 20 + 1 = \boxed{083}$ | null | 083 |
a2f17726f85ff19d52ab94b2ca2670f2 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_13 | A particle moves in the Cartesian plane according to the following rules:
How many different paths can the particle take from $(0,0)$ to $(5,5)$ | Another possibility is to use block-walking and recursion : for each vertex, the number of ways to reach it is $a + b + c$ , where $a$ is the number of ways to reach the vertex from the left (without having come to that vertex (the one on the left) from below), $b$ is the number of ways to reach the vertex from the vertex diagonally down and left, and $c$ is the number of ways to reach the vertex from below (without having come to that vertex (the one below) from the left).
Assign to each point $(i,j)$ the triplet $(a_{i,j}, b_{i,j}, c_{i,j})$ . Let $s(i,j) = a_{i,j}+ b_{i,j}+ c_{i,j}$ . Let all lattice points that contain exactly one negative coordinate be assigned to $(0,0,0)$ . This leaves the lattice points of the first quadrant, the positive parts of the $x$ and $y$ axes, and the origin unassigned. As a seed, assign to $(0,1,0)$ . (We will see how this correlates with the problem.) Then define for each lattice point $(i,j)$ its triplet thus: \begin{align*} a_{i,j} &= s(i-1,j) - c_{i-1,j}\\ b_{i,j} &= s(i-1,j-1) \\ c_{i,j} &= s(i,j-1) - a_{i,j-1}. \end{align*} It is evident that $s(i,j)$ is the number of ways to reach $(i,j)$ from $(0,0)$ . Therefore we compute vertex by vertex the triplets $(a_{i,j}, b_{i,j}, c_{i,j})$ with $0 \leq i, j \leq 5$ [asy] defaultpen(fontsize(8)+0.8+heavyblue); size(250); for(int i = 0; i<6; ++i) { draw((0,i)--(5,i)^^(i,0)--(i,5), gray+0.25); } label("$(0,0,0)$", (0,0), N); for(int i = 1; i<6; ++i) { label("$(0,0,1)$", (0,i), N); label("$(1,0,0)$", (i,0), N); label("$({"+string(i-1)+"},1,0)$", (i,1), N); label("$(0,1,{"+string(i-1)+"})$", (1,i), N);} real[] val={1,2,4,7}; for(int i = 2; i<6; ++i) { label("$(1,{"+string(i-1)+"}, {"+string(val[i-2])+"})$", (2,i), N); label("$({"+string(val[i-2])+"},{"+string(i-1)+"}, 1)$", (i,2), N);} label("$(3,3,3)$", (3,3), N); label("$(9,9,9)$", (4,4), N); label("$(28,27,28)$", (5,5), N); label("$(4,5,6)$", (3,4), N); label("$(6,5,4)$", (4,3), N); label("$(5,8,11)$", (3,5), N); label("$(11,8,5)$", (5,3), N); label("$(13,15,18)$", (4,5), N); label("$(18,15,13)$", (5,4), N); [/asy] Finally, after simple but tedious calculations, we find that $(a_{5,5}, b_{5,5}, c_{5,5}) = (28,27,28)$ , so $s(i,j)=28+27+28 = \boxed{083}$ | null | 083 |
bc7ec851b0c2f1b99e1888615598c28a | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_14 | Consider the points $A(0,12), B(10,9), C(8,0),$ and $D(-4,7).$ There is a unique square $S$ such that each of the four points is on a different side of $S.$ Let $K$ be the area of $S.$ Find the remainder when $10K$ is divided by $1000$ | Let $(a,b)$ denote a normal vector of the side containing $A$ . Note that $\overline{AC}, \overline{BD}$ intersect and hence must be opposite vertices of the square. The lines containing the sides of the square have the form $ax+by=12b$ $ax+by=8a$ $bx-ay=10b-9a$ , and $bx-ay=-4b-7a$ . The lines form a square, so the distance between $C$ and the line through $A$ equals the distance between $D$ and the line through $B$ , hence $8a+0b-12b=-4b-7a-10b+9a$ , or $-3a=b$ . We can take $a=-1$ and $b=3$ . So the side of the square is $\frac{44}{\sqrt{10}}$ , the area is $K=\frac{1936}{10}$ , and the answer to the problem is $\boxed{936}$ | null | 936 |
8a4de751e6bfac6c47983adfc5744254 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_15 | Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$ | Let $E$ $F$ and $G$ be the points of tangency of the incircle with $BC$ $AC$ and $AB$ , respectively. Without loss of generality, let $AC < AB$ , so that $E$ is between $D$ and $C$ . Let the length of the median be $3m$ . Then by two applications of the Power of a Point Theorem $DE^2 = 2m \cdot m = AF^2$ , so $DE = AF$ . Now, $CE$ and $CF$ are two tangents to a circle from the same point, so by the Two Tangent Theorem $CE = CF = c$ and thus $AC = AF + CF = DE + CE = CD = 10$ . Then $DE = AF = AG = 10 - c$ so $BG = BE = BD + DE = 20 - c$ and thus $AB = AG + BG = 30 - 2c$
Now, by Stewart's Theorem in triangle $\triangle ABC$ with cevian $\overline{AD}$ , we have
\[(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10.\]
Our earlier result from Power of a Point was that $2m^2 = (10 - c)^2$ , so we combine these two results to solve for $c$ and we get
\[9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.\]
Thus $c = 2$ or $= 10$ . We discard the value $c = 10$ as extraneous (it gives us a line) and are left with $c = 2$ , so our triangle has area $\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}$ and so the answer is $24 + 14 = \boxed{038}$ | null | 038 |
8a4de751e6bfac6c47983adfc5744254 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_15 | Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$ | WLOG let E be be between C & D (as in solution 1). Assume $AD = 3m$ . We use power of a point to get that $AG = DE = \sqrt{2}m$ and $AB = AG + GB = AG + BE = 10+2\sqrt{2} m$
Since now we have $AC = 10$ $BC = 20, AB = 10+2\sqrt{2} m$ in triangle $\triangle ABC$ and cevian $AD = 3m$ . Now, we can apply Stewart's Theorem
\[2000 + 180 m^2 = 10(10+2\sqrt{2}m)^{2} + 1000\] \[1000 + 180 m^2 = 1000 + 400\sqrt{2}m + 80 m^{2}\] \[100 m^2 = 400\sqrt{2}m\]
$m = 4\sqrt{2}$ or $m = 0$ if $m = 0$ , we get a degenerate triangle, so $m = 4\sqrt{2}$ , and thus $AB = 26$ . You can now use Heron's Formula to finish. The answer is $24 \sqrt{14}$ , or $\boxed{038}$ | null | 038 |
8a4de751e6bfac6c47983adfc5744254 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_15 | Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$ | Let $E, F$ , and $G$ be the point of tangency (as stated in Solution 1). We can now let $AD$ be $3m$ . By using Power of a Point Theorem on A to the incircle, you get that $AG^2 = 2m^2$ . We can use it again on point D to the incircle to get the equation $(10 - CE)^2 = 2m^2$ . Setting the two equations equal to each other gives $(10 - CE)^2 = AG^2$ , and it can be further simplified to be $10 - CE = AG$
Let lengths $AC$ and $AB$ be called $b$ and $c$ , respectively. We can write $AG$ as $\frac{b + c - 20}{2}$ and $CE$ as $\frac{b + 20 - c}{2}$ . Plugging these into the equation, you get:
\[10 - \frac{b + 20 - c}{2} = \frac{b + c - 20}{2}\] \[b + c - 20 + b + 20 - c = 20 \rightarrow b = 10\]
Additionally, by Median of a triangle formula, you get that $3m = \frac{\sqrt{2c^2 - 200}}{2}$
Refer back to the fact that $AG^2 = 2m^2$ . We can now plug in our variables.
\[\left(\frac{b + c - 20}{2}\right)^2 = 2m^2 \rightarrow (c - 10)^2 = 8m^2\] \[c^2 - 20c + 100 = 8 \cdot \frac{2c^2 - 200}{36}\] \[9c^2 - 180c + 900 = 4c^2 - 400\] \[5c^2 - 180c + 1300 = 0\] \[c^2 - 36c + 260 = 0\]
Solving, you get that $c = 26$ or $10$ , but the latter will result in a degenerate triangle, so $c = 26$ .
Finally, you can use Heron's Formula to get that the area is $24\sqrt{14}$ , giving an answer of $\boxed{038}$ | null | 038 |
27a137abafd6d5c403c84ac80c10ec3c | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_2 | A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is $\frac mn,$ where $m$ and $n$ are relatively prime integers , find $m+n.$ | Use construction . We need only calculate the probability the first and second person all get a roll of each type, since then the rolls for the third person are determined.
Our answer is thus $\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70}$ , and $m + n = \boxed{79}$ | null | 79 |
27a137abafd6d5c403c84ac80c10ec3c | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_2 | A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is $\frac mn,$ where $m$ and $n$ are relatively prime integers , find $m+n.$ | Call the three different types of rolls as A, B, and C. We need to arrange 3As, 3Bs, and 3Cs in a string such that A, B, and C appear in the first three, second three, and the third three like ABCABCABC or BCABACCAB. This can occur in $\left(\frac{3!}{1!1!1!}\right)^3 = 6^3 = 216$ different manners. The total number of possible strings is $\frac{9!}{3!3!3!} = 1680$ . The solution is therefore $\frac{216}{1680} = \frac{9}{70}$ , and $m + n = \boxed{79}$ | null | 79 |
27a137abafd6d5c403c84ac80c10ec3c | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_2 | A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is $\frac mn,$ where $m$ and $n$ are relatively prime integers , find $m+n.$ | The denominator of m/n is equal to the total amount of possible roll configurations given to the three people. This is equal to ${9 \choose 3}{6 \choose3}$ as the amount of ways to select three rolls out of 9 to give to the first person is ${9 \choose 3}$ , and three rolls out of 6 is ${6 \choose3}$ . After that, the three remaining rolls have no more configurations.
The numerator is the amount of ways to give one roll of each type to each of the three people, which can be done by defining the three types of rolls as x flavored, y flavored, and z flavored.
xxx, yyy, zzz
So you have to choose one x, one y, and one z to give to the first person. There are 3 xs, 3 ys, and 3 zs to select from, giving $3^3$ combinations. Multiply that by the combinations of xs, ys, and zs for the second person, which is evidently $2^3$ since there are two of each letter left.
$(27*8)/{9 \choose 3}{6 \choose3}$ simplifies down to our fraction m/n, which is $9/70$ . Adding them up gives $9 + 70 = \boxed{79}$ | null | 79 |
4746f19f83b52450626bfc7abe33e834 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_3 | An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is $\frac mn$ where $m$ and $n$ are relatively prime integers . Find $m+n.$ | Let's call the first term of the original geometric series $a$ and the common ratio $r$ , so $2005 = a + ar + ar^2 + \ldots$ . Using the sum formula for infinite geometric series, we have $\;\;\frac a{1 -r} = 2005$ . Then we form a new series, $a^2 + a^2 r^2 + a^2 r^4 + \ldots$ . We know this series has sum $20050 = \frac{a^2}{1 - r^2}$ . Dividing this equation by $\frac{a}{1-r}$ , we get $10 = \frac a{1 + r}$ . Then $a = 2005 - 2005r$ and $a = 10 + 10r$ so $2005 - 2005r = 10 + 10r$ $1995 = 2015r$ and finally $r = \frac{1995}{2015} = \frac{399}{403}$ , so the answer is $399 + 403 = \boxed{802}$ | null | 802 |
4746f19f83b52450626bfc7abe33e834 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_3 | An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is $\frac mn$ where $m$ and $n$ are relatively prime integers . Find $m+n.$ | We can write the sum of the original series as $a + a\left(\dfrac{m}{n}\right) + a\left(\dfrac{m}{n}\right)^2 + \ldots = 2005$ , where the common ratio is equal to $\dfrac{m}{n}$ . We can also write the sum of the second series as $a^2 + a^2\left(\dfrac{m}{n}\right)^2 + a^2\left(\left(\dfrac{m}{n}\right)^2\right)^2 + \ldots = 20050$ . Using the formula for the sum of an infinite geometric series $S=\dfrac{a}{1-r}$ , where $S$ is the sum of the sequence, $a$ is the first term of the sequence, and $r$ is the ratio of the sequence, the sum of the original series can be written as $\dfrac{a}{1-\frac{m}{n}}=\dfrac{a}{\frac{n-m}{n}}=\dfrac{a \cdot n}{n-m}=2005\;\text{(1)}$ , and the second sequence can be written as $\dfrac{a^2}{1-\frac{m^2}{n^2}}=\dfrac{a^2}{\frac{n^2-m^2}{n^2}}=\dfrac{a^2\cdot n^2}{(n+m)(n-m)}=20050\;\text{(2)}$ . Dividing $\text{(2)}$ by $\text{(1)}$ , we obtain $\dfrac{a\cdot n}{m+n}=10$ , which can also be written as $a\cdot n=10(m+n)$ . Substitute this value for $a\cdot n$ back into $\text{(1)}$ , we obtain $10\cdot \dfrac{n+m}{n-m}=2005$ . Dividing both sides by 10 yields $\dfrac{n+m}{n-m}=\dfrac{401}{2}$ we can now write a system of equations with $n+m=401$ and $n-m=2$ , but this does not output integer solutions. However, we can also write $\dfrac{n+m}{n-m}=\dfrac{401}{2}$ as $\dfrac{n+m}{n-m}=\dfrac{802}{4}$ . This gives the system of equations $m+n=802$ and $n-m=4$ , which does have integer solutions. Our answer is therefore $m+n=\boxed{802}$ (Solving for $m$ and $n$ gives us $399$ and $403$ , respectively, which are co-prime). | null | 802 |
7e494e687a8a95e4fe19730dbd259855 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_4 | Find the number of positive integers that are divisors of at least one of $10^{10},15^7,18^{11}.$ | $10^{10} = 2^{10}\cdot 5^{10}$ so $10^{10}$ has $11\cdot11 = 121$ divisors
$15^7 = 3^7\cdot5^7$ so $15^7$ has $8\cdot8 = 64$ divisors.
$18^{11} = 2^{11}\cdot3^{22}$ so $18^{11}$ has $12\cdot23 = 276$ divisors.
Now, we use the Principle of Inclusion-Exclusion . We have $121 + 64 + 276$ total potential divisors so far, but we've overcounted those factors which divide two or more of our three numbers. Thus, we must subtract off the divisors of their pair-wise greatest common divisors
$\gcd(10^{10},15^7) = 5^7$ which has 8 divisors.
$\gcd(15^7, 18^{11}) = 3^7$ which has 8 divisors.
$\gcd(18^{11}, 10^{10}) = 2^{10}$ which has 11 divisors.
So now we have $121 + 64 + 276 - 8 -8 -11$ potential divisors. However, we've now undercounted those factors which divide all three of our numbers. Luckily, we see that the only such factor is 1, so we must add 1 to our previous sum to get an answer of $121 + 64 + 276 - 8 - 8 - 11 + 1 = \boxed{435}$ | null | 435 |
7e494e687a8a95e4fe19730dbd259855 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_4 | Find the number of positive integers that are divisors of at least one of $10^{10},15^7,18^{11}.$ | We can rewrite the three numbers as $10^{10} = 2^{10}\cdot 5^{10}$ $15^7 = 3^7\cdot5^7$ , and $18^{11} = 2^{11}\cdot3^{22}$ .
Assume that $n$ (a positive integer) is a divisor of one of the numbers. Therefore, $n$ can be expressed as ${p_1}^{e_1}$ or as ${p_2}^{e_2}{p_3}^{e_3}$ where $p_1$ $p_2$ are in $\{2,3,5\}$ and $e_1$ $e_2$ are positive integers.
If $n$ is the power of a single prime, then there are 11 possibilities ( $2^1$ to $2^{11}$ ) for $p_1=2$ , 22 possibilities ( $3^1$ to $3^{22}$ ) for $p_1=3$ , 10 possibilities ( $5^1$ to $5^{10}$ ) for $p_1=5$ , and 1 possibility if $n=1$ . From this case, there are $11+22+10+1=44$ possibilities.
If $n$ is the product of the powers of two primes, then we can just multiply the exponents of each rewritten product to get the number of possibilities, since each exponent of the product must be greater than 0. From this case, there are $10*10+11*22+7*7=100+242+49=391$ possibilities.
Adding up the two cases, there are $44+391=\boxed{435}$ positive integers. | null | 435 |
35ce17c70c2ab593e16600368d4d9bae | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_5 | Determine the number of ordered pairs $(a,b)$ of integers such that $\log_a b + 6\log_b a=5, 2 \leq a \leq 2005,$ and $2 \leq b \leq 2005.$ | The equation can be rewritten as $\frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a \log b}=5$ Multiplying through by $\log a \log b$ and factoring yields $(\log b - 3\log a)(\log b - 2\log a)=0$ . Therefore, $\log b=3\log a$ or $\log b=2\log a$ , so either $b=a^3$ or $b=a^2$
There are $44-2+1=43$ possibilities for the square case and $12-2+1=11$ possibilities for the cube case. Thus, the answer is $43+11= \boxed{054}$ | null | 054 |
35ce17c70c2ab593e16600368d4d9bae | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_5 | Determine the number of ordered pairs $(a,b)$ of integers such that $\log_a b + 6\log_b a=5, 2 \leq a \leq 2005,$ and $2 \leq b \leq 2005.$ | Let $k=\log_a b$ . Then our equation becomes $k+\frac{6}{k}=5$ . Multiplying through by $k$ and solving the quadratic gives us $k=2$ or $k=3$ . Hence $a^2=b$ or $a^3=b$
For the first case $a^2=b$ $a$ can range from 2 to 44, a total of 43 values.
For the second case $a^3=b$ $a$ can range from 2 to 12, a total of 11 values.
Thus the total number of possible values is $43+11=\boxed{54}$ | null | 54 |
35ce17c70c2ab593e16600368d4d9bae | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_5 | Determine the number of ordered pairs $(a,b)$ of integers such that $\log_a b + 6\log_b a=5, 2 \leq a \leq 2005,$ and $2 \leq b \leq 2005.$ | Using the change of base formula on the second equation to change to base $a$ , we get $\log_a(b) + \frac{6 \log_a(a)}{\log_a(b)}$ . If we substitute $x$ for $\log_a(b)$ , we get $x + \frac{6}{x}$ . Multiplying by $x$ on both sides and solving, we get $x=3,2$ . Substituting back in, we get $\log_a(b) = 3,2$ . That means $a^3 = b$ or $a^2 = b$ . Since $b \leq 2005$ , we can see that for the cubed case, the maximum $a$ can be without exceeding 2005 is 12(because $13^3 = 2197$ ) and for the squared case it can be a maximum of 44. Since $a \neq 1$ , the number of values is $(44-1)+(12-1) = \boxed{54}$ | null | 54 |
a5c529bdcc51c7acdde6bc2f34fecd4a | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_6 | The cards in a stack of $2n$ cards are numbered consecutively from 1 through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A.$ The remaining cards form pile $B.$ The cards are then restacked by taking cards alternately from the tops of pile $B$ and $A,$ respectively. In this process, card number $(n+1)$ becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles $A$ and $B$ are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named magical. For example, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in the magical stack in which card number 131 retains its original position. | Since a card from B is placed on the bottom of the new stack, notice that cards from pile B will be marked as an even number in the new pile, while cards from pile A will be marked as odd in the new pile. Since 131 is odd and retains its original position in the stack, it must be in pile A. Also to retain its original position, exactly $131 - 1 = 130$ numbers must be in front of it. There are $\frac{130}{2} = 65$ cards from each of piles A, B in front of card 131. This suggests that $n = 131 + 65 = 196$ ; the total number of cards is $196 \cdot 2 = \boxed{392}$ | null | 392 |
a5c529bdcc51c7acdde6bc2f34fecd4a | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_6 | The cards in a stack of $2n$ cards are numbered consecutively from 1 through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A.$ The remaining cards form pile $B.$ The cards are then restacked by taking cards alternately from the tops of pile $B$ and $A,$ respectively. In this process, card number $(n+1)$ becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles $A$ and $B$ are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named magical. For example, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in the magical stack in which card number 131 retains its original position. | If you index the final stack $1,2,\dots,2n$ , you notice that pile A resides only in the odd indices and has maintained its original order aside from flipping over. The same has happened to pile B except replace odd with even. Thus, if 131 is still at index 131, an odd number, then 131 must be from pile A. The numbers in pile A are the consecutive integers $1,2,\dots, n$ . This all leads us to the following equation. \[131=2n-2(131)+1\implies2n=\boxed{392}\] | null | 392 |
08b15eab85b8057b1408da3f437b06fb | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_7 | Let $x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}.$ Find $(x+1)^{48}$ | We note that in general,
It now becomes apparent that if we multiply the numerator and denominator of $\frac{4}{ (\sqrt{5}+1) (\sqrt[4]{5}+1) (\sqrt[8]{5}+1) (\sqrt[16]{5}+1) }$ by $(\sqrt[16]{5} - 1)$ , the denominator will telescope to $\sqrt[1]{5} - 1 = 4$ , so
It follows that $(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = \boxed{125}$ | null | 125 |
08b15eab85b8057b1408da3f437b06fb | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_7 | Let $x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}.$ Find $(x+1)^{48}$ | Like Solution $2$ , let $z=\sqrt[16]{5}$ Then, the expression becomes
$x=\frac{4}{(z+1)(z^2+1)(z^4+1)(z^8+1)}$ Now, multiplying by the conjugate of each binomial in the denominator, we obtain...
$x=\frac{4(z-1)(z^2-1)(z^4-1)(z^8-1)}{(z^2-1)(z^4-1)(z^8-1)(z^{16}-1)}\implies x=\frac{4(z-1)}{z^{16}-1}$ Plugging back in,
$x=\frac{4({\sqrt[16]{5}-1)}}{5-1}\implies x=\sqrt[16]{5}-1$
Hence, after some basic exponent rules, we find the answer is $\boxed{125}$ | null | 125 |
cd77e9ab14615039bbe58bef14c09de1 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_8 | Circles $C_1$ and $C_2$ are externally tangent , and they are both internally tangent to circle $C_3.$ The radii of $C_1$ and $C_2$ are 4 and 10, respectively, and the centers of the three circles are all collinear . A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2.$ Given that the length of the chord is $\frac{m\sqrt{n}}p$ where $m,n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime , and $n$ is not divisible by the square of any prime , find $m+n+p.$ | Let $O_1, O_2, O_3$ be the centers and $r_1 = 4, r_2 = 10,r_3 = 14$ the radii of the circles $C_1, C_2, C_3$ . Let $T_1, T_2$ be the points of tangency from the common external tangent of $C_1, C_2$ , respectively, and let the extension of $\overline{T_1T_2}$ intersect the extension of $\overline{O_1O_2}$ at a point $H$ . Let the endpoints of the chord/tangent be $A,B$ , and the foot of the perpendicular from $O_3$ to $\overline{AB}$ be $T$ . From the similar right triangles $\triangle HO_1T_1 \sim \triangle HO_2T_2 \sim \triangle HO_3T$
\[\frac{HO_1}{4} = \frac{HO_1+14}{10} = \frac{HO_1+10}{O_3T}.\]
It follows that $HO_1 = \frac{28}{3}$ , and that $O_3T = \frac{58}{7}\dagger$ . By the Pythagorean Theorem on $\triangle ATO_3$ , we find that
\[AB = 2AT = 2\left(\sqrt{r_3^2 - O_3T^2}\right) = 2\sqrt{14^2 - \frac{58^2}{7^2}} = \frac{8\sqrt{390}}{7}\]
and the answer is $m+n+p=\boxed{405}$ | null | 405 |
cd77e9ab14615039bbe58bef14c09de1 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_8 | Circles $C_1$ and $C_2$ are externally tangent , and they are both internally tangent to circle $C_3.$ The radii of $C_1$ and $C_2$ are 4 and 10, respectively, and the centers of the three circles are all collinear . A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2.$ Given that the length of the chord is $\frac{m\sqrt{n}}p$ where $m,n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime , and $n$ is not divisible by the square of any prime , find $m+n+p.$ | Call our desired length $x$ . Note for any $X$ on $\overline{AB}$ and $Y$ on $\overline{O_1O_2}$ such that $\overline{XY}\perp\overline{AB}$ that the function $f$ such that $f(\overline{O_1Y})=\overline{XY}$ is linear. Since $(0,4)$ and $(14,10)$ , we can quickly interpolate that $f(10)=\overline{O_3T}=\frac{58}{7}$ . Then, extend $\overline{O_3T}$ until it reaches the circle on both sides; call them $P,Q$ . By Power of a Point, $\overline{PT}\cdot\overline{TQ}=\overline{AT}\cdot\overline{TB}$ .
Since $\overline{AT}=\overline{TB}=\frac{1}{2}x$ \[(\overline{PO_3}-\overline{O_3T})(\overline{QO_3}+\overline{O_3T})=\frac{1}{4}x^2\] \[\left(14+\frac{58}{7}\right)\left(14-\frac{58}{7}\right)=\frac{1}{4}x^2\] After solving for $x$ , we get $x=\frac{8\sqrt{390}}{7}$ , so our answer is $8+390+7=\boxed{405}$ | null | 405 |
3cb50186fb1fc4e62f51a1a3bd8d7828 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | We know by De Moivre's Theorem that $(\cos t + i \sin t)^n = \cos nt + i \sin nt$ for all real numbers $t$ and all integers $n$ . So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem.
Recall the trigonometric identities $\cos \left(\frac{\pi}2 - u\right) = \sin u$ and $\sin \left(\frac{\pi}2 - u\right) = \cos u$ hold for all real $u$ . If our original equation holds for all $t$ , it must certainly hold for $t = \frac{\pi}2 - u$ . Thus, the question is equivalent to asking for how many positive integers $n \leq 1000$ we have that $\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \sin n \left(\frac\pi2 -u \right) + i\cos n \left(\frac\pi2 - u\right)$ holds for all real $u$
$\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \left(\cos u + i \sin u\right)^n = \cos nu + i\sin nu$ . We know that two complex numbers are equal if and only if both their real part and imaginary part are equal. Thus, we need to find all $n$ such that $\cos n u = \sin n\left(\frac\pi2 - u\right)$ and $\sin nu = \cos n\left(\frac\pi2 - u\right)$ hold for all real $u$
$\sin x = \cos y$ if and only if either $x + y = \frac \pi 2 + 2\pi \cdot k$ or $x - y = \frac\pi2 + 2\pi\cdot k$ for some integer $k$ . So from the equality of the real parts we need either $nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k$ , in which case $n = 1 + 4k$ , or we need $-nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k$ , in which case $n$ will depend on $u$ and so the equation will not hold for all real values of $u$ . Checking $n = 1 + 4k$ in the equation for the imaginary parts, we see that it works there as well, so exactly those values of $n$ congruent to $1 \pmod 4$ work. There are $\boxed{250}$ of them in the given range. | null | 250 |
3cb50186fb1fc4e62f51a1a3bd8d7828 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | This problem begs us to use the familiar identity $e^{it} = \cos(t) + i \sin(t)$ . Notice, $\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}$ since $\sin(-t) = -\sin(t)$ . Using this, $(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)$ is recast as $(i e^{-it})^n = i e^{-itn}$ . Hence we must have $i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}$ . Thus since $1000$ is a multiple of $4$ exactly one quarter of the residues are congruent to $1$ hence we have $\boxed{250}$ | null | 250 |
3cb50186fb1fc4e62f51a1a3bd8d7828 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | We can rewrite $\sin(t)$ as $\cos\left(\frac{\pi}{2}-t\right)$ and $\cos(t)$ as $\sin\left(\frac{\pi}{2}-t\right)$ . This means that $\sin t + i\cos t = e^{i\left(\frac{\pi}{2}-t\right)}=\frac{e^{\frac{\pi i}{2}}}{e^{it}}$ . This theorem also tells us that $e^{\frac{\pi i}{2}}=i$ , so $\sin t + i\cos t = \frac{i}{e^{it}}$ . By the same line of reasoning, we have $\sin nt + i\cos nt = \frac{i}{e^{int}}$
For the statement in the question to be true, we must have $\left(\frac{i}{e^{it}}\right)^n=\frac{i}{e^{int}}$ . The left hand side simplifies to $\frac{i^n}{e^{int}}$ . We cancel the denominators and find that the only thing that needs to be true is that $i^n=i$ . This is true if $n\equiv1\pmod{4}$ , and there are $\boxed{250}$ such numbers between $1$ and $1000$ . Solution by Zeroman | null | 250 |
3cb50186fb1fc4e62f51a1a3bd8d7828 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | We are using degrees in this solution instead of radians. I just process stuff better that way.
We can see that the LHS is $cis(n(90^{\circ}-t))$ , and the RHS is $cis(90^{\circ}-nt)$ So, $n(90-t) \equiv 90-nt \mod 360$ Expanding and canceling the nt terms, we will get $90n \equiv 90 \mod 360$ . Canceling gets $n \equiv 1 \mod 4$ , and thus there are $\boxed{250}$ values of n. | null | 250 |
3cb50186fb1fc4e62f51a1a3bd8d7828 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | Let $t=0$ . Then, we have $i^n=i$ which means $n\equiv 1\pmod{4}$ . Thus, the answer is $\boxed{250}$ | null | 250 |
3cb50186fb1fc4e62f51a1a3bd8d7828 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | We factor out $i^n$ from $(\sin t + i \cos t)^n = i^n (\cos(t) - i \sin t)= i^n(\cos(nt) - i\sin nt).$ We know the final expression must be the same as $\sin nt + i \cos nt$ so we must have $i^n(\cos(nt) - i\sin nt) = \sin nt + i \cos nt$ in which testing yields $n \equiv 1 \pmod{4}$ is the only mod that works, so we have a total of $1000 \cdot\frac{1}{4} = \boxed{250}$ integers $n$ that work. | null | 250 |
3cb50186fb1fc4e62f51a1a3bd8d7828 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | Note that this looks like de Moivre's except switched around. Using de Moivre's as motivation we try to convert the given expression into de Moivre's. Note that $\sin t = \cos(90 - t)$ and $\cos t = \sin(90 - t)$ . So we rewrite the expression and setting it equal to the given expression in the problem, we get $\cos(90 - nt) + i\sin(90 - nt) = \cos(90n - nt) + i\sin(90n - nt)$ . Now we can just look at the imaginary parts. Doing so and simplifying, we see that $1 + 4k= n$ . From this we see that $n \equiv 1\pmod{4}$ . So there are $\boxed{250}$ solutions. | null | 250 |
3cb50186fb1fc4e62f51a1a3bd8d7828 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_9 | For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$ | $(\sin\theta + i\cos\theta)^{n} = i^{n} (\cos\theta - i\sin\theta)^n$
Hence the required condition is just $i^{n} = i$ which is true for exactly 1 in 4 consecutive numbers. Thus $\boxed{250}$ | null | 250 |
96a45eb0235aa4b66f2450140033d0b6 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_10 | Given that $O$ is a regular octahedron , that $C$ is the cube whose vertices are the centers of the faces of $O,$ and that the ratio of the volume of $O$ to that of $C$ is $\frac mn,$ where $m$ and $n$ are relatively prime integers, find $m+n.$ | Let the side of the octahedron be of length $s$ . Let the vertices of the octahedron be $A, B, C, D, E, F$ so that $A$ and $F$ are opposite each other and $AF = s\sqrt2$ . The height of the square pyramid $ABCDE$ is $\frac{AF}2 = \frac s{\sqrt2}$ and so it has volume $\frac 13 s^2 \cdot \frac s{\sqrt2} = \frac {s^3}{3\sqrt2}$ and the whole octahedron has volume $\frac {s^3\sqrt2}3$
Let $M$ be the midpoint of $BC$ $N$ be the midpoint of $DE$ $G$ be the centroid of $\triangle ABC$ and $H$ be the centroid of $\triangle ADE$ . Then $\triangle AMN \sim \triangle AGH$ and the symmetry ratio is $\frac 23$ (because the medians of a triangle are trisected by the centroid), so $GH = \frac{2}{3}MN = \frac{2s}3$ $GH$ is also a diagonal of the cube, so the cube has side-length $\frac{s\sqrt2}3$ and volume $\frac{2s^3\sqrt2}{27}$ . The ratio of the volumes is then $\frac{\left(\frac{2s^3\sqrt2}{27}\right)}{\left(\frac{s^3\sqrt2}{3}\right)} = \frac29$ and so the answer is $\boxed{011}$ | null | 011 |
96a45eb0235aa4b66f2450140033d0b6 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_10 | Given that $O$ is a regular octahedron , that $C$ is the cube whose vertices are the centers of the faces of $O,$ and that the ratio of the volume of $O$ to that of $C$ is $\frac mn,$ where $m$ and $n$ are relatively prime integers, find $m+n.$ | Let the octahedron have vertices $(\pm 3, 0, 0), (0, \pm 3, 0), (0, 0, \pm 3)$ . Then the vertices of the cube lie at the centroids of the faces, which have coordinates $(\pm 1, \pm 1, \pm 1)$ . The cube has volume 8. The region of the octahedron lying in each octant is a tetrahedron with three edges mutually perpendicular and of length 3. Thus the octahedron has volume $8 \cdot \left(\frac 16 \cdot3^3\right) = 36$ , so the ratio is $\frac 8{36} = \frac 29$ and so the answer is $\boxed{011}$ | null | 011 |
fb0aa8696cef3e1e18cd065da0d7acc2 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_11 | Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals such that $a_0 = 37, a_1 = 72, a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac 3{a_k}$ for $k = 1,2,\ldots, m-1.$ Find $m.$ | For $0 < k < m$ , we have
Thus the product $a_{k}a_{k+1}$ is a monovariant : it decreases by 3 each time $k$ increases by 1. For $k = 0$ we have $a_{k}a_{k+1} = 37\cdot 72$ , so when $k = \frac{37 \cdot 72}{3} = 888$ $a_{k}a_{k+1}$ will be zero for the first time, which implies that $m = \boxed{889}$ , our answer. | null | 889 |
fb0aa8696cef3e1e18cd065da0d7acc2 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_11 | Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals such that $a_0 = 37, a_1 = 72, a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac 3{a_k}$ for $k = 1,2,\ldots, m-1.$ Find $m.$ | Plugging in $k = m-1$ to the given relation, we get $0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}$ . Inspecting the value of $a_{k}a_{k+1}$ for small values of $k$ , we see that $a_{k}a_{k+1} = 37\cdot 72 - 3k$ . Setting the RHS of this equation equal to $3$ , we find that $m$ must be $\boxed{889}$ | null | 889 |
fb0aa8696cef3e1e18cd065da0d7acc2 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_11 | Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals such that $a_0 = 37, a_1 = 72, a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac 3{a_k}$ for $k = 1,2,\ldots, m-1.$ Find $m.$ | Note that $a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3$ . Then, we can generate a sum of series of equations such that $\sum_{k=1}^{m-1} a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3(m-1)$ . Then, note that all but the first and last terms on the LHS cancel out, leaving us with $a_m\cdot a_{m-1} - a_1\cdot a_0 = -3(m-1)$ . Plugging in $a_m=0$ $a_0=37$ $a_1=72$ , we have $-37\cdot 72 = -3(m-1)$ . Solving for $m$ gives $m=\boxed{889}$ .
~sigma | null | 889 |
f90fccd1844e5708e797b3d0a8c6cdf7 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$ | Let $G$ be the foot of the perpendicular from $O$ to $AB$ . Denote $x = EG$ and $y = FG$ , and $x > y$ (since $AE < BF$ and $AG = BG$ ). Then $\tan \angle EOG = \frac{x}{450}$ , and $\tan \angle FOG = \frac{y}{450}$
By the tangent addition rule $\left( \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \right)$ , we see that \[\tan 45 = \tan (EOG + FOG) = \frac{\frac{x}{450} + \frac{y}{450}}{1 - \frac{x}{450} \cdot \frac{y}{450}}.\] Since $\tan 45 = 1$ , this simplifies to $1 - \frac{xy}{450^2} = \frac{x + y}{450}$ . We know that $x + y = 400$ , so we can substitute this to find that $1 - \frac{xy}{450^2} = \frac 89 \Longrightarrow xy = 150^2$
Substituting $x = 400 - y$ again, we know have $xy = (400 - y)y = 150^2$ . This is a quadratic with roots $200 \pm 50\sqrt{7}$ . Since $y < x$ , use the smaller root, $200 - 50\sqrt{7}$
Now, $BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}$ . The answer is $250 + 50 + 7 = \boxed{307}$ | null | 307 |
f90fccd1844e5708e797b3d0a8c6cdf7 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$ | [asy] size(3inch); pair A, B, C, D, M, O, X, Y; A = (0,900); B = (900,900); C = (900,0); D = (0,0); M = (450,900); O = (450,450); X = (250 - 50*sqrt(7),900); Y = (650 - 50*sqrt(7),900); draw(A--B--C--D--cycle); draw(X--O--Y); draw(M--O--A); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",X,N); label("$F$",Y,NNE); label("$O$",O,S); label("$M$",M,N); [/asy] Let the midpoint of $\overline{AB}$ be $M$ and let $FB = x$ , so then $MF = 450 - x$ and $AF = 900 - x$ . Drawing $\overline{AO}$ , we have $\triangle OEF\sim\triangle AOF$ , so \[\frac{OF}{EF} = \frac{AF}{OF}\Rightarrow (OF)^2 = 400(900 - x).\] By the Pythagorean Theorem on $\triangle OMF$ \[(OF)^2 = 450^2 + (450 - x)^2.\] Setting these two expressions for $(OF)^2$ equal and solving for $x$ (it is helpful to scale the problem down by a factor of 50 first), we get $x = 250\pm 50\sqrt{7}$ . Since $BF > AE$ , we want the value $x = 250 + 50\sqrt{7}$ , and the answer is $250 + 50 + 7 = \boxed{307}$ | null | 307 |
f90fccd1844e5708e797b3d0a8c6cdf7 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$ | Let lower case letters be the complex numbers correspond to their respective upper case points in the complex plane, with $o = 0, a = -450 + 450i, b = 450 + 450i$ , and $f = x + 450i$ . Since $EF$ = 400, $e = (x-400) + 450i$ . From $\angle{EOF} = 45^{\circ}$ , we can deduce that the rotation of point $F$ 45 degrees counterclockwise, $E$ , and the origin are collinear. In other words, \[\dfrac{e^{i \frac{\pi}{4}} \cdot (x + 450i)}{(x - 400) + 450i}\] is a real number. Simplyfying using the fact that $e^{i \frac{\pi}{4}} = \dfrac{\sqrt{2}}{2} + i \dfrac{\sqrt{2}}{2}$ , clearing the denominator, and setting the imaginary part equal to $0$ , we eventually get the quadratic \[x^2 - 400x + 22500 = 0\] which has solutions $x = 200 \pm 50\sqrt{7}$ . It is given that $AE < BF$ , so $x = 200 - 50\sqrt{7}$ and \[BF = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7} \Rightarrow \boxed{307}.\] | null | 307 |
f90fccd1844e5708e797b3d0a8c6cdf7 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$ | [asy] size(250); pair A,B,C,D,O,E,F,G,H,K; A = (0,0); B = (900,0); C = (900,900); D = (0,900); O = (450,450); E = (600,0); F = (150,0); G = (-600,0); H = (450,0); K = (0,270); draw(A--B--C--D--cycle); draw(O--E); draw(O--F); draw(O--G); draw(A--G); draw(O--H); label("O",O,N); label("A",A,S); label("B",B,SE); label("C",C,NE); label("D",D,NW); label("E",E,SE); label("F",F,S); label("H",H,SW); label("G",G,SW); label("x",H--E,S); label("K",K,NW); [/asy] Let G be a point such that it lies on AB, and GOE is 90 degrees. Let H be foot of the altitude from O to AB.
Since $\triangle GOE \sim \triangle OHE$ $\frac{GO}{OE} = \frac{450}{x}$ , and by Angle Bisector Theorem $\frac{GF}{FE} = \frac{450}{x}$ . Thus, $GF = \frac{450 \cdot 400}{x}$ $AF = AH-FH = 50+x$ , and $KA = EB$ (90 degree rotation), and now we can bash on 2 similar triangles $\triangle GAK \sim \triangle GHO$
\[\frac{GA}{AK} = \frac{GH}{OH}\] \[\frac{\frac{450 \cdot 400}{x}-50-x}{450-x} = \frac{\frac{450 \cdot 400}{x}+400-x}{450}\] I hope you like expanding \[x^2 - 850x + \frac{81000000}{x} = -450x - 22500 + \frac{81000000}{x}\] \[x^2 - 400x + 22500 = 0\] Quadratic formula gives us \[x = 200 \pm 50 \sqrt{7}\] Since AE < BF \[x = 200 - 50 \sqrt{7}\] Thus, \[BF = 250 + 50 \sqrt{7}\] So, our answer is $\boxed{307}$ | null | 307 |
f90fccd1844e5708e797b3d0a8c6cdf7 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$ | We know that G is on the perpendicular bisector of $EF$ , which means that $EJ=JF=200$ $EG=GF=200\sqrt{2}$ and $GH=250$ . Now, let $HO$ be equal to $x$ . We can set up an equation with the Pythagorean Theorem:
\begin{align*} \sqrt{x^2+250^2}&=(200\sqrt{2})^2 \\ x^2+62500&=80000 \\ x^2&=17500 \\ x&=50\sqrt{7} \end{align*}
Now, since $IO=450$
\begin{align*} HI&=450-x \\ &=450-50\sqrt{7} \\ \end{align*} \\
Since $HI=AJ$ , we now have:
\begin{align*} BF&=AB-AJ-JF \\ &=900-(450-50\sqrt{7})-200 \\ &=250+50\sqrt{7} \\ \end{align*}
This means that our answer would be $250+50+7=\boxed{307}$ | null | 307 |
f90fccd1844e5708e797b3d0a8c6cdf7 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$ | Construct $BO, AO.$ Let $\angle{FOB} = \alpha.$ Also let $FB = x$ then $AE = 500-x.$ We then have from simple angle-chasing: \begin{align*} \angle{BFO} = 135 - \alpha \\ \angle{OFE} = 45 + \alpha \\ \angle{EOA} = 45 - \alpha \\ \angle{AEO} = 90 + \alpha \\ \angle{OEF} = 90 - \alpha. \end{align*} From AA similarity we have \[\triangle{EOB} \sim \triangle{EFO}.\] This gives the ratios, \[\dfrac{400 + x}{EO} = \dfrac{450\sqrt{2}}{FO}.\] Similarly from AA similarity \[\triangle{FOA} \sim \triangle{FEO}.\] So we get the ratios \[\dfrac{EO}{450\sqrt{2}} = \dfrac{FO}{900-x}.\] We can multiply to get \[\dfrac{400 + x}{450\sqrt{2}} = \dfrac{450\sqrt{2}}{900 - x}.\] Cross-multiplying reveals \[360000 + 500x - x^2 = 405000.\] Bringing everything to one side we have \[x^2 - 500x + 45000 = 0.\] By the quadratic formula we get \[x = \dfrac{500 + \sqrt{500^2 - 4\cdot45000}}{2} = \dfrac{500 + \sqrt{70000}}{2} = \dfrac{500 + 100\sqrt{7}}{2} = 250 + 50\sqrt{7}.\] Therefore \[p+q+r = 250 + 50 + 7 = \boxed{307}.\] ~aa1024 | null | 307 |
f90fccd1844e5708e797b3d0a8c6cdf7 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_12 | Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime , find $p+q+r.$ | Draw AO, OB, and extend OB to D. Let $\angle{FOB} = \alpha.$ Then, after angle chasing, we find that \[\angle{AEB} = 90 + \alpha\] .
Using this, we draw a line perpendicular to $AB$ at $E$ to meet $BD$ at $M$ . Since $\angle{MEO} = \alpha$ and $\angle{EMO} = 45$ , we have that \[\triangle{EMO} \sim \triangle{OBF}\] Let $FB = x$ . Then $EM = 400+x$ . Since $FB/BO = \frac{x}{450\sqrt{2}}$ , and $MO/EM = FB/OB$ , we have \[MO = \frac{(400+x)x}{450\sqrt{2}}\] Since $\triangle{EBM}$ is a $45-45-90$ triangle, \[(400+x)\sqrt{2} = 450 \sqrt{2} + \frac{(400+x)x}{450\sqrt{2}}\] Solving for $x$ , we get that $x=250 +- 50s\sqrt{7}$ , but since $FB>AE$ $FB = 250+50\sqrt{7}$ , thus \[p+q+r=\boxed{307}\] -dchang0524 | null | 307 |
009bb6054e35e55f6b0817e1624f6f18 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_13 | Let $P(x)$ be a polynomial with integer coefficients that satisfies $P(17)=10$ and $P(24)=17.$ Given that $P(n)=n+3$ has two distinct integer solutions $n_1$ and $n_2,$ find the product $n_1\cdot n_2.$ | We define $Q(x)=P(x)-x+7$ , noting that it has roots at $17$ and $24$ . Hence $P(x)-x+7=A(x-17)(x-24)$ . In particular, this means that $P(x)-x-3=A(x-17)(x-24)-10$ . Therefore, $x=n_1,n_2$ satisfy $A(x-17)(x-24)=10$ , where $A$ $(x-17)$ , and $(x-24)$ are integers. This cannot occur if $x\le 17$ or $x\ge 24$ because the product $(x-17)(x-24)$ will either be too large or not be a divisor of $10$ . We find that $x=19$ and $x=22$ are the only values that allow $(x-17)(x-24)$ to be a factor of $10$ . Hence the answer is $19\cdot 22=\boxed{418}$ | null | 418 |
009bb6054e35e55f6b0817e1624f6f18 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_13 | Let $P(x)$ be a polynomial with integer coefficients that satisfies $P(17)=10$ and $P(24)=17.$ Given that $P(n)=n+3$ has two distinct integer solutions $n_1$ and $n_2,$ find the product $n_1\cdot n_2.$ | We know that $P(n)-(n+3)=0$ so $P(n)$ has two distinct solutions so $P(x)$ is at least quadratic. Let us first try this problem out as if $P(x)$ is a quadratic polynomial. Thus $P(n)-(n+3)= an^2+(b-1)n+(c-3)=0$ because $P(n)=an^2+bn+c$ where $a,b,c$ are all integers. Thus $P(x)=ax^2+bx+c$ where $a,b,c$ are all integers. We know that $P(17)$ or $289a+17b+c=10$ and $P(24)$ or $576a+24b+c=17$ . By doing $P(24)-P(17)$ we obtain that $287a+7b=7$ or $41a+b=1$ or $-41a=b-1$ . Thus $P(n)=an^2- (41a)n+(c-3)=0$ . Now we know that $b=-41a+1$ , we have $289a+17(-41a+1)+c=10$ or $408a=7+c$ which makes $408a-10=c-3$ . Thus $P(n)=an^2-(41a)n+(408a-10)=0$ . By Vieta's formulas, we know that the sum of the roots( $n$ ) is equal to 41 and the product of the roots( $n$ ) is equal to $408-\frac{10}{a}$ . Because the roots are integers $\frac{10}{a}$ has to be an integer, so $a=1,2,5,10,-1,-2,-5,-10$ . Thus the product of the roots is equal to one of the following: $398,403,406,407,409,410,413,418$ . Testing every potential product of the roots, we find out that the only product that can have divisors that sum up to $41$ is $\boxed{418}$ | null | 418 |
bc30162e80d10e3de36f768dff414606 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | By the Law of Sines and since $\angle BAE = \angle CAD, \angle BAD = \angle CAE$ , we have
\begin{align*} \frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} \\ &= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} \\ &= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2} \end{align*}
Substituting our knowns, we have $\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}$ . The answer is $q = \boxed{463}$ | null | 463 |
bc30162e80d10e3de36f768dff414606 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively.
From here, we can use Heron's Formula to find the altitude. The area of the triangle is $\sqrt{21*6*7*8} = 84$ . We can then use similar triangles with triangle $AQC$ and triangle $DSC$ to find $DS=\frac{24}{5}$ . Consequently, from Pythagorean theorem, $SC = \frac{18}{5}$ and $AS = 14-SC = \frac{52}{5}$ . We can also use the Pythagorean theorem on triangle $AQB$ to determine that $BQ = \frac{33}{5}$
Label $AR$ as $y$ and $RE$ as $x$ $RB$ then equals $13-y$ . Then, we have two similar triangles.
Firstly: $\triangle ARE \sim \triangle ASD$ . From there, we have $\frac{x}{y} = \frac{\frac{24}{5}}{\frac{53}{5}}$
Next: $\triangle BRE \sim \triangle BQA$ . From there, we have $\frac{x}{13-y} = \frac{\frac{56}{5}}{\frac{33}{5}}$
Solve the system to get $x = \frac{2184}{463}$ and $y = \frac{4732}{463}$ . Notice that 463 is prime, so even though we use the Pythagorean theorem on $x$ and $13-y$ , the denominator won't change. The answer we desire is $\boxed{463}$ | null | 463 |
bc30162e80d10e3de36f768dff414606 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | Let $\angle CAD = \angle BAE = \theta$ . Note by Law of Sines on $\triangle BEA$ we have \[\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{\angle BEA}}\] As a result, our goal is to find $\sin{\angle BEA}$ and $\sin{\theta}$ (we already know $AB$ ).
Let the foot of the altitude from $A$ to $BC$ be $H$ . By law of cosines on $\triangle ABC$ we have \[169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5}\] It follows that $AH = \frac{56}{5}$ and $HC = \frac{42}{5} \Rightarrow HD = \frac{12}{5}$
Note that by PT on $\triangle AHD$ we have that $AD^2 = \left(\frac{56}{5}\right)^2 + \left(\frac{12}{5}\right)^2 = \frac{656}{5}$ . By Law of Sines on $\triangle ADC$ (where we square everything to avoid taking the square root) we see \[\frac{36}{\sin^2{\theta}} = \frac{656}{5 \cdot \frac{16}{25}} \Rightarrow \sin^2{\theta} = \frac{36}{205}.\] How are we going to find $\sin{\angle BEA}$ though? $\angle BEA$ and $\theta$ are in the same triangle. Applying Law of Sines on $\triangle ABC$ we see that \[\frac{13}{\frac{4}{5}} = \frac{14}{\sin{\angle B}} \Rightarrow \sin{\angle B} = \frac{56}{65} \Rightarrow \cos{\angle B} = \frac{33}{65}.\] $\theta$ $\angle B$ , and $\angle BEA$ are all in the same triangle. We know they add up to $180^{\circ}$ . There's a good chance we can exploit this using the identity $\sin{p} = \sin{180^{\circ}-p}$
We have that $\sin{(180^{\circ} - (\theta + \angle B))} = \sin{\angle BEA} = \sin{(\theta + \angle B)}$ . Success! We know $\sin{\theta}$ and $\sin{\angle B}$ already. Applying the $\sin$ addition formula we see \[\sin{\theta + \angle B} = \sin{\theta} \cos{\angle B} + \sin{\angle B} \cos{\theta} = \frac{6}{\sqrt{205}} \cdot \frac{33}{65} + \frac{56}{65} \cdot \frac{13}{\sqrt{205}}=\frac{1}{65 \cdot \sqrt{205}} (198 + 728) = \frac{926}{65 \sqrt{205}}.\] This is the last stretch! Applying Law of Sines a final time on $\triangle BEA$ we see \[\frac{BE}{\sin{\theta}} = \frac{13}{\sin{BEA}} \Rightarrow \frac{BE}{\frac{6}{\sqrt{205}}} = \frac{13}{\frac{926}{65\sqrt{205}}} \Rightarrow \frac{BE}{6} = \frac{13 \cdot 65}{926} \Rightarrow \frac{13 \cdot 65 \cdot 6}{926} = BE = \frac{2535}{463}.\] It follows that the answer is $\boxed{463}$ | null | 463 |
bc30162e80d10e3de36f768dff414606 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | Since $AE$ and $AD$ are isogonal with respect to the $A$ angle bisector, we have \[\frac{BE}{EC}\cdot \frac{BD}{DC}=(\frac{AB}{AC})^2.\] To prove this, let $\angle BAE=\angle DAC=x$ and $\angle BAD=\angle CAE=y.$ Then, by the Ratio Lemma, we have \[\frac{BD}{DC}=\frac{AB\sin y}{AC\sin x}\] \[\frac{BE}{EC}=\frac{AB\sin x}{AC\sin y}\] and multiplying these together proves the formula for isogonal lines. Hence, we have \[\frac{BE}{15-BE}\cdot \frac{9}{6}=\frac{169}{196}\implies BE=\frac{2535}{463}\] so our desired answer is $\boxed{463}.$ | null | 463 |
bc30162e80d10e3de36f768dff414606 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | Let $ED = x$ , such that $BE = 9-x$ . Since $\overline{AE}$ and $\overline{AD}$ are isogonal, we get $\frac{9-x}{6+x} \cdot \frac{9}{6} = \frac{13^2}{14^2} \Rightarrow 588(9 - x) = 338(6 + x)$ , and we can solve to get $x = \frac{1632}{463}$ (and $BE = \frac{2535}{463}$ ). Hence, our answer is $\boxed{463}$ . - Spacesam | null | 463 |
bc30162e80d10e3de36f768dff414606 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | Diagram borrowed from Solution 1.
Applying Law of Cosines on $\bigtriangleup ABC$ with respect to $\angle C$ we have \[AB^2=AC^2+BC^2-2(AC)(BC)\cos C\] Solving gets $\cos C=\frac{3}{5}$ , which implies that \[\sin C=\sqrt{1-\cos C}=\frac{4}{5}\] Applying Stewart's Theorem with cevian $AD$ we have \[(BC)(AD)^2+(BC)(BD)(CD)=(CD)(AB)^2+(BD)(AC)^2\] Solving gets $AD=\frac{4\sqrt{205}}{5}$
Applying Law of Sines on $\bigtriangleup ACD$ to solve for $\sin CAD$ we have \[\frac{AD}{\sin C}=\frac{CD}{\sin CAD}\] Solving gets $\sin CAD=\frac{6\sqrt{205}}{205}$ . Thus $\sin BAE=\sin CAD=\frac{6\sqrt{205}}{205}$
Applying Law of Sines on $\bigtriangleup ABC$ we have \[\frac{AC}{\sin B}=\frac{AB}{\sin C}\] Solving gets $\sin B=\frac{56}{65}$
Applying Stewart's Theorem with cevian $AE$ we have \[(BC)(AE)^2+(BC)(BE)(CE)=(CE)(AB)^2+(BE)(AC)^2\] \[(BC)(AE)^2+(BC)(BE)(BC-BE)=(BC-BE)(AB)^2+(BE)(AC)^2\] Solving gets $AE=\sqrt{\frac{15BE^2-198BE+2535}{15}}$
Finally, applying Law of Sines on $\bigtriangleup BAE$ we have \[\frac{AE}{\sin B}=\frac{BE}{\sin BAE}\] \[\frac{\sqrt{\frac{15BE^2-198BE+2535}{15}}}{\frac{56}{65}}=\frac{BE}{\frac{6\sqrt{205}}{205}}\] \[7605BE^2-32342BE+2535=0\] Solving the easy quadratic equation gets $BE=\frac{1632}{463}\Longrightarrow q=\boxed{463}$ | null | 463 |
bc30162e80d10e3de36f768dff414606 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_14 | In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | Making perpendicular lines from $D$ to $AC$ , meeting at $N$ ; from $E$ to $AB$ , meeting at $J$ . According to LOC, we can get that $\cos\angle{C}=\frac{3}{5}$ . So we get that $CN=\frac{18}{5};DN=\frac{24}{5};AN=AC-CN=\frac{52}{5}$ . Now we can see that $\tan\angle{DAN}=\frac{DN}{AN}=\frac{6}{13}$ . Now we see that in $\triangle{EJA}$ , assuming $EJ=6x;AJ=13x$ since $\tan\angle{JAE}=\tan\angle{DAN}$ . Now we need to find the tangent of $\angle{B}$ . Making a perpendicular line from $C$ to $AB$ at $M$ . We can see that $CM=\frac{168}{13};AM=\frac{70}{13};BM=AB-AM=13-\frac{70}{13}=\frac{99}{13}$ . We get that $\tan\angle{B}=\frac{56}{33}$ so $BJ=\frac{33}{56}*6x=\frac{99x}{28}$ .
After getting AJ and BJ, we can get that $AB=AJ+BJ=\frac{463x}{28}=13$ , which means that $x=\frac{364}{463}$ . According to the similarity, $\frac{CM}{BC}=\frac{JE}{BE};BE=\frac{2535}{463}$ which $\boxed{463}$ is our answer
~bluesoul | null | 463 |
9bc0df4473ff44371b4ccd3acd7357c7 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_15 | Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$ | Rewrite the given equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$
Let $w_3$ have center $(x,y)$ and radius $r$ . Now, if two circles with radii $r_1$ and $r_2$ are externally tangent, then the distance between their centers is $r_1 + r_2$ , and if they are internally tangent, it is $|r_1 - r_2|$ . So we have
\begin{align*} r + 4 &= \sqrt{(x-5)^2 + (y-12)^2} \\ 16 - r &= \sqrt{(x+5)^2 + (y-12)^2} \end{align*}
Solving for $r$ in both equations and setting them equal, then simplifying, yields
\begin{align*} 20 - \sqrt{(x+5)^2 + (y-12)^2} &= \sqrt{(x-5)^2 + (y-12)^2} \\ 20+x &= 2\sqrt{(x+5)^2 + (y-12)^2} \end{align*}
Squaring again and canceling yields $1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.$
So the locus of points that can be the center of the circle with the desired properties is an ellipse
Since the center lies on the line $y = ax$ , we substitute for $y$ and expand: \[1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75} \Longrightarrow (3+4a^2)x^2 - 96ax + 276 = 0.\]
We want the value of $a$ that makes the line $y=ax$ tangent to the ellipse, which will mean that for that choice of $a$ there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is $0$ , so $(-96a)^2 - 4(3+4a^2)(276) = 0$
Solving yields $a^2 = \frac{69}{100}$ , so the answer is $\boxed{169}$ | null | 169 |
9bc0df4473ff44371b4ccd3acd7357c7 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_15 | Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$ | As above, we rewrite the equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$ . Let $F_1=(-5,12)$ and $F_2=(5,12)$ . If a circle with center $C=(a,b)$ and radius $r$ is externally tangent to $w_2$ and internally tangent to $w_1$ , then $CF_1=16-r$ and $CF_2=4+r$ . Therefore, $CF_1+CF_2=20$ . In particular, the locus of points $C$ that can be centers of circles must be an ellipse with foci $F_1$ and $F_2$ and major axis $20$
Clearly, the minimum value of the slope $a$ will occur when the line $y=ax$ is tangent to this ellipse. Suppose that this point of tangency is denoted by $T$ , and the line $y=ax$ is denoted by $\ell$ . Then we reflect the ellipse over $\ell$ to a new ellipse with foci $F_1'$ and $F_2'$ as shown below.
By the reflection property of ellipses (i.e., the angle of incidence to a tangent line is equal to the angle of reflection for any path that travels between the foci), we know that $F_1$ $T$ , and $F_2'$ are collinear, and similarly, $F_2$ $T$ and $F_1'$ are collinear. Therefore, $OF_1F_2F_2'F_1'$ is a pentagon with $OF_1=OF_2=OF_1'=OF_2'=13$ $F_1F_2=F_1'F_2'=10$ , and $F_1F_2'=F_1'F_2=20$ . Note that $\ell$ bisects $\angle F_1'OF_1$ . We can bisect this angle by bisecting $\angle F_1'OF_2$ and $F_2OF_1$ separately.
We proceed using complex numbers. Triangle $F_2OF_1'$ is isosceles with side lengths $13,13,20$ . The height of this from the base of $20$ is $\sqrt{69}$ . Therefore, the complex number $\sqrt{69}+10i$ represents the bisection of $\angle F_1'OF_2$
Similarly, using the 5-12-13 triangles, we easily see that $12+5i$ represents the bisection of the angle $F_2OF_1$ . Therefore, we can add these two angles together by multiplying the complex numbers, finding \[\text{arg}\left((\sqrt{69}+10i)(12+5i)\right)=\frac{1}{2}\angle F_1'OF_1.\] Now the point $F_1$ is given by the complex number $-5+12i$ . Therefore, to find a point on line $\ell$ , we simply subtract $\frac{1}{2}\angle F_1'OF_1$ , which is the same as multiplying $-5+12i$ by the conjugate of $(\sqrt{69}+10i)(12+5i)$ . We find \[(-5+12i)(\sqrt{69}-10i)(12-5i)=169(10+i\sqrt{69}).\] In particular, note that the tangent of the argument of this complex number is $\sqrt{69}/10$ , which must be the slope of the tangent line. Hence $a^2=69/100$ , and the answer is $\boxed{169}$ | null | 169 |
9bc0df4473ff44371b4ccd3acd7357c7 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_15 | Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$ | We use the same reflection as in Solution 2. As $OF_1'=OF_2=13$ , we know that $\triangle OF_1'F_2$ is isosceles. Hence $\angle F_2F_1'O=\angle F_1'F_2O$ . But by symmetry, we also know that $\angle OF_1T=\angle F_2F_1'O$ . Hence $\angle OF_1T=\angle F_1'F_2O$ . In particular, as $\angle OF_1T=\angle OF_2T$ , this implies that $O, F_1, F_2$ , and $T$ are concyclic.
Let $X$ be the intersection of $F_2F_1'$ with the $x$ -axis. As $F_1F_2$ is parallel to the $x$ -axis, we know that \[\angle TXO=180-\angle F_1F_2T.\tag{1}\] But \[180-\angle F_1F_2T=\angle F_2F_1T+\angle F_1TF_2.\tag{2}\] By the fact that $OF_1F_2T$ is cyclic, \[\angle F_2F_1T=\angle F_2OT\qquad\text{and}\qquad \angle F_1TF_2=\angle F_1OF_2.\tag{3}\] Therefore, combining (1), (2), and (3), we find that \[\angle TXO=\angle F_2OT+\angle F_1OF_2=\angle F_1OT.\tag{4}\]
By symmetry, we also know that \[\angle F_1TO=\angle OTF_1'.\tag{5}\] Therefore, (4) and (5) show by AA similarity that $\triangle F_1OT\sim \triangle OXT$ . Therefore, $\angle XOT=\angle OF_1T$
Now as $OF_1=OF_2'=13$ , we know that $\triangle OF_1F_2'$ is isosceles, and as $F_1F_2'=20$ , we can drop an altitude to $F_1F_2'$ to easily find that $\tan \angle OF_1T=\sqrt{69}/10$ . Therefore, $\tan\angle XOT$ , which is the desired slope, must also be $\sqrt{69}/10$ . As before, we conclude that the answer is $\boxed{169}$ | null | 169 |
9bc0df4473ff44371b4ccd3acd7357c7 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_15 | Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$ | First, rewrite the equations for the circles as $(x+5)^2+(y-12)^2=16^2$ and $(x-5)^2+(y-12)^2=4^2$ .
Then, choose a point $(a,b)$ that is a distance of $x$ from both circles. Use the distance formula between $(a,b)$ and each of $A$ and $C$ (in the diagram above). The distances, as can be seen in the diagram above are $16-x$ and $4+x$ , respectively. \[(a-5)^2+(b-12)^2=(4+x)^2\] \[(a+5)^2+(b-12)^2=(16-x)^2\] Subtracting the first equation from the second gives \[20a=240-40x\rightarrow a=12-2x\rightarrow x=6-\frac a2\] Substituting this into the first equation gives \[a^2-10a+25+b^2-24b+144=100-10a+\frac{a^2}4\] \[b^2-24b+69+\frac{3a^2}4=0\] Now, instead of converting this to the equation of an eclipse, solve for $b$ and then divide by $a$ \[b=\frac{24\pm\sqrt{300-3a^2}}{2}\] We take the smaller root to minimize $\frac b a$ \[\frac b a=\frac{24-\sqrt{300-3a^2}}{2a}=\frac{24-\sqrt3\cdot\sqrt{100-a^2}}{2a}=\frac{12}a-\frac{\sqrt3}{2a}\sqrt{100-a^2}\] Now, let $10\cos\theta=a$ . This way, $\sqrt{100-a^2}=10\sin\theta$ .
Substitute this in. $\frac{b}{a}=\frac{12}{10\cos\theta}-\frac{\sqrt3\cdot\sin\theta}{2\cos\theta}=\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta$ Then, take the derivative of this and set it to 0 to find the minimum value. $\frac{6}{5}\sec\theta\tan\theta-\frac{\sqrt3}{2}\sec^2\theta=0\rightarrow\frac{6}{5}\sin\theta-frac{\sqrt3}{2}=0\rightarrow\sin\theta=\frac{5\sqrt3}{12}$ Then, use this value of $\sin\theta$ to find the minimum of $\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta$ to get $\frac{\sqrt{69}}{10}\rightarrow\left(\frac{\sqrt{69}}{10}\right)^2=\frac{69}{100}\rightarrow69+100=\boxed{169}$ | null | 169 |
9bc0df4473ff44371b4ccd3acd7357c7 | https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_15 | Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$ | First, obtain the equation of the ellipse as laid out in previous solutions. We now scale the coordinate plane in the $x$ direction by a factor of $\frac{\sqrt{3}}{2}$ centered at $x=0.$ This takes the ellipse to a circle centered at $(0,12)$ with radius $5\sqrt{3}$ and takes the line $y=ax$ to $y=\left( \frac{\sqrt{3}}{2} \right)^{-1} ax.$ The tangent point of our line to the circle with positive slope forms a right triangle with the origin and the center of the circle. Thus, the distance from this tangent point to the origin is $\sqrt{69}.$ By similar triangles, the slope of this line is then $\frac{\sqrt{69}}{5\sqrt{3}}.$ We multiply this by $\frac{\sqrt{3}}{2}$ to get $a=\frac{\sqrt{69}}{10},$ so our final answer is $\boxed{169.}$ | null | 169. |
f99f8c9df0bc9ef24d615c7404c48d7e | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_1 | The digits of a positive integer $n$ are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when $n$ is divided by $37$ | A brute-force solution to this question is fairly quick, but we'll try something slightly more clever: our numbers have the form ${\underline{(n+3)}}\,{\underline{(n+2)}}\,{\underline{( n+1)}}\,{\underline {(n)}}$ $= 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n$ , for $n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace$
Now, note that $3\cdot 37 = 111$ so $30 \cdot 37 = 1110$ , and $90 \cdot 37 = 3330$ so $87 \cdot 37 = 3219$ . So the remainders are all congruent to $n - 9 \pmod{37}$ . However, these numbers are negative for our choices of $n$ , so in fact the remainders must equal $n + 28$
Adding these numbers up, we get $(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}$ | null | 217 |
cd97092dcb5d5604040a8c009469f3ce | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2 | Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$ | Note that since set $A$ has $m$ consecutive integers that sum to $2m$ , the middle integer (i.e., the median) must be $2$ . Therefore, the largest element in $A$ is $2 + \frac{m-1}{2}$
Further, we see that the median of set $B$ is $0.5$ , which means that the "middle two" integers of set $B$ are $0$ and $1$ . Therefore, the largest element in $B$ is $1 + \frac{2m-2}{2} = m$ $2 + \frac{m-1}{2} > m$ if $m < 3$ , which is clearly not possible, thus $2 + \frac{m-1}{2} < m$
Solving, we get \begin{align*} m - 2 - \frac{m-1}{2} &= 99\\ m-\frac{m}{2}+\frac{1}{2}&=101\\ \frac{m}{2}&=100\frac{1}{2}.\\ m &= \boxed{201} | null | 201 |
cd97092dcb5d5604040a8c009469f3ce | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2 | Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$ | Let us give the elements of our sets names: $A = \{x, x + 1, x + 2, \ldots, x + m - 1\}$ and $B = \{y, y + 1, \ldots, y + 2m - 1\}$ . So we are given that \[2m = x + (x + 1) + \ldots + (x + m - 1) = mx + (1 + 2 + \ldots + (m - 1)) = mx + \frac{m(m -1)}2,\] so $2 = x + \frac{m - 1}2$ and $x + (m - 1) = \frac{m + 3}2$ (this is because $x = 2 - \frac{m-1}{2}$ so plugging this into $x+(m-1)$ yields $\frac{m+3}{2}$ ). Also, \[m = y + (y + 1) + \ldots + (y + 2m - 1) = 2my + \frac{2m(2m - 1)}2,\] so $1 = 2y + (2m - 1)$ so $2m = 2(y + 2m - 1)$ and $m = y + 2m - 1$
Then by the given, $99 = |(x + m - 1) - (y + 2m - 1)| = \left|\frac{m + 3}2 - m\right| = \left|\frac{m - 3}2\right|$ $m$ is a positive integer so we must have $99 = \frac{m - 3}2$ and so $m = \boxed{201}$ | null | 201 |
cd97092dcb5d5604040a8c009469f3ce | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2 | Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$ | The thing about this problem is, you have some "choices" that you can make freely when you get to a certain point, and these choices won't affect the accuracy of the solution, but will make things a lot easier for us.
First, we note that for set $A$
\[\frac{m(f + l)}{2} = 2m\]
Where $f$ and $l$ represent the first and last terms of $A$ . This comes from the sum of an arithmetic sequence.
Solving for $f+l$ , we find the sum of the two terms is $4$
Doing the same for set B, and setting up the equation with $b$ and $e$ being the first and last terms of set $B$
\[m(b+e) = m\]
and so $b+e = 1$
Now we know, assume that both sequences are increasing sequences, for the sake of simplicity. Based on the fact that set $A$ has half the number of elements as set $B$ , and the difference between the greatest terms of the two two sequences is $99$ (forget about absolute value, it's insignificant here since we can just assume both sets end with positive last terms), you can set up an equation where $x$ is the last term of set A:
\[2(x-(-x+4)+1) = 1+(x+99)-(-x-99+1)\]
Note how i basically just counted the number of terms in each sequence here. It's made a lot simpler because we just assumed that the first term is negative and last is positive for each set, it has absolutely no effect on the end result! This is a great strategy that can help significantly simplify problems. Also note how exactly i used the fact that the first and last terms of each sequence sum to $4$ and $1$ respectively (add $x$ and $(-x+4)$ to see what i mean).
Solving this equation we find $x = 102$ . We know the first and last terms have to sum to $4$ so we find the first term of the sequence is $-98$ . Now, the solution is in clear sight, we just find the number of integers between $-98$ and $102$ , inclusive, and it is $m = \boxed{201}$ | null | 201 |
cd97092dcb5d5604040a8c009469f3ce | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2 | Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$ | First, calculate the average of set $A$ and set $B$ . It's obvious that they are $2$ and $1/2$ respectively.
Let's look at both sets. Obviously, there is an odd number of integers in the set with $2$ being in the middle, which means that $m$ is an odd number and that the number of consecutive integers on each side of $2$ are equal. In set $B$ , it is clear that it contains an even number of integers, but since the number in the middle is $1/2$ , we know that the range of the consecutive numbers on both sides will be $(x$ to $0)$ and $(1$ to $-y)$
Nothing seems useful right now, but let's try plugging an odd number, $3$ , for $m$ in set $B$ . We see that there are $6$ consecutive integers and $3$ on both sides of $1/2$ . After plugging this into set $A$ , we find that the set equals \[{1,2,3}\] . From there, we find the absolute value of the difference of both of the greatest values, and get 0.
Let's try plugging in another odd number, $55$ . We see that the resulting set of numbers is $(-54$ to $0)$ , and $(1$ to $55)$ . We then plug this into set $A$ , and find that the set of numbers is $(-25$ to $-29)$ which indeed results in the average being $2$ . We then find the difference of the greatest values to be 26.
From here, we see a pattern that can be proven by more trial and error. When we make $m$ equal to $3$ , then the difference is $0$ whearas when we make it $55$ , then the difference is $26$ $55-3$ equals to $52$ and $26-0$ is just $0$ . We then see that $m$ increases twice as fast as the difference. So when the difference is $99$ , it increased $99$ from when it was $0$ , which means that $m$ increased by $99*2$ which is $198$ . We then add this to our initial $m$ of $3$ , and get $\boxed{201}$ as our answer. | null | 201 |
cd97092dcb5d5604040a8c009469f3ce | https://artofproblemsolving.com/wiki/index.php/2004_AIME_I_Problems/Problem_2 | Set $A$ consists of $m$ consecutive integers whose sum is $2m$ , and set $B$ consists of $2m$ consecutive integers whose sum is $m.$ The absolute value of the difference between the greatest element of $A$ and the greatest element of $B$ is $99$ . Find $m.$ | Let the first term of $A$ be $a$ and the first term of $B$ be $b$ . There are $m$ elements in $A$ so $A$ is $a, a+1, a+2,...,a+m-1$ . Adding these up, we get $\frac{2a+m-1}{2}\cdot m = 2m \implies 2a+m=5$ . Set $B$ contains the numbers $b, b+1, b+2,...,b+2m-1$ . Summing these up, we get $\frac{2b+2m-1}{2}\cdot 2m =m \implies 2b+2m=2$ . The problem gives us that the absolute value of the difference of the largest terms in $A$ and $B$ is $99$ . The largest term in $A$ is $a+m-1$ and the largest term in $B$ is $b+2m-1$ so $|b-a+m|=99$ . From the first two equations we get, we can get that $2(b-a)+m=-3$ . Now, we make a guess and assume that $b-a+m=99$ (if we get a negative value for $m$ , we can try $b-a+m=-99$ ). From here we get that $b-a=-102$ . Solving for $m$ , we get that the answer is $\boxed{201}$ | null | 201 |