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fc958d16f44e146b2b2253153b47ede8 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_12 | Pyramid $OABCD$ has square base $ABCD,$ congruent edges $\overline{OA}, \overline{OB}, \overline{OC},$ and $\overline{OD},$ and $\angle AOB=45^\circ.$ Let $\theta$ be the measure of the dihedral angle formed by faces $OAB$ and $OBC.$ Given that $\cos \theta=m+\sqrt{n},$ where $m_{}$ and $n_{}$ are integers, find $m+n.$ | Without loss of generality, place the pyramid in a 3-dimensional coordinate system such that $A = (1,0,0),$ $B = (0,1,0),$ $C = ( - 1,0,0),$ $D = (0, - 1,0),$ and $O = (0,0,z),$ where $z$ is unknown.
We first find $z.$ Note that
\[\overrightarrow{OA}\cdot \overrightarrow{OB} = \parallel \overrightarrow{OA}\parallel \parallel \overrightarrow{OB}\parallel \cos 45^\circ.\]
Since $\overrightarrow{OA} =\, <1,0, - z>$ and $\overrightarrow{OB} =\, <0,1, - z> ,$ this simplifies to
\[z^{2}\sqrt {2} = 1 + z^{2}\implies z^{2} = 1 + \sqrt {2}.\]
Now let's find $\cos \theta.$ Let $\vec{u}$ and $\vec{v}$ be normal vectors to the planes containing faces $OAB$ and $OBC,$ respectively. From the definition of the dot product as $\vec{u}\cdot \vec{v} = \parallel \vec{u}\parallel \parallel \vec{v}\parallel \cos \theta$ , we will be able to solve for $\cos \theta.$ A cross product yields (alternatively, it is simple to find the equation of the planes $OAB$ and $OAC$ , and then to find their normal vectors)
\[\vec{u} = \overrightarrow{OA}\times \overrightarrow{OB} = \left| \begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & - z \\ 0 & 1 & - z \end{array}\right| =\, < z,z,1 > .\]
Similarly,
\[\vec{v} = \overrightarrow{OB}\times \overrightarrow{OC} - \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & - z \\ - 1 & 0 & - z \end{array}\right| =\, < - z,z,1 > .\]
Hence, taking the dot product of $\vec{u}$ and $\vec{v}$ yields
\[\cos \theta = \frac{ \vec{u} \cdot \vec{v} }{ \parallel \vec{u} \parallel \parallel \vec{v} \parallel } = \frac{- z^{2} + z^{2} + 1}{(\sqrt {1 + 2z^{2}})^{2}} = \frac {1}{3 + 2\sqrt {2}} = 3 - 2\sqrt {2} = 3 - \sqrt {8}.\]
Flipping the signs (we found the cosine of the supplement angle) yields $\cos \theta = - 3 + \sqrt {8},$ so the answer is $\boxed{005}$ | null | 005 |
fc958d16f44e146b2b2253153b47ede8 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_12 | Pyramid $OABCD$ has square base $ABCD,$ congruent edges $\overline{OA}, \overline{OB}, \overline{OC},$ and $\overline{OD},$ and $\angle AOB=45^\circ.$ Let $\theta$ be the measure of the dihedral angle formed by faces $OAB$ and $OBC.$ Given that $\cos \theta=m+\sqrt{n},$ where $m_{}$ and $n_{}$ are integers, find $m+n.$ | Similar to Solution 1, $\angle APC$ is the dihedral angle we want. WLOG, we will let $AB=1,$ meaning $AC=\sqrt{2}$
Because $\triangle OAB,\triangle OBC$ are isosceles, $\angle ABP = 67.5^{\circ}$ $PC=PA=\cos(\angle PAB)=\cos(22.5^{\circ})$
Thus by the half-angle identity,
\[PA=\cos\left(\frac{45}{2}\right) = \sqrt{\frac{1+\cos(45^{\circ})}{2}}\] \[= \sqrt{\frac{2+\sqrt{2}}{4}}.\]
Now looking at triangle $\triangle PAC,$ we drop the perpendicular from $P$ to $AC$ , and call the foot $H$ . Then $\angle CPH = \theta / 2.$ By Pythagoreas, \[PH=\sqrt{\frac{2+\sqrt{2}}{4}-\frac{1}{2}}=\frac{\sqrt[4]{2}}{2}.\]
We have that \[\cos\left(\frac{\theta}{2}\right)=\frac{\sqrt[4]{2}}{\sqrt{2+\sqrt{2}}},\text{ so}\] \[\cos(\theta)=2\cos^{2}\left(\frac{\theta}{2}\right)-1\] \[=2\left(\frac{\sqrt{2}}{2+\sqrt{2}}\right)-1\] \[=2(\frac{2\sqrt{2}-2}{2})-1\] \[=-3+\sqrt{8}.\]
Because $m$ and $n$ can be negative integers, our answer is $(-3)+8=\boxed{005.}$ | null | 005. |
728780b05b876993d069bd0aef7c1dd2 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_13 | Let $f(n)$ be the integer closest to $\sqrt[4]{n}.$ Find $\sum_{k=1}^{1995}\frac 1{f(k)}.$ | This is a pretty easy problem just to bash. Since the max number we can get is $7$ , we just need to test $n$ values for $1.5,2.5,3.5,4.5,5.5$ and $6.5$ . Then just do how many numbers there are times $\frac{1}{\lfloor n \rfloor}$ , which should be $5+17+37+65+101+145+30 = \boxed{400}$ | null | 400 |
54f59ae2a0bb3cfe4681f5b027954131 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_14 | In a circle of radius $42$ , two chords of length $78$ intersect at a point whose distance from the center is $18$ . The two chords divide the interior of the circle into four regions. Two of these regions are bordered by segments of unequal lengths, and the area of either of them can be expressed uniquely in the form $m\pi-n\sqrt{d},$ where $m, n,$ and $d_{}$ are positive integers and $d_{}$ is not divisible by the square of any prime number. Find $m+n+d.$ | Let the center of the circle be $O$ , and the two chords be $\overline{AB}, \overline{CD}$ and intersecting at $E$ , such that $AE = CE < BE = DE$ . Let $F$ be the midpoint of $\overline{AB}$ . Then $\overline{OF} \perp \overline{AB}$
By the Pythagorean Theorem $OF = \sqrt{OB^2 - BF^2} = \sqrt{42^2 - 39^2} = 9\sqrt{3}$ , and $EF = \sqrt{OE^2 - OF^2} = 9$ . Then $OEF$ is a $30-60-90$ right triangle , so $\angle OEB = \angle OED = 60^{\circ}$ . Thus $\angle BEC = 60^{\circ}$ , and by the Law of Cosines
It follows that $\triangle BCO$ is an equilateral triangle , so $\angle BOC = 60^{\circ}$ . The desired area can be broken up into two regions, $\triangle BCE$ and the region bounded by $\overline{BC}$ and minor arc $\stackrel{\frown}{BC}$ . The former can be found by Heron's formula to be $[BCE] = \sqrt{60(60-48)(60-42)(60-30)} = 360\sqrt{3}$ . The latter is the difference between the area of sector $BOC$ and the equilateral $\triangle BOC$ , or $\frac{1}{6}\pi (42)^2 - \frac{42^2 \sqrt{3}}{4} = 294\pi - 441\sqrt{3}$
Thus, the desired area is $360\sqrt{3} + 294\pi - 441\sqrt{3} = 294\pi - 81\sqrt{3}$ , and $m+n+d = \boxed{378}$ | null | 378 |
e08c15188573f494efb1fbd42f80a188 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_15 | Let $p_{}$ be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of $5$ heads before one encounters a run of $2$ tails. Given that $p_{}$ can be written in the form $m/n$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n$ | Think of the problem as a sequence of 's and 's. No two 's can occur in a row, so the sequence is blocks of $1$ to $4$ 's separated by 's and ending in $5$ 's. Since the first letter could be or the sequence could start with a block of 's, the total probability is that $3/2$ of it has to start with an
The answer to the problem is then the sum of all numbers of the form $\frac 32 \left( \frac 1{2^a} \cdot \frac 12 \cdot \frac 1{2^b} \cdot \frac 12 \cdot \frac 1{2^c} \cdots \right) \cdot \left(\frac 12\right)^5$ , where $a,b,c \ldots$ are all numbers $1-4$ , since the blocks of 's can range from $1-4$ in length. The sum of all numbers of the form $(1/2)^a$ is $1/2+1/4+1/8+1/16=15/16$ , so if there are n blocks of 's before the final five 's, the answer can be rewritten as the sum of all numbers of the form $\frac 32\left( \left(\frac {15}{16}\right)^n \cdot \left(\frac 12\right)^n \right) \cdot \left(\frac 1{32}\right)=\frac 3{64}\left(\frac{15}{32}\right)^n$ , where $n$ ranges from $0$ to $\infty$ , since that's how many blocks of 's there can be before the final five. This is an infinite geometric series whose sum is $\frac{3/64}{1-(15/32)}=\frac{3}{34}$ , so the answer is $\boxed{037}$ | null | 037 |
e08c15188573f494efb1fbd42f80a188 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_15 | Let $p_{}$ be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of $5$ heads before one encounters a run of $2$ tails. Given that $p_{}$ can be written in the form $m/n$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n$ | Let $p_H, p_T$ respectively denote the probabilities that a string beginning with 's and 's are successful. Thus,
A successful string can either start with 1 to 4 H's, start with a T and then continue with a string starting with (as there cannot be $2$ 's in a row, or be the string HHHHH.
There is a $\frac{1}{16}$ probability we roll $4$ consecutive 's, and there is a $\frac{15}{16}$ probability we roll a . Thus,
The answer is $p_H + p_T = \frac{3}{2}p_H = \frac{3}{34}$ , and $m+n = \boxed{037}$ | null | 037 |
e08c15188573f494efb1fbd42f80a188 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_15 | Let $p_{}$ be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of $5$ heads before one encounters a run of $2$ tails. Given that $p_{}$ can be written in the form $m/n$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n$ | For simplicity, let's compute the complement, namely the probability of getting to $2$ tails before $5$ heads.
Let $h_{i}$ denote the probability that we get $2$ tails before $5$ heads, given that we have $i$ consecutive heads. Similarly, let $t_{i}$ denote the probability that we get $2$ tails before $5$ heads, given that we have $i$ consecutive tails. Specifically, $h_{5} = 0$ and $t_{2} = 1$ . If we can solve for $h_{1}$ and $t_{1}$ , we are done; the answer is simply $1/2 * (h_{1} + t_{1})$ , since on our first roll, we have equal chances of getting a string with "1 consecutive head" or "1 consecutive tail."
Consider solving for $t_{1}$ . If we have 1 consecutive tail, then (a) rolling a head gets us to 1 consecutive head and (b) rolling a tail gets us to 2 consecutive tails. So, we must have:
Applying similar logic, we get the equations:
\begin{align*} h_{1} &= \frac{1}{2} h_{2} + \frac{1}{2} t_{1}\\ h_{2} &= \frac{1}{2} h_{3} + \frac{1}{2} t_{1}\\ h_{3} &= \frac{1}{2} h_{4} + \frac{1}{2} t_{1}\\ h_{4} &= \frac{1}{2} h_{5} + \frac{1}{2} t_{1} \end{align*}
Since $h_{5} = 0$ , we get $h_{4} = \frac{1}{2} t_{1}$ $\Rightarrow h_{3} = \frac{3}{4} t_{1}$ $\Rightarrow h_{2} = \frac{7}{8} t_{1}$ $\Rightarrow h_{1} = \frac{15}{16} t_{1}$ $\Rightarrow t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} \cdot \frac{15}{16} t_{1} = \frac{1}{2} + \frac{15}{32} t_{1} \Rightarrow t_{1} = \frac{16}{17}$ $\Rightarrow h_{1} = \frac{15}{16} \cdot \frac{16}{17} = \frac{15}{17}$
So, the probability of reaching $2$ tails before $5$ heads is $\frac{1}{2} (h_{1} + t_{1}) = \frac{31}{34}$ ; we want the complement, $\frac{3}{34}$ , yielding an answer of $3 + 34 = \boxed{037}$ | null | 037 |
e08c15188573f494efb1fbd42f80a188 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_15 | Let $p_{}$ be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of $5$ heads before one encounters a run of $2$ tails. Given that $p_{}$ can be written in the form $m/n$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n$ | Consider what happens in the "endgame" or what ultimately leads to the end. Let A denote that a head has been flipped, and let B denote that a tail has been flipped. The endgame outcomes are AAAAA, BAAAAA, BB, ABB, AABB, AAABB, AAAABB. The probabilities of each of these are $\frac{1}{32},\frac{1}{64},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\frac{1}{64}$ respectively. The ones where there is five heads are AAAAA and BAAAAA. The sum of these probabilities are $\frac{3}{64}$ . The sum of all these endgame outcomes are $\frac{34}{64}$ , hence the desired probability is $\frac{3}{34}$ , and in this case $m=3,n=34$ so we have $m+n=\boxed{037}$ -vsamc | null | 037 |
11129888d41e42438abb9aa36e24c22e | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_1 | The increasing sequence $3, 15, 24, 48, \ldots\,$ consists of those positive multiples of 3 that are one less than a perfect square . What is the remainder when the 1994th term of the sequence is divided by 1000? | One less than a perfect square can be represented by $n^2 - 1 = (n+1)(n-1)$ . Either $n+1$ or $n-1$ must be divisible by 3. This is true when $n \equiv -1,\ 1 \equiv 2,\ 1 \pmod{3}$ . Since 1994 is even, $n$ must be congruent to $1 \pmod{3}$ . It will be the $\frac{1994}{2} = 997$ th such term, so $n = 4 + (997-1) \cdot 3 = 2992$ . The value of $n^2 - 1 = 2992^2 - 1 \pmod{1000}$ is $\boxed{063}$ | null | 063 |
03f5b44be80f13b57db57455c8df76b5 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_2 | A circle with diameter $\overline{PQ}$ of length 10 is internally tangent at $P$ to a circle of radius 20. Square $ABCD$ is constructed with $A$ and $B$ on the larger circle, $\overline{CD}$ tangent at $Q$ to the smaller circle, and the smaller circle outside $ABCD$ . The length of $\overline{AB}$ can be written in the form $m + \sqrt{n}$ , where $m$ and $n$ are integers. Find $m + n$
1994 AIME Problem 2.png
Note: The diagram was not given during the actual contest. | 1994 AIME Problem 2 - Solution.png
Call the center of the larger circle $O$ . Extend the diameter $\overline{PQ}$ to the other side of the square (at point $E$ ), and draw $\overline{AO}$ . We now have a right triangle , with hypotenuse of length $20$ . Since $OQ = OP - PQ = 20 - 10 = 10$ , we know that $OE = AB - OQ = AB - 10$ . The other leg, $AE$ , is just $\frac 12 AB$
Apply the Pythagorean Theorem
The quadratic formula shows that the answer is $\frac{16 \pm \sqrt{16^2 + 4 \cdot 240}}{2} = 8 \pm \sqrt{304}$ . Discard the negative root, so our answer is $8 + 304 = \boxed{312}$ | null | 312 |
c9ec465de29de52b45f7994d4a2aafea | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_3 | The function $f_{}^{}$ has the property that, for each real number $x,\,$
If $f(19)=94,\,$ what is the remainder when $f(94)\,$ is divided by $1000$ | \begin{align*}f(94)&=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=\cdots \\ &= (94^2-93^2) + (92^2-91^2) +\cdots+ (22^2-21^2)+ 20^2-f(19) \\ &= 94+93+\cdots+21+400-94 \\ &= 4561 \end{align*}
So, the remainder is $\boxed{561}$ | null | 561 |
c9ec465de29de52b45f7994d4a2aafea | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_3 | The function $f_{}^{}$ has the property that, for each real number $x,\,$
If $f(19)=94,\,$ what is the remainder when $f(94)\,$ is divided by $1000$ | Those familiar with triangular numbers and some of their properties will quickly recognize the equation given in the problem. It is well-known (and easy to show) that the sum of two consecutive triangular numbers is a perfect square; that is, \[T_{n-1} + T_n = n^2,\] where $T_n = 1+2+...+n = \frac{n(n+1)}{2}$ is the $n$ th triangular number.
Using this, as well as using the fact that the value of $f(x)$ directly determines the value of $f(x+1)$ and $f(x-1),$ we conclude that $f(n) = T_n + K$ for all odd $n$ and $f(n) = T_n - K$ for all even $n,$ where $K$ is a constant real number.
Since $f(19) = 94$ and $T_{19} = 190,$ we see that $K = -96.$ It follows that $f(94) = T_{94} - (-96) = \frac{94\cdot 95}{2} + 96 = 4561,$ so the answer is $\boxed{561}$ | null | 561 |
df8dfcf765351f3b14a535dd7e4ce771 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_4 | Find the positive integer $n\,$ for which \[\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994\] (For real $x\,$ $\lfloor x\rfloor\,$ is the greatest integer $\le x.\,$ | Note that if $2^x \le a<2^{x+1}$ for some $x\in\mathbb{Z}$ , then $\lfloor\log_2{a}\rfloor=\log_2{2^{x}}=x$
Thus, there are $2^{x+1}-2^{x}=2^{x}$ integers $a$ such that $\lfloor\log_2{a}\rfloor=x$ . So the sum of $\lfloor\log_2{a}\rfloor$ for all such $a$ is $x\cdot2^x$
Let $k$ be the integer such that $2^k \le n<2^{k+1}$ . So for each integer $j<k$ , there are $2^j$ integers $a\le n$ such that $\lfloor\log_2{a}\rfloor=j$ , and there are $n-2^k+1$ such integers such that $\lfloor\log_2{a}\rfloor=k$
Therefore, $\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor= \sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1994$
Through computation: $\sum_{j=0}^{7}(j\cdot2^j)=1538<1994$ and $\sum_{j=0}^{8}(j\cdot2^j)=3586>1994$ . Thus, $k=8$
So, $\sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1538+8(n-2^8+1)=1994 \Rightarrow n = \boxed{312}$ | null | 312 |
067f0b089095b4b384e3d0f791979204 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_5 | Given a positive integer $n\,$ , let $p(n)\,$ be the product of the non-zero digits of $n\,$ . (If $n\,$ has only one digits, then $p(n)\,$ is equal to that digit.) Let
What is the largest prime factor of $S\,$ | Suppose we write each number in the form of a three-digit number (so $5 \equiv 005$ ), and since our $p(n)$ ignores all of the zero-digits, replace all of the $0$ s with $1$ s. Now note that in the expansion of
we cover every permutation of every product of $3$ digits, including the case where that first $1$ represents the replaced $0$ s. However, since our list does not include $000$ , we have to subtract $1$ . Thus, our answer is the largest prime factor of $(1+1+2+3+\cdots +9)^3 - 1 = 46^3 - 1 = (46-1)(46^2 + 46 + 1) = 3^3 \cdot 5 \cdot 7 \cdot \boxed{103}$ | null | 103 |
067f0b089095b4b384e3d0f791979204 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_5 | Given a positive integer $n\,$ , let $p(n)\,$ be the product of the non-zero digits of $n\,$ . (If $n\,$ has only one digits, then $p(n)\,$ is equal to that digit.) Let
What is the largest prime factor of $S\,$ | Note that $p(1)=p(11), p(2)=p(12), p(3)=p(13), \cdots p(19)=p(9)$ , and $p(37)=3p(7)$ . So $p(10)+p(11)+p(12)+\cdots +p(19)=46$ $p(10)+p(11)+\cdots +p(99)=46*45=2070$ . We add $p(1)+p(2)+p(3)+\cdots +p(10)=45$ to get 2115. When we add a digit we multiply the sum by that digit. Thus $2115\cdot (1+1+2+3+4+5+6+7+8+9)=2115\cdot 46=47\cdot 45\cdot 46$ . But we didn't count 100, 200, 300, ..., 900. We add another 45 to get $45\cdot 2163$ . The largest prime factor of that is $\boxed{103}$ | null | 103 |
9a5a6f1fd4069ff884ee2830fa74b6f3 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_6 | The graphs of the equations
are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side length $\tfrac{2}{\sqrt{3}}.\,$ How many such triangles are formed? | We note that the lines partition the hexagon of the six extremal lines into disjoint unit regular triangles, and forms a series of unit regular triangles along the edge of the hexagon.
[asy] size(200); picture pica, picb, picc; int i; for(i=-10;i<=10;++i){ if((i%10) == 0){draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(20/sqrt(3)+abs((0,i))/sqrt(3),i),black+0.7);} else{draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(20/sqrt(3)+abs((0,i))/sqrt(3),i));} } picb = rotate(120,origin)*pica; picc = rotate(240,origin)*pica; add(pica);add(picb);add(picc); [/asy]
Solving the above equations for $k=\pm 10$ , we see that the hexagon in question is regular, with side length $\frac{20}{\sqrt{3}}$ . Then, the number of triangles within the hexagon is simply the ratio of the area of the hexagon to the area of a regular triangle. Since the ratio of the area of two similar figures is the square of the ratio of their side lengths, we see that the ratio of the area of one of the six equilateral triangles composing the regular hexagon to the area of a unit regular triangle is just $\left(\frac{20/\sqrt{3}}{2/\sqrt{3}}\right)^2 = 100$ . Thus, the total number of unit triangles is $6 \times 100 = 600$
There are $6 \cdot 10$ equilateral triangles formed by lines on the edges of the hexagon. Thus, our answer is $600+60 = \boxed{660}$ | null | 660 |
9a5a6f1fd4069ff884ee2830fa74b6f3 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_6 | The graphs of the equations
are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side length $\tfrac{2}{\sqrt{3}}.\,$ How many such triangles are formed? | There are three types of lines: horizontal, upward-slanting diagonal, and downward-slanting diagonal. There are $21$ of each type of line, and a unit equilateral triangle is determined by exactly one of each type of line. Given an upward-slanting diagonal and a downward-slanting diagonal, they determine exactly two unit equilateral triangles as shown below. [asy] size(60); pair u=rotate(60)*(1,0),d=rotate(-60)*(1,0),h=(1,0); draw((0,0)--4*u^^-2*h+4*u--(-2*h+4*u+4*d)); draw(u--2*u+d,dotted); draw(3*u--3*u-h,dotted); [/asy] Therefore, if all horizontal lines are drawn, there will be a total of $2\cdot 21^2=882$ unit equilateral triangles. Of course, we only draw $21$ horizontal lines, so we are overcounting the triangles that are caused by the undrawn horizontal lines. In the below diagram, we draw the diagonal lines and the highest and lowest horizontal lines. [asy] size(200); pair u=rotate(60)*(2/sqrt(3),0),d=rotate(-60)*(2/sqrt(3),0),h=(2/sqrt(3),0); for (int i=0;i<21;++i) {path v=(-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d;draw(shift(0,-2*i)*v);} for (int i=0;i<21;++i) {path v=rotate(180)*((-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d);draw(shift(0,2*i)*v);} draw((-15,-10)--(15,-10)); draw((-15,10)--(15,10)); [/asy]
We see that the lines $y=-21,-20,\dots, -11$ and $y=11,12,\dots,21$ would complete several of the $882$ unit equilateral triangles. In fact, we can see that the lines $y=-21,-20,\dots,-11$ complete $1,2,(1+3),(2+4),(3+5),(4+6),\dots,(9+11)$ triangles, or $111$ triangles. The positive horizontal lines complete the same number of triangles, hence the answer is $882-2\cdot 111=\boxed{660}$ | null | 660 |
9a5a6f1fd4069ff884ee2830fa74b6f3 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_6 | The graphs of the equations
are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side length $\tfrac{2}{\sqrt{3}}.\,$ How many such triangles are formed? | Picturing the diagram in your head should give you an illustration similar to the one above. The distance from parallel sides of the center hexagon is 20 units, and by extending horizontal lines to the sides of the hexagon. This shows that for every side of the hexagon there are 10 spaces. Therefore the side length of the biggest triangle (imagine one of the overlapping triangles in the Star of David) is 30. The total number of triangles in the hexagon can be found by finding the number of triangles in the extended triangle and subtracting the 3 corner triangles. This gives us $30^2 - 10^2 - 10^2- 10^2 = 600$ . That is the number of triangles in the hexagon. The remaining triangles form in groups of 10 on the exterior of each side. $600 + 6 * 10 = \boxed{660}$ | null | 660 |
e04415018886b333e8fcfca30ac133d1 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_7 | For certain ordered pairs $(a,b)\,$ of real numbers , the system of equations
has at least one solution, and each solution is an ordered pair $(x,y)\,$ of integers. How many such ordered pairs $(a,b)\,$ are there? | The equation $x^2+y^2=50$ is that of a circle of radius $\sqrt{50}$ , centered at the origin. By testing integers until the left side becomes too big, we see that the lattice points on this circle are $(\pm1,\pm7)$ $(\pm5,\pm5)$ , and $(\pm7,\pm1)$ where the signs are all independent of each other, for a total of $3\cdot 2\cdot 2=12$ lattice points. They are indicated by the blue dots below.
[asy] size(150); draw(circle((0,0),sqrt(50))); draw((1,7)--(-1,-7),red); draw((7,1)--(5,-5), green); dot((0,0)); dot((1,7),blue); dot((1,-7),blue); dot((-1,7),blue); dot((-1,-7),blue); dot((5,5),blue); dot((5,-5),blue); dot((-5,5),blue); dot((-5,-5),blue); dot((7,1),blue); dot((7,-1),blue); dot((-7,1),blue); dot((-7,-1),blue); [/asy]
Since $(x,y)=(0,0)$ yields $a\cdot 0+b\cdot 0=0 \neq 1$ , we know that $ax+by=1$ is the equation of a line that does not pass through the origin. So, we are looking for the number of lines which pass through at least one of the $12$ lattice points on the circle, but do not pass through the origin or through any non-lattice point on the circle. An example is the green line above. It is straightforward to show that a line passes through the origin precisely when there exist two opposite points $(p,q)$ and $(-p,-q)$ through which it passes. And example is the red line above.
There are $\binom{12}{2}=66$ ways to pick two distinct lattice points, and subsequently $66$ distinct lines which pass through two distinct lattice points on the circle. Then we subtract the lines which pass through the origin by noting that the lattice points on the circle can be grouped into opposite pairs $(p,q)$ and $(-p,-q)$ , for a total of $\frac{12}{2}=6$ lines. Finally, we add the $12$ unique tangent lines to the circle at each of the lattice points.
Therefore, our final count of distinct lines which pass through one or two of the lattice points on the circle, but do not pass through the origin, is \[66-6+12=\boxed{72}.\] | null | 72 |
862e857ca1f58447636654eebabbede3 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8 | The points $(0,0)\,$ $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$ | Consider the points on the complex plane . The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, so:
\[(a+11i)\left(\mathrm{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.\]
Equating the real and imaginary parts, we have:
\begin{align*}b&=\frac{a}{2}-\frac{11\sqrt{3}}{2}\\37&=\frac{11}{2}+\frac{a\sqrt{3}}{2} \end{align*}
Solving this system, we find that $a=21\sqrt{3}, b=5\sqrt{3}$ . Thus, the answer is $\boxed{315}$ | null | 315 |
862e857ca1f58447636654eebabbede3 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8 | The points $(0,0)\,$ $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$ | Using the Pythagorean theorem with these beastly numbers doesn't seem promising. How about properties of equilateral triangles? $\sqrt{3}$ and perpendiculars inspires this solution:
First, drop a perpendicular from $O$ to $AB$ . Call this midpoint of $AB M$ . Thus, $M=\left(\frac{a+b}{2}, 24\right)$ . The vector from $O$ to $M$ is $\left[\frac{a+b}{2}, 24\right]$ . Meanwhile from point $M$ we can use a vector with $\frac{\sqrt{3}}{3}$ the distance; we have to switch the $x$ and $y$ and our displacement is $\left[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}\right]$ . (Do you see why we switched $x$ and $y$ due to the rotation of 90 degrees?)
We see this displacement from $M$ to $A$ is $\left[\frac{a-b}{2}, 13\right]$ as well. Equating the two vectors, we get $a+b=26\sqrt{3}$ and $a-b=16\sqrt{3}$ . Therefore, $a=21\sqrt{3}$ and $b=5\sqrt{3}$ . And the answer is $\boxed{315}$ | null | 315 |
862e857ca1f58447636654eebabbede3 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8 | The points $(0,0)\,$ $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$ | Plot this equilateral triangle on the complex plane.
Translate the equilateral triangle so that its centroid is located at the origin. (The centroid can be found by taking the average of the three vertices of the triangle, which gives $\left(\frac{a+b}{3}, 16i\right)$ . The new coordinates of the equilateral triangle are $\left(-\frac{a+b}{3}-16i\right), \left(a-\frac{a+b}{3}-5i\right), \left(b-\frac{a+b}{3}+21i\right)$ . These three vertices are solutions of a cubic polynomial of form $x^3 + C$ . By Vieta's Formulas, the sum of the paired roots of the cubic polynomial are zero. (Or for the three roots $r_1, r_2,$ and $r_3,$ $\, r_1r_2 + r_2r_3 + r_3r_1 = 0$ .) The vertices of the equilateral triangle represent the roots of a polynomial, so the vertices can be plugged into the above equation. Because both the real and complex components of the equation have to sum to zero, you really have two equations. Multiply out the equation given by Vieta's Formulas and isolate the ones with imaginary components. Simplify that equation, and that gives the equation $5a = 21b.$ Now use the equation with only real parts. This should give you a quadratic $a^2 - ab + b^2 = 1083$ . Use your previously obtained equation to plug in for $a$ and solve for $b$ , which should yield $5\sqrt{3}$ $a$ is then $\frac{21}{5}\sqrt{3}$ . Multiplying $a$ and $b$ yields $\boxed{315}$ | null | 315 |
862e857ca1f58447636654eebabbede3 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_8 | The points $(0,0)\,$ $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$ | Just using the Pythagorean Theorem, we get that $a^2 + 11^2 = (b-a)^2 + 26^2 = b^2 = 37^2$
$a^2 + 121 = b^2 + 1369 ==> a^2 = b^2 + 1248$ . Expanding the second and subtracting the first equation from it we get $b^2 = 2ab - 555$ $b^2 = 2ab - 555 ==> a^2 = 2ab + 693$
We have $b^2 + 1248 = 2b\sqrt{b^2+1248} + 693$
Moving the square root to one side and non square roots to the other we eventually get $b^4 + 1110b^2 + 308025 = 4b^4 + 4992b^2$
$3b^4 + 3882b^2 - 308025 = 0$
This factors to $(3b^2-225)(b^2+1369)$ , so $3b^2 = 225, b = 5\sqrt 3$
Plugging it back in, we find that a = $\sqrt{1323}$ which is $21\sqrt3$ , so the product $ab$ is $\boxed{315}$ | null | 315 |
85aa89f6a26ea296e807ef8b837b98a8 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_9 | A solitaire game is played as follows. Six distinct pairs of matched tiles are placed in a bag. The player randomly draws tiles one at a time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand. The game ends if the player ever holds three tiles, no two of which match; otherwise the drawing continues until the bag is empty. The probability that the bag will be emptied is $p/q,\,$ where $p\,$ and $q\,$ are relatively prime positive integers. Find $p+q.\,$ | Let $P_k$ be the probability of emptying the bag when it has $k$ pairs in it. Let's consider the possible draws for the first three cards:
Therefore, we obtain the recursion $P_k = \frac {3}{2k - 1}P_{k - 1}$ . Iterating this for $k = 6,5,4,3,2$ (obviously $P_1 = 1$ ), we get $\frac {3^5}{11*9*7*5*3} = \frac {9}{385}$ , and $p+q=\boxed{394}$ | null | 394 |
85aa89f6a26ea296e807ef8b837b98a8 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_9 | A solitaire game is played as follows. Six distinct pairs of matched tiles are placed in a bag. The player randomly draws tiles one at a time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand. The game ends if the player ever holds three tiles, no two of which match; otherwise the drawing continues until the bag is empty. The probability that the bag will be emptied is $p/q,\,$ where $p\,$ and $q\,$ are relatively prime positive integers. Find $p+q.\,$ | Call the case that we begin with [ABCDEF]. It doesn't matter what letter we choose at first, so WLOG assume we choose A. Now there is BCDEFABCDEF remaining in the bag. We have two cases to consider here.
1. We pick the other A. There's a $\frac{1}{11}$ chance for this to happen. We remain with the case [BCDEF] if this is the case.
2. We pick any other letter that is not an A. There's a $\frac{10}{11}$ chance for this to happen. WLOG, assume we pick the letter B. Now in order for us to continue the game, we must choose either the other A or B. There's a $\frac{2}{10}$ chance for this to happen. WLOG, assume we choose A. Now we have BCDEFCDEF left.
Notice however that in the first case, the probability of emptying the bag with [BCDEF] is the same thing as with BCDEFCDEF, as the only difference is you've removed one of the letters (and it doesn't matter which you chose).
Hence for this case, there is a $\frac{1}{11} + \frac{10}{11}*\frac{2}{10} = \frac{3}{11}$ * [BCDEF] chance to empty the bag.
Continuing this process, we get that:
[BCDEF] = $\frac{3}{9}$ * [CDEF]
[CDEF] = $\frac{3}{7}$ * [DEF]
[DEF] = $\frac{3}{5}$ * [EF]
[EF] = 1 (clearly, since if we are only left with EFEF then we are going to empty the bag).
And hence [ABCDEF] = $1*\frac{3}{5}*\frac{3}{7}*\frac{3}{9}*\frac{3}{11} = \frac{9}{385}$ so our answer is $9+385=\boxed{394}$ | null | 394 |
f44d9a06c349f917921cef4710f583e7 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_10 | In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. Find $m+n.\,$ | Since $\triangle ABC \sim \triangle CBD$ , we have $\frac{BC}{AB} = \frac{29^3}{BC} \Longrightarrow BC^2 = 29^3 AB$ . It follows that $29^2 | BC$ and $29 | AB$ , so $BC$ and $AB$ are in the form $29^2 x$ and $29 x^2$ , respectively, where x is an integer.
By the Pythagorean Theorem , we find that $AC^2 + BC^2 = AB^2 \Longrightarrow (29^2x)^2 + AC^2 = (29 x^2)^2$ , so $29x | AC$ . Letting $y = AC / 29x$ , we obtain after dividing through by $(29x)^2$ $29^2 = x^2 - y^2 = (x-y)(x+y)$ . As $x,y \in \mathbb{Z}$ , the pairs of factors of $29^2$ are $(1,29^2)(29,29)$ ; clearly $y = \frac{AC}{29x} \neq 0$ , so $x-y = 1, x+y= 29^2$ . Then, $x = \frac{1+29^2}{2} = 421$
Thus, $\cos B = \frac{BC}{AB} = \frac{29^2 x}{29x^2} = \frac{29}{421}$ , and $m+n = \boxed{450}$ | null | 450 |
f44d9a06c349f917921cef4710f583e7 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_10 | In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. Find $m+n.\,$ | We will solve for $\cos B$ using $\triangle CBD$ , which gives us $\cos B = \frac{29^3}{BC}$ . By the Pythagorean Theorem on $\triangle CBD$ , we have $BC^2 - DC^2 = (BC + DC)(BC - DC) = 29^6$ . Trying out factors of $29^6$ , we can either guess and check or just guess to find that $BC + DC = 29^4$ and $BC - DC = 29^2$ (The other pairs give answers over 999). Adding these, we have $2BC = 29^4 + 29^2$ and $\frac{29^3}{BC} = \frac{2*29^3}{29^2 (29^2 +1)} = \frac{58}{842} = \frac{29}{421}$ , and our answer is $\boxed{450}$ | null | 450 |
f44d9a06c349f917921cef4710f583e7 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_10 | In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. Find $m+n.\,$ | Using similar right triangles, we identify that $CD = \sqrt{AD \cdot BD}$ . Let $AD$ be $29 \cdot k^2$ , to avoid too many radicals, getting $CD = k \cdot 29^2$ . Next we know that $AC = \sqrt{AB \cdot AD}$ and that $BC = \sqrt{AB \cdot BD}$ . Applying the logic with the established values of k, we get $AC = 29k \cdot \sqrt{29^2 + k^2}$ and $BC = 29^2 \cdot \sqrt{29^2 + k^2}$ . Next we look to the integer requirement. Since $k$ is both outside and inside square roots, we know it must be an integer to keep all sides as integers. Let $y$ be $\sqrt{29^2 + k^2}$ , thus $29^2 = y^2 - k^2$ , and $29^2 = (y + k)(y - k)$ . Since $29$ is prime, and $k$ cannot be zero, we find $k = 420$ and $y = 421$ as the smallest integers satisfying this quadratic Diophantine equation. Then, since $cos B$ $\frac{29}{\sqrt{29^2 + k^2}}$ . Plugging in we get $cos B = \frac{29}{421}$ , thus our answer is $\boxed{450}$ | null | 450 |
644c8e9f0765aa73b3f1dc7ca7cb92bc | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_11 | Ninety-four bricks, each measuring $4''\times10''\times19'',$ are to be stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes $4''\,$ or $10''\,$ or $19''\,$ to the total height of the tower. How many different tower heights can be achieved using all ninety-four of the bricks? | We have the smallest stack, which has a height of $94 \times 4$ inches. Now when we change the height of one of the bricks, we either add $0$ inches, $6$ inches, or $15$ inches to the height. Now all we need to do is to find the different change values we can get from $94$ $0$ 's, $6$ 's, and $15$ 's. Because $0$ $6$ , and $15$ are all multiples of $3$ , the change will always be a multiple of $3$ , so we just need to find the number of changes we can get from $0$ 's, $2$ 's, and $5$ 's.
From here, we count what we can get:
\[0, 2 = 2, 4 = 2+2, 5 = 5, 6 = 2+2+2, 7 = 5+2, 8 = 2+2+2+2, 9 = 5+2+2, \ldots\]
It seems we can get every integer greater or equal to four; we can easily deduce this by considering parity or using the Chicken McNugget Theorem , which says that the greatest number that cannot be expressed in the form of $2m + 5n$ for $m,n$ being positive integers is $5 \times 2 - 5 - 2=3$
But we also have a maximum change ( $94 \times 5$ ), so that will have to stop somewhere. To find the gaps, we can work backwards as well. From the maximum change, we can subtract either $0$ 's, $3$ 's, or $5$ 's. The maximum we can't get is $5 \times 3-5-3=7$ , so the numbers $94 \times 5-8$ and below, except $3$ and $1$ , work. Now there might be ones that we haven't counted yet, so we check all numbers between $94 \times 5-8$ and $94 \times 5$ $94 \times 5-7$ obviously doesn't work, $94 \times 5-6$ does since 6 is a multiple of 3, $94 \times 5-5$ does because it is a multiple of $5$ (and $3$ ), $94 \times 5-4$ doesn't since $4$ is not divisible by $5$ or $3$ $94 \times 5-3$ does since $3=3$ , and $94 \times 5-2$ and $94 \times 5-1$ don't, and $94 \times 5$ does.
Thus the numbers $0$ $2$ $4$ all the way to $94 \times 5-8$ $94 \times 5-6$ $94 \times 5-5$ $94 \times 5-3$ , and $94\times 5$ work. That's $2+(94 \times 5 - 8 - 4 +1)+4=\boxed{465}$ numbers. That's the number of changes you can make to a stack of bricks with dimensions $4 \times 10 \times 19$ , including not changing it at all. | null | 465 |
644c8e9f0765aa73b3f1dc7ca7cb92bc | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_11 | Ninety-four bricks, each measuring $4''\times10''\times19'',$ are to be stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes $4''\,$ or $10''\,$ or $19''\,$ to the total height of the tower. How many different tower heights can be achieved using all ninety-four of the bricks? | Using bricks of dimensions $4''\times10''\times19''$ is comparable to using bricks of dimensions $0''\times6''\times15''$ which is comparable to using bricks of dimensions $0''\times2''\times5''$ . Using 5 bricks of height $2''$ can be replaced by using 2 bricks of height $5''$ and 3 bricks of height $0''$
It follows that all tower heights can be made by using 4 or fewer bricks of height $2''$ . There are $95+94+93+92+91=465$ ways to build a tower using 4 or fewer bricks of height $2''$ . Taking the heights $\mod 5$ , we see that towers using a different number of bricks of height $2''$ have unequal heights. Thus, all of the $\boxed{465}$ tower heights are different. | null | 465 |
b6975a99467e2c794d22ce9f6b4cfab2 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_12 | A fenced, rectangular field measures $24$ meters by $52$ meters. An agricultural researcher has 1994 meters of fence that can be used for internal fencing to partition the field into congruent, square test plots. The entire field must be partitioned, and the sides of the squares must be parallel to the edges of the field. What is the largest number of square test plots into which the field can be partitioned using all or some of the 1994 meters of fence? | Suppose there are $n$ squares in every column of the grid, so there are $\frac{52}{24}n = \frac {13}6n$ squares in every row. Then $6|n$ , and our goal is to maximize the value of $n$
Each vertical fence has length $24$ , and there are $\frac{13}{6}n - 1$ vertical fences; each horizontal fence has length $52$ , and there are $n-1$ such fences. Then the total length of the internal fencing is $24\left(\frac{13n}{6}-1\right) + 52(n-1) = 104n - 76 \le 1994 \Longrightarrow n \le \frac{1035}{52} \approx 19.9$ , so $n \le 19$ . The largest multiple of $6$ that is $\le 19$ is $n = 18$ , which we can easily verify works, and the answer is $\frac{13}{6}n^2 = \boxed{702}$ | null | 702 |
b6975a99467e2c794d22ce9f6b4cfab2 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_12 | A fenced, rectangular field measures $24$ meters by $52$ meters. An agricultural researcher has 1994 meters of fence that can be used for internal fencing to partition the field into congruent, square test plots. The entire field must be partitioned, and the sides of the squares must be parallel to the edges of the field. What is the largest number of square test plots into which the field can be partitioned using all or some of the 1994 meters of fence? | Assume each partitioned square has a side length of $1$ (just so we can get a clear image of what the formula will look like). The amount of internal fencing that is required to partition the field is clearly $52*(24+1) + 24(52+1)$ . (If you are confused, just draw the square out). This is clearly greater than $1994$ , so the actual side length that we are looking for is greater than $1$
Now we can convert this into an equation. The equation is simply $(\frac{52}{x})(\frac{24}{x}+1) + (\frac{24}{x})(\frac{52}{x}+1)$ (The intutition comes from considering partioning the field into side lengths of $1$ and then partitioning those squares). This is equivalent to $\frac{2496}{x^2} + \frac{76}{x}$ , which should be less than or equal to $1994$
Now we can just find possible lengths of the square that are greater than $1$ and test them out. A viable side length would mean that $\frac{24}{x}, \frac{52}{x} \in$ N. Since $\gcd(24,52) = 4$ , then the smallest value greater than $1$ that we satisfies the conditions has $4$ in the numerator, and hence $3$ in the denominator. Test out $x=\frac{4}{3}$ . This will equate to something less than $1994$ , and hence the smallest square length that is plausible is $\frac{4}{3}$
Now the rest is elementary, we do $\frac{52}{\frac{4}{3}} * \frac{24}{\frac{4}{3}} \Rightarrow 39*18 = \boxed{702}$ | null | 702 |
e4028acce874a195c37e48c60930845d | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_13 | The equation
has 10 complex roots $r_1, \overline{r_1}, r_2, \overline{r_2}, r_3, \overline{r_3}, r_4, \overline{r_4}, r_5, \overline{r_5},\,$ where the bar denotes complex conjugation. Find the value of | Let $t = 1/x$ . After multiplying the equation by $t^{10}$ $1 + (13 - t)^{10} = 0\Rightarrow (13 - t)^{10} = - 1$
Using DeMoivre, $13 - t = \text{cis}\left(\frac {(2k + 1)\pi}{10}\right)$ where $k$ is an integer between $0$ and $9$
$t = 13 - \text{cis}\left(\frac {(2k + 1)\pi}{10}\right) \Rightarrow \bar{t} = 13 - \text{cis}\left(-\frac {(2k + 1)\pi}{10}\right)$
Since $\text{cis}(\theta) + \text{cis}(-\theta) = 2\cos(\theta)$ $t\bar{t} = 170 - 26\cos \left(\frac {(2k + 1)\pi}{10}\right)$ after expanding. Here $k$ ranges from 0 to 4 because two angles which sum to $2\pi$ are involved in the product.
The expression to find is $\sum t\bar{t} = 850 - 26\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}$
But $\cos \frac {\pi}{10} + \cos \frac {9\pi}{10} = \cos \frac {3\pi}{10} + \cos \frac {7\pi}{10} = \cos \frac {\pi}{2} = 0$ so the sum is $\boxed{850}$ | null | 850 |
e4028acce874a195c37e48c60930845d | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_13 | The equation
has 10 complex roots $r_1, \overline{r_1}, r_2, \overline{r_2}, r_3, \overline{r_3}, r_4, \overline{r_4}, r_5, \overline{r_5},\,$ where the bar denotes complex conjugation. Find the value of | Divide both sides by $x^{10}$ to get \[1 + \left(13-\dfrac{1}{x}\right)^{10}=0\]
Rearranging: \[\left(13-\dfrac{1}{x}\right)^{10} = -1\]
Thus, $13-\dfrac{1}{x} = \omega$ where $\omega = e^{i(\pi n/5+\pi/10)}$ where $n$ is an integer.
We see that $\dfrac{1}{x}=13-\omega$ . Thus, \[\dfrac{1}{x\overline{x}}=(13\, -\, \omega)(13\, -\, \overline{\omega})=169-13(\omega\, +\, \overline{\omega})\, +\, \omega\overline{\omega}=170\, -\, 13(\omega\, +\, \overline{\omega})\]
Summing over all terms: \[\dfrac{1}{r_1\overline{r_1}}+\cdots + \dfrac{1}{r_5\overline{r_5}} = 5\cdot 170 - 13(e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)})\]
However, note that $e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)}=0$ from drawing the numbers on the complex plane, our answer is just \[5\cdot 170=\boxed{850}\] | null | 850 |
045d232f164e12aaaaf3d63f18e87e22 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_14 | A beam of light strikes $\overline{BC}\,$ at point $C\,$ with angle of incidence $\alpha=19.94^\circ\,$ and reflects with an equal angle of reflection as shown. The light beam continues its path, reflecting off line segments $\overline{AB}\,$ and $\overline{BC}\,$ according to the rule: angle of incidence equals angle of reflection. Given that $\beta=\alpha/10=1.994^\circ\,$ and $AB=BC,\,$ determine the number of times the light beam will bounce off the two line segments. Include the first reflection at $C\,$ in your count.
AIME 1994 Problem 14.png | At each point of reflection, we pretend instead that the light continues to travel straight. [asy] pathpen = linewidth(0.7); size(250); real alpha = 28, beta = 36; pair B = MP("B",(0,0),NW), C = MP("C",D((1,0))), A = MP("A",expi(alpha * pi/180),N); path r = C + .4 * expi(beta * pi/180) -- C - 2*expi(beta * pi/180); D(A--B--(1.5,0));D(r);D(anglemark(C,B,D(A)));D(anglemark((1.5,0),C,C+.4*expi(beta*pi/180)));MP("\beta",B,(5,1.2),fontsize(9));MP("\alpha",C,(4,1.2),fontsize(9)); for(int i = 0; i < 180/alpha; ++i){ path l = B -- (1+i/2)*expi(-i * alpha * pi / 180); D(l, linetype("4 4")); D(IP(l,r)); } D(B); [/asy] Note that after $k$ reflections (excluding the first one at $C$ ) the extended line will form an angle $k \beta$ at point $B$ . For the $k$ th reflection to be just inside or at point $B$ , we must have $k\beta \le 180 - 2\alpha \Longrightarrow k \le \frac{180 - 2\alpha}{\beta} = 70.27$ . Thus, our answer is, including the first intersection, $\left\lfloor \frac{180 - 2\alpha}{\beta} \right\rfloor + 1 = \boxed{071}$ | null | 071 |
c29bedc35453e71703582a7061dd3026 | https://artofproblemsolving.com/wiki/index.php/1994_AIME_Problems/Problem_15 | Given a point $P^{}_{}$ on a triangular piece of paper $ABC,\,$ consider the creases that are formed in the paper when $A, B,\,$ and $C\,$ are folded onto $P.\,$ Let us call $P_{}^{}$ a fold point of $\triangle ABC\,$ if these creases, which number three unless $P^{}_{}$ is one of the vertices, do not intersect. Suppose that $AB=36, AC=72,\,$ and $\angle B=90^\circ.\,$ Then the area of the set of all fold points of $\triangle ABC\,$ can be written in the form $q\pi-r\sqrt{s},\,$ where $q, r,\,$ and $s\,$ are positive integers and $s\,$ is not divisible by the square of any prime. What is $q+r+s\,$ | Let $O_{AB}$ be the intersection of the perpendicular bisectors (in other words, the intersections of the creases) of $\overline{PA}$ and $\overline{PB}$ , and so forth. Then $O_{AB}, O_{BC}, O_{CA}$ are, respectively, the circumcenters of $\triangle PAB, PBC, PCA$ . According to the problem statement, the circumcenters of the triangles cannot lie within the interior of the respective triangles, since they are not on the paper. It follows that $\angle APB, \angle BPC, \angle CPA > 90^{\circ}$ ; the locus of each of the respective conditions for $P$ is the region inside the (semi)circles with diameters $\overline{AB}, \overline{BC}, \overline{CA}$
We note that the circle with diameter $AC$ covers the entire triangle because it is the circumcircle of $\triangle ABC$ , so it suffices to take the intersection of the circles about $AB, BC$ . We note that their intersection lies entirely within $\triangle ABC$ (the chord connecting the endpoints of the region is in fact the altitude of $\triangle ABC$ from $B$ ). Thus, the area of the locus of $P$ (shaded region below) is simply the sum of two segments of the circles. If we construct the midpoints of $M_1, M_2 = \overline{AB}, \overline{BC}$ and note that $\triangle M_1BM_2 \sim \triangle ABC$ , we see that thse segments respectively cut a $120^{\circ}$ arc in the circle with radius $18$ and $60^{\circ}$ arc in the circle with radius $18\sqrt{3}$
[asy] pair project(pair X, pair Y, real r){return X+r*(Y-X);} path endptproject(pair X, pair Y, real a, real b){return project(X,Y,a)--project(X,Y,b);} pathpen = linewidth(1); size(250); pen dots = linetype("2 3") + linewidth(0.7), dashes = linetype("8 6")+linewidth(0.7)+blue, bluedots = linetype("1 4") + linewidth(0.7) + blue; pair B = (0,0), A=(36,0), C=(0,36*3^.5), P=D(MP("P",(6,25), NE)), F = D(foot(B,A,C)); D(D(MP("A",A)) -- D(MP("B",B)) -- D(MP("C",C,N)) -- cycle); fill(arc((A+B)/2,18,60,180) -- arc((B+C)/2,18*3^.5,-90,-30) -- cycle, rgb(0.8,0.8,0.8)); D(arc((A+B)/2,18,0,180),dots); D(arc((B+C)/2,18*3^.5,-90,90),dots); D(arc((A+C)/2,36,120,300),dots); D(B--F,dots); D(D((B+C)/2)--F--D((A+B)/2),dots); D(C--P--B,dashes);D(P--A,dashes); pair Fa = bisectorpoint(P,A), Fb = bisectorpoint(P,B), Fc = bisectorpoint(P,C); path La = endptproject((A+P)/2,Fa,20,-30), Lb = endptproject((B+P)/2,Fb,12,-35); D(La,bluedots);D(Lb,bluedots);D(endptproject((C+P)/2,Fc,18,-15),bluedots);D(IP(La,Lb),blue); [/asy]
The diagram shows $P$ outside of the grayed locus; notice that the creases [the dotted blue] intersect within the triangle, which is against the problem conditions. The area of the locus is the sum of two segments of two circles; these segments cut out $120^{\circ}, 60^{\circ}$ angles by simple similarity relations and angle-chasing.
Hence, the answer is, using the $\frac 12 ab\sin C$ definition of triangle area, $\left[\frac{\pi}{3} \cdot 18^2 - \frac{1}{2} \cdot 18^2 \sin \frac{2\pi}{3} \right] + \left[\frac{\pi}{6} \cdot \left(18\sqrt{3}\right)^2 - \frac{1}{2} \cdot (18\sqrt{3})^2 \sin \frac{\pi}{3}\right] = 270\pi - 324\sqrt{3}$ , and $q+r+s = \boxed{597}$ | null | 597 |
83cd1ad378c8268971e509dbe974b8a1 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_1 | How many even integers between 4000 and 7000 have four different digits? | The thousands digit is $\in \{4,5,6\}$
Case $1$ : Thousands digit is even
$4, 6$ , two possibilities, then there are only $\frac{10}{2} - 1 = 4$ possibilities for the units digit. This leaves $8$ possible digits for the hundreds and $7$ for the tens places, yielding a total of $2 \cdot 8 \cdot 7 \cdot 4 = 448$
Case $2$ : Thousands digit is odd
$5$ , one possibility, then there are $5$ choices for the units digit, with $8$ digits for the hundreds and $7$ for the tens place. This gives $1 \cdot 8 \cdot 7 \cdot 5= 280$ possibilities.
Together, the solution is $448 + 280 = \boxed{728}$ | null | 728 |
83cd1ad378c8268971e509dbe974b8a1 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_1 | How many even integers between 4000 and 7000 have four different digits? | Firstly, we notice that the thousands digit could be $4$ $5$ or $6$ .
Since the parity of the possibilities are different, we cannot cover all cases in one operation. Therefore, we must do casework.
Case $1$
Here we let thousands digit be $4$
4 _ _ _
We take care of restrictions first, and realize that there are 4 choices for the last digit, namely $2$ $6$ $8$ and $0$ .
Now that there are no restrictions we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \cdot 7 \cdot 4 = 224$ numbers that satisfy the conditions posed by the problem.
Case $2$
Here we let thousands digit be $5$
5 _ _ _
Again, we take care of restrictions first. This time there are 5 choices for the last digit, which are all the even numbers because 5 is odd.
Now that there are no restrictions we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \cdot 7 \cdot 5 = 280$ numbers that satisfy the conditions posed by the problem.
Case $3$
Here we let thousands digit be $6$
6 _ _ _
Once again, we take care of restrictions first. Since 6 is even, there are 4 choices for the last digit, namely $2$ $6$ $8$ and $0$
Now we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \cdot 7 \cdot 4 = 224$ numbers that satisfy the conditions posed by the problem.
Now that we have our answers for each possible thousands place digit, we add up our answers and get $224+280+224$ $\boxed{728}$ ~PEKKA | null | 728 |
6e6392f54f3f43496bf5b43175ad8a64 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_2 | During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went $n^{2}_{}/2$ miles on the $n^{\mbox{th}}_{}$ day of this tour, how many miles was he from his starting point at the end of the $40^{\mbox{th}}_{}$ day? | On the first day, the candidate moves $[4(0) + 1]^2/2\ \text{east},\, [4(0) + 2]^2/2\ \text{north},\, [4(0) + 3]^2/2\ \text{west},\, [4(0) + 4]^2/2\ \text{south}$ , and so on. The E/W displacement is thus $1^2 - 3^2 + 5^2 \ldots +37^2 - 39^2 = \left|\sum_{i=0}^9 \frac{(4i+1)^2}{2} - \sum_{i=0}^9 \frac{(4i+3)^2}{2}\right|$ . Applying difference of squares , we see that \begin{align*} \left|\sum_{i=0}^9 \frac{(4i+1)^2 - (4i+3)^2}{2}\right| &= \left|\sum_{i=0}^9 \frac{(4i+1+4i+3)(4i+1-(4i+3))}{2}\right|\\ &= \left|\sum_{i=0}^9 -(8i+4) \right|. \end{align*} The N/S displacement is \[\left|\sum_{i=0}^9 \frac{(4i+2)^2}{2} - \sum_{i=0}^9 \frac{(4i+4)^2}{2}\right| = \left|\sum_{i=0}^9 -(8i+6) \right|.\] Since $\sum_{i=0}^{9} i = \frac{9(10)}{2} = 45$ , the two distances evaluate to $8(45) + 10\cdot 4 = 400$ and $8(45) + 10\cdot 6 = 420$ . By the Pythagorean Theorem , the answer is $\sqrt{400^2 + 420^2} = 29 \cdot 20 = \boxed{580}$ | null | 580 |
53e112d374764867391a4f889396fd50 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_3 | The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught $n\,$ fish for various values of $n\,$
In the newspaper story covering the event, it was reported that
What was the total number of fish caught during the festival? | Suppose that the number of fish is $x$ and the number of contestants is $y$ . The $y-(9+5+7)=y-21$ fishers that caught $3$ or more fish caught a total of $x - \left(0\cdot(9) + 1\cdot(5) + 2\cdot(7)\right) = x - 19$ fish. Since they averaged $6$ fish,
Similarily, those who caught $12$ or fewer fish averaged $5$ fish per person, so
Solving the two equation system, we find that $y = 175$ and $x = \boxed{943}$ , the answer. | null | 943 |
53e112d374764867391a4f889396fd50 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_3 | The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught $n\,$ fish for various values of $n\,$
In the newspaper story covering the event, it was reported that
What was the total number of fish caught during the festival? | Let $f$ be the total number of fish caught by the contestants who didn't catch $0, 1, 2, 3, 13, 14$ , or $15$ fish and let $a$ be the number of contestants who didn't catch $0, 1, 2, 3, 13, 14$ , or $15$ fish. From $\text{(b)}$ , we know that $\frac{69+f+65+28+15}{a+31}=6\implies f=6a+9$ . From $\text{(c)}$ we have $\frac{f+69+14+5}{a+44}=5\implies f=5a+132$ . Using these two equations gets us $a=123$ . Plug this back into the equation to get $f=747$ . Thus, the total number of fish caught is $5+14+69+f+65+28+15=\boxed{943}$ - Heavytoothpaste | null | 943 |
2bca157a6152983a5c8ba7479224a87a | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_4 | How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$ | Let $k = a + d = b + c$ so $d = k-a, b=k-c$ . It follows that $(k-c)c - a(k-a) = (a-c)(a+c-k) = (c-a)(d-c) = 93$ . Hence $(c - a,d - c) = (1,93),(3,31),(31,3),(93,1)$
Solve them in terms of $c$ to get $(a,b,c,d) = (c - 93,c - 92,c,c + 1),$ $(c - 31,c - 28,c,c + 3),$ $(c - 1,c + 92,c,c + 93),$ $(c - 3,c + 28,c,c + 31)$ . The last two solutions don't follow $a < b < c < d$ , so we only need to consider the first two solutions.
The first solution gives us $c - 93\geq 1$ and $c + 1\leq 499$ $\implies 94\leq c\leq 498$ , and the second one gives us $32\leq c\leq 496$
So the total number of such quadruples is $405 + 465 = \boxed{870}$ | null | 870 |
2bca157a6152983a5c8ba7479224a87a | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_4 | How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$ | Let $b = a + m$ and $c = a + m + n$ . From $a + d = b + c$ $d = b + c - a = a + 2m + n$
Substituting $b = a + m$ $c = a + m + n$ , and $d = b + c - a = a + 2m + n$ into $bc - ad = 93$ \[bc - ad = (a + m)(a + m + n) - a(a + 2m + n) = m(m + n). = 93 = 3(31)\] Hence, $(m,n) = (1,92)$ or $(3,28)$
For $(m,n) = (1,92)$ , we know that $0 < a < a + 1 < a + 93 < a + 94 < 500$ , so there are $405$ four-tuples. For $(m,n) = (3,28)$ $0 < a < a + 3 < a + 31 < a + 34 < 500$ , and there are $465$ four-tuples. In total, we have $405 + 465 = \boxed{870}$ four-tuples. | null | 870 |
2bca157a6152983a5c8ba7479224a87a | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_4 | How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$ | Square both sides of the first equation in order to get $bc$ and $ad$ terms, which we can plug $93$ in for. \begin{align*} (a+d)^2 = (b+c)^2 &\implies a^2 + 2ad + d^2 = b^2 + 2bc + c^2 \\ &\implies 2bc-2ad = a^2-b^2 + d^2-c^2 \\ &\implies 2(bc-ad) = (a-b)(a+b)+(d-c)(d+c) \end{align*} We can plug $93$ in for $bc - ad$ to get $186$ on the left side, and also observe that $a-b = c-d$ after rearranging the first equation. Plug in $c-d$ for $a-b$
$186 = (c-d)(a+b) + (d-c)(d+c) \implies 186 = -(d-c)(a+b) + (d-c)(d+c) \implies 186 = (d-c)(d+c-a-b)$
Now observe the possible factors of $186$ , which are $1 \cdot 186, 2\cdot 93, 3 \cdot 62, 6\cdot 31$ $(d-c)$ and $(d+c-a-b)$ must be factors of $186$ , and $(d+c-a-b)$ must be greater than $(d-c)$
$1 \cdot 186$ work, and yields $405$ possible solutions. $2 \cdot 93$ does not work, because if $c-d = 2$ , then $a+b$ must differ by 2 as well, but an odd number $93$ can only result from two numbers of different parity. $c-d$ will be even, and $a+b$ will be even, so $c+d - (a+b)$ must be even. $3 \cdot 62$ works, and yields $465$ possible solutions, while $6 \cdot 31$ fails for the same reasoning above.
Thus, the answer is $405 + 465 = \boxed{870}$ | null | 870 |
2bca157a6152983a5c8ba7479224a87a | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_4 | How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$ | Add the two conditions together to get $a+d+ad+93=b+c+bc$ . Rearranging and factorising with SFFT, $(a+1)(d+1)+93=(b+1)(c+1)$ . This implies that for every quadruple $(a,b,c,d)$ , we can replace $a\longrightarrow a+1$ $b\longrightarrow b+1$ , etc. and this will still produce a valid quadruple. This means, that we can fix $a=1$ , and then just repeatedly add $1$ to get the other quadruples.
Now, our conditions are $b+c=d+1$ and $bc=d+93$ . Replacing $d$ in the first equation, we get $bc-b-c=92$ . Factorising again with SFFT gives $(b-1)(c-1)=93$ . Since $b<c$ , we have two possible cases to consider.
Case 1: $b=2$ $c=94$ . This produces the quadruple $(1,2,94,95)$ , which indeed works.
Case 2: $b=4$ $c=32$ . This produces the quadruple $(1,4,32,35)$ , which indeed works.
Now, for case 1, we can add $1$ to each term exactly $404$ times (until we get the quadruple $(405,406,498,499)$ ), until we violate $d<500$ . This gives $405$ quadruples for case 1.
For case 2, we can add $1$ to each term exactly $464$ times (until we get the quadruple $(465,468,496,499)$ ). this gives $465$ quadruples for case 2.
In conclusion, having exhausted all cases, we can finish. There are hence $405+465=\boxed{870}$ possible quadruples. | null | 870 |
2bca157a6152983a5c8ba7479224a87a | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_4 | How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$ | Let $r = d-c$ . From the equation $a+d = b+c$ , we have \[r = d-c = b-a ,\] so $b = a+r$ and $c = d-r$ . We then have \[93 = (a+r)(d-r) - ad = rd - ra - r^2 = r(d-a-r) .\] Since $c > b$ $d-r > a+r$ , or $d-a-r > r$ . Since the prime factorization of 93 is $3 \cdot 31$ , we must either have $r=1$ and $d-a-r = 93$ , or $r=3$ and $d-a-r = 31$ . We consider these cases separately.
If $r=1$ , then $d-a = 94$ $b= a+1$ , and $c = d-1$ . Thus $d$ can be any integer between 95 and 499, inclusive, and our choice of $d$ determines the four-tuple $(a,b,c,d)$ . We therefore have $499-95+1 = 405$ possibilities in this case.
If $r=3$ , then $d-a = 34$ $b = a+3$ , and $c=d-3$ . Thus $d$ can be any integer between 35 and 499, inclusive, and our choice of $d$ determines the four-tuple $(a,b,c,d)$ , as before. We therefore have $499-35+1 = 465$ possibilities in this case.
Since there are 405 possibilities in the first case and 465 possibilities in the second case, in total there are $405 + 465 = \boxed{870}$ four-tuples. | null | 870 |
2bca157a6152983a5c8ba7479224a87a | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_4 | How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$ | Assume $d = x+m, a = x-m, c = x+n$ , and $b = x-n$ . This clearly satisfies the condition that $a+d = b+c$ since ( $2x = 2x$ ) . Now plug this into $bc-ad = 93$ . You get $(x+n)(x-n) - (x+m)(x-m) = 93 \Rightarrow m^2 - n^2 = 93 \Rightarrow (m-n)(m+n) = 93$
Since $m>n$ (as given by the condition that $a<b<c<d$ ), $m+n>m-n$ and $m$ and $n$ are integers, there are two cases we have to consider since $93 = 3\cdot 31$ . We first have to consider $m-n = 1, m+n = 93$ , and then consider $m-n=3, m+n = 31$
In the first case, we get $m=47, n=46$ and in the second case we get $m=17, n=14$ . Now plug these values (in separate cases) back into $a,b,c,d$ . Since the only restriction is that all numbers have to be greater than $0$ or less than $500$ , we can write two inequalities. Either $x+47 < 500, x-47 > 0$ , or $x+17 < 500, x-17 > 0$ (using the inequalities given by $d$ and $a$ , and since $b$ and $c$ are squeezed in between $d$ and $a$ , we only have to consider these two inequalities).
This gives us either $47 < x < 453$ or $17 < x < 483$ , and using simple counting, there are $405$ values for $x$ in the first case and $465$ values for $x$ in the second case, and hence our answer is $405+465 = \boxed{870}$ | null | 870 |
4e7e77b0e0ae821246063b13cd27511a | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_5 | Let $P_0(x) = x^3 + 313x^2 - 77x - 8\,$ . For integers $n \ge 1\,$ , define $P_n(x) = P_{n - 1}(x - n)\,$ . What is the coefficient of $x\,$ in $P_{20}(x)\,$ | Notice that \begin{align*}P_{20}(x) &= P_{19}(x - 20)\\ &= P_{18}((x - 20) - 19)\\ &= P_{17}(((x - 20) - 19) - 18)\\ &= \cdots\\ &= P_0(x - (20 + 19 + 18 + \ldots + 2 + 1)).\end{align*}
Using the formula for the sum of the first $n$ numbers, $1 + 2 + \cdots + 20 = \frac{20(20+1)}{2} = 210$ . Therefore, \[P_{20}(x) = P_0(x - 210).\]
Substituting $x - 210$ into the function definition, we get $P_0(x-210) = (x - 210)^3 + 313(x - 210)^2 - 77(x - 210) - 8$ . We only need the coefficients of the linear terms, which we can find by the binomial theorem
Adding up the coefficients, we get $630 \cdot 210 - 626 \cdot 210 - 77 = \boxed{763}$ | null | 763 |
4e7e77b0e0ae821246063b13cd27511a | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_5 | Let $P_0(x) = x^3 + 313x^2 - 77x - 8\,$ . For integers $n \ge 1\,$ , define $P_n(x) = P_{n - 1}(x - n)\,$ . What is the coefficient of $x\,$ in $P_{20}(x)\,$ | Notice the transformation of $P_{n-1}(x)\to P_n(x)$ adds $n$ to the roots. Thus, all these transformations will take the roots and add $1+2+\cdots+20=210$ to them. (Indeed, this is very easy to check in general.)
Let the roots be $r_1,r_2,r_3.$ Then $P_{20}(x)=(x-r_1-210)(x-r_2-210)(x-r_3-210).$ By Vieta's/expanding/common sense, you see the coefficient of $x$ is $(r_1+210)(r_2+210)+(r_2+210)(r_3+210)+(r_3+210)(r_1+210).$ Expanding yields $r_1r_2+r_2r_3+r_3r_1+210\cdot 2(r_1+r_2+r_3)+3\cdot 210^2.$ Using Vieta's (again) and plugging stuff in yields $-77+210\cdot 2\cdot -313+3\cdot 210^2=\boxed{763}.$ | null | 763 |
68fa04f8c7142d287d8b3884ea7a7a0b | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_6 | What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers? | Let the desired integer be $n$ . From the information given, it can be determined that, for positive integers $a, \ b, \ c$
$n = 9a + 36 = 10b + 45 = 11c + 55$
This can be rewritten as the following congruences:
$n \equiv 0 \pmod{9}$
$n \equiv 5 \pmod{10}$
$n \equiv 0 \pmod{11}$
Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is $\boxed{495}$ | null | 495 |
68fa04f8c7142d287d8b3884ea7a7a0b | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_6 | What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers? | Let $n$ be the desired integer. From the given information, we have \begin{align*}9x &= a \\ 11y &= a \\ 10z + 5 &= a, \end{align*} here, $x,$ and $y$ are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have $z$ as the 4th term of the sequence. Since, $a$ is a multiple of $9$ and $11,$ it is also a multiple of $\text{lcm}[9,11]=99.$ Hence, $a=99m,$ for some $m.$ So, we have $10z + 5 = 99m.$ It follows that $99(5) = \boxed{495}$ is the smallest integer that can be represented in such a way. | null | 495 |
68fa04f8c7142d287d8b3884ea7a7a0b | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_6 | What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers? | By the method in Solution 1, we find that the number $n$ can be written as $9a+36=10b+45=11c+55$ for some integers $a,b,c$ . From this, we can see that $n$ must be divisible by 9, 5, and 11. This means $n$ must be divisible by 495. The only multiples of 495 that are small enough to be AIME answers are 495 and 990. From the second of the three expressions above, we can see that $n$ cannot be divisible by 10, so $n$ must equal $\boxed{495}$ . Solution by Zeroman. | null | 495 |
68fa04f8c7142d287d8b3884ea7a7a0b | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_6 | What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers? | First note that the integer clearly must be divisible by $9$ and $11$ since we can use the "let the middle number be x" trick. Let the number be $99k$ for some integer $k.$ Now let the $10$ numbers be $x,x+1, \cdots x+9.$ We have $10x+45 = 99k.$ Taking mod $5$ yields $k \equiv 0 \pmod{5}.$ Since $k$ is positive, we take $k=5$ thus obtaining $99 \cdot 5 = \boxed{495}$ as our answer. | null | 495 |
be489f3abb4e054e32d7dbe3b02ce482 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_7 | Three numbers, $a_1, a_2, a_3$ , are drawn randomly and without replacement from the set $\{1, 2, 3,\ldots, 1000\}$ . Three other numbers, $b_1, b_2, b_3$ , are then drawn randomly and without replacement from the remaining set of $997$ numbers. Let $p$ be the probability that, after suitable rotation, a brick of dimensions $a_1 \times a_2 \times a_3$ can be enclosed in a box of dimension $b_1 \times b_2 \times b_3$ , with the sides of the brick parallel to the sides of the box. If $p$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator? | There is a total of $P(1000,6)$ possible ordered $6$ -tuples $(a_1,a_2,a_3,b_1,b_2,b_3).$
There are $C(1000,6)$ possible sets $\{a_1,a_2,a_3,b_1,b_2,b_3\}.$ We have five valid cases for the increasing order of these six elements:
Note that the $a$ 's are different from each other, as there are $3!=6$ ways to permute them as $a_1,a_2,$ and $a_3.$ Similarly, the $b$ 's are different from each other, as there are $3!=6$ ways to permute them as $b_1,b_2,$ and $b_3.$
So, there is a total of $C(1000,6)\cdot5\cdot6^2$ valid ordered $6$ -tuples. The requested probability is \[p=\frac{C(1000,6)\cdot5\cdot6^2}{P(1000,6)}=\frac{C(1000,6)\cdot5\cdot6^2}{C(1000,6)\cdot6!}=\frac14,\] from which the answer is $1+4=\boxed{005}.$ | null | 005 |
be489f3abb4e054e32d7dbe3b02ce482 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_7 | Three numbers, $a_1, a_2, a_3$ , are drawn randomly and without replacement from the set $\{1, 2, 3,\ldots, 1000\}$ . Three other numbers, $b_1, b_2, b_3$ , are then drawn randomly and without replacement from the remaining set of $997$ numbers. Let $p$ be the probability that, after suitable rotation, a brick of dimensions $a_1 \times a_2 \times a_3$ can be enclosed in a box of dimension $b_1 \times b_2 \times b_3$ , with the sides of the brick parallel to the sides of the box. If $p$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator? | Call the six numbers selected $x_1 > x_2 > x_3 > x_4 > x_5 > x_6$ . Clearly, $x_1$ must be a dimension of the box, and $x_6$ must be a dimension of the brick.
The total number of arrangements is ${6\choose3} = 20$ ; therefore, $p = \frac{3 + 2}{20} = \frac{1}{4}$ , and the answer is $1 + 4 = \boxed{005}$ | null | 005 |
be489f3abb4e054e32d7dbe3b02ce482 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_7 | Three numbers, $a_1, a_2, a_3$ , are drawn randomly and without replacement from the set $\{1, 2, 3,\ldots, 1000\}$ . Three other numbers, $b_1, b_2, b_3$ , are then drawn randomly and without replacement from the remaining set of $997$ numbers. Let $p$ be the probability that, after suitable rotation, a brick of dimensions $a_1 \times a_2 \times a_3$ can be enclosed in a box of dimension $b_1 \times b_2 \times b_3$ , with the sides of the brick parallel to the sides of the box. If $p$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator? | As in Solutions 2 and 3, we let $x_1>x_2>x_3>x_4>x_5>x_6$ where each $x_i$ is a number selected. It is clear that when choosing whether each number must be in the set with larger dimensions (the box) or the set with smaller dimensions (the brick) there must always be at least as many numbers in the former set as the latter. We realize that this resembles Catalan numbers, where the indices of the numbers in the first set can be replaced with rising sections of a mountain, and the other indices representing falling sections of a mountain. The formula for the $n$ th Catalan number (where $n$ is the number of pairs of rising and falling sections) is \[\frac{\binom{2n}{n}}{n+1}\] Thus, there are $\frac{\binom{6}{3}}{4}$ ways to pick which of $x_1,x_2,x_3,x_4,x_5,$ and $x_6$ are the dimensions of the box, and which are the dimensions of the brick, such that the condition is fulfilled. There are $\binom{6}{3}$ total ways to choose which numbers make up the brick and box, so the probability of the condition being fulfilled is $\frac{\binom{6}{3}/4}{\binom{6}{3}}=\frac14\Longrightarrow \boxed{005}$ | null | 005 |
d3aed27916ed9e05d7b9ac77aee6e484 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_8 | Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ? The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\},\{a, c\}.$ | Call the two subsets $m$ and $n.$ For each of the elements in $S,$ we can assign it to either $m,n,$ or both. This gives us $3^6$ possible methods of selection. However, because the order of the subsets does not matter, each possible selection is double counted, except the case where both $m$ and $n$ contain all $6$ elements of $S.$ So our final answer is then $\frac {3^6 - 1}{2} + 1 = \boxed{365}.$ | null | 365 |
d3aed27916ed9e05d7b9ac77aee6e484 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_8 | Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ? The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\},\{a, c\}.$ | Given one of ${6 \choose k}$ subsets with $k$ elements, the other also has $2^k$ possibilities; this is because it must contain all of the "missing" $n - k$ elements and thus has a choice over the remaining $k.$ We want $\sum_{k = 0}^6 {6 \choose k}2^k = (2 + 1)^6 = 729$ by the Binomial Theorem. But the order of the sets doesn't matter, so we get $\dfrac{729 - 1}{2} + 1 = \boxed{365}.$ | null | 365 |
d3aed27916ed9e05d7b9ac77aee6e484 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_8 | Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ? The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\},\{a, c\}.$ | We evaluate $f(6)$ recursively: \begin{alignat*}{6} f(0)&=1, \\ f(1)&=3f(0)-1&&=2, \\ f(2)&=3f(1)-1&&=5, \\ f(3)&=3f(2)-1&&=14, \\ f(4)&=3f(3)-1&&=41, \\ f(5)&=3f(4)-1&&=122, \\ f(6)&=3f(5)-1&&=\boxed{365} ~MRENTHUSIASM | null | 365 |
d3aed27916ed9e05d7b9ac77aee6e484 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_8 | Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ? The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\},\{a, c\}.$ | For all $n\geq1,$ we have \begin{align*} f(n) &= 3f(n-1)-1 \\ &= 3\left(3f(n-2)-1\right)-1 \\ &= 3^2f(n-2)-3-1 \\ &= 3^2\left(3f(n-3)-1\right)-3-1 \\ &= 3^3f(n-3)-3^2-3-1 \\ & \ \vdots \\ &= 3^nf(0)-3^{n-1}-3^{n-2}-3^{n-3}-\cdots-1 \\ &= 3^n-\left(3^{n-1}+3^{n-2}+3^{n-3}+\cdots+1\right) \\ &= 3^n-\frac{3^n-1}{2} \\ &= \frac{3^n+1}{2} \\ &= \frac{3^n-1}{2}+1, \end{align*} which resembles the result in Solutions 1 and 2. The answer is $f(6)=\boxed{365}.$ | null | 365 |
d3aed27916ed9e05d7b9ac77aee6e484 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_8 | Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ? The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\},\{a, c\}.$ | We can perform casework based on the number of overlapping elements. If no elements overlap, there is $\binom60=1$ way to choose the overlapping elements, and $2^{6-0}$ ways to distribute the remaining elements--each element can go in one subset or the other. We must also divide by $2$ because the order of the subsets does not matter. Proceeding similarly for the other cases, our sum is \[\dbinom{6}0 \cdot 2^5+\dbinom{6}1 \cdot 2^4+\cdots +\dbinom{6}4 \cdot 2+\dbinom{6}5 \cdot 1+\dbinom{6}6 \cdot 1.\] (Note that we have to be especially careful with the last case, as it does not follow the pattern of the other cases.) Adding these up gives a total of $\boxed{365}.$ | null | 365 |
d3aed27916ed9e05d7b9ac77aee6e484 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_8 | Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ? The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\},\{a, c\}.$ | If we wanted to, we could use casework, being very careful not to double count cases. Let's first figure out what cases we need to look at. The notation I will be using is $x\rightarrow y,$ which implies that we pick $x$ numbers for the first set which then the second set can have $y$ numbers.
Clearly: \begin{align*} 0&\rightarrow6 \\ 1&\rightarrow5\mid6 \\ 2&\rightarrow4\mid5\mid6 \\ 3&\rightarrow3\mid4\mid5\mid6 \\ 4&\rightarrow2\mid3\mid4\mid5\mid6 \\ 5&\rightarrow1\mid2\mid3\mid4\mid5\mid6 \\ 6&\rightarrow0\mid1\mid2\mid3\mid4\mid5\mid6 \end{align*} However notice that many of the cases are double counted as direction does not matter, e.g. $2\rightarrow4$ is the same as $4\rightarrow2.$ Get rid of those cases so we are just left with: \begin{align*} 0&\rightarrow6 \\ 1&\rightarrow5\mid6 \\ 2&\rightarrow4\mid5\mid6 \\ 3&\rightarrow3\mid4\mid5\mid6 \\ 4&\rightarrow4\mid5\mid6 \\ 5&\rightarrow5\mid6 \\ 6&\rightarrow6 \end{align*} Now we start computing, $0\rightarrow6$ is just $1$ case, $1\rightarrow5\mid6$ is just $\binom{6}{1}\cdot2$ cases, $2\rightarrow4\mid5\mid6$ is just $\binom{6}{2}\cdot2^2$ cases, and $3\rightarrow3\mid4\mid5\mid6$ is just $\binom{6}{3}\cdot2^3$ cases (If you have trouble understanding this, write down the six letters and then try to understand what $x\rightarrow y$ really means.). Now what you can do is continue on this same pattern like Solution 2 does, and then use simple symmetry to figure out the double counted cases. However, the purpose of this solution is to bash out the double counted cases, so we will do exactly that.
One quick thing though. We have a double counted case with the $3\rightarrow3,$ as choosing ABC and DEF is the same thing as choosing DEF and then ABC. There are $\frac{\binom{6}{3}}{2} = 10$ cases of this.
For computing $4\rightarrow4\mid5\mid6,$ we use the same process as before. We have $\binom{6}{4}\cdot(3+4+1)$ (Note, the $3$ comes from $\frac{\binom{4}{2}}{2}$ ), and for $5\rightarrow5\mid6$ we have $\binom{6}{5}\cdot\left(\frac{5}{2}+1\right),$ and then for $6\rightarrow6$ we just have $\binom{6}{6}$ (there is no double counted case since ABCDEF, ABCDEF is only counted once).
Summing case by case, we have $1+12+60+150+120+21+1 = \boxed{365}.$ | null | 365 |
79288ceb7287ce8d70bb97a7e4e4a1ab | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_9 | Two thousand points are given on a circle . Label one of the points $1$ . From this point, count $2$ points in the clockwise direction and label this point $2$ . From the point labeled $2$ , count $3$ points in the clockwise direction and label this point $3$ . (See figure.) Continue this process until the labels $1,2,3\dots,1993\,$ are all used. Some of the points on the circle will have more than one label and some points will not have a label. What is the smallest integer that labels the same point as $1993$
AIME 1993 Problem 9.png | The label $1993$ will occur on the $\frac12(1993)(1994) \pmod{2000}$ th point around the circle. (Starting from 1) A number $n$ will only occupy the same point on the circle if $\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}$
Simplifying this expression, we see that $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\equiv 0\pmod{2000}$ . Therefore, one of $1993 - n$ or $1994 + n$ is odd, and each of them must be a multiple of $125$ or $16$
For $1993 - n$ to be a multiple of $125$ and $1994 + n$ to be a multiple of $16$ $n \equiv 118 \pmod {125}$ and $n\equiv 6 \pmod {16}$ . The smallest $n$ for this case is $118$
In order for $1993 - n$ to be a multiple of $16$ and $1994 + n$ to be a multiple of $125$ $n\equiv 9\pmod{16}$ and $n\equiv 6\pmod{125}$ . The smallest $n$ for this case is larger than $118$ , so $\boxed{118}$ is our answer. | null | 118 |
79288ceb7287ce8d70bb97a7e4e4a1ab | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_9 | Two thousand points are given on a circle . Label one of the points $1$ . From this point, count $2$ points in the clockwise direction and label this point $2$ . From the point labeled $2$ , count $3$ points in the clockwise direction and label this point $3$ . (See figure.) Continue this process until the labels $1,2,3\dots,1993\,$ are all used. Some of the points on the circle will have more than one label and some points will not have a label. What is the smallest integer that labels the same point as $1993$
AIME 1993 Problem 9.png | Two labels $a$ and $b$ occur on the same point if $\ a(a+1)/2\equiv \ b(b+1)/2\pmod{2000}$ . If we assume the final answer be $n$ , then we have $\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}$
Multiply $2$ on both side we have $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\equiv 0\pmod{4000}$ . As they have different parities, the even one must be divisible by $32$ . As $(1993 - n)+(1994 + n)\equiv 2\pmod{5}$ , one of them is divisible by $5$ , which indicates it's divisible by $125$
Which leads to four different cases: $1993-n\equiv 0\pmod{4000}$ $1994+n\equiv 0\pmod{4000}$ $1993-n\equiv 0\pmod{32}$ and $1994+n\equiv 0\pmod{125}$ $1993-n\equiv 0\pmod{125}$ and $1994+n\equiv 0\pmod{32}$ . Which leads to $n\equiv 1993,2006,3881$ and $118\pmod{4000}$ respectively, and only $n=118$ satisfied.Therefore answer is $\boxed{118}$ .(by ZJY) | null | 118 |
4a4a9f8e28c94ed3039285e00e38a9a2 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_10 | Euler's formula states that for a convex polyhedron with $V$ vertices $E$ edges , and $F$ faces $V-E+F=2$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon . At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$ | The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with $12$ equilateral pentagons) in which the $20$ vertices have all been truncated to form $20$ equilateral triangles with common vertices. The resulting solid has then $p=12$ smaller equilateral pentagons and $t=20$ equilateral triangles yielding a total of $t+p=F=32$ faces. In each vertex, $T=2$ triangles and $P=2$ pentagons are concurrent. Now, the number of edges $E$ can be obtained if we count the number of sides that each triangle and pentagon contributes: $E=\frac{3t+5p}{2}$ , (the factor $2$ in the denominator is because we are counting twice each edge, since two adjacent faces share one edge). Thus, $E=60$ . Finally, using Euler's formula we have $V=E-30=30$
In summary, the solution to the problem is $100P+10T+V=\boxed{250}$ | null | 250 |
4a4a9f8e28c94ed3039285e00e38a9a2 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_10 | Euler's formula states that for a convex polyhedron with $V$ vertices $E$ edges , and $F$ faces $V-E+F=2$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon . At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$ | As seen above, $E=V+30$ . Every vertex $V$ , there is a triangle for every $T$ and a pentagon for every $P$ by the given. However, there are three times every triangle will be counted and five times every pentagon will be counted because of their numbers of vertices. From this observation, $\frac{VT}3+\frac{VP}5=32\implies V(5T+3P)=480$ . Also, at every vertex $V$ , there are $T+P$ edges coming out from that vertex (one way to see this is to imagine the leftmost segment of each triangle and pentagon that is connected to the given vertex, and note that it includes every one of the edges exactly once), so $\frac{V(T+P)}2=E\implies V(T+P)=2E\implies V(5T+5P)=10E$ , and subtracting the other equation involving the vertices from this gives $2VP=10E-480\implies VP=5E-240=5(V+30)-240=5V-90$ $\implies V(5-P)=90$ .
Since $V|480$ from the first vertex-related observation and $P>0\implies5-P<5$ , and it quickly follows that $V=30\implies E=60\implies P=2\implies T=2\implies100P+10T+V=\boxed{250}$ | null | 250 |
4a4a9f8e28c94ed3039285e00e38a9a2 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_10 | Euler's formula states that for a convex polyhedron with $V$ vertices $E$ edges , and $F$ faces $V-E+F=2$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon . At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$ | Notice that at each vertex, we must have the sum of the angles be less than $360$ degrees or we will not be able to fold the polyhedron. Therefore, we have $60T + 108P < 360.$ Now, let there be $t$ triangles and $p$ pentagons total such that $t+p = 32.$ From the given, we know that $E = V + 30.$ Lastly, we see that $E = \frac{3t+5p}{2}$ and $V = \frac{3t}{T}=\frac{5p}{P}.$
Now, we do casework on what $P$ is.
Case 1: $P = 2$ Notice that we must have $t$ and $p$ integral. Trying $T = 1, 2$ yields a solution with $t=2.$ Trying other cases of $P$ and $T$ yields no solutions. Therefore, $T=2, P=2$ and after solving for $t, p$ we get $V=30.$ Finally, we have $100P+10T+V = \boxed{250}$ | null | 250 |
4a4a9f8e28c94ed3039285e00e38a9a2 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_10 | Euler's formula states that for a convex polyhedron with $V$ vertices $E$ edges , and $F$ faces $V-E+F=2$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon . At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$ | We know that $V-E = -30 \implies V = E-30$ based off the problem condition. Furthermore, if we draw out a few pentagons as well as triangles on each of side of the pentagons, it's clear that each vertex has 4 edges connected to it, with two triangles and two pentagons for each vertex. However, each edge is used for two vertices and thus counted twice. Therefore, we have that $E = \frac{4V}{2} = 2V$ . Plugging this in, we have that $V=30$ and so our answer is $200+20+30 = \boxed{250}$ | null | 250 |
81676e7bfc491c56ca8e61967afa7c61 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_11 | Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is $m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. What are the last three digits of $m+n\,$ | In order to begin this problem, we need to calculate the probability that Alfred will win on the first round.
Because he goes first, Alfred has a $\frac{1}{2}$ chance of winning (getting heads) on his first flip.
Then, Bonnie, who goes second, has a $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ , chance of winning on her first coin toss.
Therefore Alfred’s chance of winning on his second flip is $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$
From this, we can see that Alfred’s (who goes first) chance of winning the first round is: $\frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \cdots = \frac{2}{3}$
Bonnie’s (who goes second) chance of winning the first round is then $1 - \frac{2}{3} = \frac{1}{3}$
This means that the person who goes first has a $\frac{2}{3}$ chance of winning the round, while the person who goes second has a $\frac{1}{3}$ chance of winning.
Now, through casework, we can calculate Alfred’s chance of winning the second round.
Case 1: Alfred wins twice; $\frac{2}{3} \times \frac{1}{3}$ (Bonnie goes first this round) $=\frac{2}{9}$
Case 2: Alfred loses the first round, but wins the second; $\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}$
Adding up the cases, we get $\frac{2}{9} + \frac{2}{9} = \frac{4}{9}$
Alfred, therefore, has a $\frac{4}{9}$ of winnning the second round, and Bonnie has a $1-\frac{4}{9} = \frac{5}{9}$ chance of winning this round.
From here, it is not difficult to see that the probabilities alternate in a pattern.
Make $A$ the probability that Alfred wins a round.
The chances of Alfred and Bonnie, respectively, winning the first round are $A$ and $A - \frac{1}{3}$ , which can be written as $2A - \frac{1}{3} = 1$
The second round’s for chances are $A$ and $A + \frac{1}{9}$ , which can also be written as $2A + \frac{1}{9}$
From this, we can conclude that for the $n$ th even round, the probability that Alfred ( $A$ ) wins can be calculated through the equation $2A + \frac{1}{3^n} = 1$
Solving the equation for $n = 6$ , we get $A = \frac{364}{729}$
$364 + 729 = 1093$
So our answer is $\boxed{093}$ | null | 093 |
81676e7bfc491c56ca8e61967afa7c61 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_11 | Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is $m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. What are the last three digits of $m+n\,$ | Rather than categorizing games as wins or losses, we can categorize them as starters (S), where Alfred starts, and non-starters (NS), where Bonnie starts. Game 1 is a starter, and since Alfred must win Game 6, Game 7 is a non-starter.
As shown in Solution 1, if a player starts a certain game, the probability $P(NS)$ that they will not start the next game (i.e. win the current game) is $\frac{2}{3}$ , and the probability $P(S)$ that they will start the next game (i.e. lose the current game) is $\frac{1}{3}$ . Similarly, if a player does not start a certain game, $P(NS) = \frac{1}{3}$ and $P(S) = \frac{2}{3}$ . We conclude that the probability of switching from S to NS or vice versa is always $\frac{2}{3}$ , and the probability of staying the same is always $\frac{1}{3}$
Listing out all the games from Game 1 to Game 7, we get: S, ?, ?, ?, ?, ?, NS. Between the 7 games, there are 6 opportunities to either switch or stay the same. We need to eventually switch from S to NS, so there must be an odd number of switches. Furthermore, this number is less than 6, so it must be 1, 3, or 5.
1 switch: There are ${6 \choose 1} = 6$ ways to place the switch, so $P = 6\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right) = \frac{12}{729}$
3 switches: There are ${6 \choose 3} = 20$ ways to place the switches, so $P = 20\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^3 = \frac{160}{729}$
5 switches: There are ${6 \choose 5} = 6$ ways to place the switch, so $P = 6\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^5 = \frac{192}{729}$
Add up all these probabilities to get $\frac{12+160+192}{729} = \frac{364}{729}$ $364+729=1093$ , so the answer is $\boxed{093}$ | null | 093 |
8c847d4a261f33e055f55e83c0796b7d | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_12 | The vertices of $\triangle ABC$ are $A = (0,0)\,$ $B = (0,420)\,$ , and $C = (560,0)\,$ . The six faces of a die are labeled with two $A\,$ 's, two $B\,$ 's, and two $C\,$ 's. Point $P_1 = (k,m)\,$ is chosen in the interior of $\triangle ABC$ , and points $P_2\,$ $P_3\,$ $P_4, \dots$ are generated by rolling the die repeatedly and applying the rule: If the die shows label $L\,$ , where $L \in \{A, B, C\}$ , and $P_n\,$ is the most recently obtained point, then $P_{n + 1}^{}$ is the midpoint of $\overline{P_n L}$ . Given that $P_7 = (14,92)\,$ , what is $k + m\,$ | If we have points $(p,q)$ and $(r,s)$ and we want to find $(u,v)$ so $(r,s)$ is the midpoint of $(u,v)$ and $(p,q)$ , then $u=2r-p$ and $v=2s-q$ . So we start with the point they gave us and work backwards. We make sure all the coordinates stay within the triangle. We have \[P_{n-1}=(x_{n-1},y_{n-1}) = (2x_n\bmod{560},\ 2y_n\bmod{420})\] Then $P_7=(14,92)$ , so $x_7=14$ and $y_7=92$ , and we get \[\begin{array}{c||ccccccc} n & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\ \hline\hline x_n & 14 & 28 & 56 & 112 & 224 & 448 & 336 \\ \hline y_n & 92 & 184 & 368 & 316 & 212 & 4 & 8 \end{array}\]
So the answer is $\boxed{344}$ | null | 344 |
8c847d4a261f33e055f55e83c0796b7d | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_12 | The vertices of $\triangle ABC$ are $A = (0,0)\,$ $B = (0,420)\,$ , and $C = (560,0)\,$ . The six faces of a die are labeled with two $A\,$ 's, two $B\,$ 's, and two $C\,$ 's. Point $P_1 = (k,m)\,$ is chosen in the interior of $\triangle ABC$ , and points $P_2\,$ $P_3\,$ $P_4, \dots$ are generated by rolling the die repeatedly and applying the rule: If the die shows label $L\,$ , where $L \in \{A, B, C\}$ , and $P_n\,$ is the most recently obtained point, then $P_{n + 1}^{}$ is the midpoint of $\overline{P_n L}$ . Given that $P_7 = (14,92)\,$ , what is $k + m\,$ | Let $L_1$ be the $n^{th}$ roll that directly influences $P_{n + 1}$
Note that $P_7 = \cfrac{\cfrac{\cfrac{P_1 + L_1}2 + L_2}2 + \cdots}{2\ldots} = \frac {(k,m)}{64} + \frac {L_1}{64} + \frac {L_2}{32} + \frac {L_3}{16} + \frac {L_4}8 + \frac {L_5}4 + \frac {L_6}2 = (14,92)$
Then quickly checking each addend from the right to the left, we have the following information (remembering that if a point must be $(0,0)$ , we can just ignore it!):
for $\frac {L_6}2,\frac {L_5}4$ , since all addends are nonnegative, a non- $(0,0)$ value will result in a $x$ or $y$ value greater than $14$ or $92$ , respectively, and we can ignore them,
for $\frac {L_4}8,\frac {L_3}{16},\frac {L_2}{32}$ in a similar way, $(0,0)$ and $(0,420)$ are the only possibilities,
and for $\frac {L_1}{64}$ , all three work.
Also, to be in the triangle, $0\le k\le560$ and $0\le m\le420$
Since $L_1$ is the only point that can possibly influence the $x$ coordinate other than $P_1$ , we look at that first.
If $L_1 = (0,0)$ , then $k = 2^6\cdot14 = 64\cdot14 > 40\cdot14 = 560$
so it can only be that $L_1 = (560,0)$ , and $k + 560 = 2^6\cdot14$
$\implies k = 64\cdot14 - 40\cdot14 = 24\cdot14 = 6\cdot56 = 336$
Now, considering the $y$ coordinate, note that if any of $L_2,L_3,L_4$ are $(0,0)$ $L_2$ would influence the least, so we test that),
then $\frac {L_2}{32} + \frac {L_3}{16} + \frac {L_4}8 < \frac {420}{16} + \frac {420}8 = 79\pm\epsilon < 80$
which would mean that $P_1 > 2^6\cdot(92 - 80) = 64\cdot12 > 42\cdot10 = 420\ge m$ , so $L_2,L_3,L_4 = (0,420)$
and now $\frac {P_1}{64} + \frac {420}{2^5} + \frac {420}{2^4} + \frac {420}{2^3} = 92$
$\implies P_1$
$= 64\cdot92 - 420(2 + 4 + 8)$
$= 64\cdot92 - 420\cdot14= 64(100 - 8) - 14^2\cdot30$
$= 6400 - 512 - (200 - 4)\cdot30$
$= 6400 - 512 - 6000 + 120$
$= - 112 + 120$
$= 8$
and finally, $k + m = 336 + 8 = \boxed{344}$ | null | 344 |
20a88d5b9e4171ba5e0ffc6851ebe9da | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_13 | Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let $t\,$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t\,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator? | Consider the unit cicle of radius 50. Assume that they start at points $(-50,100)$ and $(-50,-100).$ Then at time $t$ , they end up at points $(-50+t,100)$ and $(-50+3t,-100).$ The equation of the line connecting these points and the equation of the circle are \begin{align}y&=-\frac{100}{t}x+200-\frac{5000}{t}\\50^2&=x^2+y^2\end{align}. When they see each other again, the line connecting the two points will be tangent to the circle at the point $(x,y).$ Since the radius is perpendicular to the tangent we get \[-\frac{x}{y}=-\frac{100}{t}\] or $xt=100y.$ Now substitute \[y= \frac{xt}{100}\] into $(2)$ and get \[x=\frac{5000}{\sqrt{100^2+t^2}}.\] Now substitute this and \[y=\frac{xt}{100}\] into $(1)$ and solve for $t$ to get \[t=\frac{160}{3}.\] Finally, the sum of the numerator and denominator is $160+3=\boxed{163}.$ | null | 163 |
20a88d5b9e4171ba5e0ffc6851ebe9da | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_13 | Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let $t\,$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t\,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator? | Let $A$ and $B$ be Kenny's initial and final points respectively and define $C$ and $D$ similarly for Jenny. Let $O$ be the center of the building. Also, let $X$ be the intersection of $AC$ and $BD$ . Finaly, let $P$ and $Q$ be the points of tangency of circle $O$ to $AC$ and $BD$ respectively.
From the problem statement, $AB=3t$ , and $CD=t$ . Since $\Delta ABX \sim \Delta CDX$ $CX=AC\cdot\left(\frac{CD}{AB-CD}\right)=200\cdot\left(\frac{t}{3t-t}\right)=100$
Since $PC=100$ $PX=200$ . So, $\tan(\angle OXP)=\frac{OP}{PX}=\frac{50}{200}=\frac{1}{4}$
Since circle $O$ is tangent to $BX$ and $AX$ $OX$ is the angle bisector of $\angle BXA$
Thus, $\tan(\angle BXA)=\tan(2\angle OXP)=\frac{2\tan(\angle OXP)}{1- \tan^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}=\frac{8}{15}$
Therefore, $t = CD = CX\cdot\tan(\angle BXA) = 100 \cdot \frac{8}{15} = \frac{160}{3}$ , and the answer is $\boxed{163}$ | null | 163 |
20a88d5b9e4171ba5e0ffc6851ebe9da | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_13 | Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let $t\,$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t\,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator? | [asy] size(8cm); defaultpen(linewidth(0.7)); pair A,B,C,D,P,Q,O,R,S; A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); R=(100,106.667); S=(0,53.333); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(R); dot(S); draw(A--B--D--C--cycle); draw(P--O); draw(D--S); draw(O--Q--R--cycle); draw(Circle(O,50)); label("$A$",A,SW); label("$B$",B,NNW); label("$C$",(200,-205),S); label("$D$",D,NE); label("$P$",(100,-205),S); label("$Q$",Q,NE); label("$O$",O,SW); label("$R$",R,NE); label("$S$",S,W); [/asy]
Let $t$ be the time they walk. Then $CD=t$ and $AB=3t$
Draw a line from point $O$ to $Q$ such that $OQ$ is perpendicular to $BD$ . Further, draw a line passing through points $O$ and $P$ , so $OP$ is parallel to $AB$ and $CD$ and is midway between those two lines. Then $PR=\dfrac{AB+CD}{2}=\dfrac{3t+t}{2}=2t$ . Draw another line passing through point $D$ and parallel to $AC$ , and call the point of intersection of this line with $AB$ as $S$ . Then $SB=AB-AS=3t-t=2t$
We see that $m\angle SBD=m\angle ORQ$ since they are corresponding angles, and thus by angle-angle similarity, $\triangle QOR\sim\triangle SDB$
Then \begin{align*} \dfrac{OQ}{DS}=\dfrac{RO}{BD}&\implies\dfrac{50}{200}=\dfrac{RO}{\sqrt{200^2+4t^2}}\\ &\implies RO=\dfrac{1}{4}\left(\sqrt{200^2+4t^2}\right)\\ &\implies RO=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right) \end{align*}
And we obtain \begin{align*} PR-OP&=RO\\ 2t-50&=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right)\\ 4t-100&=\sqrt{100^2+t^2}\\ (4t-100)^2&=\left(\sqrt{100^2+t^2}\right)^2\\ 16t^2-800t+100^2&=t^2+100^2\\ 15t^2&=800t\\ t&=\dfrac{800}{15} \end{align*}
so we have $t=\frac{160}{3}$ , and our answer is thus $160+3=\boxed{163}$ | null | 163 |
ec196b7d665851565e1f1a2442a27372 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_14 | A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form $\sqrt{N}\,$ , for a positive integer $N\,$ . Find $N\,$ | Put the rectangle on the coordinate plane so its vertices are at $(\pm4,\pm3)$ , for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, $O$
Note that such a rectangle is unstuck if its four vertices are in or on the edge of all four quadrants, and it is not the same rectangle as the big one. Let the four vertices of this rectangle be $A(4,y)$ $B(-x,3)$ $C(-4,-y)$ and $D(x,-3)$ for nonnegative $x,y$ . Then this is a rectangle, so $OA=OB$ , or $16+y^2=9+x^2$ , so $x^2=y^2+7$
Reflect $D$ across the side of the rectangle containing $C$ to $D'(-8-x,-3)$ . Then $BD'=\sqrt{(-8-x-(-x))^2+(3-(-3))^2}=10$ is constant, and the perimeter of the rectangle is equal to $2(BC+CD')$ . The midpoint of $\overline{BD'}$ is $(-4-x,0)$ , and since $-4>-4-x$ and $-y\le0$ $C$ always lies below $\overline{BD'}$
If $y$ is positive, it can be decreased to $y'<y$ . This causes $x$ to decrease as well, to $x'$ , where $x'^2=y'^2+7$ and $x'$ is still positive. If $B$ and $D'$ are held in place as everything else moves, then $C$ moves $(y-y')$ units up and $(x-x')$ units left to $C'$ , which must lie within $\triangle BCD'$ . Then we must have $BC'+C'D'<BC+CD'$ , and the perimeter of the rectangle is decreased. Therefore, the minimum perimeter must occur with $y=0$ , so $x=\sqrt7$
By the distance formula, this minimum perimeter is \[2\left(\sqrt{(4-\sqrt7)^2+3^2}+\sqrt{(4+\sqrt7)^2+3^2}\right)=4\left(\sqrt{8-2\sqrt7}+\sqrt{8+2\sqrt7}\right)\] \[=4(\sqrt7-1+\sqrt7+1)=8\sqrt7=\sqrt{448}.\] Therefore $N$ would equal $\boxed{448}.$ | null | 448 |
ec196b7d665851565e1f1a2442a27372 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_14 | A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form $\sqrt{N}\,$ , for a positive integer $N\,$ . Find $N\,$ | Note that the diagonal of the rectangle with minimum perimeter must have the diagonal along the middle segment of length 8 of the rectangle (any other inscribed rectangle can be rotated a bit, then made smaller; this one can't because then the rectangle cannot be inscribed since its longest diagonal is less than 8 in length). Then since a rectangle must have right angles, we draw a circle of radius 4 around the center of the rectangle. Picking the two midpoints on the sides of length 6 and opposite intersection points on the segments of length 8, we form a rectangle. Let $a$ and $b$ be the sides of the rectangle. Then $ab = 3(8) = 24$ since both are twice the area of the same right triangle, and $a^2+b^2 = 64$ . So $(a+b)^2 = 64+2(24) = 112$ , so $2(a+b) = \sqrt{\boxed{448}$ | null | 448 |
d7f61bc2ccb412295bb3763d401acb36 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_15 | Let $\overline{CH}$ be an altitude of $\triangle ABC$ . Let $R\,$ and $S\,$ be the points where the circles inscribed in the triangles $ACH\,$ and $BCH^{}_{}$ are tangent to $\overline{CH}$ . If $AB = 1995\,$ $AC = 1994\,$ , and $BC = 1993\,$ , then $RS\,$ can be expressed as $m/n\,$ , where $m\,$ and $n\,$ are relatively prime integers. Find $m + n\,$ | [asy] unitsize(48); pair A,B,C,H; A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,N); label("$H$",H,NE); draw(circle((2,1),1)); pair [] x=intersectionpoints(C--H,circle((2,1),1)); dot(x[0]); label("$S$",x[0],SW); draw(circle((4.29843788128,1.29843788128),1.29843788128)); pair [] y=intersectionpoints(C--H,circle((4.29843788128,1.29843788128),1.29843788128)); dot(y[0]); label("$R$",y[0],NE); label("$1993$",(1.5,2),NW); label("$1994$",(5.5,2),NE); label("$1995$",(4,0),S); [/asy]
From the Pythagorean Theorem $AH^2+CH^2=1994^2$ , and $(1995-AH)^2+CH^2=1993^2$
Subtracting those two equations yields $AH^2-(1995-AH)^2=3987$
After simplification, we see that $2*1995AH-1995^2=3987$ , or $AH=\frac{1995}{2}+\frac{3987}{2*1995}$
Note that $AH+BH=1995$
Therefore we have that $BH=\frac{1995}{2}-\frac{3987}{2*1995}$
Therefore $AH-BH=\frac{3987}{1995}$
Now note that $RS=|HR-HS|$ $RH=\frac{AH+CH-AC}{2}$ , and $HS=\frac{CH+BH-BC}{2}$
Therefore we have $RS=\left| \frac{AH+CH-AC-CH-BH+BC}{2} \right|=\frac{|AH-BH-1994+1993|}{2}$
Plugging in $AH-BH$ and simplifying, we have $RS=\frac{1992}{1995*2}=\frac{332}{665} \rightarrow 332+665=\boxed{997}$ | null | 997 |
b3efdc28d7aa9eb5c447623a6ed3eaab | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_1 | Find the sum of all positive rational numbers that are less than 10 and that have denominator 30 when written in lowest terms | There are 8 fractions which fit the conditions between 0 and 1: $\frac{1}{30},\frac{7}{30},\frac{11}{30},\frac{13}{30},\frac{17}{30},\frac{19}{30},\frac{23}{30},\frac{29}{30}$
Their sum is 4. Note that there are also 8 terms between 1 and 2 which we can obtain by adding 1 to each of our first 8 terms. For example, $1+\frac{19}{30}=\frac{49}{30}.$ Following this pattern, our answer is $4(10)+8(1+2+3+\cdots+9)=\boxed{400}.$ | null | 400 |
b3efdc28d7aa9eb5c447623a6ed3eaab | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_1 | Find the sum of all positive rational numbers that are less than 10 and that have denominator 30 when written in lowest terms | Note that if $x$ is a solution, then $(300-x)$ is a solution. We know that $\phi(300) = 80.$ Therefore the answer is $\frac{80}{2} \cdot\frac{300}{30} = \boxed{400}.$ | null | 400 |
2cc1ff361b05ac72a53e65b26dce0f0d | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_2 | positive integer is called ascending if, in its decimal representation , there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are there? | Note that an ascending number is exactly determined by its digits : for any set of digits (not including 0, since the only position for 0 is at the leftmost end of the number, i.e. a leading 0), there is exactly one ascending number with those digits.
So, there are nine digits that may be used: $1,2,3,4,5,6,7,8,9.$ Note that each digit may be present or may not be present. Hence, there are $2^9=512$ potential ascending numbers, one for each subset of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$
However, we've counted one-digit numbers and the empty set , so we must subtract them off to get our answer, $512-10=\boxed{502}.$ | null | 502 |
f31fb7045abc2d37976c951d3aaa4158 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_3 | A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly $.500$ . During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than $.503$ . What's the largest number of matches she could've won before the weekend began? | Let $n$ be the number of matches won, so that $\frac{n}{2n}=\frac{1}{2}$ , and $\frac{n+3}{2n+4}>\frac{503}{1000}$
Cross multiplying $1000n+3000>1006n+2012$ , so $n<\frac{988}{6}=164 \dfrac {4}{6}=164 \dfrac{2}{3}$ . Thus, the answer is $\boxed{164}$ | null | 164 |
f31fb7045abc2d37976c951d3aaa4158 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_3 | A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly $.500$ . During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than $.503$ . What's the largest number of matches she could've won before the weekend began? | Let $n$ be the number of matches she won before the weekend began. Since her win ratio started at exactly . $500 = \tfrac{1}{2},$ she must have played exactly $2n$ games total before the weekend began. After the weekend, she would have won $n+3$ games out of $2n+4$ total. Therefore, her win ratio would be $(n+3)/(2n+4).$ This means that \[\frac{n+3}{2n+4} > .503 = \frac{503}{1000}.\] Cross-multiplying, we get $1000(n+3) > 503(2n+4),$ which is equivalent to $n < \frac{988}{6} = 164.\overline{6}.$ Since $n$ must be an integer, the largest possible value for $n$ is $\boxed{164}.$ | null | 164 |
7566d50acf6505d9f6df3cc96825b89d | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_4 | In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below.
\[\begin{array}{c@{\hspace{8em}} c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} c@{\hspace{6pt}}c@{\hspace{6pt}}c} \vspace{4pt} \text{Row 0: } & & & & & & & 1 & & & & & & \\\vspace{4pt} \text{Row 1: } & & & & & & 1 & & 1 & & & & & \\\vspace{4pt} \text{Row 2: } & & & & & 1 & & 2 & & 1 & & & & \\\vspace{4pt} \text{Row 3: } & & & & 1 & & 3 & & 3 & & 1 & & & \\\vspace{4pt} \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt} \text{Row 5: } & & 1 & & 5 & &10& &10 & & 5 & & 1 & \\\vspace{4pt} \text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1 \end{array}\] In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3 :4 :5$ | Consider what the ratio means. Since we know that they are consecutive terms, we can say \[\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.\]
Taking the first part, and using our expression for $n$ choose $k$ \[\frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}\] \[\frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!}\] \[\frac{1}{3(n-k+1)} = \frac{1}{4k}\] \[n-k+1 = \frac{4k}{3}\] \[n = \frac{7k}{3} - 1\] \[\frac{3(n+1)}{7} = k\] Then, we can use the second part of the equation. \[\frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!}\] \[\frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!}\] \[\frac{1}{4(n-k)} = \frac{1}{5(k+1)}\] \[\frac{4(n-k)}{5} = k+1\] \[\frac{4n}{5}-\frac{4k}{5} = k+1\] \[\frac{4n}{5} = \frac{9k}{5} +1.\] Since we know $k = \frac{3(n+1)}{7}$ we can plug this in, giving us \[\frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1\] \[4n = 9\left(\frac{3(n+1)}{7}\right)+5\] \[7(4n - 5) = 27n+27\] \[28n - 35 = 27n+27\] \[n = 62\] We can also evaluate for $k$ , and find that $k = \frac{3(62+1)}{7} = 27.$ Since we want $n$ , however, our final answer is $\boxed{062.}$ $\LaTeX$ by ciceronii | null | 062. |
7566d50acf6505d9f6df3cc96825b89d | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_4 | In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below.
\[\begin{array}{c@{\hspace{8em}} c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} c@{\hspace{6pt}}c@{\hspace{6pt}}c} \vspace{4pt} \text{Row 0: } & & & & & & & 1 & & & & & & \\\vspace{4pt} \text{Row 1: } & & & & & & 1 & & 1 & & & & & \\\vspace{4pt} \text{Row 2: } & & & & & 1 & & 2 & & 1 & & & & \\\vspace{4pt} \text{Row 3: } & & & & 1 & & 3 & & 3 & & 1 & & & \\\vspace{4pt} \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt} \text{Row 5: } & & 1 & & 5 & &10& &10 & & 5 & & 1 & \\\vspace{4pt} \text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1 \end{array}\] In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3 :4 :5$ | Call the row $x=t+k$ , and the position of the terms $t-1, t, t+1$ . Call the middle term in the ratio $N = \dbinom{t+k}{t} = \frac{(t+k)!}{k!t!}$ . The first term is $N \frac{t}{k+1}$ , and the final term is $N \frac{k}{t+1}$ . Because we have the ratio $3:4:5$
$\frac{t}{k+1} = \frac{3}{4}$ and $\frac{k}{t+1} = \frac{5}{4}$
$4t = 3k+3$ and $4k= 5t+5$
$4t-3k=3$ $5t-4k=-5$
Solve the equations to get $t= 27, k=35$ and $x = t+k = \boxed{062}$ | null | 062 |
24419649e24d083a5883025a6e5aa867 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_5 | Let $S^{}_{}$ be the set of all rational numbers $r^{}_{}$ $0^{}_{}<r<1$ , that have a repeating decimal expansion in the form $0.abcabcabc\ldots=0.\overline{abc}$ , where the digits $a^{}_{}$ $b^{}_{}$ , and $c^{}_{}$ are not necessarily distinct. To write the elements of $S^{}_{}$ as fractions in lowest terms, how many different numerators are required? | We consider the method in which repeating decimals are normally converted to fractions with an example:
$x=0.\overline{176}$
$\Rightarrow 1000x=176.\overline{176}$
$\Rightarrow 999x=1000x-x=176$
$\Rightarrow x=\frac{176}{999}$
Thus, let $x=0.\overline{abc}$
$\Rightarrow 1000x=abc.\overline{abc}$
$\Rightarrow 999x=1000x-x=abc$
$\Rightarrow x=\frac{abc}{999}$
If $abc$ is not divisible by $3$ or $37$ , then this is in lowest terms. Let us consider the other multiples: $333$ multiples of $3$ $27$ of $37$ , and $9$ of both $3$ and $37$ , so $999-333-27+9 = 648$ , which is the amount that are neither. The $12$ numbers that are multiples of $81$ reduce to multiples of $3$ . We have to count these since it will reduce to a multiple of $3$ which we have removed from $999$ , but, this cannot be removed since the numerator cannot cancel the $3$ .There aren't any numbers which are multiples of $37^2$ , so we can't get numerators which are multiples of $37$ . Therefore $648 + 12 = \boxed{660}$ | null | 660 |
80d57b864147b627c2741c5e8830a270 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_6 | For how many pairs of consecutive integers in $\{1000,1001,1002,\ldots,2000\}$ is no carrying required when the two integers are added? | For one such pair of consecutive integers, let the smaller integer be $\underline{1ABC},$ where $A,B,$ and $C$ are digits from $0$ through $9.$
We wish to count the ordered triples $(A,B,C).$ By casework, we consider all possible forms of the larger integer, as shown below. \[\begin{array}{c|c|c|c|c|c|c} & & & & & & \\ [-2.5ex] \textbf{Case} & \textbf{Thousands} & \textbf{Hundreds} & \textbf{Tens} & \textbf{Ones} & \textbf{Conditions for No Carrying} & \boldsymbol{\#}\textbf{ of Ordered Triples} \\ [0.5ex] \hline & & & & & & \\ [-2ex] 0\leq C\leq 8 & 1 & A & B & C+1 & 0\leq A,B,C\leq 4 & 5^3 \\ 0\leq B\leq 8; \ C=9 & 1 & A & B+1 & 0 & 0\leq A,B\leq 4; \ C=9 & 5^2 \\ 0\leq A\leq 8; \ B=C=9 & 1 & A+1 & 0 & 0 & 0\leq A\leq 4; \ B=C=9 & 5 \\ A=B=C=9 & 2 & 0 & 0 & 0 & A=B=C=9 & 1 \end{array}\] Together, the answer is $5^3+5^2+5+1=\boxed{156}.$ | null | 156 |
80d57b864147b627c2741c5e8830a270 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_6 | For how many pairs of consecutive integers in $\{1000,1001,1002,\ldots,2000\}$ is no carrying required when the two integers are added? | Consider what carrying means: If carrying is needed to add two numbers with digits $abcd$ and $efgh$ , then $h+d\ge 10$ or $c+g\ge 10$ or $b+f\ge 10$ . 6. Consider $c \in \{0, 1, 2, 3, 4\}$ $1abc + 1ab(c+1)$ has no carry if $a, b \in \{0, 1, 2, 3, 4\}$ . This gives $5^3=125$ possible solutions.
With $c \in \{5, 6, 7, 8\}$ , there obviously must be a carry. Consider $c = 9$ $a, b \in \{0, 1, 2, 3, 4\}$ have no carry. This gives $5^2=25$ possible solutions. Considering $b = 9$ $a \in \{0, 1, 2, 3, 4, 9\}$ have no carry. Thus, the solution is $125 + 25 + 6=\boxed{156}$ | null | 156 |
80d57b864147b627c2741c5e8830a270 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_6 | For how many pairs of consecutive integers in $\{1000,1001,1002,\ldots,2000\}$ is no carrying required when the two integers are added? | Consider the ordered pair $(1abc , 1abc - 1)$ where $a,b$ and $c$ are digits. We are trying to find all ordered pairs where $(1abc) + (1abc - 1)$ does not require carrying. For the addition to require no carrying, $2a,2b < 10$ , so $a,b < 5$ unless $1abc$ ends in $00$ , which we will address later. Clearly, if $c \in \{0, 1, 2, 3, 4 ,5\}$ , then adding $(1abc) + (1abc - 1)$ will require no carrying. We have $5$ possibilities for the value of $a$ $5$ for $b$ , and $6$ for $c$ , giving a total of $(5)(5)(6) = 150$ , but we are not done yet.
We now have to consider the cases where $b,c = 0$ , specifically when $1abc \in \{1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000\}$ . We can see that $1100, 1200, 1300, 1400, 1500$ , and $2000$ all work, giving a grand total of $150 + 6 = \boxed{156}$ ordered pairs. | null | 156 |
46fe5a0c8f236f3f4fc5ebd101b6f6fa | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_7 | Faces $ABC^{}_{}$ and $BCD^{}_{}$ of tetrahedron $ABCD^{}_{}$ meet at an angle of $30^\circ$ . The area of face $ABC^{}_{}$ is $120^{}_{}$ , the area of face $BCD^{}_{}$ is $80^{}_{}$ , and $BC=10^{}_{}$ . Find the volume of the tetrahedron. | Since the area $BCD=80=\frac{1}{2}\cdot10\cdot16$ , the perpendicular from $D$ to $BC$ has length $16$
The perpendicular from $D$ to $ABC$ is $16 \cdot \sin 30^\circ=8$ . Therefore, the volume is $\frac{8\cdot120}{3}=\boxed{320}$ | null | 320 |
b681ced455ae01672932a3a2b4ac4f0b | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_8 | For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1^{}$ | Note that the $\Delta$ s are reminiscent of differentiation; from the condition $\Delta(\Delta{A}) = 1$ , we are led to consider the differential equation \[\frac{d^2 A}{dn^2} = 1\] This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; \[a_{n} = \frac{1}{2}(n-19)(n-92)\] as we must have roots at $n = 19$ and $n = 92$
Thus, $a_1=\frac{1}{2}(1-19)(1-92)=\boxed{819}$ | null | 819 |
b681ced455ae01672932a3a2b4ac4f0b | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_8 | For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1^{}$ | Let $\Delta^1 A=\Delta A$ , and $\Delta^n A=\Delta(\Delta^{(n-1)}A)$
Note that in every sequence of $a_i$ $a_n=\binom{n-1}{1}\Delta a_n + \binom{n-1}{2}\Delta^2 a_n +\binom{n-1}{3}\Delta^3 a_n + ...$
Then $a_n=a_1 +\binom{n-1}{1}\Delta a_1 +\binom{n-1}{2}\Delta^2 a_1 +\binom{n-1}{3}\Delta^3 a_1 + ...$
Since $\Delta a_1 =a_2 -a_1$ $a_n=a_1 +\binom{n-1}{1}(a_2-a_1) +\binom{n-1}{2}\cdot 1=a_1 +n(a_2-a_1) +\binom{n-1}{2}$
$a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153$
$a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095$
Solving, $a_1=\boxed{819}$ | null | 819 |
b681ced455ae01672932a3a2b4ac4f0b | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_8 | For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1^{}$ | The sequence $\Delta(\Delta A)$ is the second finite difference sequence, and the first $k-1$ terms of this sequence can be computed in terms of the original sequence as shown below.
$\begin{array}{rcl} a_3+a_1-2a_2&=&1\\ a_4+a_2-2a_3&=&1\\ &\vdots\\ a_k + a_{k-2} - 2a_{k-1} &= &1\\ a_{k+1} + a_{k-1} - 2a_k &=& 1.\\ \end{array}$
Adding the above $k-1$ equations we find that
\[(a_{k+1} - a_k) = k-1 + (a_2-a_1).\tag{1}\]
We can sum equation $(1)$ from $k=1$ to $18$ , finding \[18(a_1-a_2) - a_1 = 153.\tag{2}\]
We can also sum equation $(1)$ from $k=1$ to $91$ , finding \[91(a_1-a_2) - a_1 = 4095.\tag{3}\] Finally, $18\cdot (3) - 91\cdot(2)$ gives $a_1=\boxed{819}$ | null | 819 |
b681ced455ae01672932a3a2b4ac4f0b | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_8 | For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1^{}$ | Since all terms of $\Delta(\Delta A)$ are 1, we know that $\Delta A$ looks like $(k,k+1,k+2,...)$ for some $k$ . This means $A$ looks like $(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)$ . More specifically, $A_n=a_1+k(n-1)+\frac{(n-1)(n-2)}{2}$ . Plugging in $a_{19}=a_{92}=0$ , we have the following linear system: \[a_1+91k=-4095\] \[a_1+18k=-153\] From this, we can easily find that $k=-54$ and $a_1=\boxed{819}$ .
Solution by Zeroman | null | 819 |
b681ced455ae01672932a3a2b4ac4f0b | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_8 | For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1^{}$ | Since the result of two finite differences of some sequence is a constant sequence, we know that sequence is a quadratic. Furthermore, we know that $f(19) = f(92) = 0$ so the quadratic is $f(x) = a(x-19)(x-92)$ for some constant $a.$ Now we use the conditions that the finite difference is $1$ to find $a.$ We know $f(19) = 0$ and $f(20) = -72a$ and $f(18) = 74a.$ Therefore applying finite differences once yields the sequence $-74a,-72a$ and then applying finite differences one more time yields $2a$ so $a =\frac{1}{2}.$ Therefore $f(1) = 9 \cdot 91 = \boxed{819}.$ | null | 819 |
b681ced455ae01672932a3a2b4ac4f0b | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_8 | For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1^{}$ | Let $a_1=a,a_2=b.$ From the conditions, we have \[a_{n-1}+a_{n+1}=2a_n+1,\] for all $n>1.$ From this, we find that \begin{align*} a_3&=2b+1-a \\ a_4&=3b+3-2a\\ a_5&=4b+6-3a, \end{align*} or, in general, \[a_n=(n-1)b+\frac{(n-2)(n-1)}{2}-(n-2)a.\] Note: we can easily prove this by induction. Now, substituting $n=19,92,$ we find that \begin{align*} 0=&18b+\frac{17\cdot18}{2}-17a\\ 0=&91b+\frac{90\cdot91}{2}-90a\\ b=&\frac{17a-\frac{17\cdot18}{2}}{18}=\frac{90a-\frac{90\cdot91}{2}}{91}. \end{align*} Now, cross multiplying, we find that \begin{align*} 91\left(17a-\frac{17\cdot18}{2}\right)&=18\left(90a-\frac{90\cdot91}{2}\right)\\ 1547a-\frac{(18\cdot91)\cdot17}{2}&=1620-\frac{(18\cdot91)\cdot90}{2}\\ 73a&=\frac{18\cdot91}{2}\cdot(90-17=73)\\ a=\frac{18\cdot91}{2}=9\cdot91=\boxed{819} | null | 819 |
509fff5796fec190c8431af92f3678e7 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_9 | Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$ $BC=50^{}_{}$ $CD=19^{}_{}$ , and $AD=70^{}_{}$ , with $AB^{}_{}$ parallel to $CD^{}_{}$ . A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$ . Given that $AP^{}_{}=\frac mn$ , where $m^{}_{}$ and $n^{}_{}$ are relatively prime positive integers, find $m+n^{}_{}$ | Let $AP=x$ so that $PB=92-x.$ Extend $AD, BC$ to meet at $X,$ and note that $XP$ bisects $\angle AXB;$ let it meet $CD$ at $E.$ Using the angle bisector theorem, we let $XB=y(92-x), XA=xy$ for some $y.$
Then $XD=xy-70, XC=y(92-x)-50,$ thus \[\frac{xy-70}{y(92-x)-50} = \frac{XD}{XC} = \frac{ED}{EC}=\frac{AP}{PB} = \frac{x}{92-x},\] which we can rearrange, expand and cancel to get $120x=70\cdot 92,$ hence $AP=x=\frac{161}{3}$ . This gives us a final answer of $161+3=\boxed{164}$ | null | 164 |
509fff5796fec190c8431af92f3678e7 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_9 | Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$ $BC=50^{}_{}$ $CD=19^{}_{}$ , and $AD=70^{}_{}$ , with $AB^{}_{}$ parallel to $CD^{}_{}$ . A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$ . Given that $AP^{}_{}=\frac mn$ , where $m^{}_{}$ and $n^{}_{}$ are relatively prime positive integers, find $m+n^{}_{}$ | From $(1)$ above, $x = \frac{70r}{h}$ and $92-x = \frac{50r}{h}$ . Adding these equations yields $92 = \frac{120r}{h}$ . Thus, $x = \frac{70r}{h} = \frac{7}{12}\cdot\frac{120r}{h} = \frac{7}{12}\cdot92 = \frac{161}{3}$ , and $m+n = \boxed{164}$ | null | 164 |
509fff5796fec190c8431af92f3678e7 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_9 | Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$ $BC=50^{}_{}$ $CD=19^{}_{}$ , and $AD=70^{}_{}$ , with $AB^{}_{}$ parallel to $CD^{}_{}$ . A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$ . Given that $AP^{}_{}=\frac mn$ , where $m^{}_{}$ and $n^{}_{}$ are relatively prime positive integers, find $m+n^{}_{}$ | The area of the trapezoid is $\frac{(19+92)h}{2}$ , where $h$ is the height of the trapezoid.
Draw lines $CP$ and $BP$ . We can now find the area of the trapezoid as the sum of the areas of the three triangles $BPC$ $CPD$ , and $PBA$
$[BPC] = \frac{1}{2} \cdot 50 \cdot r$ (where $r$ is the radius of the tangent circle.)
$[CPD] = \frac{1}{2} \cdot 19 \cdot h$
$[PBA] = \frac{1}{2} \cdot 70 \cdot r$
$[BPC] + [CPD] + [PBA] = 60r + \frac{19h}{2} = [ABCD] = \frac{(19+92)h}{2}$
$60r = 46h$
$r = \frac{23h}{30}$
From Solution 1 above, $\frac{h}{70} = \frac{r}{x}$
Substituting $r = \frac{23h}{30}$ , we find $x = \frac{161}{3}$ , hence the answer is $\boxed{164}$ | null | 164 |
8df169eeae51f41b750e3478f6910555 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_10 | Consider the region $A$ in the complex plane that consists of all points $z$ such that both $\frac{z}{40}$ and $\frac{40}{\overline{z}}$ have real and imaginary parts between $0$ and $1$ , inclusive. What is the integer that is nearest the area of $A$ | Let $z=a+bi \implies \frac{z}{40}=\frac{a}{40}+\frac{b}{40}i$ . Since $0\leq \frac{a}{40},\frac{b}{40}\leq 1$ we have the inequality \[0\leq a,b \leq 40\] which is a square of side length $40$
Also, $\frac{40}{\overline{z}}=\frac{40}{a-bi}=\frac{40a}{a^2+b^2}+\frac{40b}{a^2+b^2}i$ so we have $0\leq a,b \leq \frac{a^2+b^2}{40}$ , which leads to: \[(a-20)^2+b^2\geq 20^2\] \[a^2+(b-20)^2\geq 20^2\]
We graph them:
To find the area outside the two circles but inside the square, we want to find the unique area of the two circles. We can do this by adding the area of the two circles and then subtracting out their overlap. There are two methods of finding the area of overlap:
1. Consider that the area is just the quarter-circle with radius $20$ minus an isosceles right triangle with base length $20$ , and then doubled (to consider the entire overlapped area)
2. Consider that the circles can be converted into polar coordinates, and their equations are $r = 40sin\theta$ and $r = 40cos\theta$ . Using calculus with the appropriate bounds, we can compute the overlapped area.
Using either method, we compute the overlapped area to be $200\pi + 400$ , and so the area of the intersection of those three graphs is $40^2-(200\pi + 400) \Rightarrow 1200 - 200\pi \approx 571.68$
$\boxed{572}$ | null | 572 |