id
int64 1
7.14k
| link
stringlengths 75
84
| no
int64 1
14
| problem
stringlengths 14
5.33k
| solution
stringlengths 21
6.43k
| answer
int64 0
999
|
|---|---|---|---|---|---|
5,901
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_9
| 2
|
A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$
[asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy]
|
Visualizing the corner which is submerged in the cylinder, we can see that it's like slicing off a corner, where they slice an equal part off every edge, to make a tetrahedron where there are 3 right isosceles triangles and one equilateral triangle. With this out of the way, we can now just find the area of that equilateral triangle, using the fact that the circle of radius $4$ is the circumcircle of the equilateral triangle. Using equilateral triangle properties, you can find that the height of the triangle is $6$ , and the side length is $\frac{6}{\sqrt{3}}=2\sqrt{3} \cdot 2=4\sqrt{3}$ . As the other faces are right isosceles triangles, they are $\frac{4\sqrt{3}}{\sqrt{2}}=2\sqrt{6}$ . Therefore the volume of this tetrahedron is
\[\left(\frac{2\sqrt{6}}{2}\right)^2=12 \ \cdot \ (2\sqrt{6})=24\sqrt{6} \implies \frac{24\sqrt{6}}{3}=8\sqrt{6} \implies (8\sqrt{6})^2=\boxed{384}\]
| 384
|
5,902
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_9
| 3
|
A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$
[asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy]
|
We can use the same method as in Solution 2 to find the side length of the equilateral triangle, which is $4\sqrt3$ . From here, its area is \[\dfrac{\bigl(4\sqrt3\bigr)^2\sqrt3}4=12\sqrt3.\] The leg of the isosceles right triangle is $\dfrac{4\sqrt3}{\sqrt2}=2\sqrt6$ , and the horizontal distance from the vertex to the base of the tetrahedron is $4$ (the radius of the cylinder), so we can find the height, as shown in the diagram. [asy] import olympiad; pair V, T, B; V = (-4, 0); B = origin; T = (0, 2*sqrt(2)); draw(V--B--T--cycle); draw(rightanglemark(V, B, T)); label("Vertex", V, W); label("Tip", T, N); label("Base", B, SE); label("$4$", V--B, S); label("$2\sqrt6$", V--T, NW); [/asy] The height from the tip to the base is $2\sqrt2$ , so the volume is $\dfrac{12\sqrt3\cdot2\sqrt2}3=8\sqrt6$ , and thus the answer is $\boxed{384}$
| 384
|
5,903
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_9
| 4
|
A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$
[asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy]
|
Since the diagonal is perpendicular to the base of the cylinder, all three edges and faces can be treated symmetrically.
The cross-section of the cube with the top face of the cylinder is an equilateral triangle inscribed in a circle with radius $4$ . This means the medians of the triangle have length $\frac{3}{2} \cdot 4=6$ , because the circumcenter is also the centroid, and the centroid divides the medians into lengths of ratio $2:1$ . Using $30-60-90$ triangles, the side length of the triangle is $4\sqrt{3}$ , and its area is $\frac{(4\sqrt{3})^2\sqrt{3}}{4}=12\sqrt{3}$
Next, consider the submerged triangular sections of the faces. Each is a $45-45-90$ triangle with leg length $x$ . The area of each is then $\frac{x^2}{2}$ . By De Gua's Theorem on the submerged pyramid (which we can apply because it has a right-angled corner), $3\left( \frac{x^2}{2} \right) ^2=(12\sqrt{3})^2$ . Solving yields $x=2\sqrt{6}$
The height of the pyramid is then $\sqrt{(2\sqrt{6})^2-4^2}=2\sqrt{2}$ , by the Pythagorean Theorem (using the slant height and circumradius). The volume is then $v=\frac{1}{3}\cdot 12\sqrt{3} \cdot 2\sqrt{2}=8\sqrt{6}$ , and the requested answer is $v^2=(8\sqrt{6})^2=\boxed{384}$ . ~bad_at_mathcounts
| 384
|
5,904
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_10
| 1
|
Call a permutation $a_1, a_2, \ldots, a_n$ of the integers $1, 2, \ldots, n$ quasi-increasing if $a_k \leq a_{k+1} + 2$ for each $1 \leq k \leq n-1$ . For example, 53421 and 14253 are quasi-increasing permutations of the integers $1, 2, 3, 4, 5$ , but 45123 is not. Find the number of quasi-increasing permutations of the integers $1, 2, \ldots, 7$
|
The simple recurrence can be found.
When inserting an integer $n$ into a string with $n - 1$ integers, we notice that the integer $n$ has 3 spots where it can go: before $n - 1$ , before $n - 2$ , and at the very end.
EXAMPLE:
Putting 4 into the string 123:
4 can go before the 2: 1423,
Before the 3: 1243,
And at the very end: 1234.
Only the addition of the next number, n, will change anything.
Thus the number of permutations with n elements is three times the number of permutations with $n-1$ elements.
Start with $n=3$ since all $6$ permutations work. And go up: $18, 54, 162, 486$
Thus for $n=7$ there are $2*3^5=\boxed{486}$ permutations.
| 486
|
5,905
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_11
| 5
|
The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
$\textit{Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.}$
Denote the circumradius of $ABC$ to be $R$ , the circumcircle of $ABC$ to be $O$ , and the shortest distance from $Q$ to circle $O$ to be $x$
Using Power of a Point on $Q$ relative to circle $O$ , we get that $x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}$ . Using Pythagorean Theorem on triangle $QOB$ to get $(x + r)^2 + r^2 = \frac{81}{4}$ . Subtracting the first equation from the second, we get that $2r^2 = 18$ and therefore $r = 3$ . Now, set $\cos{ABC} = y$ . Using law of cosines on $ABC$ to find $AC$ in terms of $y$ and plugging that into the extended law of sines, we get $\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6$ . Squaring both sides and cross multiplying, we get $36x^2 - 40x + 5 = 0$ . Now, we get $x = \frac{10 \pm \sqrt{55}}{18}$ using quadratic formula. If you drew a decent diagram, $B$ is acute and therefore $x = \frac{10 + \sqrt{55}}{18}$ (You can also try plugging in both in the end and seeing which gives a rational solution). Note that $BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.$ Using the cosine addition formula and then plugging in what we know about $QBO$ , we get that $BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}$ . Now, the hard part is to find what $\sin{B}$ is. We therefore want $\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}$ . For the numerator, by inspection $(a + b\sqrt{55})^2$ will not work for integers $a$ and $b$ . The other case is if there is $(a\sqrt{5} + b\sqrt{11})^2$ . By inspection, $5\sqrt{5} - 2\sqrt{11}$ works. Therefore, plugging all this in yields the answer, $\frac{18}{5} \rightarrow \boxed{23}$ . Solution by hyxue
| 23
|
5,906
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12
| 1
|
There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.
|
Let $a_{n}$ be the number of ways to form $n$ -letter strings made up of As and Bs such that no more than $3$ adjacent letters are identical.
Note that, at the end of each $n$ -letter string, there are $3$ possibilities for the last letter chain: it must be either $1$ $2$ , or $3$ letters long. Removing this last chain will make a new string that is $n-1$ $n-2$ , or $n-3$ letters long, respectively.
Therefore we can deduce that $a_{n}=a_{n-1}+a_{n-2}+a_{n-3}$
We can see that \[a_{1}=2\] \[a_{2}=2^{2}=4\] \[a_{3}=2^{3}=8\] so using our recursive relation we find \[a_{4}=14\] \[a_{5}=26\] \[a_{6}=48\] \[a_{7}=88\] \[a_{8}=162\] \[a_{9}=298\] \[a_{10}=\boxed{548}\]
| 548
|
5,907
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12
| 2
|
There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.
|
The solution is a simple recursion:
We have three cases for the ending of a string: three in a row, two in a row, and a single:
...AAA $(1)$ ...BAA $(2)$ ...BBA or ...ABA $(3)$
(Here, WLOG each string ends with A. This won't be the case when we actually solve for values in recursion.)
For case $(1)$ , we could only add a B to the end, making it a case $(3)$ .
For case $(2)$ , we could add an A or a B to the end, making it a case $(1)$ if you add an A, or a case $(3)$ if you add a B.
For case $(3)$ , we could add an A or a B to the end, making it a case $(2)$ or a case $(3)$
Let us create three series to represent the number of permutations for each case: $\{a\}$ $\{b\}$ , and $\{c\}$ representing case $(1)$ $(2)$ , and $(3)$ respectively.
The series have the following relationship:
$a_n=b_{n-1}$ $b_n=c_{n-1}$ $c_n=c_{n-1}+a_{n-1}+b_{n-1}$
For $n=3$ $a_3$ and $b_3$ both equal $2$ $c_3=4$ . With some simple math, we have: $a_{10}=88$ $b_{10}=162$ , and $c_{10}=298$ .
Summing the three up we have our solution: $88+162+298=\boxed{548}$
| 548
|
5,908
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12
| 3
|
There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.
|
This is a recursion problem. Let $a_n$ be the number of valid strings of $n$ letters, where the first letter is $A$ . Similarly, let $b_n$ be the number of valid strings of $n$ letters, where the first letter is $B$
Note that $a_n=b_{n-1}+b_{n-2}+b_{n-3}$ for all $n\ge4$
Similarly, we have $b_n=a_{n-1}+a_{n-2}+a_{n-3}$ for all $n\ge4$
Here is why: every valid strings of $n$ letters $(n\ge4)$ where the first letter is $A$ must begin with one of the following:
$AAAB$ - and the number of valid ways is $b_{n-3}$
$AAB$ - and the number of valid ways is $b_{n-2}$
$AB$ - and there are $b_{n-1}$ ways.
We know that $a_1=1$ $a_2=2$ , and $a_3=4$ . Similarly, we have $b_1=1$ $b_2=2$ , and $b_3=4$ . We can quickly check our recursion to see if our recursive formula works. By the formula, $a_4=b_3+b_2+b_1=7$ , and listing out all $a_4$ , we can quickly verify our formula.
Therefore, we have the following:
$\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c}Value of n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\\hline a & 1 & 2 & 4 & 7 & 13 & 24 & 44 & 81 & 149 & 274\\b & 1 & 2 & 4 & 7 & 13 & 24 & 44 & 81 & 149 & 274\end{tabular}$
The total number of valid $10$ letter strings is equal to $a_{10}+b_{10}=274+274=\boxed{548}$
| 548
|
5,909
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12
| 4
|
There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.
|
Playing around with strings gives this approach: We have a certain number of As, then Bs, and so on. Therefore, what if we denoted each solution with numbers like $(3,3,3,1)$ to denote AAABBBAAAB or vice versa (starting with Bs)? Every string can be represented like this!
We can have from 4 to 10 numbers in our parentheses. For each case, we will start with the largest number possible, usually a bunch of 3s, then go down systematically. Realize also that if we are left with just 2s and 1s, there is only one number of 2s and 1s that adds up to the leftover amount. Our final answer is the sum of all of these parenthetical sets [each set multiplied by its permutations, as order matters] multiplied by two [starting with either A or B, and alternating as we go along].
$4 \rightarrow (3,3,3,1) = 4, (3,3,2,2) = 6$
$5 \rightarrow (3, 3, 2, 1, 1) = 30, (3, 2, 2, 2, 1) = 20, (2,2,2,2,2)=1$
$6 \rightarrow (3,3,1,1,1,1) = 15, (3,2,2,1,1,1) = 60, (2,2,2,2,1,1) = 15$
$7 \rightarrow (3,2,1,1,1,1,1) = 42, (2,2,2,1,1,1,1) = 35$
$8 \rightarrow (3,1...1) = 8, (2,2,1...1) = 28$
$9 \rightarrow (2,1...1) = 9$
$10 \rightarrow (1,1....1) =1$
Adding them all up gives you 274; multiplying by 2 gives $\boxed{548}$
| 548
|
5,910
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12
| 5
|
There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.
|
We are going to build the string, 1 character at a time. And, we are going to only care about the streak of letters at the end of the string.
Let $a_n$ be the number of strings of length n that satisfy the problem statement and also has a "streak" of length 1 at the end. ABABBA has a streak of length 1.
Let $b_n$ be the number of strings of length n that satisfy the problem statement and also has of length 2 at the end. ABABBAA has a streak of length 2. There are 2 "A" s at the end of the string.
Let $c_n$ be the number of string of length n that that satisfy the problem statement and also has a "streak" of length 3 at the end. ABABBAAA has a streak of length 3. There are 3 "A" s at the end of the string.
Let's establish a recursive relationship. $a_{n+1} = a_n+b_n+c_n$ , since you can simply break the streak. $b_{n+1} = a_n$ , and $c_{n+1} = b_n$ Since you can just add to the streak.
We can log everything using a table.
$\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c}Value of n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\\hline a & 2 & 2 & 4 & 8 & 14 & 26 & 48 & 88 & 162 & 298\\b & 0 & 2 & 2 & 4 & 8 & 14 & 26 & 48 & 88 & 162 \\c & 0 & 0 & 2 & 2 &4 & 8 & 14 & 26 & 48 & 88\end{tabular}$
Adding $a_{10}$ $b_{10}$ $c_{10}$ gets the total number of numbers that doesn't have more than 3 concecutive letters.
That gets a total of $298+162+88 = \boxed{548}$
| 548
|
5,911
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_12
| 6
|
There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.
|
Let $S_n$ be the number of n-letter strings satisfying the problem criteria. Then we can easily see $S_1 = 2, S_2 = 4, S_3 = 8, S_4 = 16-2 =14$ . For $n \ge 4$ consider the last three elements of the list. For example to get $S_5$ we could try to take $S_4$ and multiply by 2 since for each $S_4$ string we have two choices for the fifth character. However we have to be careful if the last three characters of $S_4$ are all the same, as in that case we would be overcounting, (i.e if we have a string $\dots aaa$ , we can only add $b$ to the end so that the problem criteria is satisfied. Additionally strings such as $\dots aaa$ or $\dots bbb$ should only be counted once as we only have one choice for the $n$ th character to add ( $b$ and $a$ respectively).
Thus to compute $S_n$ we start with $S_{n-1}$ then take out the strings ending in $\dots bbb$ or $\dots aaa$ . There are $S_{n-4}$ remaining valid strings (i.e if we pick any of the $S_{n-4}$ valid strings, examine the last character, if it is a $b$ then we append $aaa$ to it, and if if it an $a$ we append $bbb$ to it, hence the number of $S_{n-1}$ strings are in one-to-one correspondence with the number of strings in $S_{n-4}$ . Thus we have the recursion $S_n = 2(S_{n-1} - S_{n-4}) + S_{n-4}$ (where we are first taking away $S_{n-4}$ , doubling the result, then adding $S_{n-4}$ back in signifying that we only count it once, as described above).
The recursion simplifies to $S_n = 2S_{n-1} - S_{n-4}$ and we can now quickly compute the remaining values: \[S_5 = 2S_4 - S_1 = 26, S_6 = 2(26)-4 = 48, S_7 = 2(48) - 8 = 88, S_8 = 2(88) - 14 = 162, S_9 = 2(162) - 26 = 298, S_{10} = 2S_9 - 48 = \boxed{548}\]
| 548
|
5,912
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_13
| 1
|
Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$
|
If $n = 1$ $a_n = \sin(1) > 0$ . Then if $n$ satisfies $a_n < 0$ $n \ge 2$ , and \[a_n = \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{2\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{2\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].\] Since $2\sin 1$ is positive, it does not affect the sign of $a_n$ . Let $b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)$ . Now since $\cos(0) + \cos(1) = 2\cos\left(\cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)$ and $\cos(n) + \cos(n + 1) = 2\cos\left(n + \cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)$ $b_n$ is negative if and only if $\cos\left(\cfrac{1}{2}\right) < \cos\left(n + \cfrac{1}{2}\right)$ , or when $n \in [2k\pi - 1, 2k\pi]$ . Since $\pi$ is irrational, there is always only one integer in the range, so there are values of $n$ such that $a_n < 0$ at $2\pi, 4\pi, \cdots$ . Then the hundredth such value will be when $k = 100$ and $n = \lfloor 200\pi \rfloor = \lfloor 628.318 \rfloor = \boxed{628}$
| 628
|
5,913
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_13
| 2
|
Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$
|
Notice that $a_n$ is the imaginary part of $\sum_{k=1}^n e^{ik}$ , by Euler's formula. Using the geometric series formula, we find that this sum is equal to \[\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}\] We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have \[\frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}\] We only need to look at the imaginary part, which is \[\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}\] Since $\cos 1 < 1$ $2-2 \cos 1 > 0$ , so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have $\sin (n+1) - \sin n > \sin 1 \implies 2 \cos \left(n + \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( n + \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),$ by sum to product. This only holds when $n$ is between $2\pi k - 1$ and $2\pi k$ for integer $k$ [continuity proof here], and since this has exactly one integer solution for every such interval, the $100$ th such $n$ is $\lfloor 200\pi \rfloor = \boxed{628}$
| 628
|
5,914
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_13
| 3
|
Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$
|
Similar to solution 2, we set a complex number $z=\cos 1+i\sin 1$ . We start from $z$ instead of $1$ because $k$ starts from $1$ : be careful.
The sum of $z+z^2+z^3+z^4+z^5\dots=\frac{z-z^{n+1}}{1-z}=\frac{z}{z-1}\left(z^n-1\right)$
We are trying to make $n$ so that the imaginary part of this expression is negative.
The argument of $z$ is $1$ . The argument of $z-1$ , however, is a little more tricky. $z-1$ is on a circle centered on $(-1,0)$ with radius $1$ . The change in angle due to $z$ is $1$ with respect to the center, but the angle that $z-1$ makes with the $y$ -axis is $half$ the change, due to Circle Theorems (this intercepted arc is the argument of $z$ ), because the $y$ - axis is tangent to the circle at the origin. So $\text{arg}(z-1)=\frac{\pi+1}{2}$ . Dividing $z$ by $z-1$ subtracts the latter argument from the former, so the angle of the quotient with the $x$ -axis is $\frac{1-\pi}{2}$
We want the argument of the whole expression $-\pi<\theta<0$ . This translates into $\frac{-\pi-1}{2}<\text{arg}\left(z^n-1\right)<\frac{\pi-1}{2}$ $z^n-1$ also consists of points on the circle centered at $(-1,0)$ , so we deal with this argument similarly: the argument of $z^n$ is twice the angle $z^n-1$ makes with the $y$ -axis. Since $z^n-1$ is always negative, $\frac{-3\pi}{2}<\text{arg}\left(z^n-1\right)<\frac{-\pi}{2}$ , and the left bound is the only one that is important. Either way, the line (the line consists of both bounds) makes a $\frac{\pi}{2}-\frac{\pi-1}{2}=\frac{-1}{2}$ angle with the $y$ -axis both ways.
So the argument of $z^n$ must be in the bound $-1<\theta<0$ by doubling, namely the last $z^n$ negative before another rotation. Since there is always one $z^n$ in this category per rotation because $\pi$ is irrational, $n_{100}\equiv z^{628}$ and the answer is $\boxed{628}$
| 628
|
5,915
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14
| 7
|
Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$
|
Take $w=x+y$ and $z=xy$ . Remark that \begin{align*} ~&2x^{3}+(xy)^{3}+2y^{3} \\ =~&2(x^{3}+y^{3})+(xy)^{3} \\ =~&2\left[(x+y)^{3}-3xy(x+y)\right]+(xy)^{3} \\ =~&2(w^{3}-3wz)+z^{3} \\ =~& 2w^{3}-6wz+z^{3}.\end{align*} The given equations imply that \[wz^{4}=810~~~\text{and}~~~(wz)^{3}-3wz^{4}=945.\] Substituting the first equation into the second, we have that $(wz)^{3}=945+3\cdot 810=3375$ , thus $wz=\sqrt[3]{3375}=15$ . Now \[z^{3}=\frac{wz^{4}}{wz}=\frac{810}{15}=54\] and \[w^{3}=\frac{3375}{54}=\frac{125}{2}.\] Thus \begin{align*}2w^{3}-6wz+z^{3}&=2\left(\frac{125}{2}\right)-6(15)+54 \\ &=125-90+54 \\ &=\boxed{89}
| 89
|
5,916
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14
| 8
|
Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$
|
$x^4y^4(x+y)=810; x^3y^3(x^3+y^3)=945, \frac{x^2-xy+y^2}{xy}=\frac{7}{6}, \frac{x^2+y^2}{xy}=\frac{13}{6}$
Let $x^2+y^2=13k; xy=6k$ , then we can see $(x+y)^2-12k=13k, x+y=5\sqrt{k}$ , now, we see $x^4y^4\cdot (x+y)=1296k^4\cdot 5\sqrt{k}, k=\frac{1}{\sqrt[3]{4}}$
The rest is easy, $2(x^3+y^3)+x^3y^3=216k^3+2[(x+y)^3-3xy(x+y)]=216k^3+2\cdot 35k^{\frac{3}{2}}=\boxed{89}$
| 89
|
5,917
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_14
| 9
|
Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$
|
Let's first put the left hand sides of the equations in factored forms. Doing this we obtain $(xy)^4 \cdot (x + y) = 810$ and $(xy)(x + y)(x^2 - xy + y^2) = 945$ . Now, we will subtract and add the equations to gather information on $x$ and $y$ . When we subtract the equations and clean it up via factoring, we yield $(xy)^3 \cdot (x + y) \cdot (x - y)^2 = 115$ , and when we add them, we yield $(xy)^3 \cdot (x + y) \cdot (x^2 + y^2) = 1755$ . Now with some intuition, you should divide the equations to obtain $\frac{(x^2 + y^2)}{(x - y)^2} = 13$ . Now, we clean this up to obtain the following factoring of $0 = 2 \cdot (2x - 3y) \cdot (3x - 2y)$ . This implies that $x = \frac{3y}{2}$ . We plug that into the target expression to reduce it down to one variable, and get that target expression is $2x^3 + (xy)^3 + 2y^3 = \frac{27}{4} \cdot y^3 + \frac{27}{8} \cdot y^6 + 2y^3$ . This means that if we can find a way to get $y^3$ , then the rest is trivial. We get $y^3$ by plugging in $x = \frac{3y}{2}$ into $x^3 \cdot y^6 + y^3 \cdot x^6 = 945$ . However, this time we only factor as $(xy)^3 \cdot (x^3 + y^3)$ because we particularly want a cubic degree on $y$ . Plugging in $x = \frac{3y}{2}$ we get $y^3 = 4$ . Now lets plug this into our target expression to get $\frac{27}{4} \cdot 4 + \frac{27}{8} \cdot 4^2 + 2 \cdot 4 = \boxed{89}$
| 89
|
5,918
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15
| 1
|
Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy]
|
Let $M$ be the intersection of $\overline{BC}$ and the common internal tangent of $\mathcal P$ and $\mathcal Q.$ We claim that $M$ is the circumcenter of right $\triangle{ABC}.$ Indeed, we have $AM = BM$ and $BM = CM$ by equal tangents to circles, and since $BM = CM, M$ is the midpoint of $\overline{BC},$ implying that $\angle{BAC} = 90.$ Now draw $\overline{PA}, \overline{PB}, \overline{PM},$ where $P$ is the center of circle $\mathcal P.$ Quadrilateral $PAMB$ is cyclic, and by Pythagorean Theorem $PM = \sqrt{5},$ so by Ptolemy on $PAMB$ we have \[AB \sqrt{5} = 2 \cdot 1 + 2 \cdot 1 = 4 \iff AB = \dfrac{4 \sqrt{5}}{5}.\] Do the same thing on cyclic quadrilateral $QAMC$ (where $Q$ is the center of circle $\mathcal Q$ and get $AC = \frac{8 \sqrt{5}}{5}.$
Let $\angle A = \angle{DAB}.$ By Law of Sines, $BD = 2R \sin A = 2 \sin A.$ Note that $\angle{D} = \angle{ABC}$ from inscribed angles, so \begin{align*} [ABD] &= \dfrac{1}{2} BD \cdot AB \cdot \sin{\angle B} \\ &= \dfrac{1}{2} \cdot \dfrac{4 \sqrt{5}}{5} \cdot 2 \sin A \sin{\left(180 - \angle A - \angle D\right)} \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \sin{\left(\angle A + \angle D\right)} \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\sin A \cos D + \cos A \sin D\right) \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\sin A \cos{\angle{ABC}} + \cos A \sin{\angle{ABC}}\right) \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\dfrac{\sqrt{5} \sin A}{5} + \dfrac{2 \sqrt{5} \cos A}{5}\right) \\ &= \dfrac{4}{5} \cdot \sin A \left(\sin A + 2 \cos A\right) \end{align*} after angle addition identity.
Similarly, $\angle{EAC} = 90 - \angle A,$ and by Law of Sines $CE = 8 \sin{\angle{EAC}} = 8 \cos A.$ Note that $\angle{E} = \angle{ACB}$ from inscribed angles, so \begin{align*} [ACE] &= \dfrac{1}{2} AC \cdot CE \sin{\angle C} \\ &= \dfrac{1}{2} \cdot \dfrac{8 \sqrt{5}}{5} \cdot 8 \cos A \sin{\left[180 - \left(90 - \angle A\right) - \angle E\right]} \\ &= \dfrac{32 \sqrt{5}}{5} \cdot \cos A \sin{\left[\left(90 - \angle A\right) + \angle{ACB}\right]} \\ &= \dfrac{32 \sqrt{5}}{5} \cdot \cos A \left(\dfrac{2 \sqrt{5} \cos A}{5} + \dfrac{\sqrt{5} \sin A}{5}\right) \\ &= \dfrac{32}{5} \cdot \cos A \left(\sin A + 2 \cos A\right) \end{align*} after angle addition identity.
Setting the two areas equal, we get \[\tan A = \frac{\sin A}{\cos A} = 8 \iff \sin A = \frac{8}{\sqrt{65}}, \cos A = \frac{1}{\sqrt{65}}\] after Pythagorean Identity. Now plug back in and the common area is $\frac{64}{65} \iff \boxed{129}.$
| 129
|
5,919
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15
| 2
|
Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy]
|
[asy] unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label("$A$",A,NE);label("$O_1$",O_1,NE);label("$O_2$",O_2,NE);label("$B$",B,SW);label("$C$",C,SW);label("$D$",D,NE);label("$E$",E,NE);label("$N$",N,W);label("$K$",(-24/15,0.2));label("$L$",(24/15,0.2));label("$n$",(-0.8,-0.12));label("$p$",((29/15,-48/15)));label("$\mathcal{P}$",(-1.6,1.1));label("$\mathcal{Q}$",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); [/asy]
Call $O_1$ and $O_2$ the centers of circles $\mathcal{P}$ and $\mathcal{Q}$ , respectively, and extend $CB$ and $O_2O_1$ to meet at point $N$ . Call $K$ and $L$ the feet of the altitudes from $B$ to $O_1N$ and $C$ to $O_2N$ , respectively. Using the fact that $\triangle{O_1BN} \sim \triangle{O_2CN}$ and setting $NO_1 = k$ , we have that $\frac{k+5}{k} = \frac{4}{1} \implies k=\frac{5}{3}$ . We can do some more length chasing using triangles similar to $O_1BN$ to get that $AK = AL = \frac{24}{15}$ $BK = \frac{12}{15}$ , and $CL = \frac{48}{15}$ . Now, consider the circles $\mathcal{P}$ and $\mathcal{Q}$ on the coordinate plane, where $A$ is the origin. If the line $\ell$ through $A$ intersects $\mathcal{P}$ at $D$ and $\mathcal{Q}$ at $E$ then $4 \cdot DA = AE$ . To verify this, notice that $\triangle{AO_1D} \sim \triangle{EO_2A}$ from the fact that both triangles are isosceles with $\angle{O_1AD} \cong \angle{O_2AE}$ , which are corresponding angles. Since $O_2A = 4\cdot O_1A$ , we can conclude that $4 \cdot DA = AE$
Hence, we need to find the slope $m$ of line $\ell$ such that the perpendicular distance $n$ from $B$ to $AD$ is four times the perpendicular distance $p$ from $C$ to $AE$ . This will mean that the product of the bases and heights of triangles $ACE$ and $DBA$ will be equal, which in turn means that their areas will be equal. Let the line $\ell$ have the equation $y = -mx \implies mx + y = 0$ , and let $m$ be a positive real number so that the negative slope of $\ell$ is preserved. Setting $A = (0,0)$ , the coordinates of $B$ are $(x_B, y_B) = \left(\frac{-24}{15}, \frac{-12}{15}\right)$ , and the coordinates of $C$ are $(x_C, y_C) = \left(\frac{24}{15}, \frac{-48}{15}\right)$ . Using the point-to-line distance formula and the condition $n = 4p$ , we have \[\frac{|mx_B + 1(y_B) + 0|}{\sqrt{m^2 + 1}} = \frac{4|mx_C + 1(y_C) + 0|}{\sqrt{m^2 + 1}}\] \[\implies |mx_B + y_B| = 4|mx_C + y_C| \implies \left|\frac{-24m}{15} + \frac{-12}{15}\right| = 4\left|\frac{24m}{15} + \frac{-48}{15}\right|.\] If $m > 2$ , then clearly $B$ and $C$ would not lie on the same side of $\ell$ . Thus since $m > 0$ , we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have \[\frac{6m}{15} + \frac{3}{15} = \frac{48}{15} - \frac{24m}{15} \implies 2m = 3 \implies m = \frac{3}{2}.\] Thus, the equation of $\ell$ is $y = -\frac{3}{2}x$
Then we can find the coordinates of $D$ by finding the point $(x,y)$ other than $A = (0,0)$ where the circle $\mathcal{P}$ intersects $\ell$ $\mathcal{P}$ can be represented with the equation $(x + 1)^2 + y^2 = 1$ , and substituting $y = -\frac{3}{2}x$ into this equation yields $x = 0, -\frac{8}{13}$ as solutions. Discarding $x = 0$ , the $y$ -coordinate of $D$ is $-\frac{3}{2} \cdot -\frac{8}{13} = \frac{12}{13}$ . The distance from $D$ to $A$ is then $\frac{4}{\sqrt{13}}.$ The perpendicular distance from $B$ to $AD$ or the height of $\triangle{DBA}$ is $\frac{|\frac{3}{2}\cdot\frac{-24}{15} + \frac{-12}{15} + 0|}{\sqrt{\frac{3}{2}^2 + 1}} = \frac{\frac{48}{15}}{\frac{\sqrt{13}}{2}} = \frac{32}{5\sqrt{13}}.$ Finally, the common area is $\frac{1}{2}\left(\frac{32}{5\sqrt{13}} \cdot \frac{4}{\sqrt{13}}\right) = \frac{64}{65}$ , and $m + n = 64 + 65 = \boxed{129}$
| 129
|
5,920
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15
| 3
|
Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy]
|
By homothety , we deduce that $AE = 4 AD$ . (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of $P$ and $Q$ to $l$ .) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from $B$ to $l$ is four times that from $C$ to $l$ . Let the distance from $C$ be $x$ and the distance from $B$ be $4x$
Let $P$ and $Q$ be the centers of their respective circles. Then dropping a perpendicular from $P$ to $Q$ creates a $3-4-5$ right triangle, from which $BC = 4$ and, if $\alpha = \angle{AQC}$ , that $\cos \alpha = \dfrac{3}{5}$ . Then $\angle{BPA} = 180^\circ - \alpha$ , and the Law of Cosines on triangles $APB$ and $AQC$ gives $AB = \dfrac{4}{\sqrt{5}}$ and $AC = \dfrac{8}{\sqrt{5}}.$
Now, using the Pythagorean Theorem to express the length of the projection of $BC$ onto line $l$ gives \[\sqrt{\frac{16}{5} - 16x^2} + \sqrt{\frac{64}{5} - x^2} = \sqrt{16 - 9x^2}.\] Squaring and simplifying gives \[\sqrt{\left(\frac{1}{5} - x^2\right)\left(\frac{64}{5} - x^2\right)} = x^2,\] and squaring and solving gives $x = \dfrac{8}{5\sqrt{13}}.$
By the Law of Sines on triangle $ABD$ , we have \[\frac{BD}{\sin A} = 2.\] But we know $\sin A = \dfrac{4x}{AB}$ , and so a small computation gives $BD = \dfrac{16}{\sqrt{65}}.$ The Pythagorean Theorem now gives \[AD = \sqrt{BD^2 - (4x)^2} + \sqrt{AB^2 - (4x)^2} = \frac{4}{\sqrt{13}},\] and so the common area is $\dfrac{1}{2} \cdot \frac{4}{\sqrt{13}} \cdot \frac{32}{5\sqrt{13}} = \frac{64}{65}.$ The answer is $\boxed{129}.$
| 129
|
5,921
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15
| 4
|
Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy]
|
$DE$ goes through $A$ , the point of tangency of both circles. So $DE$ intercepts equal arcs in circle $P$ and $Q$ homothety . Hence, $AE=4AD$ . We will use such similarity later.
The diagonal distance between the centers of the circles is $4+1=5$ . The difference in heights is $4-1=3$ . So $BC=\sqrt{5^2-3^2}=4$
The triangle connecting the centers with a side parallel to $BC$ is a $3-4-5$ right triangle. Since $O_PA=1$ , the height of $A$ is $1+3/5=8/5$ . Drop an altitude from $A$ to $BC$ and call it $I$ $IB=4/5$ and $IC=4-4/5=32/5$ . Since right $\triangle AIB\sim\triangle CIB$ $ABC$ is a right triangle also; $IB:IA:IC$ form a geometric progression $\times 2$
Extend $BA$ through $A$ to a point $G$ on the other side of $\circ Q$ . By homothety $\triangle DAB\sim\triangle EAG$ . By angle chasing $\triangle DAB$ through right triangle $ABC$ , we deduce that $\angle CEG$ is a right angle. Since $ACEG$ is cyclic, $\angle GAC$ is also right. So $CG$ is a diameter of $\circ G$ . Because of this, $CG \perp BC$ , the tangent line. $\triangle BCG$ is right and $\triangle BCG\sim\triangle ABC\sim\triangle CAG$
$AC=\sqrt{(8/5)^2+(16/5)^2}=8\sqrt{5}/5$ so $AG=2AC=16\sqrt{5}/5$ and $[\triangle CAG]=64/5$
Since $[\triangle DAB]=[\triangle ACE]$ , the common area is $[ACEG]/17$ $16[\triangle DAB]=[\triangle GAE]$ because the triangles are similar with a ratio of $1:4$ . So we only need to find $[\triangle CEG]$ now.
Extend $DE$ through $E$ to intersect the tangent at $F$ . Because $4DA=AE$ , the altitude from $B$ to $AD$ is $4$ times the height from $C$ to $EA$ . So $BC=3/4BF$ and $BF=16/3$ . We look at right triangle $\triangle AIF$ $IF=68/15$ and $AI=8/5$ $\triangle AIF$ is a $17-6-5\sqrt{13}$ right triangle. Hypotenuse $AF$ intersects $CG$ at a point, we call it $H$ $CH=4/3\div 68/15\cdot 8/5=8/17$ . So $HG=8-8/17=128/17$
By Power of a Point $CH\cdot HG=AH\cdot HE$ $AH=16/5\cdot 5\sqrt{13}/17=16\sqrt{13}/17.$ So $HE=1024/289\cdot 17/(16\sqrt{13})=64/(17\sqrt{13})$ . The height from $E$ to $CG$ is $17/(5\sqrt{13})\cdot 64/(17\sqrt{13})=64/65$
Thus, $[\triangle CEG]=64/65\cdot 8\div 2=256/65$ . The area of the whole cyclic quadrilateral is $64/5+256/65=(832+256)/65=1088/65$ . Lastly, the common area is $1/17$ the area of the quadrilateral, or $64/65$ . So $64+65=\boxed{129}$
| 129
|
5,922
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15
| 5
|
Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy]
|
Let $K$ be the intersection of $BC$ and $AE$ . Since the radii of the two circles are 1:4, so we have $AD:AE=1:4$ , and the distance from $B$ to line $l$ and the distance from $C$ to line $l$ are in a ratio of 4:1, so $BK:CK=4:1$ . We can easily calculate the length of $BC$ to be 4, so $CK=\frac{4}{3}$ . Let $J$ be the foot of perpendicular line from $A$ to $BC$ , we can know that $BJ:CJ=1:4$ , so $BJ = 0.8$ $CJ=3.2$ $AJ=1.6$ , and $AK=\sqrt{1.6^2+\left(3.2+\frac{4}{3}\right)^2}=\frac{4}{3}\sqrt{13}$ . Since $CK^2 = EK\cdot AK$ , so $EK=\frac{4}{39}\sqrt{13}$ , and $AE = \frac{4}{3}\sqrt{13} - \frac{4}{39}\sqrt{13}= \frac{16}{13}\sqrt{13}$ $\sin\angle AKB=\frac{AJ}{AK} = \frac{1.6}{\frac{4}{3}\sqrt{13}}=\frac{1.2}{\sqrt{13}}$ , so the distance from $C$ to line $l$ is $d=CK\cdot \sin\angle AKB = \frac{4}{3}\cdot \frac{1.2}{\sqrt{13}}=\frac{1.6}{\sqrt{13}}$ . so the area is \[[ACE] = \frac{1}{2}\cdot AE\cdot d = \frac{1}{2}\cdot\frac{16}{13}\sqrt{13}\frac{1.6}{\sqrt{13}} = \frac{64}{65}\] The final answer is $\boxed{129}$
| 129
|
5,923
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15
| 6
|
Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy]
|
Consider the common tangent from $A$ to both circles. Let this intersect $BC$ at point $K$ . From equal tangents, we have $BK=AK=CK$ , which implies that $\angle BAC = 90^\circ$
Let the center of $\mathcal{P}$ be $O_1$ , and the center of $\mathcal{Q}$ be $O_2$ . Angle chasing, we find that $\triangle O_1DA \sim \triangle O_2EA$ with a ratio of $1:4$ . Hence $4AD = AE$
We can easily deduce that $BC=4$ by dropping an altitude from $O_1$ to $O_2C$ . Let $\angle ABC = \theta$ . By some simple angle chasing, we obtain that $\angle BO_1A = 2\angle BDA = 2\angle ABC = 2\theta,$ and similarly $\angle CO_2A = 180 - 2\theta$
Using LoC, we get that $AB = \sqrt{2-2\cos2\theta}$ and $AC = \sqrt{32+32\cos2\theta}$ . From Pythagorean theorem, we have \[AB^2 + AC^2 = BC^2 \implies \cos 2\theta = -\frac{3}{5} \implies \cos \theta = \frac{1}{\sqrt5}, \sin \theta = \frac{2}{\sqrt 5}\] In other words, $AB = \frac{4}{\sqrt5}, AC = \frac{8}{\sqrt5}$
Using the area condition, we have: \begin{align*} \frac12 AD*AB \sin \angle DAB &= \frac 12 AE*AC \sin(90-\angle DAB) \\ AD*\frac{4}{\sqrt5} \sin \angle DAB &= 4AD*\frac{8}{\sqrt5} \cos \angle DAB \\ \sin \angle DAB &= 8 \cos \angle DAB \\ \implies \sin \angle DAB &= \frac{8}{\sqrt{65}} \end{align*}
Now, for brevity, let $\angle D = \angle ADB$ and $\angle A = \angle DAB$
From Law of Sines on $\triangle ABD$ , we have \begin{align*} \frac{AB}{\sin \angle D} &= \frac{AD}{\sin (180-\angle A - \angle D)} \\ \frac{\frac{4}{\sqrt5}}{\frac{2}{\sqrt5}} &= \frac{AD}{\sin \angle A \cos\angle D + \sin \angle D\cos\angle A} \\ 2 &= \frac{AD}{\frac{2}{\sqrt{13}}} \\ AD &= \frac{4}{\sqrt{13}} \end{align*}
It remains to find the area of $\triangle ABD$ . This is just \[\frac12 AD*AB*\sin \angle A = \frac12 * \frac{4}{\sqrt{13}}*\frac{4}{\sqrt5}*\frac{8}{\sqrt{65}} = \frac{64}{65}\] for an answer of $\boxed{129}.$
| 129
|
5,924
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15
| 7
|
Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy]
|
Add in the line $k$ as the internal tangent between the two circles. Let $M$ be the midpoint of $BC$ ; It is well-known that $M$ is on $k$ and because $k$ is the radical axis of the two circles, $AM=BM=CM$ . Therefore because $M$ is the circumcenter of $\triangle{BAC}$ $\angle{BAC}=90^{\circ}$ . Let $O_P$ be the center of circle $\mathcal{P}$ and likewise let $O_Q$ be the center of circle $\mathcal{Q}$ . It is well known that by homothety $O_P, A,$ and $O_Q$ are collinear. It is well-known that $\angle{ABC}=\angle{ADB}=b$ , and likewise $\angle{ACB}=\angle{AEC}=c$ . By homothety, $AD=4AE$ , therefore since the two triangles mentioned in the problem, the length of the altitude from $B$ to $AD$ is four times the length of the altitude from $C$ to $AE$ . Using the Pythagorean Theorem, $BC = 4$ . By angle-chasing, $O_{P}AMB$ is cyclic, and likewise $O_{Q}CMA$ is cyclic. Use the Pythagorean Theorem for $\triangle{O_{P}BM}$ to get $O_{P}M=\sqrt{5}$ . Then by Ptolemy's Theorem $AB=\frac{4\sqrt{5}}{5} \implies AC=\frac{8\sqrt{5}}{5}$ . Now to compute the area, using what we know about the length of the altitude from $B$ to $AD$ is four times the length of the altitude from $C$ to $AE$ , letting $x$ be the length of the altitude from $C$ to $AE$ $x=\frac{8}{5\sqrt{13}}$ . From the Law of Sines, $\frac{BD}{\sin{A}}=2 \implies \sin{A}=\frac{4x}{AB} \implies BD=\frac{8x}{AB}=\frac{16\sqrt{65}}{65}$ . Then use the Pythagorean Theorem twice and add up the lengths to get $AD=\frac{4}{\sqrt{13}}$ . Use the formula $\frac{b \times h}{2}$ to get $\frac{64}{65} = \boxed{129}$ as the answer.
| 129
|
5,925
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15
| 8
|
Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy]
|
Let $P$ and $Q$ be the centers of circles $\mathcal{P}$ and $\mathcal{Q}$ , respectively.
Let $M$ be midpoint $BC, \beta = \angle ACB.$
Upper diagram shows that
$\sin 2\beta = \frac {4}{5}$ and $AC = 2 AB.$ Therefore $\cos 2\beta = \frac {3}{5}.$
Let $CH\perp l, BH'\perp l.$ Lower diagram shows that
$\angle CAE = \angle ABH' = \alpha$ (perpendicular sides)
and $\angle CQE = 2\alpha$ (the same intersept $\overset{\Large\frown} {CE}).$ \[\tan\alpha = \frac {1}{8}, \sin2\alpha = \frac{2 \tan \alpha}{1 + \tan^2 \alpha} = \frac {16}{65}, \cos2\alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} = \frac {63}{65}.\] The area \[[ACE] = [AQC]+[CQE]– [AQE].\] Hence \[[ACE] =\frac{AQ^2}{2} \left(\sin 2\alpha + \sin 2\beta - \sin(2\alpha + 2\beta)\right),\] \[[ACE] = 8\left( \frac{16}{65}+\frac{4}{5} - \frac{4}{5}\cdot \frac{63}{65} - \frac{3}{5}\cdot \frac{16}{65}\right) = \frac{64}{65}\implies \boxed{129}.\] vladimir.shelomovskii@gmail.com, vvsss
| 129
|
5,926
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15
| 9
|
Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy]
|
We begin by extending $\overline{AB}$ upwards until it intersects Circle $\mathcal{Q}$ . We can call this point of intersection $F$ . Connect $F$ with $E$ $C$ , and $A$ for future use.
Create a trapezoid with points $B$ $C$ , and the origins of Circles $\mathcal{P}$ and $\mathcal{Q}$ . After quick inspection, we can conclude that the distance between the origins is 5 and that $\overline{BC}$ is 4.
(Note: It is important to understand that we could simply use the formula for the distance between the tangent points on a line and two different circles, which is $2\cdot \sqrt{a\cdot b}$ , where $a$ and $b$ are the two respective radii of the circles. In our case, we get $2\cdot \sqrt{4} = 4$
Using similar triangles or homotheties, $AE=4\cdot AD$ and $AF=4\cdot BA$ $BC^2 = BA\cdot BF$ \[16 = BA\cdot (5\cdot BA)\] \[BA = \frac{4}{\sqrt{5}}\] \[AF = \frac{16}{\sqrt{5}}\] \[BF = 4\cdot \sqrt{5}\]
Inspecting $\triangle{BFC}$ , we recognize that it is a right triangle ( $\angle{BCF} = 90$ ) as the final length ( $\overline{FC}$ ) being 8 would allow for an $x-2x-x\sqrt{5}$ triangle. Hence, the diameter of circle $\mathcal{Q}$ $CF$ . This also means that $\angle{CAF} = \angle{CEF} = 90$
From the fact that $\triangle{ABC}$ is a right triangle: \[AB^2 + AC^2 = BC^2\] \[AC = \frac{8}{\sqrt{5}}\] (Note: We could have also used $\triangle{FAC}$ .)
Our next step is to start angle chasing to find any other similar triangles or shared angles. Label the intersection between $\overline{FC}$ and $\overline{DE}$ as $M$ . Label $\angle{CAE} = \angle{CFE} = \theta$ . Since $\angle{BAC} = 90$ $\angle{EAF} = \angle{DAB} = 90-\theta$ . Now, use the sine formula and the fact that the areas of $\triangle{DBA}$ and $\triangle{ACE}$ are equal to get: \[\frac{1}{2}\cdot AC\cdot AE\cdot \sin{\theta} = \frac{1}{2}\cdot AD\cdot AB\cdot \sin{(90-\theta)}\] Since $\sin{(90-\theta)} = \cos{\theta}$ \[\tan{\theta} = \frac{AD}{2\cdot AE}\]
Using right $\triangle{FCE}$ , since $\angle{ACM} = \angle{CFE}$ $\tan{\theta} = \frac{CE}{FE}$ . Hence, plugging into the previous equation: \[\frac{CE}{FE} = \frac{AD}{2\cdot AE}\] Using the Pythagorean theorem on $\triangle{FCE}$ $FE = \sqrt{(64 - CE^2)}$ . We also know that $AE=4\cdot AD$ Plugging back in: \[\frac{CE}{\sqrt{(64 - CE^2)}} = \frac{AD}{2\cdot (4\cdot AD)}\] \[\frac{CE}{\sqrt{(64 - CE^2)}} = \frac{1}{8}\] .
From here, we can square both sides and bring everything to one side to get: \[EF^2 + 64EF - 64 = 0\] \[EF = \frac{64}{\sqrt{65}}\] \[CE = \frac{8}{\sqrt{65}}\] We should also return to the fact that $\sin{\theta} = \frac{CE}{CF}$ from $\triangle{FCE}$ , so \[\sin{\theta} = \frac{1}{\sqrt{65}}\]
From the fact that $\angle{CAF} = \angle{CEF} = 90$ , we can use Ptolemy's Theorem on quadrilateral $ACEF$ $AC\cdot EF + CE\cdot FA = CF\cdot AE$ . Plugging in and solving, we get that $AE = \frac{16}{\sqrt{13}}$
We now have all of our pieces to use the Sine Formula on $\triangle{ACE}$ \[\frac{1}{2}\cdot AC\cdot AF\cdot \sin{\theta}\] \[\frac{1}{2}\cdot \frac{8}{\sqrt{5}}\cdot \frac{16}{\sqrt{13}}\cdot \frac{1}{\sqrt{65}} = \frac{64}{65} = \boxed{129}\]
| 129
|
5,927
|
https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15
| 10
|
Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy]
|
Let the center of the larger circle be $O,$ and the center of the smaller circle be $P.$ It is not hard to find the areas of $ACO$ and $ABP$ using pythagorean theorem, which are $\frac{32}{5}$ and $\frac{2}{5}$ respectively. Assign $\angle AOC=a,\angle COE=c,\angle DPB=d,\angle BPA=b.$ We can figure out that $\angle AOE=\angle DOA=\theta$ using vertical angles and isosceles triangles. Now, using $[ABC]=\frac{1}{2}ab\sin C$ \[[ACE]=[AOC]+[COE]-[AOC]=\dfrac{32}{5}-8\sin c-8\sin \theta,\] \[[DAB]=[APD]+[APB]+[BPD]=\dfrac{2}{5}+\dfrac{1}{2}\sin \theta+\dfrac{1}{2}\sin d.\] We can also figure out that $\sin a=\dfrac{4}{5},\cos a=\dfrac{3}{5},\sin b=\dfrac{4}{5}, \cos b=-\dfrac{3}{5}.$ Also, $c=\theta-a$ and $d=360-\theta-b.$ Using sum and difference identities: \[\sin c=\dfrac{3}{5}\sin \theta-\dfrac{4}{5}\cos \theta,\] \[\sin d=\dfrac{3}{5}\sin\theta-\dfrac{4}{5}\cos\theta.\] (We can also notice that $c+d=360-\theta-b+\theta-a=360-(a+b)=180$ which means that $\sin c=\sin d.$ )
Substituting in the equations for $\sin c$ and $\sin d$ into the equations for $[ACE]$ and $[DAB],$ setting them equal, and simplifying: \[3=2\sin\theta+3\cos\theta.\] Solving this equation we get that $\sin\theta=\frac{12}{13}$ and $\cos\theta=\frac{5}{13}.$ Doing a lot of substitution gives us \[[ACE]=[DAB]=\dfrac{64}{65},\] which means the answer is $64+65=\boxed{129}.$
| 129
|
5,928
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_1
| 1
|
The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.
[asy] size(200); defaultpen(linewidth(0.7)); path laceL=(-20,-30)..tension 0.75 ..(-90,-135)..(-102,-147)..(-152,-150)..tension 2 ..(-155,-140)..(-135,-40)..(-50,-4)..tension 0.8 ..origin; path laceR=reflect((75,0),(75,-240))*laceL; draw(origin--(0,-240)--(150,-240)--(150,0)--cycle,gray); for(int i=0;i<=3;i=i+1) { path circ1=circle((0,-80*i),5),circ2=circle((150,-80*i),5); unfill(circ1); draw(circ1); unfill(circ2); draw(circ2); } draw(laceL--(150,-80)--(0,-160)--(150,-240)--(0,-240)--(150,-160)--(0,-80)--(150,0)^^laceR,linewidth(1));[/asy]
|
The rectangle is divided into three smaller rectangles with a width of 50 mm and a length of $\dfrac{80}{3}$ mm. According to the Pythagorean Theorem (or by noticing the 8-15-17 Pythagorean triple), the diagonal of the rectangle is $\sqrt{50^2+\left(\frac{80}{3}\right)^2}=\frac{170}{3}$ mm. Since that on the lace, there are 6 of these diagonals, a width, and an extension of at least 200 mm on each side. Therefore, the minimum of the lace in millimeters is \[6\times \dfrac{170}{3}+50+200\times 2=\boxed{790}.\]
| 790
|
5,929
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_2
| 1
|
An urn contains $4$ green balls and $6$ blue balls. A second urn contains $16$ green balls and $N$ blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is $0.58$ . Find $N$
|
First, we find the probability both are green, then the probability both are blue, and add the two probabilities. The sum should be equal to $0.58$
The probability both are green is $\frac{4}{10}\cdot\frac{16}{16+N}$ , and the probability both are blue is $\frac{6}{10}\cdot\frac{N}{16+N}$ , so \[\frac{4}{10}\cdot\frac{16}{16+N}+\frac{6}{10}\cdot\frac{N}{16+N}=\frac{29}{50}\] Solving this equation, \[20\left(\frac{16}{16+N}\right)+30\left(\frac{N}{16+N}\right)=29\] Multiplying both sides by $16+N$ , we get
\begin{align*} 20\cdot16+30\cdot N&=29(16+N)\\ 320+30N&=464+29N\\ N&=\boxed{144}
| 144
|
5,930
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3
| 1
|
Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.
|
Let the numerator and denominator $x,y$ with $\gcd(x,y)=1$ and $x+y = 1000.$ Now if $\gcd(x,y) = 1$ then $\gcd(x,y) =\gcd(x,1000-x)= \gcd(x,1000-x-(-1)x)=\gcd(x,1000)=1.$ Therefore any pair that works satisfies $\gcd(x,1000)= 1.$ By Euler's totient theorem, there are $\phi(1000) = 400$ numbers relatively prime to 1000 from 1 to 1000. Recall that $r=\frac{x}{y}<1$ and note by Euclidean algorithm $\gcd(1000,1000-x)=1$ , so we want $x<y=1000-x.$ Thus the $400$ relatively prime numbers can generate $\boxed{200}$ desired fractions.
| 200
|
5,931
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3
| 2
|
Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.
|
If the initial manipulation is not obvious, consider the Euclidean Algorithm. Instead of using $\frac{n}{m}$ as the fraction to use the Euclidean Algorithm on, we can rewrite this as $\frac{500-x}{500+x}$ or \[\gcd(500+x,500-x)=\gcd((500+x)+(500-x),500-x)=\gcd(1000,500-x).\] Thus, we want $\gcd(1000,500-x)=1$ . You can either proceed as Solution $1$ , or consider that no even numbers work, limiting us to $250$ choices of numbers and restricting $x$ to be odd. If $x$ is odd, $500-x$ is odd, so the only possible common factors $1000$ and $500-x$ can share are multiples of $5$ . Thus, we want to avoid these. There are $50$ odd multiples of $5$ less than $500$ , so the answer is $250-50=\boxed{200}$
| 200
|
5,932
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3
| 3
|
Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.
|
Say $r=\frac{d}{1000-d}$ ; then $1\leq d\leq499$ . If this fraction is reducible, then the modulus of some number for $d$ is the same as the modulus for $1000-d$ . Since $1000=2^3\cdot5^3$ , that modulus can only be $2$ or $5$ . This implies that if $d\mid2$ or $d\mid5$ , the fraction is reducible. There are $249$ cases where $d\mid2$ $99$ where $d\mid5$ , and $49$ where $d\mid(2\cdot5=10)$ , so by PIE, the number of fails is $299$ , so our answer is $\boxed{200}$
| 200
|
5,933
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3
| 4
|
Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.
|
We know that the numerator of the fraction cannot be even, because the denominator would also be even. We also know that the numerator cannot be a multiple of $5$ because the denominator would also be a multiple of $5$ . Proceed by listing out all the other possible fractions and we realize that the numerator and denominator are always relatively prime. We have $499$ fractions to start with, and $250$ with odd numerators. Subtract $50$ to account for the multiples of $5$ , and we get $\boxed{200}$
| 200
|
5,934
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3
| 5
|
Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.
|
We know that the set of these rational numbers is from $\dfrac{1}{999}$ to $\dfrac{499}{501}$ where each each element $\dfrac{n}{m}$ has $n+m =1000$ and $\dfrac{n}{m}$ is irreducible.
We note that $\dfrac{n}{m} =\dfrac{1000-m}{m}=\dfrac{1000}{m}-1$ .
Hence, $\dfrac{n}{m}$ is irreducible if $\dfrac{1000}{m}$ is irreducible, and $\dfrac{1000}{m}$ is irreducible if $m$ is not divisible by $2$ or $5$ . Thus, the answer to the question is the number of integers between $501$ and $999$ inclusive that are not divisible by $2$ or $5$
We note there are $499$ numbers between $501$ and $999$ , and
Using the Principle of Inclusion and Exclusion, we get that there are $499-249-99+49=200$ numbers between $501$ and $999$ are not divisible by either $2$ or $5$ , so our answer is $\boxed{200}$
| 200
|
5,935
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3
| 6
|
Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.
|
We notice that there are a total of $400$ fractions that are in simplest form where the numerator and denominator add up to $1000$ . Because the numerator and denominator have to be relatively prime, there are $\varphi(1000)=400$ fractions. Half of these are greater than $1$ , so the answer is $400\div2=\boxed{200}$
| 200
|
5,936
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3
| 7
|
Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.
|
Our fraction can be written in the form $\frac{1000 - a}{a} = \frac{1000}{a} - 1.$ Thus the fraction is reducible when $a$ divides $1000.$ We also want $500 < a < 1000.$ By PIE , the total values of $a$ that make the fraction reducible is, \[249 + 99 - 49 = 299.\] By complementary counting , the answer we want is $499 - 299 = \boxed{200}.$
| 200
|
5,937
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3
| 8
|
Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.
|
Suppose our fraction is $\frac{a}{b}$ . The given condition means $a+b=1000$ . Now, if $a$ and $b$ share a common factor greater than $1$ , then the expression $a+b$ must also contain that common factor. This means our fraction cannot have a factor of $5$ or be even.
There are $250$ fractions that aren’t even. From this, $50$ are divisible by $5$ , which means the answer is $250-50=\boxed{200}$
| 200
|
5,938
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_5
| 1
|
Let the set $S = \{P_1, P_2, \dots, P_{12}\}$ consist of the twelve vertices of a regular $12$ -gon. A subset $Q$ of $S$ is called "communal" if there is a circle such that all points of $Q$ are inside the circle, and all points of $S$ not in $Q$ are outside of the circle. How many communal subsets are there? (Note that the empty set is a communal subset.)
|
By looking at the problem and drawing a few pictures, it quickly becomes obvious that one cannot draw a circle that covers $2$ disjoint areas of the $12$ -gon without including all the vertices in between those areas. In other words, in order for a subset to be communal, all the vertices in the subset must be adjacent to one another. We now count the number of ways to select a row of adjacent vertices. We notice that for any subset size between $1$ and $11$ , there are $12$ possible subsets like this (this is true because we can pick any of the $12$ vertices as a "starting" vertex, include some number of vertices counterclockwise from that vertex, and generate all possible configurations). However, we also have to include the set of all $12$ vertices, as well as the empty set. Thus, the total number is $12\cdot11 + 2 = \boxed{134}$
| 134
|
5,939
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_6
| 5
|
The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$
|
We have the equation $y=3(x-h)^2 + j.$
We know: $(x,y):(0,2013)$ , so $h^2=2013/3 - j/3$ after plugging in the values and isolating $h^2$ . Therefore, $h^2=671-j/3$
Lets call the x-intercepts $x_1$ $x_2$ . Since both $x_1$ and $x_2$ are positive there is a relationship between $x_1$ $x_2$ and $h$ . Namely, $x_1+x_2=2h$ . The is because: $x_1-h=-(x_2-h)$
Similarly, we know: $(x,y):(x_1,0)$ , so $j=-3(x_1-h)^2$ . Combining the two equations gives us \[h^2=671+(x_1-h)^2\] \[h^2=671+x_1^2-2x_1h+h^2\] \[h=(671+x_1^2)/2x_1.\]
Now since we have this relationship, $2h=x_1+x_2$ , we can just multiply the last equation by 2(so that we get $2h$ on the left side) which gives us \[2h=671/x_1+x_1^2/x^1\] \[2h=671/x_1+x_1\] \[x_1+x_2=671/x_1+x_1\] \[x_2=671/x_1\] \[x_1x_2=671.\] Prime factorization of 671 gives 11 and 61. So now we know $x_1=11$ and $x_2=61$ . Lastly, we plug in the numbers,11 and 61, into $x_1+x_2=2h$ , so $\boxed{36}$
| 36
|
5,940
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_6
| 6
|
The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$
|
First, we start of exactly like solutions above and we find out that $j=2013-3h^2$ and $k=2014-2h^2$ We then plug j and k into $3(x-h)^2+j$ and $y=2(x-h)^2+k$ respectively. After that, we get two equations, $y=3x^2-6xh+2013$ and $y=2x^2-4xh+2014$ . We can apply Vieta's. Let the roots of the first equation be $a, b$ and the roots of the second equation be $c, d$ . Thus, we have that $a\cdot b=1007$ $a+b=2h$ and $c\cdot d=671$ $c+d=2h$ . Simple evaluations finds that $\boxed{36}$
| 36
|
5,941
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_7
| 1
|
Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$ . Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$ . The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . (Note that $\arg(w)$ , for $w \neq 0$ , denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane)
|
Let $w = \operatorname{cis}{(\alpha)}$ and $z = 10\operatorname{cis}{(\beta)}$ . Then, $\dfrac{w - z}{z} = \dfrac{\operatorname{cis}{(\alpha)} - 10\operatorname{cis}{(\beta)}}{10\operatorname{cis}{\beta}}$
Multiplying both the numerator and denominator of this fraction by $\operatorname{cis}{(-\beta)}$ gives us:
$\dfrac{w - z}{z} = \dfrac{1}{10}\operatorname{cis}{(\alpha - \beta)} - 1 = \dfrac{1}{10}\cos{(\alpha - \beta)} + \dfrac{1}{10}i\sin{(\alpha - \beta)} - 1$
We know that $\tan{\theta}$ is equal to the imaginary part of the above expression divided by the real part. Let $x = \alpha - \beta$ . Then, we have that:
$\tan{\theta} = \dfrac{\sin{x}}{\cos{x} - 10}.$
We need to find a maximum of this expression, so we take the derivative:
Note (not from author): To take the derivative, we need to use the Quotient Rule . In this case, \[\frac{d}{dx}\left(\frac{\sin x}{\cos x-10}\right)=\frac{\cos x(\cos x-10)-(-\sin x)\sin x}{(\cos x-10)^2}=\dfrac{1 - 10\cos{x}}{(\cos{x} - 10)^2}\]
Thus, we see that the maximum occurs when $\cos{x} = \dfrac{1}{10}$ . Therefore, $\sin{x} = \pm\dfrac{\sqrt{99}}{10}$ , and $\tan{\theta} = \pm\dfrac{\sqrt{99}}{99}$ . Thus, the maximum value of $\tan^2{\theta}$ is $\dfrac{99}{99^2}$ , or $\dfrac{1}{99}$ , and our answer is $1 + 99 = \boxed{100}$
| 100
|
5,942
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_7
| 2
|
Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$ . Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$ . The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . (Note that $\arg(w)$ , for $w \neq 0$ , denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane)
|
Without the loss of generality one can let $z$ lie on the positive x axis and since $arg(\theta)$ is a measure of the angle if $z=10$ then $arg(\dfrac{w-z}{z})=arg(w-z)$ and we can see that the question is equivalent to having a triangle $OAB$ with sides $OA =10$ $AB=1$ and $OB=t$ and trying to maximize the angle $BOA$ [asy] pair O = (0,0); pair A = (100,0); pair B = (80,30); pair D = (sqrt(850),sqrt(850)); draw(A--B--O--cycle); dotfactor = 3; dot("$A$",A,dir(45)); dot("$B$",B,dir(45)); dot("$O$",O,dir(135)); dot("$ \theta$",O,(7,1.2)); label("$1$", ( A--B )); label("$10$",(O--A)); label("$t$",(O--B)); [/asy]
using the Law of Cosines we get: $1^2=10^2+t^2-t*10*2\cos\theta$ rearranging: \[20t\cos\theta=t^2+99\] solving for $\cos\theta$ we get:
\[\frac{99}{20t}+\frac{t}{20}=\cos\theta\] if we want to maximize $\theta$ we need to minimize $\cos\theta$ , using AM-GM inequality we get that the minimum value for $\cos\theta= 2\left(\sqrt{\dfrac{99}{20t}\dfrac{t}{20}}\right)=2\sqrt{\dfrac{99}{400}}=\dfrac{\sqrt{99}}{10}$ hence using the identity $\tan^2\theta=\sec^2\theta-1$ we get $\tan^2\theta=\frac{1}{99}$ and our answer is $1 + 99 = \boxed{100}$
| 100
|
5,943
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_7
| 3
|
Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$ . Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$ . The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . (Note that $\arg(w)$ , for $w \neq 0$ , denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane)
|
Note that $\frac{w-z}{z}=\frac{w}{z}-1$ , and that $\left|\frac{w}{z}\right|=\frac{1}{10}$ . Thus $\frac{w}{z}-1$ is a complex number on the circle with radius $\frac{1}{10}$ and centered at $-1$ on the complex plane. Let $\omega$ denote this circle.
Let $A$ and $C$ be the points that represent $\frac{w}{z}-1$ and $-1$ respectively on the complex plane. Let $O$ be the origin. In order to maximize $\tan^2(\theta)$ , we need to maximize $\angle{AOC}$ . This angle is maximized when $AO$ is tangent to $\omega$ . Using the Pythagorean Theorem, we get
\[AO^2=1^2-\left(\frac{1}{10}\right)^2=\frac{99}{100}\]
Thus
\[\tan^2(\theta)=\frac{AC^2}{AO^2}=\frac{1/100}{99/100}=\frac{1}{99}\]
And the answer is $1+99=\boxed{100}$
| 100
|
5,944
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_8
| 1
|
The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$
|
We have that $N^2 - N = N(N - 1)\equiv 0\mod{10000}$
Thus, $N(N-1)$ must be divisible by both $5^4$ and $2^4$ . Note, however, that if either $N$ or $N-1$ has both a $5$ and a $2$ in its factorization, the other must end in either $1$ or $9$ , which is impossible for a number that is divisible by either $2$ or $5$ . Thus, one of them is divisible by $2^4 = 16$ , and the other is divisible by $5^4 = 625$ . Noting that $625 \equiv 1\mod{16}$ , we see that $625$ would work for $N$ , except the thousands digit is $0$ . The other possibility is that $N$ is a multiple of $16$ and $N-1$ is a multiple of $625$ . In order for this to happen, \[N-1 \equiv -1 \pmod {16}.\] Since $625 \equiv 1 \pmod{16}$ , we know that $15 \cdot 625 = 9375 \equiv 15 \equiv -1 \mod{16}$ . Thus, $N-1 = 9375$ , so $N = 9376$ , and our answer is $\boxed{937}$
| 937
|
5,945
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_8
| 2
|
The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$
|
let $N= 10000t+1000a+100b+10c+d$ for positive integer values $t,a,b,c,d$ .
When we square $N$ we get that \begin{align*} N^2 &=(10000t+1000a+100b+10c+d)^2\\ &=10^8t^2+10^6a^2+10^4b^2+10^2c^2+d^2+2(10^7ta+10^6tb+10^5tc+10^4td+10^5ab+10^4ac+10^3bc+10^ad+10^2bd+10cd) \end{align*}
However, we don't have to deal with this whole expression but only with its last 4 digits so it is suffices to consider only: \[2000ad+2000bc+100c^2+200bd+20cd+d^2.\] Now we need to compare each decimal digit with $1000a+100b+10c+d$ and see whether the digits are congruent in base 10.
we first consider the ones digits:
$d^2\equiv d \pmod{10}.$
This can happen for only 3 values : 1, 5 and 6.
We can try to solve each case:
Considering the tenths place,
we have that:
$20cd=20c\equiv 10c \pmod {100}$ so $c= 0$
Considering the hundreds place we have that
$200bd+100c^2= 200b \equiv 100b \pmod{1000}$ so again $b=0$
now considering the thousands place we have that
$2000ad+2000bc = 2000a \equiv 1000a \pmod {10000}$ so we get $a=0$ but $a$ cannot be equal to $0$ so we consider $d=5.$
considering the tenths place
we have that:
$20cd+20=100c+20\equiv 20 \equiv 10c \mod {100}$ ( the extra $20$ is carried from $d^2$ which is equal to $25$ )
so $c=2$
considering the hundreds place we have that
$200bd+100c^2+100c= 1000b+600 \equiv600\equiv 100b \pmod{1000}$ ( the extra $100c$ is carried from the tenths place)
so $b=6$
now considering the thousands place we have that
$2000ad+2000bc +1000b= 10000a+24000+ 6000\equiv0\equiv 1000a \pmod {10000}$ ( the extra $1000b$ is carried from the hundreds place)
so a is equal 0 again
considering the tenths place
we have that:
$20cd+30=120c+30\equiv 30+20c \equiv 10c \pmod {100}$ ( the extra $20$ is carried from $d^2$ which is equal to $25$ )
if $c=7$ then we have
$30+20 \cdot 7 \equiv 70\equiv7 \cdot 10 \pmod{100}$
so $c=7$
considering the hundreds place we have that
$200bd+100c^2+100c+100= 1200b+4900+800 \equiv200b+700\equiv 100b \pmod{1000}$ ( the extra $100c+100$ is carried from the tenths place)
if $b=3$ then we have
$700+200 \cdot 3 \equiv 300\equiv3 \cdot 100 \pmod {1000}$
so $b=3$
now considering the thousands place we have that
$2000ad+2000bc +1000b+5000+1000= 12000a+42000+ 3000+6000\equiv0\equiv 2000a+1000\equiv 1000a \pmod {10000}$ ( the extra $1000b+6000$ is carried from the hundreds place)
if $a=9$ then we have
$2000 \cdot 9+1000 \equiv 9000\equiv9 \cdot 1000 \pmod {1000}$
so $a=9$
so we have that the last 4 digits of $N$ are $9376$ and $abc$ is equal to $\boxed{937}$
| 937
|
5,946
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_8
| 3
|
The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$
|
By the Chinese Remainder Theorem, the equation $N(N-1)\equiv 0\pmod{10000}$ is equivalent to the two equations: \begin{align*} N(N-1)&\equiv 0\pmod{16},\\ N(N-1)&\equiv 0\pmod{625}. \end{align*} Since $N$ and $N-1$ are coprime, the only solutions are when $(N\mod{16},N\mod{625})\in\{(0,0),(0,1),(1,0),(1,1)\}$
Let \[\varphi:\mathbb Z/10000\mathbb Z\to\mathbb Z/16\mathbb Z\times\mathbb Z/625\mathbb Z,\] \[x\mapsto (x\mod{16},x\mod{625}).\] The statement of the Chinese Remainder theorem is that $\varphi$ is an isomorphism between the two rings. In this language, the solutions are $\varphi^{-1}(0,0)$ $\varphi^{-1}(0,1)$ $\varphi^{-1}(1,0)$ , and $\varphi^{-1}(1,1)$ . Now we easily see that \[\varphi^{-1}(0,0)=0\] and \[\varphi^{-1}(1,1)=1.\] Noting that $625\equiv 1\pmod{16}$ , it follows that \[\varphi^{-1}(1,0)=625.\] To compute $\varphi^{-1}(0,1)$ , note that \[(0,1)=15(1,0)+(1,1)\] in \[\mathbb Z/16\mathbb Z\times\mathbb Z/625\mathbb Z,\] so since $\varphi^{-1}$ is linear in its arguments (by virtue of being an isomorphism), \[\varphi^{-1}(0,1)=15\varphi^{-1}(1,0)+\varphi^{-1}(1,1)=15\times 625+1=9376.\]
The four candidate digit strings $abcd$ are then $0000,0001,0625,9376$ . Of those, only $9376$ has nonzero first digit, and therefore the answer is $\boxed{937}$
| 937
|
5,947
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_8
| 4
|
The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$
|
WLOG, we can assume that $N$ is a 4-digit integer $\overline{abcd}$ . Note that the only $d$ that will satisfy $N$ will also satisfy $d^2\equiv d\pmod{10}$ , as the units digit of $\overline{abcd}^2$ is affected only by $d$ , regardless of $a$ $b$ , or $c$
By checking the numbers 0-9, we see that the only possible values of $d$ are $d=0, 1, 5, 6$
Now, we seek to find $c$ . Note that the only $\overline{cd}$ that will satisfy $N$ will also satisfy $\overline{cd}^2 \equiv \overline{cd}\pmod{100}$ , by the same reasoning as above - the last two digits of $\overline{abcd}^2$ are only affected by $c$ and $d$ . As we already have narrowed choices for $d$ , we start reasoning out.
First, we note that if $d=0$ , then $c=0$ , as a number ending in 0, and therefore divisible by 10, is squared, the result is divisible by 100, meaning it ends in two 0's. However, if $N$ ends in $00$ , then recursively, $a$ and $b$ must be $0$ , as a number divisible by 100 squared ends in four zeros. As $a$ cannot be 0, we throw out this possibility, as the only solution in this case is $0$
Now, let's assume that $d=1$ $\overline{cd}$ is equal to $10c + d = 10c + 1$ . Squaring this gives $100c^2 + 20c + 1$ , and when modulo 100 is taken, it must equal $10c + 1$ . As $c$ is an integer, $100c^2$ must be divisible by 100, so $100c^2+20c+1 \equiv 20c + 1\pmod{100}$ , which must be equivalent to $10c + 1$ . Note that this is really $\overline{(2c)1}$ and $\overline{c1}$ , and comparing the 10's digits. So really, we're just looking for when the units digit of $2c$ and $c$ are equal, and a quick check reveals that this is only true when $c=0$ .However, if we extend this process to find $b$ and $a$ , we'd find that they are also 0. The only solution in this case is $1$ , and since $a=0$ here, this is not our solution. Therefore, there are no valid solutions in this case.
Let's assume that $d=5$ . Note that $(10c + 5)^2 = 100c^2 + 100c + 25$ , and when modulo $100$ is taken, $25$ is the remainder. So all cases here have squares that end in 25, so $\overline{cd}=25$ is our only case here. A quick check reveals that $25^2=625$ , which works for now.
Now, let $d=6$ . Note that $(10c + 6)^2 = 100c^2 + 120c + 36$ . Taking modulo 100, this reduces to $20c+36$ , which must be equivalent to $10c+6$ . Again, this is similar to $\overline{(2c+3)6}$ and $\overline{c6}$ , so we see when the units digits of $2c+3$ and $c$ are equal. To make checking faster, note that $2c$ is necessarily even, so $2c+3$ is necessarily odd, so $c$ must be odd. Checking all the odds reveals that only $c=3$ works, so this case gives $76$ . Checking quickly $76^2 = 5776$ , which works for now.
Now, we find $b$ , given two possibilities for $\overline{cd}$
Start with $\overline{cd} = 25$ $\overline{bcd} = 100b + \overline{cd} = 100b + 25.$ Note that if we square this, we get $10000b^2 + 5000b + 625$ , which should be equivalent to $100b + 25$ modulo 1000. Note that, since $b$ is an integer, $10000b^2 + 5000 + 625$ simplifies modulo 1000 to $625$ . Therefore, the only $\overline{bcd}$ that works here is $625$ $625^2 = 390625$
Now, assume that $\overline{cd}=76$ . We have $100b + 76$ , and when squared, becomes $10000b^2 + 15200b + 5776$ , which, modulo 1000, should be equivalent to $100b+76$ . Reducing $10000b^2 + 15200b + 5776$ modulo $1000$ gives $200b + 776$ . Using the same technique as before, we must equate the hundreds digit of $\overline{(2b+7)76}$ to $\overline{b76}$ , or equate the units digit of $2b+7$ and $b$ . Since $2b+7$ is necessarily odd, any possible $b$ 's must be odd. A quick check reveals that $b=3$ is the only solution, so we get a solution of $376$ $376^2 = 141376$
Finally, we solve for $a$ . Start with $\overline{bcd}=625$ . We have $1000a + 625$ , which, squared, gives \[1000000a^2 + 1250000a + 390625,\] and reducing modulo 10000 gives simply 625. So $\overline{abcd}=625$ . However, that makes $a=0$ . Therefore, no solutions exist in this case.
We turn to our last case, $\overline{bcd}=376$ . We have \[1000a + 376^2 = 1000000a^2 + 752000a + 141376,\] and reducing modulo $10000$ gives $2000a + 1376$ , which must be equivalent to $1000a + 376$ . So we must have $\overline{(2a+1)376}$ being equivalent to $\overline{a376}$ modulo 1000. So, the units digit of $2a+1$ must be equal to $a$ . Since $2a+1$ is odd, $a$ must be odd. Lo and behold, the only possibility for $a$ is $a=3$ . Therefore, $\overline{abcd}=9376$ , so our answer is $\boxed{937}$
| 937
|
5,948
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_11
| 1
|
A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Perform the coordinate transformation $(x, y)\rightarrow (x+y, x-y)$ . Then we can see that a movement up, right, left, or down in the old coordinates adds the vectors $\langle 1, -1 \rangle$ $\langle 1, 1 \rangle$ $\langle -1, -1 \rangle$ $\langle -1, 1 \rangle$ respectively. Moreover, the transformation takes the equation $|y| = |x|$ to the union of the x and y axis. Exactly half of the moves go up in the new coordinates, and half of them go down. In order to end up on the x axis, we need to go up thrice and down thrice. The number of possible sequences of up and down moves is the number of permutations of $UUUDDD$ , which is just $\binom63 = 20$ . The probability of any of these sequences happening is $\left(\frac12\right)^6$ . Thus, the probability of ending on the x axis is $\frac{20}{2^6}$ . Similarly, the probability of ending on the y axis is the same.
However, we overcount exactly one case: ending at $(0, 0)$ . Since ending on the x axis and ending on the y axis are independent events, the probability of both is simply $\left(\frac{20}{2^6}\right)^2 = \frac{25}{256}$ . Using PIE, the total probability is $\frac{20}{64} + \frac{20}{64} - \frac{25}{256} = \frac{135}{256}$ , giving an answer of $\boxed{391}$
| 391
|
5,949
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_11
| 2
|
A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
We have 4 possible moves: U, D, R, and L. The total number of paths that could be taken is $4^6$ , or $4096$ . There are 4 possible cases that land along the line $y = x$ $x,y = \pm 1; x,y = \pm 2; x,y = \pm 3;$ or $x = y = 0$ . We will count the number of ways to end up at $(1,1), (2,2),$ and $(3,3)$ , multiply them by 4 to account for the other quadrants, and add this to the number of ways to end up at $(0,0)$
Thus, the total number of ways to end up at $(1,1)$ is $300$
Thus, the total number of ways to end up at $(0,0)$ is $400$
Adding these cases together, we get that the total number of ways to end up on $y = x$ is $4(20 + 120 + 300) + 400 = 2160$ . Thus, our probability is $\frac{2160}{4096}$ . When this fraction is fully reduced, it is $\frac{135}{256}$ , so our answer is $135 + 256 = \boxed{391}.$
| 391
|
5,950
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_11
| 3
|
A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
We split this into cases by making a chart. In each row, the entries $(\pm1)$ before the dividing line represent the right or left steps (without regard to order), and the entries after the dividing line represent the up or down steps (again, without regard to order). This table only represents the cases where the ending position $(x,y)$ satisfies $x=y$ \[\begin{array}{ccccccccccccl} \multicolumn{5}{c}{R (+)\qquad L (-)}& |&\multicolumn{5}{c}{U (+)\qquad D (-)}\\ +1&& +1&& +1&| & +1&& +1&& +1\\ +1&& +1&& -1& | & +1&& +1&& -1\\ +1&& -1&& -1& | & +1&& -1&& -1\\ -1 && -1&& -1& | & -1&& -1&& -1\\ \\ +1&& +1&& +1&& -1 &|& +1&& +1\\ +1&& +1&& -1 && -1 &|& +1 && -1\\ +1&& -1&& -1 && -1 &|& -1 && -1 &&(\times 2 \text{ for symmetry by swapping }R-L\text{ and }U-D)\\ \\ +1&& +1 &&+1 &&-1&& -1& |& +1\\ +1&& +1 &&-1&& -1&& -1 &|& -1&& (\times 2\text{ symmetry})\\ \\ +1&& +1 &&+1&& -1&& -1 &&-1&| & (\times2 \text{ symmetry})\\ \end{array}\] Note that to account for the cases when $x=-y$ , we can simply multiply the $U-D$ steps by $-1$ , so for example, the first row would become \[+1 \qquad+1\qquad +1 \ \ \ \ |\ \ \ -1\qquad -1\qquad -1.\] Therefore, we must multiply the number of possibilities in each case by $2$ , except for when $x=y=0$
Now, we compute the number of possibilities for each case. In particular, we must compute the number of $RLUD$ words, where $R$ represents $+1$ to the left of $|$ $L$ represents $-1$ to the left of $|$ $U$ represents $+1$ to the right of $|$ , and $D$ represents $-1$ to the right of $|$ . Using multinomials, we compute the following numbers of possibilities for each case. \[{6\choose 3}\cdot 2+ \frac{6!}{2!2!}\cdot 2 + \frac{6!}{2!2!} \cdot 2 + {6\choose 3} \cdot 2 = 2(20 + 180 + 180 + 20) = 800\] \[\frac{6!}{3!2!}\cdot 2 + \frac{6!}{2!2!} + \frac{6!}{3!2!}\cdot 2 = 120 + 180 + 120 = 420\ (\times2\text{ for symmetry})\] \[\frac{6!}{3!2!} \cdot 2 + \frac{6!}{3!2!} \cdot 2 = 120 + 120 = 240\ (\times2\text{ for symmetry})\] \[{6\choose 3} = 20\ (\times 2\text{ for symmetry})\]
Thus, there are $800 + 840 + 480 + 40 = 2160$ possibilities where $|x|=|y|$ . Because there are $4^6$ total possibilities, the probability is $\frac{2160}{4^6} = \frac{135}{256}$ , so the answer is $\boxed{391}.$
| 391
|
5,951
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_11
| 4
|
A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Denote $(x, y)_n$ the probability that starting from point $(x, y)$ , the token reaches a point on the graph of $|y| = |x|$ in exactly $n$ moves. The problem asks us to find $(0, 0)_6$ . We start by breaking this down: \[(0, 0)_6 = \frac14 \cdot ((0, 1)_5 + (0, -1)_5 + (1, 0)_5 + (-1, 0)_5)\] Notice that by symmetry, $(0, 1)_5 = (0, -1)_5 = (1, 0)_5 = (-1, 0)_5$ , so the equation simplifies to \[(0, 0)_6 = (0, 1)_5\] We now expand $(0, 1)_5$ \[(0, 1)_5 = \frac14 \cdot ((0, 0)_4 + (0, 2)_4 + 2(1, 1)_4)\] First, we find $(0, 0)_4$ \[(0, 0)_4 = (0, 1)_3\] \[(0, 1)_3 = \frac14 \cdot ((0, 0)_2 + (0, 2)_2 + 2(1, 1)_2)\] At this point, we can just count the possibilities to find $(0, 0)_2 = \frac34$ $(0, 2)_2 = \frac{7}{16}$ , and $(1, 1)_2 = \frac58$ . Therefore, \[(0, 1)_3 = \frac14 \cdot (\frac34 + \frac{7}{16} + 2 \cdot \frac58)\] \[(0, 1)_3 = \frac{39}{64}\] Next, we find $(0, 2)_4$ \[(0, 2)_4 = \frac14 \cdot ((0, 1)_3 + (0, 3)_3 + 2(1, 2)_3)\] We already calculated $(0, 1)_3$ , so we just need to find $(0, 3)_3$ and $(1, 2)_3$ \[(0, 3)_3 = \frac14 \cdot ((0, 2)_2 + (0, 4)_2 + 2(1, 3)_2)\] \[(0, 3)_3 = \frac14 \cdot (\frac{7}{16} + 0 + 2 \cdot \frac{1}{4})\] \[(0, 3)_3 = \frac{15}{64}\] \[(1, 2)_3 = \frac14 \cdot ((1, 3)_2 + (1, 1)_2 + (0, 2)_2 + (2, 2)_2)\] \[(1, 2)_3 = \frac14 \cdot (\frac14 + \frac58 + \frac{7}{16} + \frac12)\] \[(1, 2)_3 = \frac{29}{64}\] Therefore, \[(0, 2)_4 = \frac14 \cdot (\frac{39}{64} + \frac{15}{64} + 2 \cdot \frac{29}{64})\] \[(0, 2)_4 = \frac{7}{16}\] Finally, we find $(1, 1)_4$ \[(1, 1)_4 = \frac12 \cdot ((0, 1)_3 + (1, 2)_3)\] \[(1, 1)_4 = \frac12 \cdot (\frac{39}{64} + \frac{29}{64})\] \[(1, 1)_4 = \frac{17}{32}\] Putting it all together, \[(0, 0)_6 = (0, 1)_5 =\frac14 \cdot (\frac{39}{64} + \frac{7}{16} + 2 \cdot \frac{17}{32})\] \[(0, 0)_6 = \frac{135}{256}\] Thus, the answer is $135 + 256 = \boxed{391}$
| 391
|
5,952
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_12
| 1
|
Let $A=\{1,2,3,4\}$ , and $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A$ . The probability that the range of $f$ and the range of $g$ are disjoint is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$
|
The natural way to go is casework. And the natural process is to sort $f$ and $g$ based on range size! Via Pigeonhole Principle, we see that the only real possibilities are: $f 1 g 1; f 1 g 2; f 1 g 3; f 2 g 2$ . Note that the $1, 2$ and $1, 3$ cases are symmetrical and we need just a $*2$ . Note also that the total number of cases is $4^4*4^4=4^8$
$f 1 g 1$ : clearly, we choose one number as the range for $f$ , one for $g$ , yielding $12$ possibilities.
$f 1 g 2$ with symmetry (WLOG $f$ has 1 element): start by selecting numbers for the ranges. This yields $4$ for the one number in $f$ , and $3$ options for the two numbers for $g$ . Afterwards, note that the function with 2 numbers in the range can have $4+6+4=14$ arrangements of these two numbers (1 of one, 3 of the other *2 and 2 of each). Therefore, we have $2*12*14$ possibilities, the 2 from symmetry.
$f 2 g 2$ : no symmetry, still easy! Just note that we have $6$ choices of which numbers go to $f$ and $g$ , and within each, $14*14=196$ choices for the orientation of each of the two numbers. That's $6*196$ possibilities.
$f 1 g 3$ : again, symmetrical (WLOG $f$ has one element): $4$ ways to select the single element for $f$ , and then find the number of ways to distribute the $3$ distinct numbers in the range for $g$ . The only arrangement for the frequency of each number is ${1, 1, 2}$ in some order. Therefore, we have $3$ ways to choose which number is the one represented twice, and then note that there are $12$ ways to arrange these! The number of possibilities in this situation is $2 * 4 * 3 * 12$
Total, divided by $4^8$ , gets $\frac{3 * (1 + 2 * 7^2 + 2^2 * 7 + 2^3 * 3)}{4^7}$ , with numerator $\boxed{453}$
| 453
|
5,953
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_12
| 2
|
Let $A=\{1,2,3,4\}$ , and $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A$ . The probability that the range of $f$ and the range of $g$ are disjoint is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$
|
We note there are $4^4 = 256$ possibilities for each of $f$ and $g$ from $A$ to $A$ since the input of the four values of each function has four options each for an output value.
We proceed with casework to determine the number of possible $f$ with range 1, 2, etc.
There are 4 possibilities: all elements output to 1, 2, 3, or 4.
We have ${{4}\choose {2}} = 6$ ways to choose the two output elements for $f$ . At this point we have two possibilities: either $f$ has 3 of 1 element and 1 of the other, or 2 of each element. In the first case, there are 2 ways to pick the element which there are 3 copies of, and ${{4}\choose {1}} = 4$ ways to rearrange the 4 elements, for a total of $6 * 4 * 2 = 48$ ways for this case. For the second case, there are ${{4}\choose {2}} = 6$ ways to rearrange the 4 elements, for a total of $6 * 6 = 36$ ways for this case. Adding these two, we get a total of $36 + 48 = 84$ total possibilities.
We have ${{4}\choose {3}} = 4$ ways to choose the three output elements for $f$ . We know we must have 2 of 1 element and 1 of each of the others, so there are 3 ways to pick this element. Finally, there are ${{4}\choose{1}}*{{3}\choose{1}} = 12$ ways to rearrange these elements (since we can pick the locations of the 2 single elements in this many ways), and our total is $4 * 3 * 12 = 144$ ways.
Since we know the elements present, we have $4!$ ways to arrange them, or 24 ways.
(To check, $4 + 84 + 144 + 24 = 256$ , which is the total number of possibilities).
We now break $f$ down by cases, and count the number of $g$ whose ranges are disjoint from $f$ 's.
We know that there are 3 possibilities for $g$ with 1 element. Since half the possibilities for $g$ with two elements will contain the element in $f$ , there are $84/2 = 42$ possibilities for $g$ with 2 elements. Since $3/4$ the possibilities for $g$ with 3 elements will contain the element in $f$ , there are $144/4 = 36$ possibilities for $g$ with 3 elements. Clearly, no 4-element range for $g$ is possible, so the total number of ways for this case to happen is $4(3 + 42 + 36) = 324$
We know that there are 2 possibilities for $g$ with 1 element. If $g$ has 2 elements in its range, they are uniquely determined, so the total number of sets with a range of 2 elements that work for $g$ is $84/6 = 14$ . No 3-element or 4-element ranges for $g$ are possible. Thus, the total number of ways for this to happen is $84(2 + 14) = 1344$
In this case, there is only 1 possibility for $g$ - all the output values are the element that does not appear in $f$ 's range. Thus, the total number of ways for this to happen is $144$
We find that the probability of $f$ and $g$ having disjoint ranges is equal to:
$\dfrac{324 + 1344 + 144}{256^2}=\dfrac{1812}{2^{16}}= \dfrac{453}{2^{14}}$
Thus, our final answer is $\boxed{453}$
| 453
|
5,954
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_13
| 1
|
On square $ABCD$ , points $E,F,G$ , and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$ . Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$ , and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$
[asy] pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", intersectionpoint( A--P, E--H )); label("$x$", intersectionpoint( B--P, E--F )); label("$y$", intersectionpoint( C--P, G--F )); label("$z$", intersectionpoint( D--P, G--H ));[/asy]
|
$269+275+405+411=1360$ , a multiple of $17$ . In addition, $EG=FH=34$ , which is $17\cdot 2$ .
Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to $EG$ and $FH$ must be a multiple of $17$ . All of these triples are primitive:
\[17=1^2+4^2\] \[34=3^2+5^2\] \[51=\emptyset\] \[68=\emptyset\text{ others}\] \[85=2^2+9^2=6^2+7^2\] \[102=\emptyset\] \[119=\emptyset \dots\]
The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting $EG=FH=34$ \[\sqrt{17}\rightarrow 34\implies 8\sqrt{17}\implies A=\textcolor{red}{1088}\] \[\sqrt{34}\rightarrow 34\implies 5\sqrt{34}\implies A=850\] \[\sqrt{85}\rightarrow 34\implies \{18\sqrt{85}/5,14\sqrt{85}/5\}\implies A=\textcolor{red}{1101.6,666.4}\]
Thus, $\boxed{850}$ is the only valid answer.
| 850
|
5,955
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_13
| 2
|
On square $ABCD$ , points $E,F,G$ , and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$ . Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$ , and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$
[asy] pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", intersectionpoint( A--P, E--H )); label("$x$", intersectionpoint( B--P, E--F )); label("$y$", intersectionpoint( C--P, G--F )); label("$z$", intersectionpoint( D--P, G--H ));[/asy]
|
Continue in the same way as solution 1 to get that $POK$ has area $3a$ , and $OK = \frac{d}{10}$ . You can then find $PK$ has length $\frac 32$
Then, if we drop a perpendicular from $H$ to $BC$ at $L$ , We get $\triangle HLF \sim \triangle OPK$
Thus, $LF = \frac{15\cdot 34}{d}$ , and we know $HL = d$ , and $HF = 34$ . Thus, we can set up an equation in terms of $d$ using the Pythagorean theorem.
\[\frac{15^2 \cdot 34^2}{d^2} + d^2 = 34^2\]
\[d^4 - 34^2 d^2 + 15^2 \cdot 34^2 = 0\]
\[(d^2 - 34 \cdot 25)(d^2 - 34 \cdot 9) = 0\]
$d^2 = 34 \cdot 9$ is extraneous, so $d^2 = 34 \cdot 25$ . Since the area is $d^2$ , we have it is equal to $34 \cdot 25 = \boxed{850}$
| 850
|
5,956
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_14
| 1
|
Let $m$ be the largest real solution to the equation
$\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4$
There are positive integers $a$ $b$ , and $c$ such that $m=a+\sqrt{b+\sqrt{c}}$ . Find $a+b+c$
|
The first step is to notice that the 4 on the right hand side can simplify the terms on the left hand side. If we distribute 1 to $\frac{3}{x-3}$ , then the fraction becomes of the form $\frac{x}{x - 3}$ . A similar cancellation happens with the other four terms. If we assume $x = 0$ is not the highest solution (which is true given the answer format) we can cancel the common factor of $x$ from both sides of the equation.
$\frac{1}{x - 3} + \frac{1}{x - 5} + \frac{1}{x - 17} + \frac{1}{x - 19} = x - 11$
Then, if we make the substitution $y = x - 11$ , we can further simplify.
$\frac{1}{y + 8} + \frac{1}{y + 6} + \frac{1}{y - 6} + \frac{1}{y - 8} =y$
If we group and combine the terms of the form $y - n$ and $y + n$ , we get this equation:
$\frac{2y}{y^2 - 64} + \frac{2y}{y^2 - 36} = y$
Then, we can cancel out a $y$ from both sides, knowing that $x = 11$ is not a possible solution given the answer format.
After we do that, we can make the final substitution $z = y^2$
$\frac{2}{z - 64} + \frac{2}{z - 36} = 1$
$2z - 128 + 2z - 72 = (z - 64)(z - 36)$
$4z - 200 = z^2 - 100z + 64(36)$
$z^2 - 104z + 2504 = 0$
Using the quadratic formula, we get that the largest solution for $z$ is $z = 52 + 10\sqrt{2}$ . Then, repeatedly substituting backwards, we find that the largest value of $x$ is $11 + \sqrt{52 + \sqrt{200}}$ . The answer is thus $11 + 52 + 200 = \boxed{263}$
| 263
|
5,957
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_1
| 1
|
Abe can paint the room in 15 hours, Bea can paint 50 percent faster than Abe, and Coe can paint twice as fast as Abe. Abe begins to paint the room and works alone for the first hour and a half. Then Bea joins Abe, and they work together until half the room is painted. Then Coe joins Abe and Bea, and they work together until the entire room is painted. Find the number of minutes after Abe begins for the three of them to finish painting the room.
|
From the given information, we can see that Abe can paint $\frac{1}{15}$ of the room in an hour, Bea can paint $\frac{1}{15}\times\frac{3}{2} = \frac{1}{10}$ of the room in an hour, and Coe can paint the room in $\frac{1}{15}\times 2 = \frac{2}{15}$ of the room in an hour. After $90$ minutes, Abe has painted $\frac{1}{15}\times\frac{3}{2}=\frac{1}{10}$ of the room. Working together, Abe and Bea can paint $\frac{1}{15}+\frac{1}{10}=\frac{1}{6}$ of the room in an hour, so it takes then $\frac{2}{5}\div \frac{1}{6}= \frac{12}{5}$ hours to finish the first half of the room. All three working together can paint $\frac{1}{6}+\frac{2}{15}=\frac{3}{10}$ of the room in an hour, and it takes them $\frac{1}{2}\div \frac{3}{10}=\frac{5}{3}$ hours to finish the room. The total amount of time they take is \[90+\frac{12}{5}\times 60+\frac{5}{3}\times 60 = 90+ 144 + 100 = \boxed{334}\]
| 334
|
5,958
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_3
| 1
|
A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find $a^2$
[asy] pair A,B,C,D,E,F,R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); A = (0,0); B = (0,18); C = (0,36); // don't look here D = (12*2.236, 36); E = (12*2.236, 18); F = (12*2.236, 0); draw(A--B--C--D--E--F--cycle); dot(" ",A,NW); dot(" ",B,NW); dot(" ",C,NW); dot(" ",D,NW); dot(" ",E,NW); dot(" ",F,NW); //don't look here R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236[/asy]
|
When we squish the rectangle, the hexagon is composed of a rectangle and two isosceles triangles with side lengths 18, 18, and 24 as shown below.
[asy] pair R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); draw(T--R,red); draw(X--Z,red); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236[/asy]
By Heron's Formula, the area of each isosceles triangle is $\sqrt{(30)(12)(12)(6)}=\sqrt{180\times 12^2}=72\sqrt{5}$ . So the area of both is $144\sqrt{5}$ . From the rectangle, our original area is $36a$ . The area of the rectangle in the hexagon is $24a$ . So we have \[24a+144\sqrt{5}=36a\implies 12a=144\sqrt{5}\implies a=12\sqrt{5}\implies a^2=\boxed{720}.\]
| 720
|
5,959
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_3
| 2
|
A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find $a^2$
[asy] pair A,B,C,D,E,F,R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); A = (0,0); B = (0,18); C = (0,36); // don't look here D = (12*2.236, 36); E = (12*2.236, 18); F = (12*2.236, 0); draw(A--B--C--D--E--F--cycle); dot(" ",A,NW); dot(" ",B,NW); dot(" ",C,NW); dot(" ",D,NW); dot(" ",E,NW); dot(" ",F,NW); //don't look here R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236[/asy]
|
Alternatively, use basic geometry. First, scale everything down by dividing everything by 6. Let $a/6=p$ . Then, the dimensions of the central rectangle in the hexagon is p x 4, and the original rectangle is 6 x p. By Pythagorean theorem and splitting the end triangles of the hexagon into two right triangles, the altitude of the end triangles is $\sqrt{3^2-2^2}=\sqrt{5}$ given 2 as the base of the constituent right triangles. The two end triangles form a large rectangle of area $\sqrt{5}$ $4$ . Then, the area of the hexagon is $4p+4\sqrt{5}$ , and the area of the rectangle is $6p$ . Equating them, $p=2\sqrt{5}$ . Multiply by scale factor of 6 and square it to get $36(20)= 720 \implies a^2=\boxed{720}$
| 720
|
5,960
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_4
| 1
|
The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
\[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\]
where $a$ $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$
|
Notice repeating decimals can be written as the following:
$0.\overline{ab}=\frac{10a+b}{99}$
$0.\overline{abc}=\frac{100a+10b+c}{999}$
where a,b,c are the digits. Now we plug this back into the original fraction:
$\frac{10a+b}{99}+\frac{100a+10b+c}{999}=\frac{33}{37}$
Multiply both sides by $999*99.$ This helps simplify the right side as well because $999=111*9=37*3*9$
$9990a+999b+9900a+990b+99c=33/37*37*3*9*99=33*3*9*99$
Dividing both sides by $9$ and simplifying gives:
$2210a+221b+11c=99^2=9801$
At this point, seeing the $221$ factor common to both a and b is crucial to simplify. This is because taking $mod 221$ to both sides results in:
$2210a+221b+11c \equiv 9801 \mod 221 \iff 11c \equiv 77 \mod 221$
Notice that we arrived to the result $9801 \equiv 77 \mod 221$ by simply dividing $9801$ by $221$ and seeing $9801=44*221+77.$ Okay, now it's pretty clear to divide both sides by $11$ in the modular equation but we have to worry about $221$ being multiple of $11.$ Well, $220$ is a multiple of $11$ so clearly, $221$ couldn't be. Also, $221=13*17.$ Now finally we simplify and get:
$c \equiv 7 \mod 221$
But we know $c$ is between $0$ and $9$ because it is a digit, so $c$ must be $7.$ Now it is straightforward from here to find $a$ and $b$
$2210a+221b+11(7)=9801 \iff 221(10a+b)=9724 \iff 10a+b=44$
and since a and b are both between $0$ and $9$ , we have $a=b=4$ . Finally we have the $3$ digit integer $\boxed{447}$
| 447
|
5,961
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_4
| 2
|
The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
\[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\]
where $a$ $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$
|
Note that $\frac{33}{37}=\frac{891}{999} = 0.\overline{891}$ . Also note that the period of $0.abab\overline{ab}+0.abcabc\overline{abc}$ is at most $6$ . Therefore, we only need to worry about the sum $0.ababab+ 0.abcabc$ . Adding the two, we get \[\begin{array}{ccccccc}&a&b&a&b&a&b\\ +&a&b&c&a&b&c\\ \hline &8&9&1&8&9&1\end{array}\] From this, we can see that $a=4$ $b=4$ , and $c=7$ , so our desired answer is $\boxed{447}$
| 447
|
5,962
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_4
| 3
|
The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
\[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\]
where $a$ $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$
|
Noting as above that $0.\overline{ab} = \frac{10a + b}{99}$ and $0.\overline{abc} = \frac{100a + 10b + c}{999}$ , let $u = 10a + b$ .
Then \[\frac{u}{99} + \frac{10u + c}{999} = \frac{33}{37}\]
\[\frac{u}{11} + \frac{10u + c}{111} = \frac{9\cdot 33}{37}\]
\[\frac{221u + 11c}{11\cdot 111} = \frac{9\cdot 33}{37}\]
\[221u + 11c = \frac{9\cdot 33\cdot 11\cdot 111}{37}\]
\[221u + 11c = 9\cdot 33^2.\]
Solving for $c$ gives
\[c = 3\cdot 9\cdot 33 - \frac{221u}{11}\]
\[c = 891 - \frac{221u}{11}\]
Because $c$ must be integer, it follows that $u$ must be a multiple of $11$ (because $221$ clearly is not). Inspecting the equation, one finds that only $u = 44$ yields a digit $c, 7$ . Thus $abc = 10u + c = \boxed{447}.$
| 447
|
5,963
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_4
| 4
|
The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
\[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\]
where $a$ $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$
|
We note as above that $0.\overline{ab} = \frac{10a + b}{99}$ and $0.\overline{abc} = \frac{100a + 10b + c}{999},$ so
\[\frac{10a + b}{99} + \frac{100a + 10b + c}{999} = \frac{33}{37} = \frac{891}{999}.\]
As $\frac{10a + b}{99}$ has a factor of $11$ in the denominator while the other two fractions don't, we need that $11$ to cancel, so $11$ divides $10a + b.$ It follows that $a = b,$ so $\frac{10a + b}{99} = \frac{11a}{99} = \frac{111a}{999},$ so
\[\frac{111a}{999} + \frac{110a+c}{999} = \frac{891}{999}.\]
Then $111a + 110a + c = 891,$ or $221a + c = 891.$ Thus $a = b = 4$ and $c = 7,$ so the three-digit integer $abc$ is $\boxed{447}.$
| 447
|
5,964
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_5
| 1
|
Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$
|
Because the coefficient of $x^2$ in both $p(x)$ and $q(x)$ is 0, the remaining root of $p(x)$ is $-(r+s)$ , and the remaining root of $q(x)$ is $-(r+s+1)$ . The coefficients of $x$ in $p(x)$ and $q(x)$ are both equal to $a$ , and equating the two coefficients gives \[rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2\] from which $s = \tfrac 12 (5r+13)$ . The product of the roots of $p(x)$ differs from that of $q(x)$ by $240$ , so \[(r+4)\cdot \tfrac 12 (5r+7)\cdot \tfrac 12(7r+15)- r\cdot \tfrac 12 (5r+13)\cdot \tfrac 12(7r+13)=240\] from which $r^2+4r-5 =0$ , with roots $r=1$ and $r=-5$
If $r = 1$ , then the roots of $p(x)$ are $r=1$ $s=9$ , and $-(r+s)=-10$ , and $b=-rst=90$
If $r = -5$ , then the roots of $p(x)$ are $r=-5$ $s=-6$ , and $-(r+s)=11$ , and $b=-rst=-330$
Therefore the requested sum is $|- 330| + |90| = \boxed{420}$
| 420
|
5,965
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_5
| 2
|
Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$
|
Let $r$ $s$ , and $-r-s$ be the roots of $p(x)$ (per Vieta's). Then $r^3 + ar + b = 0$ and similarly for $s$ . Also, \[q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0\]
Set up a similar equation for $s$
\[q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0.\]
Simplifying and adding the equations gives \begin{align}\tag{*} r^2 - s^2 + 4r + 3s + 49 &= 0 \end{align} Now, let's deal with the $ax$ terms. Plugging the roots $r$ $s$ , and $-r-s$ into $p(x)$ yields a long polynomial, and plugging the roots $r+4$ $s-3$ , and $-1-r-s$ into $q(x)$ yields another long polynomial. Equating the coefficients of $x$ in both polynomials, we get: \[rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),\] which eventually simplifies to \[s = \frac{13 + 5r}{2}.\] Substitution into (*) should give $r = -5$ and $r = 1$ , corresponding to $s = -6$ and $s = 9$ , and $|b| = 330, 90$ , for an answer of $\boxed{420}$
| 420
|
5,966
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_5
| 3
|
Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$
|
The roots of $p(x)$ are $r$ $s$ , and $-r-s$ since they sum to $0$ by Vieta's Formula (co-efficient of $x^2$ term is $0$ ).
Similarly, the roots of $q(x)$ are $r + 4$ $s - 3$ , and $-r-s-1$ , as they too sum to $0$
Then:
$a = rs+r(-r-s)+s(-r-s) = rs-(r+s)^2$ and $-b = rs(-r-s)$ from $p(x)$ and
$a=(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2$ and $-(b+240)=(r+4)(s-3)(-r-s-1)$ from $q(x)$
From these equations, we can write that \[rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2 = a\] and simplifying gives \[2s-5r-13=0 \Rightarrow s = \frac{5r+13}{2}.\]
We now move to the other two equations regarding the product of the roots. We see that we can cancel a negative from both sides to get \[rs(r+s) = b\] \[(r+4)(s-3)(r+s+1)=b + 240.\] Subtracting the first equation from the second equation gives us $(r+4)(s-3)(r+s+1) - rs(r+s) = 240$
Expanding, simplifying, substituting $s = \frac{5r+13}{2}$ , and simplifying some more yields the simple quadratic $r^2 + 4r - 5 = 0$ , so $r = -5, 1$ . Then $s = -6, 9$
Finally, we substitute back into $b=rs(r+s)$ to get $b = (-5)(-6)(-5-6) = -330$ , or $b = (1)(9)(1 + 9) = 90$
The answer is $|-330|+|90| = \boxed{420}$
| 420
|
5,967
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_5
| 4
|
Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$
|
By Vieta's, we know that the sum of roots of $p(x)$ is $0$ . Therefore,
the roots of $p$ are $r, s, -r-s$ . By similar reasoning, the roots of $q(x)$ are $r + 4, s - 3, -r - s - 1$ . Thus, $p(x) = (x - r)(x - s)(x + r + s)$ and $q(x) = (x - r - 4)(x - s + 3)(x + r + s + 1)$
Since $p(x)$ and $q(x)$ have the same coefficient for $x$ , we can go ahead
and match those up to get \begin{align*} rs - r(r + s) - s(r + s) &= (r + 4)(s - 3) - (r + 4)(r + s + 1) - (s - 3)(r + s + 1) \\ 0 &= -13 - 5r + 2s \\ s &= \frac{5r + 13}{2} \end{align*}
At this point, we can go ahead and compare the constant term in $p(x)$ and $q(x)$ . Doing so is certainly valid, but we can actually do this another way. Notice that $p(s) = 0$ . Therefore, $q(s) = 240$ . If we plug that into
our expression, we get that \begin{align*} q(s) &= 3(s - r - 4)(r + 2s + 1) \\ 240 &= 3(s - r - 4)(r + 2s + 1) \\ 240 &= 3\left( \frac{3r + 5}{2} \right)(6r + 14) \\ 80 &= (3r + 5)(3r + 7) \\ 0 &= r^2 + 4r - 5 \end{align*} This tells us that $(r, s) = (1, 9)$ or $(-5, -6)$ . Since $-b$ is the product of the roots, we have that the two possibilities are $1 \cdot 9 \cdot (-10) = -90$ and $(-5)(-6)11 = 330$ . Adding the absolute values of these gives us $\boxed{420}$
| 420
|
5,968
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_6
| 1
|
Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$ . Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$
|
The probability that he rolls a six twice when using the fair die is $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$ . The probability that he rolls a six twice using the biased die is $\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}=\frac{16}{36}$ . Given that Charles rolled two sixes, we can see that it is $16$ times more likely that he chose the second die. Therefore the probability that he is using the fair die is $\frac{1}{17}$ , and the probability that he is using the biased die is $\frac{16}{17}$ . The probability of rolling a third six is
\[\frac{1}{17}\times \frac{1}{6} + \frac{16}{17} \times \frac{2}{3} = \frac{1}{102}+\frac{32}{51}=\frac{65}{102}\] Therefore, our desired $p+q$ is $65+102= \boxed{167}$
| 167
|
5,969
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_6
| 2
|
Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$ . Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$
|
This is an incredibly simple problem if one is familiar with conditional probability (SO BE FAMILIAR WITH IT)!
The conditional probability that the third roll will be a six given that the first two rolls are sixes, is the conditional probability that Charles rolls three sixes given that his first two rolls are sixes. This is thus $\frac{\frac{1}{2}(\frac{2}{3})^3+\frac{1}{2}(\frac{1}{6})^3}{\frac{1}{2}(\frac{2}{3})^2+\frac{1}{2}(\frac{1}{6})^2}= \frac{65}{102}$ . The answer is therefore $65+102= \boxed{167}$
| 167
|
5,970
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_8
| 1
|
Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle D is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in the form $\sqrt{m}-n$ , where $m$ and $n$ are positive integers. Find $m+n$
|
Using the diagram above, let the radius of $D$ be $3r$ , and the radius of $E$ be $r$ . Then, $EF=r$ , and $CE=2-r$ , so the Pythagorean theorem in $\triangle CEF$ gives $CF=\sqrt{4-4r}$ . Also, $CD=CA-AD=2-3r$ , so \[DF=DC+CF=2-3r+\sqrt{4-4r}.\] Noting that $DE=4r$ , we can now use the Pythagorean theorem in $\triangle DEF$ to get \[(2-3r+\sqrt{4-4r})^2+r^2=16r^2.\]
Solving this quadratic is somewhat tedious, but the constant terms cancel, so the computation isn't terrible. Solving gives $3r=\sqrt{240}-14$ for a final answer of $\boxed{254}$
| 254
|
5,971
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_8
| 2
|
Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle D is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in the form $\sqrt{m}-n$ , where $m$ and $n$ are positive integers. Find $m+n$
|
Consider a reflection of circle $E$ over diameter $\overline{AB}$ . By symmetry, we now have three circles that are pairwise externally tangent and all internally tangent to a large circle. The small circles have radii $r$ $r$ , and $3r$ , and the big circle has radius $2$
Descartes' Circle Theorem gives $\left(\frac{1}{r}+\frac{1}{r}+\frac{1}{3r}-\frac12\right)^2 = 2\left(\left(\frac{1}{r}\right)^2+\left(\frac{1}{r}\right)^2+\left(\frac{1}{3r}\right)^2+\left(-\frac12\right)^2\right)$
Note that the big circle has curvature $-\frac12$ because it is internally tangent.
Solving gives $3r=\sqrt{240}-14$ for a final answer of $\boxed{254}$
| 254
|
5,972
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_8
| 3
|
Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle D is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in the form $\sqrt{m}-n$ , where $m$ and $n$ are positive integers. Find $m+n$
|
We use the notation of Solution 1 for triangle $\triangle DEC$ \[\sin \angle EDC = \frac {EF}{DE} = \frac {1}{4} \implies \cos \angle EDC = \frac {\sqrt{15}}{4}.\] We use Cosine Law for $\triangle DEC$ and get: \[(4r)^2 +(2 – 3r)^2 – 2 \cdot 4r \cdot (2 – 3r) \cdot \frac {\sqrt{15}}{4} = (2 – r)^2\] \[(24 + 6 \sqrt{15} ) r^2 = (8 + 4 \sqrt {15})r \implies 3r = 4 \sqrt{15} – 14 \implies \boxed{254}.\] vladimir.shelomovskii@gmail.com, vvsss
| 254
|
5,973
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_9
| 1
|
Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs.
|
We know that a subset with less than $3$ chairs cannot contain $3$ adjacent chairs. There are only $10$ sets of $3$ chairs so that they are all $3$ adjacent. There are $10$ subsets of $4$ chairs where all $4$ are adjacent, and $10 \cdot 5$ or $50$ where there are only $3.$ If there are $5$ chairs, $10$ have all $5$ adjacent, $10 \cdot 4$ or $40$ have $4$ adjacent, and $10 \cdot {5\choose 2}$ or $100$ have $3$ adjacent. With $6$ chairs in the subset, $10$ have all $6$ adjacent, $10(3)$ or $30$ have $5$ adjacent, $10 \cdot {4\choose2}$ or $60$ have $4$ adjacent, $\frac{10 \cdot 3}{2}$ or $15$ have $2$ groups of $3$ adjacent chairs, and $10 \cdot \left({5\choose2} - 3\right)$ or $70$ have $1$ group of $3$ adjacent chairs. All possible subsets with more than $6$ chairs have at least $1$ group of $3$ adjacent chairs, so we add ${10\choose7}$ or $120$ ${10\choose8}$ or $45$ ${10\choose9}$ or $10$ , and ${10\choose10}$ or $1.$ Adding, we get $10 + 10 + 50 + 10 + 40 + 100 + 10 + 30 + 60 + 15 + 70 + 120 + 45 + 10 + 1 = \boxed{581}.$
| 581
|
5,974
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_9
| 2
|
Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs.
|
Starting with small cases, we see that four chairs give $4 + 1 = 5$ , five chairs give $5 + 5 + 1 = 11$ , and six chairs give $6 + 6 + 6 + 6 + 1 = 25.$ Thus, n chairs should give $n 2^{n-4} + 1$ , as confirmed above. This claim can be verified by the principle of inclusion-exclusion: there are $n 2^{n-3}$ ways to arrange $3$ adjacent chairs, but then we subtract $n 2^{n-4}$ ways to arrange $4.$ Finally, we add $1$ to account for the full subset of chairs. Thus, for $n = 10$ we get a first count of $641.$
However, we overcount cases in which there are two distinct groups of three or more chairs. We have $5$ cases for two groups of $3$ directly opposite each other, $5$ for two groups of four, $20$ for two groups of $3$ not symmetrically opposite, $20$ for a group of $3$ and a group of $4$ , and $10$ for a group of $3$ and a group of $5.$ Thus, we have $641 - 60 = \boxed{581}$
| 581
|
5,975
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_9
| 3
|
Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs.
|
It is possible to use recursion to count the complement. Number the chairs $1, 2, 3, ..., 10.$ If chair $1$ is not occupied, then we have a line of $9$ chairs such that there is no consecutive group of three. If chair $1$ is occupied, then we split into more cases. If chairs $2$ and $10$ are empty, then we have a line of $7.$ If chair $2$ is empty but chair $10$ is occupied, then we have a line of $6$ chairs (because chair $9$ cannot be occupied); this is similar to when chair $2$ is occupied and chair $10$ is empty. Finally, chairs $2$ and $10$ cannot be simultaneously occupied. Thus, we have reduced the problem down to computing $T_9 + T_7 + 2T_6$ , where $T_n$ counts the ways to select a subset of chairs $\textit{in a line}$ from a group of n chairs such that there is no group of $3$ chairs in a row.
Now, we notice that $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ (representing the cases when the first, second, and/or third chair is unoccupied). Also, $T_0 = 1, T_1 = 2, T_2 = 4, T_3 = 7$ , and hence $T_4 = 13, T_5 = 24, T_6 = 44, T_7 = 81, T_8 = 149, T_9 = 274$ . Now we know the complement is $274 + 81 + 88 = 443$ , and subtracting from $2^{10} = 1024$ gives $1024 - 443 = \boxed{581}$
| 581
|
5,976
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_12
| 1
|
Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1.$ Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$
|
WLOG, let C be the largest angle in the triangle.
As above, we can see that $\cos3A+\cos3B-\cos(3A+3B)=1$
Expanding, we get
$\cos3A+\cos3B-\cos3A\cos3B+\sin3A\sin3B=1$
$\cos3A\cos3B-\cos3A-\cos3B+1=\sin3A\sin3B$
$(\cos3A-1)(\cos3B-1)=\sin3A\sin3B$
CASE 1: If $\sin 3A = 0$ or $\sin 3B = 0$
This implies one or both of A or B are 60 or 120.
If one of A or B is 120, we have a contradiction, since C must be the largest angle.
Otherwise, if one of A or B is 60, WLOG, assume A = 60, we would have $\cos(3B) + \cos(3C) = 2$ , and thus, cos(3B) and cos(3C) both equal 1, implying $B = C = 120$ , a contradiction to the fact that the sum of the angles of a triangle must be 180 degrees.
CASE 2: If $\sin 3A \neq 0$ and $\sin 3B \neq 0$
$\frac{\cos3A-1}{\sin3A}\cdot\frac{\cos3B-1}{\sin3B}=1$
$\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1$
Note that $\tan{x}=\frac{1}{\tan(90-x)}$ , or $\tan{x}\tan(90-x)=1$
Thus $\frac{3A}{2}+\frac{3B}{2}=90$ , or $A+B=60$
Now we know that $C=120$ , so we can just use the Law of Cosines to get $\boxed{399}$
| 399
|
5,977
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_12
| 2
|
Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1.$ Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$
|
\[\cos3A+\cos3B=1-\cos(3C)=1+\cos(3A+3B)\] \[2\cos\frac{3}{2}(A+B)\cos\frac{3}{2}(A-B)=2\cos^2\frac{3}{2}(A+B)\] If $\cos\frac{3}{2}(A+B) = 0$ , then $\frac{3}{2}(A+B)=90$ $A+B=60$ , so $C=120$ ; otherwise, \[2\cos\frac{3}{2}(A-B)=2cos\frac{3}{2}(A+B)\] \[\sin\frac{3}{2}A\sin\frac{3}{2}B=0\] so either $\sin\frac{3}{2}A=0$ or $\sin\frac{3}{2}B=0$ , i.e., either $A=120$ or $B=120$ . In all cases, one of the angles must be 120, which opposes the longest side. Final result follows. $\boxed{399}$
| 399
|
5,978
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_12
| 3
|
Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1.$ Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$
|
Let $BC$ be the unknown side length. By Law of Cosines we have that $BC = \sqrt{269-260\cos{A}}$ . We notice that $\cos{A}$ should be negative to optimize $BC$ so $A$ is between $90$ and $180$ degrees. We also know that the value inside the square root is an integer $m$ , so $269-260\cos{A}$ should be an integer. We can then assume that $A$ is $120$ degrees so $\cos{A} = \frac{-1}{2}$ . We do this because $120$ degrees is a "common" value and it makes the value inside the square root an integer. Plugging this into $269-260\cos{A}$ for $m$ we get that it is $\boxed{399}$
| 399
|
5,979
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_15
| 1
|
For any integer $k\geq 1$ , let $p(k)$ be the smallest prime which does not divide $k.$ Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$ , and $X(k)=1$ if $p(k)=2.$ Let $\{x_n\}$ be the sequence defined by $x_0=1$ , and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\geq 0.$ Find the smallest positive integer $t$ such that $x_t=2090.$
|
Note that for any $x_i$ , for any prime $p$ $p^2 \nmid x_i$ . This provides motivation to translate $x_i$ into a binary sequence $y_i$
Let the prime factorization of $x_i$ be written as $p_{a_1} \cdot p_{a_2} \cdot p_{a_3} \cdots$ , where $p_i$ is the $i$ th prime number. Then, for every $p_{a_k}$ in the prime factorization of $x_i$ , place a $1$ in the $a_k$ th digit of $y_i$ . This will result in the conversion $x_1 = 2, x_{2} = 3, x_{3} = 2 * 3 = 6, \cdots$
Multiplication for the sequence $x_i$ will translate to addition for the sequence $y_i$ . Thus, we see that $x_{n+1}X(x_n) = x_np(x_n)$ translates into $y_{n+1} = y_n+1$ . Since $x_0=1$ , and $y_0=0$ $x_i$ corresponds to $y_i$ , which is $i$ in binary. Since $x_{10010101_2} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090$ $t = 10010101_2$ $\boxed{149}$
| 149
|
5,980
|
https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_15
| 3
|
For any integer $k\geq 1$ , let $p(k)$ be the smallest prime which does not divide $k.$ Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$ , and $X(k)=1$ if $p(k)=2.$ Let $\{x_n\}$ be the sequence defined by $x_0=1$ , and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\geq 0.$ Find the smallest positive integer $t$ such that $x_t=2090.$
|
Let $P_k$ denote the $k$ th prime.
Lemma: $x_{n+2^{k-1}} = P_k \cdot x_{n}$ for all $0 \leq n \leq 2^{k-1} - 1.$
$\mathbf{\mathrm{Proof:}}$
We can prove this using induction. Assume that $x_{2^{k-1}-1} = \prod_{j=1}^{k-1} P_j.$ Then, using the given recursion $x_{k+1} = \frac{x_np(x_n)}{X(x_n)}$ , we would “start fresh” for $x_{2^{k-1}} = P_k.$ It is then easy to see that then $\frac{x_n}{P_k}$ just cycles through the previous $x_{2^{k-1}}$ terms of $\{ x_n \},$ since the recursion process is the same and $P_k$ being a factor of $x_n$ is not affected until $n = 2 \cdot {2^{k-1}} = 2^k,$ when given our assumption $x_{2^k - 1} = \prod_{j=1}^{k} P_j$ and $n = 2^k$ is now the least $n$ such that \[P_{k+1} = p(x_{2^{n-1}}),\] in which $P_a = p(x_n)$ where $a > k$ is the only way that the aforementioned cycle would be affected. Specifically, by cancellation according to our recursion, $x_{2^k} = P_{k+1},$ and the values of $x_n$ just starts cycling through the previous $x_{2^k}$ terms again until $x_{2^{k+1}}$ when a new prime shows up in the prime factorization of $x_n,$ when it starts cycling again, and so on. Using our base cases of $x_0$ and $x_1,$ our induction is complete.
Now, it is easy to see that $2090 = 2 \cdot 5 \cdot 11 \cdot 19 = P_1 \cdot P_3 \cdot P_5 \cdot P_8,$ and by Lemma #1, the least positive integer $n$ such that $19 | x_n$ is $2^7.$ By repeatedly applying our obtained recursion from Lemma #1, it is easy to see that our answer is just $2^7 + 2^4 + 2^2 + 2^0,$ or $10010101_2 = \boxed{149}.$
| 149
|
5,981
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_1
| 1
|
The AIME Triathlon consists of a half-mile swim, a 30-mile bicycle ride, and an eight-mile run. Tom swims, bicycles, and runs at constant rates. He runs fives times as fast as he swims, and he bicycles twice as fast as he runs. Tom completes the AIME Triathlon in four and a quarter hours. How many minutes does he spend bicycling?
|
Let $r$ represent the rate Tom swims in miles per minute. Then we have
$\frac{1/2}{r} + \frac{8}{5r} + \frac{30}{10r} = 255$
Solving for $r$ , we find $r = 1/50$ , so the time Tom spends biking is $\frac{30}{(10)(1/50)} = \boxed{150}$ minutes.
| 150
|
5,982
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_2
| 1
|
Find the number of five-digit positive integers, $n$ , that satisfy the following conditions:
|
The number takes a form of $5\text{x,y,z}5$ , in which $5|(x+y+z)$ . Let $x$ and $y$ be arbitrary digits. For each pair of $x,y$ , there are exactly two values of $z$ that satisfy the condition of $5|(x+y+z)$ . Therefore, the answer is $10\times10\times2=\boxed{200}$
| 200
|
5,983
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_4
| 1
|
In the array of 13 squares shown below, 8 squares are colored red, and the remaining 5 squares are colored blue. If one of all possible such colorings is chosen at random, the probability that the chosen colored array appears the same when rotated 90 degrees around the central square is $\frac{1}{n}$ , where is a positive integer. Find
[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--(0,0)); draw((2,0)--(2,2)--(3,2)--(3,0)--(3,1)--(2,1)--(4,1)--(4,0)--(2,0)); draw((1,2)--(1,4)--(0,4)--(0,2)--(0,3)--(1,3)--(-1,3)--(-1,2)--(1,2)); draw((-1,1)--(-3,1)--(-3,0)--(-1,0)--(-2,0)--(-2,1)--(-2,-1)--(-1,-1)--(-1,1)); draw((0,-1)--(0,-3)--(1,-3)--(1,-1)--(1,-2)--(0,-2)--(2,-2)--(2,-1)--(0,-1)); size(100);[/asy]
|
When the array appears the same after a 90-degree rotation, the top formation must look the same as the right formation, which looks the same as the bottom one, which looks the same as the right one. There are four of the same configuration. There are not enough red squares for these to be all red, nor are there enough blue squares for there to be more than one blue square in each three-square formation. Thus there are 2 reds and 1 blue in each, and a blue in the center. There are 3 ways to choose which of the squares in the formation will be blue, leaving the other two red.
There are $\binom{13}{5}$ ways to have 5 blue squares in an array of 13.
$\frac{3}{\binom{13}{5}}$ $\frac{1}{429}$ , so $n$ $\boxed{429}$
| 429
|
5,984
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_5
| 1
|
The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ $b$ , and $c$ are positive integers. Find $a+b+c$
|
We note that $8x^3 - 3x^2 - 3x - 1 = 9x^3 - x^3 - 3x^2 - 3x - 1 = 9x^3 - (x + 1)^3$ . Therefore, we have that $9x^3 = (x+1)^3$ , so it follows that $x\sqrt[3]{9} = x+1$ . Solving for $x$ yields $\frac{1}{\sqrt[3]{9}-1} = \frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}$ , so the answer is $\boxed{98}$
| 98
|
5,985
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_5
| 2
|
The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ $b$ , and $c$ are positive integers. Find $a+b+c$
|
Let $r$ be the real root of the given polynomial . Now define the cubic polynomial $Q(x)=-x^3-3x^2-3x+8$ . Note that $1/r$ must be a root of $Q$ . However we can simplify $Q$ as $Q(x)=9-(x+1)^3$ , so we must have that $(\frac{1}{r}+1)^3=9$ . Thus $\frac{1}{r}=\sqrt[3]{9}-1$ , and $r=\frac{1}{\sqrt[3]{9}-1}$ . We can then multiply the numerator and denominator of $r$ by $\sqrt[3]{81}+\sqrt[3]{9}+1$ to rationalize the denominator, and we therefore have $r=\frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}$ , and the answer is $\boxed{98}$
| 98
|
5,986
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_5
| 3
|
The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ $b$ , and $c$ are positive integers. Find $a+b+c$
|
It is clear that for the algebraic degree of $x$ to be $3$ that there exists some cubefree integer $p$ and positive integers $m,n$ such that $a = m^3p$ and $b = n^3p^2$ (it is possible that $b = n^3p$ , but then the problem wouldn't ask for both an $a$ and $b$ ). Let $f_1$ be the automorphism over $\mathbb{Q}[\sqrt[3]{a}][\omega]$ which sends $\sqrt[3]{a} \to \omega \sqrt[3]{a}$ and $f_2$ which sends $\sqrt[3]{a} \to \omega^2 \sqrt[3]{a}$ (note : $\omega$ is a cubic root of unity ).
Letting $r$ be the root, we clearly we have $r + f_1(r) + f_2(r) = \frac{3}{8}$ by Vieta's formulas. Thus it follows $c=8$ .
Now, note that $\sqrt[3]{a} + \sqrt[3]{b} + 1$ is a root of $x^3 - 3x^2 - 24x - 64 = 0$ . Thus $(x-1)^3 = 27x + 63$ so $(\sqrt[3]{a} + \sqrt[3]{b})^3 = 27(\sqrt[3]{a} + \sqrt[3]{b}) + 90$ . Checking the non-cubicroot dimension part, we get $a + b = 90$ so it follows that $a + b + c = \boxed{98}$
| 98
|
5,987
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_5
| 4
|
The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ $b$ , and $c$ are positive integers. Find $a+b+c$
|
We have $cx-1=\sqrt[3]{a}+\sqrt[3]{b}.$ Therefore $(cx-1)^3=(\sqrt[3]{a}+\sqrt[3]{b})^3=a+b+3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})=a+b+3\sqrt[3]{ab}(cx-1).$ We have \[c^3x^3-3c^2x^2-(3c\sqrt[3]{ab}-3c)x-(a+b+1-3\sqrt[3]{ab})=0.\] We will find $a,b,c$ so that the equation is equivalent to the original one. Let $\dfrac{3c^2}{c^3}=\dfrac{3}{8}, \dfrac{3c\sqrt[3]{ab}-3c}{c^3}=\dfrac{3}{8}, \dfrac{a+b+1-3\sqrt[3]{ab}}{c^3}=\dfrac{1}{8}.$ Easily, $c=8, \sqrt[3]{ab}=9,$ and $a+b=90.$ So $a + b + c = 90+8=\boxed{98}$
| 98
|
5,988
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_8
| 1
|
The domain of the function $f(x) = \arcsin(\log_{m}(nx))$ is a closed interval of length $\frac{1}{2013}$ , where $m$ and $n$ are positive integers and $m>1$ . Find the remainder when the smallest possible sum $m+n$ is divided by 1000.
|
We start with the same method as above. The domain of the arcsin function is $[-1, 1]$ , so $-1 \le \log_{m}(nx) \le 1$
\[\frac{1}{m} \le nx \le m\] \[\frac{1}{mn} \le x \le \frac{m}{n}\] \[\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}\] \[n = 2013m - \frac{2013}{m}\]
For $n$ to be an integer, $m$ must divide $2013$ , and $m > 1$ . To minimize $n$ $m$ should be as small as possible because increasing $m$ will decrease $\frac{2013}{m}$ , the amount you are subtracting, and increase $2013m$ , the amount you are adding; this also leads to a small $n$ which clearly minimizes $m+n$
We let $m$ equal $3$ , the smallest factor of $2013$ that isn't $1$ . Then we have $n = 2013*3 - \frac{2013}{3} = 6039 - 671 = 5368$
$m + n = 5371$ , so the answer is $\boxed{371}$
| 371
|
5,989
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_8
| 2
|
The domain of the function $f(x) = \arcsin(\log_{m}(nx))$ is a closed interval of length $\frac{1}{2013}$ , where $m$ and $n$ are positive integers and $m>1$ . Find the remainder when the smallest possible sum $m+n$ is divided by 1000.
|
Note that we need $-1\le f(x)\le 1$ , and this eventually gets to $\frac{m^2-1}{mn}=\frac{1}{2013}$ . From there, break out the quadratic formula and note that \[m= \frac{n+\sqrt{n^2+4026^2}}{2013\times 2}.\] Then we realize that the square root, call it $a$ , must be an integer. Then $(a-n)(a+n)=4026^2.$
Observe carefully that $4026^2 = 2\times 2\times 3\times 3\times 11\times 11\times 61\times 61$ ! It is not difficult to see that to minimize the sum, we want to minimize $n$ as much as possible. Seeing that $2a$ is even, we note that a $2$ belongs in each factor. Now, since we want to minimize $a$ to minimize $n$ , we want to distribute the factors so that their ratio is as small as possible (sum is thus minimum). The smallest allocation of $2, 61, 61$ and $2, 11, 3, 3, 11$ fails; the next best is $2, 61, 11, 3, 3$ and $2, 61, 11$ , in which $a=6710$ and $n=5368$ . That is our best solution, upon which we see that $m=3$ , thus $\boxed{371}$
| 371
|
5,990
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_9
| 1
|
A paper equilateral triangle $ABC$ has side length $12$ . The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$ . The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$
[asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy]
|
Let $M$ and $N$ be the points on $\overline{AB}$ and $\overline{AC}$ , respectively, where the paper is folded. Let $D$ be the point on $\overline{BC}$ where the folded $A$ touches it. [asy] import cse5; size(8cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", 12*dir(60), dir(90)); pair C = MP("C", (12,0), dir(-20)); pair D = MP("D", (9,0), dir(-80)); pair Y = MP("Y", midpoint(A--D), dir(-50)); pair M = MP("M", extension(A,B,Y,Y+(dir(90)*(D-A))), dir(180)); pair N = MP("N", extension(A,C,M,Y), dir(20)); pair F = MP("F", foot(A,B,C), dir(-90)); pair X = MP("X", extension(A,F,M,N), dir(-120)); draw(B--A--C--cycle, tpen); draw(M--N^^F--A--D); draw(rightanglemark(D,F,A,15)); draw(rightanglemark(A,Y,M,15)); MA("\theta",F,A,D,1.8); [/asy] We have $AF=6\sqrt{3}$ and $FD=3$ , so $AD=3\sqrt{13}$ . Denote $\angle DAF = \theta$ ; we get $\cos\theta = 2\sqrt{3}/\sqrt{13}$
In triangle $AXY$ $AY=\tfrac 12 AD = \tfrac 32 \sqrt{13}$ , and $AX=AY\sec\theta =\tfrac{13}{4}\sqrt{3}$
In triangle $AMX$ , we get $\angle AMX=60^\circ-\theta$ and then use sine-law to get $MX=\tfrac 12 AX\csc(60^\circ-\theta)$ ; similarly, from triangle $ANX$ we get $NX=\tfrac 12 AX\csc(60^\circ+\theta)$ . Thus \[MN=\tfrac 12 AX(\csc(60^\circ-\theta) +\csc(60^\circ+\theta)).\] Since $\sin(60^\circ\pm \theta) = \tfrac 12 (\sqrt{3}\cos\theta \pm \sin\theta)$ , we get \begin{align*} \csc(60^\circ-\theta) +\csc(60^\circ+\theta) &= \frac{\sqrt{3}\cos\theta}{\cos^2\theta - \tfrac 14} = \frac{24 \cdot \sqrt{13}}{35} \end{align*} Then \[MN = \frac 12 AX \cdot \frac{24 \cdot \sqrt{13}}{35} = \frac{39\sqrt{39}}{35}\]
The answer is $39 + 39 + 35 = \boxed{113}$
| 113
|
5,991
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_9
| 2
|
A paper equilateral triangle $ABC$ has side length $12$ . The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$ . The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$
[asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy]
|
Let $P$ and $Q$ be the points on $\overline{AB}$ and $\overline{AC}$ , respectively, where the paper is folded.
Let $D$ be the point on $\overline{BC}$ where the folded $A$ touches it.
Let $a$ $b$ , and $x$ be the lengths $AP$ $AQ$ , and $PQ$ , respectively.
We have $PD = a$ $QD = b$ $BP = 12 - a$ $CQ = 12 - b$ $BD = 9$ , and $CD = 3$
Using the Law of Cosines on $BPD$
$a^{2} = (12 - a)^{2} + 9^{2} - 2 \times (12 - a) \times 9 \times \cos{60}$
$a^{2} = 144 - 24a + a^{2} + 81 - 108 + 9a$
$a = \frac{39}{5}$
Using the Law of Cosines on $CQD$
$b^{2} = (12 - b)^{2} +3^{2} - 2 \times (12 - b) \times 3 \times \cos{60}$
$b^{2} = 144 - 24b + b^{2} + 9 - 36 + 3b$
$b = \frac{39}{7}$
Using the Law of Cosines on $DPQ$
$x^{2} = a^{2} + b^{2} - 2ab \cos{60}$
$x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{39}{7})$
$x = \frac{39 \sqrt{39}}{35}$
The solution is $39 + 39 + 35 = \boxed{113}$
| 113
|
5,992
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_9
| 3
|
A paper equilateral triangle $ABC$ has side length $12$ . The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$ . The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$
[asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy]
|
Proceed with the same labeling as in Solution 1.
$\angle B = \angle C = \angle A = \angle PDQ = 60^\circ$
$\angle PDB + \angle PDQ + \angle QDC = \angle QDC + \angle CQD + \angle C = 180^\circ$
Therefore, $\angle PDB = \angle CQD$
Similarly, $\angle BPD = \angle QDC$
Now, $\bigtriangleup BPD$ and $\bigtriangleup CDQ$ are similar triangles, so
$\frac{3}{12-a} = \frac{12-b}{9} = \frac{b}{a}$
Solving this system of equations yields $a = \frac{39}{5}$ and $b = \frac{39}{7}$
Using the Law of Cosines on $APQ$
$x^{2} = a^{2} + b^{2} - 2ab \cos{60}$
$x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{39}{7})$
$x = \frac{39 \sqrt{39}}{35}$
The solution is $39 + 39 + 35 = \boxed{113}$
| 113
|
5,993
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_9
| 4
|
A paper equilateral triangle $ABC$ has side length $12$ . The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$ . The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$
[asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy]
|
We let the original position of $A$ be $A$ , and the position of $A$ after folding be $D$ . Also, we put the triangle on the coordinate plane such that $A=(0,0)$ $B=(-6,-6\sqrt3)$ $C=(6,-6\sqrt3)$ , and $D=(3,-6\sqrt3)$
[asy] size(10cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = (9,0); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle); draw(M--A--N--cycle); label("$D$", A, S); pair X = (6,6*sqrt(3)); draw(B--X--C); label("$A$",X,dir(90)); draw(A--X); [/asy]
Note that since $A$ is reflected over the fold line to $D$ , the fold line is the perpendicular bisector of $AD$ . We know $A=(0,0)$ and $D=(3,-6\sqrt3)$ . The midpoint of $AD$ (which is a point on the fold line) is $(\tfrac32, -3\sqrt3)$ . Also, the slope of $AD$ is $\frac{-6\sqrt3}{3}=-2\sqrt3$ , so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of $AD$ , or $\frac{1}{2\sqrt3}=\frac{\sqrt3}{6}$ . Then, using point slope form, the equation of the fold line is \[y+3\sqrt3=\frac{\sqrt3}{6}\left(x-\frac32\right)\] \[y=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}\] Note that the equations of lines $AB$ and $AC$ are $y=\sqrt3x$ and $y=-\sqrt3x$ , respectively. We will first find the intersection of $AB$ and the fold line by substituting for $y$ \[\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}\] \[\frac{5\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=-\frac{39}{10}\] Therefore, the point of intersection is $\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)$ . Now, lets find the intersection with $AC$ . Substituting for $y$ yields \[-\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}\] \[\frac{-7\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=\frac{39}{14}\] Therefore, the point of intersection is $\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)$ . Now, we just need to use the distance formula to find the distance between $\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)$ and $\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)$ \[\sqrt{\left(\frac{39}{14}+\frac{39}{10}\right)^2+\left(-\frac{39\sqrt3}{14}+\frac{39\sqrt3}{10}\right)^2}\] The number 39 is in all of the terms, so let's factor it out: \[39\sqrt{\left(\frac{1}{14}+\frac{1}{10}\right)^2+\left(-\frac{\sqrt3}{14}+\frac{\sqrt3}{10}\right)^2}=39\sqrt{\left(\frac{6}{35}\right)^2+\left(\frac{\sqrt3}{35}\right)^2}\] \[\frac{39}{35}\sqrt{6^2+\sqrt3^2}=\frac{39\sqrt{39}}{35}\] Therefore, our answer is $39+39+35=\boxed{113}$ , and we are done.
| 113
|
5,994
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11
| 1
|
Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are three integers $0 < x < y < z < 14$ such that when $x$ $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions.
|
$N$ must be some multiple of $\text{lcm}(14, 15, 16)= 2^{4}\cdot 3\cdot 5\cdot 7$ ; this $lcm$ is hereby denoted $k$ and $N = qk$
$1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $10$ , and $12$ all divide $k$ , so $x, y, z = 9, 11, 13$
We have the following three modulo equations:
$nk\equiv 3\pmod{9}$
$nk\equiv 3\pmod{11}$
$nk\equiv 3\pmod{13}$
To solve the equations, you can notice the answer must be of the form $9\cdot 11\cdot 13\cdot m + 3$ where $m$ is an integer.
This must be divisible by $lcm$ $(14, 15, 16)$ , which is $560\cdot 3$
Therefore, $\frac{9\cdot 11\cdot 13\cdot m + 3}{560\cdot 3} = q$ , which is an integer. Factor out $3$ and divide to get $\frac{429m+1}{560} = q$ .
Therefore, $429m+1=560q$ . We can use Bezout's Identity or a Euclidean algorithm bash to solve for the least of $m$ and $q$
We find that the least $m$ is $171$ and the least $q$ is $131$
Since we want to factor $1680\cdot q$ , don't multiply: we already know that the prime factors of $1680$ are $2$ $3$ $5$ , and $7$ , and since $131$ is prime, we have $2 + 3 + 5 + 7 + 131 = \boxed{148}$
| 148
|
5,995
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11
| 2
|
Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are three integers $0 < x < y < z < 14$ such that when $x$ $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions.
|
Note that the number of play blocks is a multiple of the LCM of $16$ $15$ , and $14$ . The value of this can be found to be $(16)(15)(7) = 1680$ . This number is also divisible by $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $10$ , and $12$ , thus, the three numbers $x, y, z$ are $9, 11, 13$
Thus, $1680k\equiv 3$ when taken mod $9$ $11$ $13$ . Since $1680$ is congruent to $6$ mod $9$ and $3$ mod $13$ , and congruent to $8$ mod $11$ , the number $k$ must be a number that is congruent to $1$ mod $13$ $2$ mod $3$ (because $6$ is a multiple of $3$ , which is a factor of $9$ that can be divided out) and cause $8$ to become $3$ when multiplied under modulo $11$
Looking at the last condition shows that $k\equiv 10$ mod $11$ (after a bit of bashing) and is congruent to $1$ mod $13$ and $2$ mod $3$ as previously noted. Listing out the numbers congruent to $10$ mod $11$ and $1$ mod $13$ yield the following lists:
$10$ mod $11$ $21$ $32$ $43$ $54$ $65$ $76$ $87$ $98$ $109$ $120$ $131$ ...
$1$ mod $13$ $14$ $27$ $40$ $53$ $66$ $79$ $92$ $105$ $118$ $131$ $144$ $157$ $170$ ...
Both lists contain $x$ elements where $x$ is the modulo being taken, thus, there must be a solution in these lists as adding $11(13)$ to this solution yields the next smallest solution. In this case, $131$ is the solution for $k$ and thus the answer is $1680(131)$ . Since $131$ is prime, the sum of the prime factors is $2 + 3 + 5 + 7 + 131 = \boxed{148}$
| 148
|
5,996
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11
| 3
|
Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are three integers $0 < x < y < z < 14$ such that when $x$ $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions.
|
It is obvious that $N=a\cdot 2^4 \cdot 3\cdot 5\cdot 7$ and so the only mod $3$ number of students are $9, 11, 13$ . Therefore, $N=1287\cdot k+3$ . Try some approaches and you will see that this one is one of the few successful ones:
Start by setting the two $N$ equations together, then we get $1680a=1287k+3$ . Divide by $3$ . Note that since the RHS is $1\pmod{3}$ , and since $560$ is $2\pmod{3}$ , then $a=3b+2$ , where $b$ is some nonnegative integer, because $a$ must be $2\pmod{3}$
This reduces to $560 \cdot 3b + 1119 = 429k$ . Now, take out the $11!$ With the same procedure, $b=11c-1$ , where $c$ is some nonnegative integer.
You also get $c=13d+4$ , at which point $k=171+560d$ $d$ cannot be equal to $0$ . Therefore, $c=4, b=43, a=131$ , and we know the prime factors of $N$ are $2, 3, 5, 7, 131$ so the answer is $\boxed{148}$
| 148
|
5,997
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11
| 4
|
Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are three integers $0 < x < y < z < 14$ such that when $x$ $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions.
|
We start by noticing that $N = a\textbf{lcm}(14, 15, 16) = 1680a$ for some integer $a$ in order to satisfy the first condition.
Next, we satisfy the second condition. Since $x<y<z < 14$ must leave a remainder when dividing $1680a$ , they are not divisors of $1680x$ . Thus, we can eliminate all $y \le 14$ s.t. $\gcd(y, 1680x) \ne 1$ which leaves $(x, y, z) = (9, 11, 13)$ . Thus, $N = 1680a \equiv 3 \pmod 9 \equiv 3 \pmod {11} \equiv 3 \pmod {13}$ . Now, we seek to find the least $a$ which satisfies this set of congruences.
By Chinese Remainder Theorem on the first two congruences, we find that $a \equiv 32 \pmod {33}$ (we divide by three before proceeding in the first congruence to ensure the minimal solution). Finally, by CRT again on $a \equiv 32 \pmod {33}$ and $1680a \equiv 3 \pmod {13}$ we find that $a \equiv 131 \pmod {429}$
Thus, the minimal value of $N$ is possible at $a = 131$ . The prime factorization of this minimum value is $2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 131$ and so the answer is $2 + 3 + 5 + 7 + 131 = \boxed{148}$
| 148
|
5,998
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11
| 5
|
Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are three integers $0 < x < y < z < 14$ such that when $x$ $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions.
|
As the problem stated, the number of boxes is definitely a multiple of $lcm(14,15,16)=1680$ , so we assume total number of boxes is $1680k$
Then, according to $(b)$ statement, we get $1680k \equiv 3 \pmod x \equiv 3 \pmod {y} \equiv 3 \pmod {z}$ . So we have $lcm(x,y,z)+3+m\cdot lcm(x,y,z)=1680k$ , we just write it to be $(1+n)lcm(x,y,z)=1680k-3$ Which tells that $x,y,z$ must be all odd number. Moreover, we can see $(1+m)lcm(x,y,z)$ can't be a multiple of $3,5,7$ (as $1680$ is a multiple of $5,7$ ) which means that $lcm(x,y,z)=lcm(9,11,13)=1287$ We let $n=1+m$
Now, we write $1287n+3=1680k, 429n+1=560k$ It is true that $n\equiv 1 \pmod{10}$ , let $n=10p+1$ , it has $429p+43=56k$ Then, $p$ must be odd, let $p=2q+1$ , it indicates $429q+236=28k$ Now, $q$ must be even, $q=2s$ tells $429s+118=14k$ Eventually, $s$ must be even, $s=2y$ $858y+59=7k$ $y=1$ is the smallest. This time, $k=131$
So the number of balls is $1680\cdot 131=2^4\cdot 3\cdot 5\cdot 7 \cdot 131$ , the desired value is $2+3+5+7+131=\boxed{148}$
| 148
|
5,999
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11
| 6
|
Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are three integers $0 < x < y < z < 14$ such that when $x$ $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions.
|
From part (a), we know that $2^4\cdot3\cdot5\cdot7 | N$ . From part (b), we know that $N \equiv 3 \pmod {1287}$ . We can expand on part (a) by saying that $N = 1680k$ for some $k$ . Rather than taking the three modulos together, we take them individually. \[1680k \equiv 6k \equiv 3 \pmod 9\] \[k \equiv 2^{-1} \pmod 9\] The inverse of 2 mod 9 is easily seen to be $5$ \[k \equiv 5 \pmod 9\]
Now moving to the second modulo which we leave as follows, \[1680k \equiv 8k \equiv 3 \pmod {11}\] Now the last modulo, \[1680k \equiv 3k \equiv 3 \pmod {13}\] \[k \equiv 1 \pmod{13}\] CRT on the first and the third one results in $k \equiv 4 \pmod {117}$ . Now doing the second one and the one we just made, $k \equiv 131 \pmod{1287}$ . Thus, the smallest value that works for $k = 131$ . Thus $N = 2^4\cdot3\cdot5\cdot7\cdot131$ $2+3+5+7+131 = \boxed{148}$
| 148
|
6,000
|
https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_13
| 1
|
Triangle $AB_0C_0$ has side lengths $AB_0 = 12$ $B_0C_0 = 17$ , and $C_0A = 25$ . For each positive integer $n$ , points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$ , respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}$ . The area of the union of all triangles $B_{n-1}C_nB_n$ for $n\geq1$ can be expressed as $\tfrac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $q$
|
Note that every $B_nC_n$ is parallel to each other for any nonnegative $n$ . Also, the area we seek is simply the ratio $k=\frac{[B_0B_1C_1]}{[B_0B_1C_1]+[C_1C_0B_0]}$ , because it repeats in smaller and smaller units. Note that the area of the triangle, by Heron's formula, is 90.
For ease, all ratios I will use to solve this problem are with respect to the area of $[AB_0C_0]$ . For example, if I say some area has ratio $\frac{1}{2}$ , that means its area is 45.
Now note that $k=$ 1 minus ratio of $[B_1C_1A]$ minus ratio $[B_0C_0C_1]$ . We see by similar triangles given that ratio $[B_0C_0C_1]$ is $\frac{17^2}{25^2}$ . Ratio $[B_1C_1A]$ is $(\frac{336}{625})^2$ , after seeing that $C_1C_0 = \frac{289}{625}$ , . Now it suffices to find 90 times ratio $[B_0B_1C_1]$ , which is given by 1 minus the two aforementioned ratios. Substituting these ratios to find $k$ and clearing out the $5^8$ , we see that the answer is $90\cdot \frac{5^8-336^2-17^2\cdot 5^4}{5^8-336^2}$ , which gives $q= \boxed{961}$
| 961
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.