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 |
|---|---|---|---|---|---|
5,801 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_14 | 2 | Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$ | This is mostly a clarification to Solution 1, but let's take the diagram for the origin to $(7,3)$ . We have the origin circle and square intersected, then two squares, then the circle and square at $(7,3)$ . If we take the circle and square at the origin out of the diagram, we will be able to repeat the resulting segm... | 574 |
5,802 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_14 | 3 | Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$ | This solution is a more systematic approach for finding when the line intersects the squares and circles. Because $1001 = 7*11*13$ and $429=3*11*13$ , the slope of our line is $\frac{3}{7}$ , and we only need to consider the line in the rectangle from the origin to $(7,3)$ , and we can iterate the line $11*13=143$ time... | 574 |
5,803 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | 1 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again... | Let $Z = XY \cap AB$ . By the radical axis theorem $AD, XY, BC$ are concurrent, say at $P$ . Moreover, $\triangle DXP \sim \triangle PXC$ by simple angle chasing. Let $y = PX, x = XZ$ . Then \[\frac{y}{37} = \frac{67}{y} \qquad \implies \qquad y^2 = 37 \cdot 67.\] Now, $AZ^2 = \tfrac 14 AB^2$ , and by power of a point,... | 270 |
5,804 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | 2 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again... | By the Radical Axis Theorem $AD, XY, BC$ concur at point $E$
Let $AB$ and $EY$ intersect at $S$ . Note that because $AXDY$ and $CYXB$ are cyclic, by Miquel's Theorem $AXBE$ is cyclic as well. Thus \[\angle AEX = \angle ABX = \angle XCB = \angle XYB\] and \[\angle XEB = \angle XAB = \angle XDA = \angle XYA.\] Thus $AY \... | 270 |
5,805 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | 3 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again... | First, we note that as $\triangle XDY$ and $\triangle XYC$ have bases along the same line, $\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{DY}{YC}$ . We can also find the ratio of their areas using the circumradius area formula. If $R_1$ is the radius of $\omega_1$ and if $R_2$ is the radius of $\omega_2$ , then \[\frac... | 270 |
5,806 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | 4 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again... | [asy] size(9cm); import olympiad; real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circ... | 270 |
5,807 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | 5 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again... | Let $E = DA \cap CB$ . By Radical Axes, $E$ lies on $XY$ . Note that $EAXB$ is cyclic as $X$ is the Miquel point of $\triangle EDC$ in this configuration.
Claim. $\triangle DXE \sim \triangle EXC$ Proof. We angle chase. \[\measuredangle XEC = \measuredangle XEB = \measuredangle XAB = \measuredangle XDA = \measuredangle... | 270 |
5,808 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | 6 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again... | $AB^2 = 4 AM^2 =2x(2x+ 2 XY) =(XP - XY) (XP + XY) = XP^2 - XY^2 = XC \cdot XD - XY^2 = 67 \cdot 37 - 47^2 = \boxed{270}.$ | 270 |
5,809 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | 7 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again... | Extend $\overline{AD}$ and $\overline{BC}$ to meet at point $P$ . Let $M$ be the midpoint of segment $AB$ . Then by radical axis on $(ADY)$ $(BCY)$ and $(ABCD)$ $P$ lies on $XY$ . By the bisector lemma, $M$ lies on $XY$ . It is well-known that $P$ $A$ $X$ , and $B$ are concyclic. By Power of a point on $M$ with respect... | 270 |
5,810 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1 | 1 | Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three number... | Let $r$ be the common ratio, where $r>1$ . We then have $ar-9-(a-5)=a(r-1)-4=ar^{2}-25-(ar-9)=ar(r-1)-16$ . We now have, letting, subtracting the 2 equations, $ar^{2}+-2ar+a=12$ , so we have $3ar=432,$ or $ar=144$ , which is how much Betty had. Now we have $144+\dfrac{144}{r}+144r=444$ , or $144(r+\dfrac{1}{r})=300$ , ... | 108 |
5,811 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1 | 2 | Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three number... | Let $a$ be Alex's peanuts and $k$ the common ratio. Then we have $a(k^2+k+1)=444$ . Adding $k$ to both sides and factoring,
\[\frac{444}{a}+k=(k+1)^2\]
For the common difference, $ak=5-(a-5)=ak^2-25-(ak-9)$ . Simplifying, $k^2-2k+1=\frac{12}{a}$ . Factoring, \[(k-1)^2=\frac{12}{a}\]
\[(k+1)^2-(k-1)^2=4k \implies 4k=\fr... | 108 |
5,812 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1 | 3 | Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three number... | Let the initial numbers of peanuts Alex, Betty and Charlie had be $a$ $b$ , and $c$ respectively.
Let the final numbers of peanuts, after eating, be $a'$ $b'$ , and $c'$
We are given that $a + b + c = 444$ . Since a total of $5 + 9 + 25 = 39$ peanuts are eaten, we must have $a' + b' + c' = 444 - 39 = 405$ .
Since $a'$ ... | 108 |
5,813 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1 | 4 | Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three number... | Bashing is not difficult. All we have to consider is the first equation. We can write it as $x*(1+r+r^2) = 444$ . The variable $x$ must be an integer, and after trying all the factors of $444$ , it's clear that $r$ is a fraction smaller than $10$ . When calculating the coefficient of $x$ , we must consider that the fra... | 108 |
5,814 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1 | 5 | Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three number... | Let $b$ be the finish number of Betty's peanuts. Then \[3b = 444-(5 + 9 + 25) = 405 = 3 \cdot 135 \implies b = 135, b+ 9 = 144.\]
Let $k > 1$ be the common ratio. Then \[\frac{144}{k} + 144 + k \cdot 144 = 444 \implies \frac{144}{k} + k \cdot 144 = 300\implies \frac{12}{k} + k \cdot 12 = 25\implies k = \frac{4}{3} \im... | 108 |
5,815 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_2 | 1 | There is a $40\%$ chance of rain on Saturday and a $30\%$ chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positiv... | Let $x$ be the probability that it rains on Sunday given that it doesn't rain on Saturday. We then have $\dfrac{4}{5}x+\dfrac{2}{5}2x = \dfrac{3}{10} \implies \dfrac{7}{5}x=\dfrac{3}{10}$ $\implies x=\dfrac{3}{14}$ . Therefore, the probability that it doesn't rain on either day is $\left(1-\dfrac{3}{14}\right)\left(\df... | 107 |
5,816 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_3 | 1 | Let $x,y,$ and $z$ be real numbers satisfying the system \begin{align*} \log_2(xyz-3+\log_5 x)&=5,\\ \log_3(xyz-3+\log_5 y)&=4,\\ \log_4(xyz-3+\log_5 z)&=4.\\ \end{align*} Find the value of $|\log_5 x|+|\log_5 y|+|\log_5 z|$ | First, we get rid of logs by taking powers: $xyz-3+\log_5 x=2^{5}=32$ $xyz-3+\log_5 y=3^{4}=81$ , and $(xyz-3+\log_5 z)=4^{4}=256$ . Adding all the equations up and using the $\log {xy}=\log {x}+\log{y}$ property, we have $3xyz+\log_5{xyz} = 378$ , so we have $xyz=125$ . Solving for $x,y,z$ by substituting $125$ for $x... | 265 |
5,817 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_4 | 1 | An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Ea... | By counting the number of green cubes $2$ different ways, we have $12a=20b$ , or $a=\dfrac{5}{3} b$ . Notice that there are only $3$ possible colors for unit cubes, so for each of the $1 \times b \times c$ layers, there are $bc-21$ yellow cubes, and similarly there are $ac-45$ red cubes in each of the $1 \times a \time... | 180 |
5,818 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_4 | 2 | An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Ea... | The total number of green cubes is given by $12a=20b\Longrightarrow a=\frac{5}{3}b$
Let $r$ be the number of red cubes on each one of the $b$ layers then the total number of red cubes is $9a=br$ . Substitute $a=\frac{5}{3}b$ gives $r=15$
Repeating the procedure on the number of yellow cubes $y$ on each of the $a$ layer... | 180 |
5,819 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5 | 1 | Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The s... | Do note that by counting the area in 2 ways, the first altitude is $x = \frac{ab}{c}$ . By similar triangles, the common ratio is $\rho = \frac{a}{c}$ for each height, so by the geometric series formula, we have \begin{align} 6p=\frac{x}{1-\rho} = \frac{ab}{c-a}. \end{align} Writing $p=a+b+c$ and clearing denominat... | 182 |
5,820 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5 | 2 | Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The s... | Note that by counting the area in 2 ways, the first altitude is $\dfrac{ab}{c}$ . By similar triangles, the common ratio is $\dfrac{a}{c}$ for reach height, so by the geometric series formula, we have $6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}$ . Multiplying by the denominator and expanding, the equation becomes $\dfrac... | 182 |
5,821 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5 | 3 | Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The s... | We start by splitting the sum of all $C_{n-2}C_{n-1}$ into two parts: those where $n-2$ is odd and those where $n-2$ is even.
First consider the sum of the lengths of the segments for which $n-2$ is odd for each $n\geq2$ . The perimeters of these triangles can be expressed using $p$ and ratios that result because of si... | 182 |
5,822 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5 | 4 | Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The s... | [asy] size(10cm); // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B); // Labels label("$A$", A, S... | 182 |
5,823 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5 | 5 | Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The s... | This solution proceeds from $\frac{\frac{ab}{c}}{1-\frac{a}{c}} = \frac{\frac{ab}{c}}{\frac{c-a}{c}} = \frac{ab}{c-a} = 6(a+b+c)$ . Note the general from for a primitive pythagorean triple, $m^2-n^2, 2mn, m^2+n^2$ and after substitution, letting $a = m^2-n^2, b = 2mn, c = m^2+n^2$ into the previous equation simplifies ... | 182 |
5,824 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5 | 6 | Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The s... | For this problem, first notice that its an infinite geometric series of $6(a+b+c)=\frac{ab}{c-b}$ if $c$ is the hypotenuse. WLOG $a<b$ , we can generalize a pythagorean triple of $x^2-y^2, 2xy, x^2+y^2$ . Let $b=2xy$ , then this generalization gives $6(a+b+c)(c-b)=ab$ \[(x^2-y^2)2xy=6(2x^2+2xy)(x-y)^2\] \[(x+y)xy=6(x^2... | 182 |
5,825 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_6 | 1 | For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$ , define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$ .
Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Note that all the coefficients of odd-powered terms is an odd number of odd degree terms multiplied together, and all coefficients of even-powered terms have an even number of odd degree terms multiplied together. Since every odd degree term is negative, and every even degree term is positive, the sum is just equal to ... | 275 |
5,826 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_6 | 2 | For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$ , define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$ .
Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We are looking for the sum of the absolute values of the coefficients of $Q(x)$ . By defining $P'(x) = 1 + \frac{1}{3}x+\frac{1}{6}x^2$ , and defining $Q'(x) = P'(x)P'(x^3)P'(x^5)P'(x^7)P'(x^9)$ , we have made it so that all coefficients in $Q'(x)$ are just the positive/absolute values of the coefficients of $Q(x)$ . .... | 275 |
5,827 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_6 | 3 | For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$ , define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$ .
Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Multiply $P(x)P(x^3)$ and notice that the odd degree terms have a negative coefficient. Observing that this is probably true for all polynomials like this (including $P(x)P(x^3)P(x^5)P(x^7)P(x^9)$ ), we plug in $-1$ to get $\frac{243}{32} \implies \boxed{275}$ | 275 |
5,828 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_7 | 1 | Squares $ABCD$ and $EFGH$ have a common center and $\overline{AB} || \overline{EF}$ . The area of $ABCD$ is 2016, and the area of $EFGH$ is a smaller positive integer. Square $IJKL$ is constructed so that each of its vertices lies on a side of $ABCD$ and each vertex of $EFGH$ lies on a side of $IJKL$ . Find the differe... | Letting $AI=a$ and $IB=b$ , we have \[IJ^{2}=a^{2}+b^{2} \geq 1008\] by AM-GM inequality . Also, since $EFGH||ABCD$ , the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and $2$ adjacent vertices of a smaller square are similar. Therefore, the areas form a ... | 840 |
5,829 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_8 | 1 | Find the number of sets $\{a,b,c\}$ of three distinct positive integers with the property that the product of $a,b,$ and $c$ is equal to the product of $11,21,31,41,51,61$ | Note that the prime factorization of the product is $3^{2}\cdot 7 \cdot 11 \cdot 17 \cdot 31 \cdot 41 \cdot 61$ . Ignoring overcounting, by stars and bars there are $6$ ways to choose how to distribute the factors of $3$ , and $3$ ways to distribute the factors of the other primes, so we have $3^{6} \cdot 6$ ways. Howe... | 728 |
5,830 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_8 | 2 | Find the number of sets $\{a,b,c\}$ of three distinct positive integers with the property that the product of $a,b,$ and $c$ is equal to the product of $11,21,31,41,51,61$ | Again, notice that the prime factors of the product are $3, 3, 7, 11, 17, 31, 41, 61$ . In this problem, we are asked to partition this set of distinct(ish) factors into three smaller indistinct sets. To do this, we can use Stirling numbers of the second kind, but Stirling numbers of the second kind would assume no emp... | 728 |
5,831 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_9 | 1 | The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$ . There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$ . Find $c_k$ | Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for $b_2$ . When we get to $b_2=9$ and $a_2=91$ , we have $a_4=271$ and $b_4=729$ , which works, therefore, the answer is $b_3+a_3=81+181=\boxed{262}$ | 262 |
5,832 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_9 | 2 | The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$ . There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$ . Find $c_k$ | Using the same reasoning ( $100$ isn't very big), we can guess which terms will work. The first case is $k=3$ , so we assume the second and fourth terms of $c$ are $100$ and $1000$ . We let $r$ be the common ratio of the geometric sequence and write the arithmetic relationships in terms of $r$
The common difference is ... | 262 |
5,833 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_10 | 1 | Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. F... | [asy] import cse5; pathpen = black; pointpen = black; pointfontsize = 9; size(8cm); pair A = origin, B = (13,0), P = (4,0), Q = (7,0), T = B + 5 dir(220), C = IP(circumcircle(A,B,T),Line(T,Q,-0.1,10)), S = IP(circumcircle(A,B,C),Line(C,P,-0.1,10)); Drawing(A--B--C--cycle); D(circumcircle(A,B,C),rgb(0,0.6,1)); DrawP... | 43 |
5,834 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_10 | 3 | Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. F... | By Ptolemy's Theorem applied to quadrilateral $ASTB$ , we find \[5\cdot 7+13\cdot ST=AT\cdot BS.\] Therefore, in order to find $ST$ , it suffices to find $AT\cdot BS$ . We do this using similar triangles, which can be found by using Power of a Point theorem.
As $\triangle APS\sim \triangle CPB$ , we find \[\frac{4}{PC}... | 43 |
5,835 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_10 | 5 | Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. F... | By Ptolemy's Theorem applied to quadrilateral $ASTB$ , we find \[AS\cdot BT+AB\cdot ST=AT\cdot BS.\] Projecting through $C$ we have \[\frac{AQ \cdot PB}{PQ \cdot AB} = (A,Q; P,B)\stackrel{C}{=}(A,T; S,B)=\frac{AT \cdot BS}{ST \cdot AB}.\] Therefore \[AT \cdot BS = \frac {AQ \cdot PB}{PQ} \times ST \implies\] \[\left(\f... | 43 |
5,836 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_11 | 1 | For positive integers $N$ and $k$ , define $N$ to be $k$ -nice if there exists a positive integer $a$ such that $a^{k}$ has exactly $N$ positive divisors. Find the number of positive integers less than $1000$ that are neither $7$ -nice nor $8$ -nice. | We claim that an integer $N$ is only $k$ -nice if and only if $N \equiv 1 \pmod k$ . By the number of divisors formula, the number of divisors of $\prod_{i=1}^n p_i^{a_i}$ is $\prod_{i=1}^n (a_i+1)$ . Since all the $a_i$ s are divisible by $k$ in a perfect $k$ power, the only if part of the claim follows. To show that ... | 749 |
5,837 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | 1 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] dr... | Choose a section to start coloring. Assume, WLOG, that this section is color $1$ . We proceed coloring clockwise around the ring. Let $f(n,C)$ be the number of ways to color the first $n$ sections (proceeding clockwise) such that the last section has color $C$ . In general (except for when we complete the coloring), we... | 732 |
5,838 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | 2 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] dr... | We use complementary counting. There are $4^6$ total colorings of the ring without restriction. To count the complement, we wish to count the number of colorings in which at least one set of adjacent sections are the same color. There are six possible sets of adjacent sections that could be the same color (think of the... | 732 |
5,839 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | 3 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] dr... | We use generating functions. Suppose that the colors are $0,1,2,3$ . Then as we proceed around a valid coloring of the ring in the clockwise direction, we know that between two adjacent sections with colors $s_i$ and $s_{i+1}$ , there exists a number $d_i\in\{1,2,3\}$ such that $s_{i+1}\equiv s_i+d_i\pmod{4}$ . Theref... | 732 |
5,840 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | 4 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] dr... | Let $f(n)$ be the number of valid ways to color a ring with $n$ sections (which we call an $n$ -ring), so the answer is given by $f(6)$ . For $n=2$ , we compute $f(n)=4\cdot3=12$ . For $n \ge 3$ , we can count the number of valid colorings as follows: choose one of the sections arbitrarily, which we may color in $4$ wa... | 732 |
5,841 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | 5 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] dr... | Label the sections 1, 2, 3, 4, 5, 6 clockwise. We do casework on the colors of sections 1, 3, 5.
Case 1: the colors of the three sections are the same.
In this case, each of sections 2, 4, 6 can be one of 3 colors, so this case yields $4 \times 3^3 = 108$ ways.
Case 2: two of sections 1, 3, 5 are the same color.
Note t... | 732 |
5,842 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | 6 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] dr... | We will take a recursive approach to this problem. We can start by writing out the number of colorings for a circle with $1, 2,$ and $3$ compartments, which are $4, 12,$ and $24.$ Now we will try to find a recursive formula, $C(n)$ , for a circle with an arbitrary number of compartments $n.$ We will do this by focusing... | 732 |
5,843 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | 7 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] dr... | WLOG, color the top left section $1$ and the top right section $2$ . Then the left section can be colored $2$ $3$ , or $4$ , and the right section can be colored $1$ $3$ , or $4$ . There are $3 \cdot 3 = 9$ ways to color the left and right sections. We split this up into two cases.
Case 1: The left and right sections a... | 732 |
5,844 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | 9 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] dr... | Let's number the regions $1,2,\dots 6$ . Suppose we color regions $1,2,3$ . Then, how many ways are there to color $4,5,6$
Note: the numbers are numbered as shown:
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.... | 732 |
5,845 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | 11 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] dr... | Let's label the regions as $1,2,3,4,5,6$ in that order. We start with region $1$ . There are no restrictions on the color of region $1$ so it can be any of the four colors. We know move on the region $2$ . It can be any color but color used for region $1$ , giving us $3$ choices. Section $3$ is where it gets a bit comp... | 732 |
5,846 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | 12 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] dr... | Let the top-right segment be segment $1,$ and remaining segments are numbered $2,3,4,5,6$ in clockwise order.
We have $4$ choices for segment $1,$ and $3$ choices for segments $2,3,4,5.$ For segment $6,$ we wish to find the expected value of the number of choices for segment $6$ 's color, which depends on whether segme... | 732 |
5,847 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_13 | 1 | Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$ ; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is th... | We casework to find the number of ways to get each possible score. Note that the lowest possible score is $2$ and the highest possible score is $7$ . Let the bijective function $f(x)=\{1,2,3,4,5,6\} \to \{1,2,3,4,5,6\}$ denote the row number of the rook for the corresponding column number.
Thus, the expected sum is $\d... | 371 |
5,848 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_13 | 2 | Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$ ; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is th... | If the score is $n+1$ , then one of the rooks must appear in the $n$ th antidiagonal, and this is the first antidiagonal in which a rook can appear. To demonstrate this, we draw the following diagram when $n=4$
[asy] for (int i=0;i<7;++i) {draw((0,10*i)--(60,10*i));draw((10*i,0)--(10*i,60));} path x=(1,1)--(9,9),y=(1,9... | 371 |
5,849 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_13 | 3 | Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$ ; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is th... | So we first count the number of permutations with score $\ge 2$ . This is obviously $6!=720$ . Then, the number of permutations with score $\ge 3$ can also be computed: in the first column, there are five ways to place a rook- anywhere but the place with score $1$ . In the next column, there are $5$ ways to place a roo... | 371 |
5,850 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_13 | 4 | Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$ ; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is th... | The problem is asking us to compute $\mathbb{E}[S]$ , where $S$ is the random variable that takes an arrangement of rooks and outputs its score, which is a non-negative integer quantity. For any random variable $S$ with non-negative integer values, we have the tail sum formula \[\mathbb{E}[S] = \sum_{n = 1}^{\infty}\ma... | 371 |
5,851 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | 1 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes).... | The inradius of $\triangle ABC$ is $100\sqrt 3$ and the circumradius is $200 \sqrt 3$ . Now, consider the line perpendicular to plane $ABC$ through the circumcenter of $\triangle ABC$ . Note that $P,Q,O$ must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since $P, Q, O$ are co... | 450 |
5,852 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | 2 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes).... | Draw a good diagram. Draw $CH$ as an altitude of the triangle. Scale everything down by a factor of $100\sqrt{3}$ , so that $AB=2\sqrt{3}$ . Finally, call the center of the triangle U. Draw a cross-section of the triangle via line $CH$ , which of course includes $P, Q$ . From there, we can call $OU=h$ . There are two c... | 450 |
5,853 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | 4 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes).... | We use the diagram from solution 3. From basic angle chasing, \[180=\angle{QOC}+\angle{CO}P=2\angle{OCP}+2\angle{OCQ}=2\angle{QCP}\] so triangle QCP is a right triangle. This means that triangles $CQI$ and $CPI$ are similar. If we let $\angle{IDQ}=x$ and $\angle{PDI}=y$ , then we know $x+y=120$ and \[\frac{PG}{GC}=\fra... | 450 |
5,854 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | 5 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes).... | We use the diagram from solution 3.
Let $BP = a$ and $BQ = b$ . Then, by Stewart's on $BPQ$ , we find \[2x^3 + 2x^3 = a^2x + b^2x \implies a^2 + b^2 = 4x^2.\]
The altitude from $P$ to $ABC$ is $\sqrt{a^2 - (200\sqrt{3})^2}$ so \[PQ = 2x = \sqrt{a^2 - (200\sqrt{3})^2} + \sqrt{b^2 - (200\sqrt{3})^2}.\]
Furthermore, the a... | 450 |
5,855 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | 6 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes).... | Let $AB = a, M$ be midpoint $BC, I$ be the center of equilateral $\triangle ABC,$ $IM = b = \frac {a}{2\sqrt{3}}, O$ be the center of sphere $ABCPQ.$ Then \[AI = 2b, AO = BO = PO =QO = d.\] \[QA=QB=QC,PA=PB=PC \implies\] \[POIQ\perp ABC, \angle PMQ = 120^\circ.\] (See upper diagram).
We construct the circle PQMD, use t... | 450 |
5,856 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | 7 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes).... | Let $M$ be the midpoint of $\overline{AB}$ and $X$ the center of $\triangle ABC$ . Then \[P, O, Q, M, X, C\] all lie in the same vertical plane. We can make the following observations:
To make calculations easier, we will denote $100\sqrt{3}=m$ , so that $MX=m$ and $XC=2m$
[asy] unitsize(20); pair P = (0, 12); pair Q =... | 450 |
5,857 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | 8 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes).... | Let $Z$ be the center of $\triangle ABC$ . Let $A’$ be the midpoint of $BC$ . Let $ZA’ = c = 100\sqrt{3}$ and $ZA = 2c = 200\sqrt{3}$ . Let $PZ = a$ and $QZ = b$ . We will be working in the plane that contains the points: $A$ $P$ $A’$ $Q$ $O$ , and $Z$
Since $P$ $O$ , and $Q$ are collinear and $PO = QO = AO$ $\triangle... | 450 |
5,858 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_15 | 1 | For $1 \leq i \leq 215$ let $a_i = \dfrac{1}{2^{i}}$ and $a_{216} = \dfrac{1}{2^{215}}$ . Let $x_1, x_2, ..., x_{216}$ be positive real numbers such that $\sum_{i=1}^{216} x_i=1$ and $\sum_{1 \leq i < j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}$ . The maximum possible value of... | Note that \begin{align*}\sum_{1 \leq i < j \leq 216} x_ix_j &= \frac12\left(\left(\sum_{i=1}^{216} x_i\right)^2-\sum_{i=1}^{216} x_i^2\right)\\&=\frac12\left(1-\sum x_i^2\right).\end{align*} Substituting this into the second equation and collecting $x_i^2$ terms, we find \[\sum_{i=1}^{216}\frac{x_i^2}{1-a_i}=\frac{1}{2... | 863 |
5,859 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_4 | 1 | Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$ . Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$ . Let $M$ be the midpoint of $\overline{AE}$ , and $N$ be the midpoint of $\overline{CD}$ . The area of $\triangle BMN$ is $x$ .... | Let $A$ be the origin, so $B=(16,0)$ and $C=(20,0).$ Using equilateral triangle properties tells us that $D=(8,8\sqrt3)$ and $E=(18,2\sqrt3)$ as well. Therefore, $M=(9,\sqrt3)$ and $N=(14,4\sqrt3).$ Applying the Shoelace Theorem to triangle $BMN$ gives
\[x=\dfrac 1 2 |16\sqrt3+36\sqrt3+0-(0+14\sqrt3+64\sqrt3)| =13\sqrt... | 507 |
5,860 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_4 | 2 | Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$ . Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$ . Let $M$ be the midpoint of $\overline{AE}$ , and $N$ be the midpoint of $\overline{CD}$ . The area of $\triangle BMN$ is $x$ .... | Note that $AB=DB=16$ and $BE=BC=4$ . Also, $\angle ABE = \angle DBC = 120^{\circ}$ . Thus, $\triangle ABE \cong \triangle DBC$ by SAS.
From this, it is clear that a $60^{\circ}$ rotation about $B$ will map $\triangle ABE$ to $\triangle DBC$ .
This rotation also maps $M$ to $N$ . Thus, $BM=BN$ and $\angle MBN=60^{\circ}... | 507 |
5,861 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_4 | 3 | Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$ . Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$ . Let $M$ be the midpoint of $\overline{AE}$ , and $N$ be the midpoint of $\overline{CD}$ . The area of $\triangle BMN$ is $x$ .... | $AB = BD, BE = BC, \angle ABE = \angle CBD \implies \triangle ABE \cong \triangle DBC$
Medians are equal, so $MB = MN, \angle ABM = \angle DBN \implies$ $\angle MBN = \angle ABD - \angle ABM + \angle DBN = 60^\circ \implies$
$\triangle MNB$ is equilateral triangle.
The height of $\triangle BCE$ is $2 \sqrt{3},$ dista... | 507 |
5,862 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_5 | 1 | In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday ... | Let the fifth sock be arbitrary; the probability that the sixth sock matches in color is $\dfrac{1}{9}$
Assuming this, then let the first sock be arbitrary; the probability that the second sock does not match is $\dfrac{6}{7}.$
The only "hard" part is the third and fourth sock. But that is simple casework. If the third... | 341 |
5,863 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_5 | 2 | In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday ... | The key is to count backwards. First, choose the pair which you pick on Wednesday in $5$ ways. Then there are four pairs of socks for you to pick a pair of on Tuesday, and you don't want to pick a pair. Since there are $4$ pairs, the number of ways to do this is $\dbinom{8}{2}-4$ . Then, there are two pairs and two non... | 341 |
5,864 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_5 | 3 | In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday ... | For the first sock, note that to pick two different socks, can complementary count to get the total, $\binom{10}{2}$ minus the number of pairs (5) to get \[\frac{\binom{10}{2} - 5}{\binom{10}{2}}\] The next steps aren't quite as simple, though.
WLOG suppose the socks are (a, a, b, b, c, c, d, d, e, e) and that we chose... | 341 |
5,865 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_7 | 1 | In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$ . Points $F$ and $G$ lie on $\overline{CE}$ , and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$ , respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$ , and $M$ and $N$ lie on $\overline{AD}$ a... | Let us find the proportion of the side length of $KLMN$ and $FJGH$ . Let the side length of $KLMN=y$ and the side length of $FJGH=x$
[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot... | 539 |
5,866 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_7 | 2 | In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$ . Points $F$ and $G$ lie on $\overline{CE}$ , and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$ , respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$ , and $M$ and $N$ lie on $\overline{AD}$ a... | [asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G... | 539 |
5,867 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8 | 1 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ | You know whatever $n$ is, it has to have 3 digits, because if it had only two, the maximum of $s(n)$ is 18.
Now let $n=100a_2+10a_1+a_0$
So first we know, $a_2+a_1+a_0=20$ . Okay now we have to split into cases based on which digit gets carried. This meaning, when you add a 3 digit number to 864, we have to know when t... | 695 |
5,868 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8 | 2 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ | First, it is easy to verify that $695$ works and that no other numbers beginning with the digit 6 work (i.e. $686, 677, 668, 659$ do not work).
Suppose by contradiction that there is a smaller valid $n$ , where the leading digit of the three-digit number $n$ is 5 or less. (Two-digit $n$ obviously do not work because 9 ... | 695 |
5,869 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8 | 3 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ | First of all, notice that the smallest $n$ with $s(n) = 20$ is $299$ . Also, if $s(n + 864) = 20$ $s(n - 136) = 19$ (because subtracting $1000$ from the number removes the $1$ in the thousands place). After checking $s(n - 136)$ for various $n$ with $s(n) = 20$ , we see that we need to have a carry when subtracting $13... | 695 |
5,870 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8 | 4 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ | Observation (Lemma) : If r is the number of regroups in the addition of n+k, $S(n+k) = S(n)+S(k)-9r$
Proof : When you add two numbers, and you do a carry, you are taking away 10 from 1 column, and adding 1 to another column, giving a net loss of 9 to the total.
Thus, we can see that we need to regroup exactly twice wh... | 695 |
5,871 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8 | 5 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ | Although this solution doesn't directly solve the problem, it greatly hastens the bashing process.
Call the three digits a, b, and c.
When you add each of 8, 6, and 4 to a, b, and c the resultant will either get smaller or larger, depending on the original number.
e.g. If c is 7, then adding 4 will reduce the 7 to a 1,... | 695 |
5,872 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8 | 6 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ | First, note that to compute $s(m+n)$ (for any positive integers $m$ and $n$ ), one can simply find the sum of $s(m)$ and $s(n)$ minus 9 times the number of times one regroups when adding $m$ to $n$ . One can see why this is by noticing that if one were to "forget" to regroup, and leave, say, a 10 in the ones' place, ... | 695 |
5,873 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_8 | 7 | For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ . Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$ | Bashing out modulo $9$ and getting lucky we get that if the $8$ and $6$ carry over when adding $n$ and $864$ , that $100a+10b+c \equiv 1+100(a-1)+10(b-4)+c+4 \pmod{9}$ such that $n=100a+10b+c$ and after maximizing $b$ and $c$ such that $c<6$ to not make the $4$ carry over to minimize $a$ we get that $\boxed{695}$ is ou... | 695 |
5,874 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_9 | 1 | Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$ . Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$ . Find the number of such sequences for which $a_n=0$ for some $n$ | Let $a_1=x, a_2=y, a_3=z$ . First note that if any absolute value equals 0, then $a_n=0$ .
Also note that if at any position, $a_n=a_{n-1}$ , then $a_{n+2}=0$ .
Then, if any absolute value equals 1, then $a_n=0$ .
Therefore, if either $|y-x|$ or $|z-y|$ is less than or equal to 1, then that ordered triple meets the cri... | 494 |
5,875 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_11 | 1 | Triangle $ABC$ has positive integer side lengths with $AB=AC$ . Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$ . Suppose $BI=8$ . Find the smallest possible perimeter of $\triangle ABC$ | Let $D$ be the midpoint of $\overline{BC}$ . Then by SAS Congruence, $\triangle ABD \cong \triangle ACD$ , so $\angle ADB = \angle ADC = 90^o$
Now let $BD=y$ $AB=x$ , and $\angle IBD = \dfrac{\angle ABD}{2} = \theta$
Then $\mathrm{cos}{(\theta)} = \dfrac{y}{8}$
and $\mathrm{cos}{(2\theta)} = \dfrac{y}{x} = 2\mathrm{cos... | 108 |
5,876 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_11 | 3 | Triangle $ABC$ has positive integer side lengths with $AB=AC$ . Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$ . Suppose $BI=8$ . Find the smallest possible perimeter of $\triangle ABC$ | Let $M$ be midpoint $BC, BM = x, AB = y, \angle IBM = \alpha.$
$BI$ is the bisector of $\angle ABM$ in $\triangle ABM.$ $BI = \frac {2 xy \cos \alpha}{x+y} = 8, \cos \alpha = \frac {x}{8} \implies \frac {x^2 y}{x+y} = 32.$ \[y = \frac {32 x} {x^2 - 32}.\] $BC = 2x$ is integer, $5.5^2 < 32 \implies x \ge 6.$ $BM < BI \i... | 108 |
5,877 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_12 | 1 | Consider all 1000-element subsets of the set $\{1, 2, 3, ... , 2015\}$ . From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | Let $M$ be the desired mean. Then because $\dbinom{2015}{1000}$ subsets have 1000 elements and $\dbinom{2015 - i}{999}$ have $i$ as their least element, \begin{align*} \binom{2015}{1000} M &= 1 \cdot \binom{2014}{999} + 2 \cdot \binom{2013}{999} + \dots + 1016 \cdot \binom{999}{999} \\ &= \binom{2014}{999} + \bino... | 431 |
5,878 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_12 | 2 | Consider all 1000-element subsets of the set $\{1, 2, 3, ... , 2015\}$ . From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | Each 1000-element subset $\left\{ a_1, a_2,a_3,...,a_{1000}\right\}$ of $\left\{1,2,3,...,2015\right\}$ with $a_1<a_2<a_3<...<a_{1000}$ contributes $a_1$ to the sum of the least element of each subset. Now, consider the set $\left\{a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ . There are $a_1$ ways to choose a positive in... | 431 |
5,879 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_12 | 3 | Consider all 1000-element subsets of the set $\{1, 2, 3, ... , 2015\}$ . From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | Let $p$ be the size of the large set and $q$ be the size of the subset (i.e. in this problem, $p = 2015$ and $q = 1000$ ). We can easily find the answers for smaller values of $p$ and $q$
For $p = 2$ and $q = 2$ , the answer is $1$
For $p = 3$ and $q = 2$ , the answer is $\frac43$
For $p = 4$ and $q = 2$ , the answer i... | 431 |
5,880 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_13 | 1 | With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$ , where $m$ and $n$ are integers greater than 1. Find $m+n$ | Let $x = \cos 1^\circ + i \sin 1^\circ$ . Then from the identity \[\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},\] we deduce that (taking absolute values and noticing $|x| = 1$ \[|2\sin 1| = |x^2 - 1|.\] But because $\csc$ is the reciprocal of $\sin$ and because $\sin z = \sin (180^\circ - z)$ , if we le... | 91 |
5,881 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_14 | 1 | For each integer $n \ge 2$ , let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$ , where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$ . Find the number of values of $n$ with ... | Let $n\ge 2$ and define $a(n) = \left\lfloor \sqrt n \right\rfloor$ . For $2\le n \le 1000$ , we have $1\le a(n)\le 31$
For $a^2 \le x < (a+1)^2$ we have $y=ax$ . Thus $A(n+1)-A(n)=a(n+\tfrac 12) = \Delta_n$ (say), and $\Delta_n$ is an integer if $a$ is even; otherwise $\Delta_n$ is an integer plus $\tfrac 12$
If $a=1$... | 483 |
5,882 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_14 | 2 | For each integer $n \ge 2$ , let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$ , where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$ . Find the number of values of $n$ with ... | By considering the graph of this function, it is shown that the graph is composed of trapezoids ranging from $a^2$ to $(a+1)^2$ with the top made of diagonal line $y=ax$ . The width of each trapezoid is $3, 5, 7$ , etc. Whenever $a$ is odd, the value of $A(n)$ increases by an integer value, plus $\frac{1}{2}$ . Wheneve... | 483 |
5,883 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_14 | 3 | For each integer $n \ge 2$ , let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$ , where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$ . Find the number of values of $n$ with ... | First, draw a graph of the function. Note that it is just a bunch of line segments. Since we only need to know whether or not $A(n)$ is an integer, we can take the area of each piece from some $x$ to $x+1$ (mod 1), aka the piece from $2$ to $3$ has area $\frac{1}{2} (\mod 1)$ . There are some patterns. Every time we in... | 483 |
5,884 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_1 | 1 | Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$ | If $N$ is $22$ percent less than one integer $k$ , then $N=\frac{78}{100}k=\frac{39}{50}k$ . In addition, $N$ is $16$ percent greater than another integer $m$ , so $N=\frac{116}{100}m=\frac{29}{25}m$ . Therefore, $k$ is divisible by $50$ and $m$ is divisible by $25$ . Setting these two equal, we have $\frac{39}{50}k=\f... | 131 |
5,885 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_1 | 2 | Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$ | Continuing from Solution 1, we have $N=\frac{39}{50}k$ and $N=\frac{29}{25}m$ . It follows that $k=\frac{50}{39}N$ and $m=\frac{25}{29}N$ . Both $m$ and $k$ have to be integers, so, in order for that to be true, $N$ has to cancel the denominators of both $\frac{50}{39}$ and $\frac{25}{29}$ . In other words, $N$ is a mu... | 131 |
5,886 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_3 | 1 | Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$ | The three-digit integers divisible by $17$ , and their digit sum: \[\begin{array}{c|c} m & s(m)\\ \hline 102 & 3 \\ 119 & 11\\ 136 & 10\\ 153 & 9\\ 170 & 8\\ 187 & 16\\ 204 & 6\\ 221 & 5\\ 238 & 13\\ 255 & 12\\ 272 & 11\\ 289 & 19\\ 306 & 9\\ 323 & 8\\ 340 & 7\\ 357 & 15\\ 374 & 14\\ 391 & 13\\ 408 & 12\\ 425 & 11\\ 44... | 476 |
5,887 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_3 | 2 | Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$ | We can do the same thing as solution 1, except note the following fact: $102$ is a multiple of $17$ and its digits sum to $3$
Therefore, we can add it onto an existing multiple of $17$ that we know of to have $s(m) = 14$ , shown in the right-hand column, provided that its units digit is less than $8$ and its hundreds d... | 476 |
5,888 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_3 | 4 | Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$ | Since the sum of the digits in the base-10 representation of $m$ is $17$ , we must have $m\equiv 17 \pmod{9}$ or $m\equiv -1\pmod{9}$ .
We also know that since $m$ is divisible by 17, $m\equiv 0 \pmod{17}$
To solve this system of linear congruences, we can use the Chinese Remainder Theorem. If we set $m\equiv (-1)(17)(... | 476 |
5,889 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_3 | 5 | Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$ | We proceed by casework on the number of digits. Clearly the answer must have at least two digits, seeing as the maximum digit sum for a one-digit number is 9. The answer must also have less than 4 digits, because this is the AIME.
Case 1: The answer is a 2-digit number.
Represent the number as $10a + b$ , where $0 < a ... | 476 |
5,890 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_4 | 2 | In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$ , and the altitude to these bases has length $\log 16$ . The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$ , where $p$ and $q$ are positive integers. Find $p + q$ | Set the base of the log as 2. Then call the trapezoid $ABCD$ with $CD$ as the longer base. Then have the two feet of the altitudes be $E$ and $F$ , with $E$ and $F$ in position from left to right respectively. Then, $CF$ and $ED$ are $\log 192 - \log 3 = \log 64$ (from the log subtraction identity. Then $CF=EF=3$ (isos... | 18 |
5,891 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_4 | 3 | In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$ , and the altitude to these bases has length $\log 16$ . The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$ , where $p$ and $q$ are positive integers. Find $p + q$ | Let $ABCD$ be the trapezoid, where $\overline{AB} || \overline{CD}$ and $AB = \log 3$ and $CD = \log 192$ . Draw altitudes from $A$ and $B$ to $\overline{CD}$ with feet at $E$ and $F$ , respectively. $AB = \log 3$ , so $EF = \log 3$ . Now, we attempt to find $DE + FC$ , or what's left of $CD$ after we take out $EF$ . W... | 18 |
5,892 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_6 | 1 | Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$ . Can you tell me the values of $a$ and $c$ ?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve sa... | We call the three roots (some may be equal to one another) $x_1$ $x_2$ , and $x_3$ . Using Vieta's formulas, we get $x_1+x_2+x_3 = a$ $x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}$ , and $x_1 \cdot x_2 \cdot x_3 = \frac{c}{2}$
Squaring our first equation we get $x_1^2+x_2^2+x_3^2+2 \cdot x_1 \cdot x_2+... | 440 |
5,893 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_6 | 2 | Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$ . Can you tell me the values of $a$ and $c$ ?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve sa... | First things first. Vietas gives us the following:
\begin{align} x_1+x_2+x_3 = a\\ x_1 \cdot x_2+x_1 \cdot x_3+x_2 \cdot x_3 = \frac{a^2-81}{2}\\ x_1 \cdot x_2 \cdot x_3 = \frac{c}{2} \end{align}
From $(2)$ $a$ must have odd parity, meaning $a^2-81$ must be a multiple of $4$ , which implies that both sides of $(2)$ are... | 440 |
5,894 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_6 | 3 | Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$ . Can you tell me the values of $a$ and $c$ ?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve sa... | Since each of the roots is positive, the local maximum of the function must occur at a positive value of $x$ . Taking $\frac{d}{dx}$ of the polynomial yields $6x^2-4ax+a^2-81$ , which is equal to $0$ at the local maximum. Since this is a quadratic in $a$ , we can find an expression for $a$ in terms of $x$ . The quadrat... | 440 |
5,895 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7 | 1 | Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS... | If $\omega = 25$ , the area of rectangle $PQRS$ is $0$ , so
\[\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0\]
and $\alpha = 25\beta$ . If $\omega = \frac{25}{2}$ , we can reflect $APQ$ over $PQ$ $PBS$ over $PS$ , and $QCR$ over $QR$ to completely cover rectangle $PQRS$ , so the area of $PQRS$ is half the are... | 161 |
5,896 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7 | 2 | Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS... | [asy] unitsize(20); pair A,B,C,E,F,P,Q,R,S; A=(48/5,36/5); B=(0,0); C=(25,0); E=(48/5,0); F=(48/5,18/5); P=(24/5,18/5); Q=(173/10,18/5); S=(24/5,0); R=(173/10,0); draw(A--B--C--cycle); draw(P--Q); draw(Q--R); draw(R--S); draw(S--P); draw(A--E,dashed); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$E$",... | 161 |
5,897 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7 | 3 | Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS... | Using the diagram from Solution 2 above, label $AF$ to be $h$ . Through Heron's formula, the area of $\triangle ABC$ turns out to be $90$ , so using $AE$ as the height and $BC$ as the base yields $AE=\frac{36}{5}$ . Now, through the use of similarity between $\triangle APQ$ and $\triangle ABC$ , you find $\frac{w}{25}=... | 161 |
5,898 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7 | 4 | Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS... | Using the diagram as shown in Solution 2, let $AE=h$ and $AP=L$ Now, by Heron's formula, we find that the $[ABC]=90$ . Hence, $h=\frac{36}{5}$
Now, we see that $\sin{B}=\frac{PS}{12-L}\implies PS=\sin{B}(12-L)$ We easily find that $\sin{B}=\frac{3}{5}$
Hence, $PS=\frac{3}{5}(12-L)$
Now, we see that $[PQRS]=\frac{3}{5}(... | 161 |
5,899 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_7 | 5 | Triangle $ABC$ has side lengths $AB = 12$ $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = \omega$ , the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS... | Proceed as in solution 1. When $\omega$ is equal to zero, $\alpha - \beta\omega=\alpha$ is equal to the altitude. This means that $25\beta$ is equal to $\frac{36}{5}$ , so $\beta = \frac{36}{125}$ , yielding $\boxed{161}$ | 161 |
5,900 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_9 | 1 | 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=or... | Our aim is to find the volume of the part of the cube submerged in the cylinder.
In the problem, since three edges emanate from each vertex, the boundary of the cylinder touches the cube at three points. Because the space diagonal of the cube is vertical, by the symmetry of the cube, the three points form an equilate... | 384 |
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