Difference between revisions of "2019 AIME I Problems/Problem 15"

(Solution 1)
(Note: I put my solution as Solution 1 as I feel like it is better organized and clearer than the other one.)
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==Solution 1==
 
==Solution 1==
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[asy]
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size(8cm);
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pair O, A, B, P, O1, O2, Q, X, Y;
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O=(0, 0);
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A=dir(140); B=dir(40);
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P=(3A+5B)/8;
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O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);
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O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);
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Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1];
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X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));
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Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));
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draw(circle(O, 1));
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draw(circle(O1, length(A-O1)));
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draw(circle(O2, length(B-O2)));
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draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);
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dot("<math>O</math>", O, S);
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dot("<math>A</math>", A, A);
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dot("<math>B</math>", B, B);
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dot("<math>P</math>", P, dir(70));
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dot("<math>Q</math>", Q, dir(200));
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dot("<math>O_1</math>", O1, SW);
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dot("<math>O_2</math>", O2, SE);
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dot("<math>X</math>", X, X);
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dot("<math>Y</math>", Y, Y);
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[/asy]
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Let <math>O_1</math> and <math>O_2</math> be the centers of <math>\omega_1</math> and <math>\omega_2</math>, respectively. There is a homothety at <math>A</math> sending <math>\omega</math> to <math>\omega_1</math> that sends <math>B</math> to <math>P</math> and <math>O</math> to <math>O_1</math>, so <math>\overline{OO_2}\parallel\overline{O_1P}</math>. Similarly, <math>\overline{OO_1}\parallel\overline{O_2P}</math>, so <math>OO_1PO_2</math> is a parallelogram. Moreover, <cmath>\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,</cmath>whence <math>OO_1O_2Q</math> is cyclic. However, <cmath>OO_1=O_2P=O_2Q,</cmath>so <math>OO_1O_2Q</math> is an isosceles trapezoid. Since <math>\overline{O_1O_2}\perp\overline{XY}</math>, <math>\overline{OQ}\perp\overline{XY}</math>, so <math>Q</math> is the midpoint of <math>\overline{XY}</math>.
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By Power of a Point, <math>PX\cdot PY=PA\cdot PB=15</math>. Since <math>PX+PY=XY=11</math>, <cmath>XP=\frac{11-\sqrt{61}}2\implies PQ=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,</cmath>and the requested sum is <math>61+4=\boxed{065}</math>.
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(Solution by TheUltimate123)
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==Solution 2==
  
 
Firstly we need to notice that <math>Q</math> is the middle point of <math>XY</math>. Assume the center of circle <math>w, w_1, w_2</math> are <math>O, O_1, O_2</math>, respectively. Then <math>A, O_2, O</math> are collinear and <math>O, O_1, B</math> are collinear. Link <math>O_1P, O_2P, O_1Q, O_2Q</math>. Notice that, <math>\angle B=\angle A=\angle APO_2=\angle BPO_1</math>. As a result, <math>PO_1\parallel O_2O</math> and <math>QO_1\parallel O_2P</math>. So we have parallelogram <math>PO_2O_1O</math>. So <math>\angle O_2PO_1=\angle O</math> Notice that, <math>O_1O_2\bot PQ</math> and <math>O_1O_2</math> divide <math>PQ</math> into two equal length pieces, So we have <math>\angle O_2PO_1=\angle O_2QO_1=\angle O</math>. As a result, <math>O_2, Q, O, O_1,</math> lie on one circle. So <math>\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P</math>. Notice that <math>\angle O_1PQ+\angle O_2O_1P=90^{\circ}</math>, we have <math>\angle OQP=90^{\circ}</math>. As a result, <math>OQ\bot PQ</math>. So <math>Q</math> is the middle point of <math>XY</math>.
 
Firstly we need to notice that <math>Q</math> is the middle point of <math>XY</math>. Assume the center of circle <math>w, w_1, w_2</math> are <math>O, O_1, O_2</math>, respectively. Then <math>A, O_2, O</math> are collinear and <math>O, O_1, B</math> are collinear. Link <math>O_1P, O_2P, O_1Q, O_2Q</math>. Notice that, <math>\angle B=\angle A=\angle APO_2=\angle BPO_1</math>. As a result, <math>PO_1\parallel O_2O</math> and <math>QO_1\parallel O_2P</math>. So we have parallelogram <math>PO_2O_1O</math>. So <math>\angle O_2PO_1=\angle O</math> Notice that, <math>O_1O_2\bot PQ</math> and <math>O_1O_2</math> divide <math>PQ</math> into two equal length pieces, So we have <math>\angle O_2PO_1=\angle O_2QO_1=\angle O</math>. As a result, <math>O_2, Q, O, O_1,</math> lie on one circle. So <math>\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P</math>. Notice that <math>\angle O_1PQ+\angle O_2O_1P=90^{\circ}</math>, we have <math>\angle OQP=90^{\circ}</math>. As a result, <math>OQ\bot PQ</math>. So <math>Q</math> is the middle point of <math>XY</math>.
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Back to our problem. Assume <math>XP=x</math>, <math>PY=y</math> and <math>x<y</math>. Then we have <math>AP\cdot PB=XP\cdot PY</math>, that is, <math>xy=15</math>. Also, <math>XP+PY=x+y=XY=11</math>. Solve these above, we have <math>x=\frac{11-\sqrt{61}}{2}=XP</math>. As a result, we hav e <math>PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}</math>. So, we have <math>PQ^2=\frac{61}{4}</math>. As a result, our answer is <math>m+n=61+4=\boxed{065}</math>.
 
Back to our problem. Assume <math>XP=x</math>, <math>PY=y</math> and <math>x<y</math>. Then we have <math>AP\cdot PB=XP\cdot PY</math>, that is, <math>xy=15</math>. Also, <math>XP+PY=x+y=XY=11</math>. Solve these above, we have <math>x=\frac{11-\sqrt{61}}{2}=XP</math>. As a result, we hav e <math>PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}</math>. So, we have <math>PQ^2=\frac{61}{4}</math>. As a result, our answer is <math>m+n=61+4=\boxed{065}</math>.
  
Solution By <math>BladeRunnerAUG</math> (Fanyuchen20020715).
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Solution By BladeRunnerAUG (Fanyuchen20020715).
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=I|num-b=14|after=Last Problem}}
 
{{AIME box|year=2019|n=I|num-b=14|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:57, 15 March 2019

Problem 15

Let $\overline{AB}$ be a chord of a circle $\omega$, and let $P$ be a point on the chord $\overline{AB}$. Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$. Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$. Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$. Line $PQ$ intersects $\omega$ at $X$ and $Y$. Assume that $AP=5$, $PB=3$, $XY=11$, and $PQ^2 = \tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

[asy] size(8cm); pair O, A, B, P, O1, O2, Q, X, Y; O=(0, 0); A=dir(140); B=dir(40); P=(3A+5B)/8; O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O); O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O); Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1]; X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1)); Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));

draw(circle(O, 1)); draw(circle(O1, length(A-O1))); draw(circle(O2, length(B-O2))); draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);

dot("$O$", O, S); dot("$A$", A, A); dot("$B$", B, B); dot("$P$", P, dir(70)); dot("$Q$", Q, dir(200)); dot("$O_1$", O1, SW); dot("$O_2$", O2, SE); dot("$X$", X, X); dot("$Y$", Y, Y); [/asy] Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$, respectively. There is a homothety at $A$ sending $\omega$ to $\omega_1$ that sends $B$ to $P$ and $O$ to $O_1$, so $\overline{OO_2}\parallel\overline{O_1P}$. Similarly, $\overline{OO_1}\parallel\overline{O_2P}$, so $OO_1PO_2$ is a parallelogram. Moreover, \[\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,\]whence $OO_1O_2Q$ is cyclic. However, \[OO_1=O_2P=O_2Q,\]so $OO_1O_2Q$ is an isosceles trapezoid. Since $\overline{O_1O_2}\perp\overline{XY}$, $\overline{OQ}\perp\overline{XY}$, so $Q$ is the midpoint of $\overline{XY}$.

By Power of a Point, $PX\cdot PY=PA\cdot PB=15$. Since $PX+PY=XY=11$, \[XP=\frac{11-\sqrt{61}}2\implies PQ=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,\]and the requested sum is $61+4=\boxed{065}$.

(Solution by TheUltimate123)

Solution 2

Firstly we need to notice that $Q$ is the middle point of $XY$. Assume the center of circle $w, w_1, w_2$ are $O, O_1, O_2$, respectively. Then $A, O_2, O$ are collinear and $O, O_1, B$ are collinear. Link $O_1P, O_2P, O_1Q, O_2Q$. Notice that, $\angle B=\angle A=\angle APO_2=\angle BPO_1$. As a result, $PO_1\parallel O_2O$ and $QO_1\parallel O_2P$. So we have parallelogram $PO_2O_1O$. So $\angle O_2PO_1=\angle O$ Notice that, $O_1O_2\bot PQ$ and $O_1O_2$ divide $PQ$ into two equal length pieces, So we have $\angle O_2PO_1=\angle O_2QO_1=\angle O$. As a result, $O_2, Q, O, O_1,$ lie on one circle. So $\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P$. Notice that $\angle O_1PQ+\angle O_2O_1P=90^{\circ}$, we have $\angle OQP=90^{\circ}$. As a result, $OQ\bot PQ$. So $Q$ is the middle point of $XY$.

Back to our problem. Assume $XP=x$, $PY=y$ and $x<y$. Then we have $AP\cdot PB=XP\cdot PY$, that is, $xy=15$. Also, $XP+PY=x+y=XY=11$. Solve these above, we have $x=\frac{11-\sqrt{61}}{2}=XP$. As a result, we hav e $PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}$. So, we have $PQ^2=\frac{61}{4}$. As a result, our answer is $m+n=61+4=\boxed{065}$.

Solution By BladeRunnerAUG (Fanyuchen20020715).

See Also

2019 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Last Problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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