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

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==Problem 15==
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==Problem==
  
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Let <math>\overline{AB}</math> be a chord of a circle <math>\omega</math>, and let <math>P</math> be a point on the chord <math>\overline{AB}</math>. Circle <math>\omega_1</math> passes through <math>A</math> and <math>P</math> and is internally tangent to <math>\omega</math>. Circle <math>\omega_2</math> passes through <math>B</math> and <math>P</math> and is internally tangent to <math>\omega</math>. Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>P</math> and <math>Q</math>. Line <math>PQ</math> intersects <math>\omega</math> at <math>X</math> and <math>Y</math>. Assume that <math>AP=5</math>, <math>PB=3</math>, <math>XY=11</math>, and <math>PQ = \sqrt{\frac{m}{n}}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
+
Let <math>\overline{AB}</math> be a chord of a circle <math>\omega</math>, and let <math>P</math> be a point on the chord <math>\overline{AB}</math>. Circle <math>\omega_1</math> passes through <math>A</math> and <math>P</math> and is internally tangent to <math>\omega</math>. Circle <math>\omega_2</math> passes through <math>B</math> and <math>P</math> and is internally tangent to <math>\omega</math>. Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>P</math> and <math>Q</math>. Line <math>PQ</math> intersects <math>\omega</math> at <math>X</math> and <math>Y</math>. Assume that <math>AP=5</math>, <math>PB=3</math>, <math>XY=11</math>, and <math>PQ^2 = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
==Solution 1==
 
==Solution 1==
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(Solution by TheUltimate123)
 
(Solution by TheUltimate123)
 +
 +
===Note===
 +
One may solve for <math>PX</math> first using PoAP, <math>PX = \frac{11}{2} - \frac{\sqrt{61}}{2}</math>. Then, notice that <math>PQ^2</math> is rational but <math>PX^2</math> is not, also <math>PX = \frac{XY}{2} - \frac{\sqrt{61}}{2}</math>. The most likely explanation for this is that <math>Q</math> is the midpoint of <math>XY</math>, so that <math>XQ = \frac{11}{2}</math> and <math>PQ=\frac{\sqrt{61}}{2}</math>. Then our answer is <math>m+n=61+4=\boxed{065}</math>. One can rigorously prove this using the methods above
  
 
==Solution 2==
 
==Solution 2==
  
 
Let the tangents to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at <math>R</math>. Then, since <math>RA^2=RB^2</math>, <math>R</math> lies on the radical axis of <math>\omega_1</math> and <math>\omega_2</math>, which is <math>\overline{PQ}</math>. It follows that <cmath>-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).</cmath>
 
Let the tangents to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at <math>R</math>. Then, since <math>RA^2=RB^2</math>, <math>R</math> lies on the radical axis of <math>\omega_1</math> and <math>\omega_2</math>, which is <math>\overline{PQ}</math>. It follows that <cmath>-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).</cmath>
−
Let <math>Q'</math> denote the midpoint of <math>\overline{XY}</math>. By the Midpoint of Harmonic Bundles Lemma, <cmath>RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,</cmath>
+
Let <math>Q'</math> denote the midpoint of <math>\overline{XY}</math>. By the Midpoint of Harmonic Bundles Lemma(EGMO 9.17), <cmath>RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,</cmath>
 
whence <math>Q=Q'</math>. Like above, <math>XP=\tfrac{11-\sqrt{61}}2</math>. Since <math>XQ=\tfrac{11}2</math>, we establish that <math>PQ=\tfrac{\sqrt{61}}2</math>, from which <math>PQ^2=\tfrac{61}4</math>, and the requested sum is <math>61+4=\boxed{065}</math>.
 
whence <math>Q=Q'</math>. Like above, <math>XP=\tfrac{11-\sqrt{61}}2</math>. Since <math>XQ=\tfrac{11}2</math>, we establish that <math>PQ=\tfrac{\sqrt{61}}2</math>, from which <math>PQ^2=\tfrac{61}4</math>, and the requested sum is <math>61+4=\boxed{065}</math>.
  
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==Solution 3==
 
==Solution 3==
  
−
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_1, O</math> are collinear and <math>O, O_2, B</math> are collinear. Link <math>O_1P, O_2P, O_1Q, O_2Q</math>. Notice that, <math>\angle B=\angle A=\angle APO_1=\angle BPO_2</math>. As a result, <math>PO_1\parallel O_2O</math> and <math>OO_1\parallel O_2P</math>. So we have parallelogram <math>PO_2OO_1</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> divides <math>PQ</math> into two equal length pieces, So we have <math>\angle O_2QO_1=\angle O_2PO_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 since <math>\angle O_1PQ+\angle O_2O_1P=90^{\circ}</math>, we have <math>\angle OQP=\angle OQO_1 + \angle O_1QP = \angle O_2O_1P + O_1PQ=90^{\circ}</math>. As a result, <math>OQ\bot PQ</math>. So <math>Q</math> is the middle point of <math>XY</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 have <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 have <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>.
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−
Solution By BladeRunnerAUG (Fanyuchen20020715).
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Solution By BladeRunnerAUG (Fanyuchen20020715). Edited by bgn4493.
  
 
==Solution 4==
 
==Solution 4==
−
Note that the tangents to the circles at <math>A</math> and <math>B</math> intersect at a point <math>Z</math> on <math>XY</math> by radical center. Then, since <math>\angle ZAB = \angle ZQA</math> and <math>\angle ZBA = \angle ZQB</math>, we have  
+
[[File:AIME-I-2019-15.png|250px|right]]
 +
Note that the tangents to the circles at <math>A</math> and <math>B</math> intersect at a point <math>Z</math> on <math>XY</math> by radical axis theorem. Since <math>\angle ZAB = \angle ZQA</math> and <math>\angle ZBA = \angle ZQB</math>, we have  
 
<cmath>\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},</cmath>
 
<cmath>\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},</cmath>
−
so <math>ZAQB</math> is cyclic. But if <math>O</math> is the center of <math>\omega</math>, clearly <math>ZAOB</math> is cyclic with diameter <math>ZO</math>, so <math>\angle ZQO = 90^{\circ} \implies Q</math> is the midpoint of <math>XY</math>. Then, by Power of a Point, <math>PY \cdot PX = PA \cdot PB = 15</math> and it is given that <math>PY+PX = 11</math>. Thus <math>PY, PX = \frac{11 \pm \sqrt{61}}{2}</math> so <math>PQ = \frac{\sqrt{61}}{2} \implies PQ^2 = \frac{61}{4}</math> and the answer is <math>61+4 = \boxed{065}</math>.
+
so <math>ZAQB</math> is cyclic.  
 +
 
 +
But if <math>O</math> is the center of <math>\omega</math>, clearly <math>ZAOB</math> is cyclic with diameter <math>ZO</math>, so <math>\angle ZQO = 90^{\circ}</math> implies that <math>Q</math> is the midpoint of <math>XY</math>. Then, by power of point <math>P</math>, <cmath>PY \cdot PX = PA \cdot PB = 15,</cmath> whereas it is given that <math>PY+PX = 11</math>. Thus <cmath>PY, PX \in \left\{\tfrac 12 (11 \pm \sqrt{61})\right\}</cmath> so <math>PQ = \frac{\sqrt{61}}{2}</math>, i.e. <math>PQ^2 = \frac{61}{4}</math> and the answer is <math>61+4 = \boxed{065}</math>.
 +
 
 +
==Solution 5==
 +
 
 +
Connect <math>AQ,QB</math>, since <math>\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}</math>, so <math>\angle{AQP}=\frac{\angle{AO_1P}}{2}=\angle{BQP}=\frac{\angle{BO_2P}}{2}, \angle{AQB}=\angle{AOB}</math> then, so <math>A,O,Q,B</math> are concyclic
 +
 
 +
We let <math>\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}=2\alpha</math>, it is clear that <math>\angle{BQP}=\alpha, \angle{O_1AP}=90^{\circ}-\alpha</math>, which leads to the conclusion <math>OQ\bot XY</math> which tells <math>Q</math> is the midpoint of <math>XY</math>
 +
 
 +
Then it is clear, <math>XP\cdot PY=15, XP=\frac{11-\sqrt{61}}{2}, PQ=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}</math> , the answer is <math>\boxed{065}</math>
 +
 
 +
~bluesoul
 +
 
 +
 
 +
 
 +
 
 +
==Solution 6(lazy)==
 +
<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;
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O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);
 +
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];
 +
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,red); draw(X -- Y,green);
 +
 
 +
dot("$A$", A, A);
 +
dot("$B$", B, B);
 +
dot("$P$", P, dir(70),blue);
 +
dot("$Q$", Q, dir(200));
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dot("$X$", X, X);
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dot("$Y$", Y, Y);
 +
label("$3$", (A+P)/2, N, red);
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label("$5$", (B+P)/2, N, red);
 +
draw(brace(X,Y));
 +
label("$11$",brace(X,Y),dir(20));
 +
</asy>
 +
<math>PX \cdot PY=AP \cdot PB=5 \cdot 3=15</math> by power of a point. Also, <math>PX+PY=XY=11</math>, so <math>PX</math> and <math>PY</math> are solutions to the quadratic <math>x^2-11x+15=0</math> so <math>PX</math> and <math>PY</math> is <math>\frac{11\pm\sqrt{61}}{2}</math> in some order. Now, because we want <math>PQ^2</math> and it is known to be rational, we can guess that <math>PQ</math> is irrational or the problem would simply ask for <math>PQ</math>. We can also figure out that since <math>PQ^2</math> is rational, <math>PQ</math> is <math>\sqrt{\text{[something]}}</math>. <math>PQ=QX-PX</math>, and chances are low that <math>QX</math> is some number with a square root plus or minus <math>\frac{\sqrt{61}}{2}</math> to cancel out the <math>\frac{\sqrt{61}}{2}</math> in <math>PX</math>, so one can see that <math>PQ^2</math> is most likely to be <math>\left(\frac{\sqrt{61}}{2}\right)^2=\frac{61}{4}</math>, and our answer is <math>61+4=\boxed{065}</math>
 +
 
 +
Note : If our answer is correct, then <math>QX=\frac{11}{2}</math>, which made <math>Q</math> the midpoint of <math>XY</math>, a feature that occurs often in AIME problems, so that again made our answer probable. Midpoints have many properties and there is a lot of ways to show if a point is the midpoint of a segment. Even if the answer is wrong, it's still the same as leaving it blank and 065 is a good guess. ~[[Ddk001]]
 +
 
 +
==Solution 7==
 +
We will show that <math>Q</math> is the midpoint of <math>XY.</math> To do this, let <math>Q^{\prime}</math> be the altitude from <math>O</math> to <math>XY</math> or, equivalently, to <math>PQ.</math> Notice that <math>O_{1}00_{2}P</math> is a parallelogram. Thus, the height from <math>O</math> to <math>O_{1}O_{2}</math> is equal to the height from <math>P</math> to <math>O_{1}O_{2}.</math> Say that the line through <math>P</math> perpendicular to <math>O_{1}O_{2}</math> intersects <math>O_{1}O_{2}</math> at <math>H.</math> Then, <math>PQ</math> is perpendicular to <math>O_{1}O_{2},</math> so <math>H</math> is on <math>PQ.</math> Now, we have that the altitude from <math>O</math> to <math>O_{1}O_{2}</math> is equal to the altitude from <math>Q^{\prime}</math> to <math>O_{1}O_{2}</math> (since <math>OQ^{\prime} \parallel O_{1}O_{2}</math>). However, the altitude from <math>Q^{\prime}</math> to <math>O_{1}O_{2}</math> is just <math>Q^{\prime}H.</math> Also, the altitude from <math>P</math> to <math>O_{1}O_{2}</math> is <math>PH</math>, so <math>PH = Q^{\prime}H.</math> Thus, <math>O_{1}O_{2}</math> bisects <math>PQ^{\prime}.</math> However, this is true for <math>Q,</math> too, so <math>Q = Q^{\prime},</math> and we are done. Now, by PoP, we have
 +
<cmath>AP \cdot BP = XP \cdot YP = 15.</cmath>
 +
Also, we have <math>XY = XP+YP = 11,</math> so <math>XP = \frac{11 \pm \sqrt{61}}{2}</math>. Notice that <math>XQ = \frac{XY}{2} = \frac{11}{2},</math> so <math>PQ = \frac{\sqrt{61}}{2},</math> giving us our answer of <math>\boxed{065}.</math>
 +
 
 +
 
 +
==Solution 8==
 +
 
 +
[[File:2019_AIME_Problem_15_Diagram.png|400px|thumb|right|[https://www.overleaf.com/read/bkdngbhwcskc#415c3f Latex]]]
 +
 
 +
Like Solution 7, let <math>Q'</math> be the altitude from <math>O</math> to <math>XY</math>. And, let <math>M</math> be the intersection of <math>O_1O_2</math> and <math>PQ</math>. Construct <math>P'</math> on line <math>AO</math> such that <math>PP' \parallel O_2O_1</math>. First, because of isosceles triangles <math>OAB</math>, <math>O_1AP</math>, and <math>O_2BP</math>, we have <math>\angle{OAP} = \angle{OBA} = \angle{APO_1} = \angle{BPO_2}</math>, which means <math>OO_1PO_2</math> is a parallelogram. So, <math>O_2P = OO_1</math>. It is also clear that <math>PP'O_1O_2</math> is a parallelogram by virtue of our definition. Thus, <math>O_2P = O_1P' = OO_1</math>. Since <math>OQ' \parallel O_1O_2 \parallel P'P</math> (because of the right angles), <math>\frac{Q'M}{MP} = \frac{OO_1}{O_1P'} = 1 \implies Q'M = MP</math>. And, because <math>QM = MP</math>, <math>Q = Q'</math>. From Power of a Point on <math>P</math>, we have <math>XP(11-XP) = 15</math>, giving us <math>XP = \frac{11 - \sqrt{61}}{2}</math>. Since <math>OQ</math> is perpendicular to <math>XY</math>, <math>Q</math> is the midpoint of <math>XY</math>, so <math>XQ = \frac{11}{2}</math>. Thus, <math>PQ = \frac{11}{2} - \frac{11 - \sqrt{61}}{2} = \frac{\sqrt{61}}{2} \implies {PQ}^2 = \frac{61}{4}</math>. Therefore, our answer is <math>\boxed{65}</math>.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Crazyvideogamez CrazyVideoGamez]
 +
 
 +
<math>
 +
\newline
 +
</math>
 +
 
 +
==Video Solution by Mr. Math==
 +
 
 +
https://www.youtube.com/watch?v=X_CSRwUh0Rc
  
 
==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}}
 +
 +
[[Category:Intermediate Geometry Problems]]

Latest revision as of 10:32, 1 February 2024

Problem

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 $PQ^2 = \frac{m}{n}$, 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 $PX+PY=XY=11$ and $XQ=11/2$, \[XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,\] and the requested sum is $61+4=\boxed{065}$.

(Solution by TheUltimate123)

Note

One may solve for $PX$ first using PoAP, $PX = \frac{11}{2} - \frac{\sqrt{61}}{2}$. Then, notice that $PQ^2$ is rational but $PX^2$ is not, also $PX = \frac{XY}{2} - \frac{\sqrt{61}}{2}$. The most likely explanation for this is that $Q$ is the midpoint of $XY$, so that $XQ = \frac{11}{2}$ and $PQ=\frac{\sqrt{61}}{2}$. Then our answer is $m+n=61+4=\boxed{065}$. One can rigorously prove this using the methods above

Solution 2

Let the tangents to $\omega$ at $A$ and $B$ intersect at $R$. Then, since $RA^2=RB^2$, $R$ lies on the radical axis of $\omega_1$ and $\omega_2$, which is $\overline{PQ}$. It follows that \[-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).\] Let $Q'$ denote the midpoint of $\overline{XY}$. By the Midpoint of Harmonic Bundles Lemma(EGMO 9.17), \[RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,\] whence $Q=Q'$. Like above, $XP=\tfrac{11-\sqrt{61}}2$. Since $XQ=\tfrac{11}2$, we establish that $PQ=\tfrac{\sqrt{61}}2$, from which $PQ^2=\tfrac{61}4$, and the requested sum is $61+4=\boxed{065}$.

(Solution by TheUltimate123)

Solution 3

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 $A, O_1, O$ are collinear and $O, O_2, B$ are collinear. Link $O_1P, O_2P, O_1Q, O_2Q$. Notice that, $\angle B=\angle A=\angle APO_1=\angle BPO_2$. As a result, $PO_1\parallel O_2O$ and $OO_1\parallel O_2P$. So we have parallelogram $PO_2OO_1$. So $\angle O_2PO_1=\angle O$ Notice that, $O_1O_2\bot PQ$ and $O_1O_2$ divides $PQ$ into two equal length pieces, So we have $\angle O_2QO_1=\angle O_2PO_1=\angle O$. 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 since $\angle O_1PQ+\angle O_2O_1P=90^{\circ}$, we have $\angle OQP=\angle OQO_1 + \angle O_1QP = \angle O_2O_1P + O_1PQ=90^{\circ}$. 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 have $PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}$. So, we have $PQ^2=\frac{61}{4}$. As a result, our answer is $m+n=61+4=\boxed{065}$.


Solution By BladeRunnerAUG (Fanyuchen20020715). Edited by bgn4493.

Solution 4

AIME-I-2019-15.png

Note that the tangents to the circles at $A$ and $B$ intersect at a point $Z$ on $XY$ by radical axis theorem. Since $\angle ZAB = \angle ZQA$ and $\angle ZBA = \angle ZQB$, we have \[\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},\] so $ZAQB$ is cyclic.

But if $O$ is the center of $\omega$, clearly $ZAOB$ is cyclic with diameter $ZO$, so $\angle ZQO = 90^{\circ}$ implies that $Q$ is the midpoint of $XY$. Then, by power of point $P$, \[PY \cdot PX = PA \cdot PB = 15,\] whereas it is given that $PY+PX = 11$. Thus \[PY, PX \in \left\{\tfrac 12 (11 \pm \sqrt{61})\right\}\] so $PQ = \frac{\sqrt{61}}{2}$, i.e. $PQ^2 = \frac{61}{4}$ and the answer is $61+4 = \boxed{065}$.

Solution 5

Connect $AQ,QB$, since $\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}$, so $\angle{AQP}=\frac{\angle{AO_1P}}{2}=\angle{BQP}=\frac{\angle{BO_2P}}{2}, \angle{AQB}=\angle{AOB}$ then, so $A,O,Q,B$ are concyclic

We let $\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}=2\alpha$, it is clear that $\angle{BQP}=\alpha, \angle{O_1AP}=90^{\circ}-\alpha$, which leads to the conclusion $OQ\bot XY$ which tells $Q$ is the midpoint of $XY$

Then it is clear, $XP\cdot PY=15, XP=\frac{11-\sqrt{61}}{2}, PQ=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}$ , the answer is $\boxed{065}$

~bluesoul



Solution 6(lazy)

[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,red); draw(X -- Y,green); dot("$A$", A, A); dot("$B$", B, B); dot("$P$", P, dir(70),blue); dot("$Q$", Q, dir(200)); dot("$X$", X, X); dot("$Y$", Y, Y); label("$3$", (A+P)/2, N, red); label("$5$", (B+P)/2, N, red); draw(brace(X,Y)); label("$11$",brace(X,Y),dir(20)); [/asy] $PX \cdot PY=AP \cdot PB=5 \cdot 3=15$ by power of a point. Also, $PX+PY=XY=11$, so $PX$ and $PY$ are solutions to the quadratic $x^2-11x+15=0$ so $PX$ and $PY$ is $\frac{11\pm\sqrt{61}}{2}$ in some order. Now, because we want $PQ^2$ and it is known to be rational, we can guess that $PQ$ is irrational or the problem would simply ask for $PQ$. We can also figure out that since $PQ^2$ is rational, $PQ$ is $\sqrt{\text{[something]}}$. $PQ=QX-PX$, and chances are low that $QX$ is some number with a square root plus or minus $\frac{\sqrt{61}}{2}$ to cancel out the $\frac{\sqrt{61}}{2}$ in $PX$, so one can see that $PQ^2$ is most likely to be $\left(\frac{\sqrt{61}}{2}\right)^2=\frac{61}{4}$, and our answer is $61+4=\boxed{065}$

Note : If our answer is correct, then $QX=\frac{11}{2}$, which made $Q$ the midpoint of $XY$, a feature that occurs often in AIME problems, so that again made our answer probable. Midpoints have many properties and there is a lot of ways to show if a point is the midpoint of a segment. Even if the answer is wrong, it's still the same as leaving it blank and 065 is a good guess. ~Ddk001

Solution 7

We will show that $Q$ is the midpoint of $XY.$ To do this, let $Q^{\prime}$ be the altitude from $O$ to $XY$ or, equivalently, to $PQ.$ Notice that $O_{1}00_{2}P$ is a parallelogram. Thus, the height from $O$ to $O_{1}O_{2}$ is equal to the height from $P$ to $O_{1}O_{2}.$ Say that the line through $P$ perpendicular to $O_{1}O_{2}$ intersects $O_{1}O_{2}$ at $H.$ Then, $PQ$ is perpendicular to $O_{1}O_{2},$ so $H$ is on $PQ.$ Now, we have that the altitude from $O$ to $O_{1}O_{2}$ is equal to the altitude from $Q^{\prime}$ to $O_{1}O_{2}$ (since $OQ^{\prime} \parallel O_{1}O_{2}$). However, the altitude from $Q^{\prime}$ to $O_{1}O_{2}$ is just $Q^{\prime}H.$ Also, the altitude from $P$ to $O_{1}O_{2}$ is $PH$, so $PH = Q^{\prime}H.$ Thus, $O_{1}O_{2}$ bisects $PQ^{\prime}.$ However, this is true for $Q,$ too, so $Q = Q^{\prime},$ and we are done. Now, by PoP, we have \[AP \cdot BP = XP \cdot YP = 15.\] Also, we have $XY = XP+YP = 11,$ so $XP = \frac{11 \pm \sqrt{61}}{2}$. Notice that $XQ = \frac{XY}{2} = \frac{11}{2},$ so $PQ = \frac{\sqrt{61}}{2},$ giving us our answer of $\boxed{065}.$


Solution 8

Latex

Like Solution 7, let $Q'$ be the altitude from $O$ to $XY$. And, let $M$ be the intersection of $O_1O_2$ and $PQ$. Construct $P'$ on line $AO$ such that $PP' \parallel O_2O_1$. First, because of isosceles triangles $OAB$, $O_1AP$, and $O_2BP$, we have $\angle{OAP} = \angle{OBA} = \angle{APO_1} = \angle{BPO_2}$, which means $OO_1PO_2$ is a parallelogram. So, $O_2P = OO_1$. It is also clear that $PP'O_1O_2$ is a parallelogram by virtue of our definition. Thus, $O_2P = O_1P' = OO_1$. Since $OQ' \parallel O_1O_2 \parallel P'P$ (because of the right angles), $\frac{Q'M}{MP} = \frac{OO_1}{O_1P'} = 1 \implies Q'M = MP$. And, because $QM = MP$, $Q = Q'$. From Power of a Point on $P$, we have $XP(11-XP) = 15$, giving us $XP = \frac{11 - \sqrt{61}}{2}$. Since $OQ$ is perpendicular to $XY$, $Q$ is the midpoint of $XY$, so $XQ = \frac{11}{2}$. Thus, $PQ = \frac{11}{2} - \frac{11 - \sqrt{61}}{2} = \frac{\sqrt{61}}{2} \implies {PQ}^2 = \frac{61}{4}$. Therefore, our answer is $\boxed{65}$.

~CrazyVideoGamez

$\newline$

Video Solution by Mr. Math

https://www.youtube.com/watch?v=X_CSRwUh0Rc

See Also

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