2019 AIME I Problems/Problem 15

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. 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

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)

$PX \cdot PY=AP \cdot PB=5 \cdot 3=15$ 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 $PQ$. $PQ=QX-PX$, 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 $(\frac{\sqrt{61}}{2})^2=\frac{61}{4}$, 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

Note 2 : I looked at the history of this page, and this was the same as the once deleted solution 5.