AIME 2019 I · 第 15 题

Solution 1

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,

$$\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,$$ hence $OO_1O_2Q$ 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 (olympiad harmonic)

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 $PQ$. 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

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Video Solution 1 by StressedPineapple

https://youtube.com/watch?v=OXxtawwIw6Q

Video Solution 2 by Mr. Math

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