Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎solution 5)
(→‎Solution 5)
Line 77: Line 77:
  
 
~bluesoul
 
~bluesoul
 +
 +
==Solution 6(calculationless)==
 +
<math>PX\cdotPY=AP\cdotPB=5\cdot3=15</math> by power of a point. Also, <math>PX+PY=XY=11</math>, so PX and PY are solutions to the quadratic <math>x^2-11x+15=0</math> so PX and PY is <math>\frac{11\mp\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>. <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>(\frac{\sqrt{61}}{2})^2=\frac{61}{4}</math>, and our answer is <math>61+4=\boxed{65}</math>
 +
Note : If our answer is correct, then <math>QX=\frac{11}{2}</math>, which made Q the midpoint of XY, a feature that occurs often in AIME problems, so that again made our answer probable, and even if it's wrong, it's still the same as leaving it blank because it's a AIME problem.
  
 
==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 13:52, 4 November 2023

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(calculationless)

$PX\cdotPY=AP\cdotPB=5\cdot3=15$ (Error compiling LaTeX. Unknown error_msg) 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 $\frac{11\mp\sqrt{61}{2}}$ (Error compiling LaTeX. Unknown error_msg) 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 $61+4=\boxed{65}$ 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, and even if it's wrong, it's still the same as leaving it blank because it's a AIME problem.

See Also

2019 AIME I (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_15&oldid=200740"