a wild symmedian appears

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Vfire

1354 posts

Vfire

#1 h Mar 14, 2019, 4:32 PM • 8 Y

Y by hwl0304, Frestho, OlympusHero, icematrix2, megarnie, Adventure10, Mango247, Rounak_iitr

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 $m+n$ .

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yayups

1614 posts

yayups

#2 h Mar 14, 2019, 4:33 PM • 6 Y

Y by a000, pad, Kanep, IMUKAT, Adventure10, Mango247

Solution to 15

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Benq

3396 posts

Benq

#3 h Mar 14, 2019, 4:35 PM • 8 Y

Y by FedeX333X, yayups, Mudkipswims42, jeff10, k2005, rayfish, Adventure10, Mango247

Coordinates

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anser

572 posts

anser

#4 h Mar 14, 2019, 4:42 PM • 3 Y

Y by jeff10, spike2015, Adventure10

Solution

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froggolover

41 posts

froggolover

#5 h Mar 14, 2019, 4:43 PM • 4 Y

Y by Vfire, awesomethree, Adventure10, Mango247

My weeb friend had a good solution using trapezoids

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djmathman

7938 posts

djmathman

#6 h Mar 14, 2019, 4:51 PM • 5 Y

Y by yayups, scrabbler94, LTE, Adventure10, centslordm

This problem is incredible.

Official Solution

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yayups

1614 posts

yayups

#7 h Mar 14, 2019, 4:57 PM • 5 Y

Y by AlastorMoody, NumberX, Kanep, Adventure10, Mango247

I was so scared when I saw all the geo, because I can never do the similar triangle-parallelogram-length chasing needed for AIME geo. But I was so happy that at least this one was more of an "olympiad geo" style problem, which I could solve

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pinetree1

1207 posts

pinetree1

#8 h Mar 14, 2019, 5:02 PM • 4 Y

Y by xwang1, Kagebaka, Mathscienceclass, Adventure10

The key claim is that $Q$ is the midpoint of $\overline{XY}$ . Then some computations yield $PQ^2 = \tfrac{61}{4}$ , which gives an answer of $\boxed{065}$ .

First, observe that by Radical center $\overline{XY}$ concurs with the common tangents at $A$ and $B$ at some point $Z$ . In particular, this implies that $XAYB$ is a harmonic quadrilateral.

Redefine $Q$ to be the midpoint of $\overline{XY}$ ; we will show that $(APQ)$ and $(BPQ)$ are tangent to $(XAYB)$ .

$[asy] size(250); defaultpen(fontsize(10pt)); pair A, B, P, Q, X, Y, Z, T, O; O = (0, 0); A = dir(210); B = dir(330); X = dir(120); Z = extension(A, rotate(90, A)*O, B, rotate(90, B)*O); Y = IP(X--Z, circumcircle(A, B, X), 1); P = extension(X, Y, A, B); Q = midpoint(X--Y); T = extension(X, rotate(90, X)*O, Y, rotate(90, Y)*O); draw(X--A--Y--B--cycle, heavygreen+linewidth(0.8)); draw(circumcircle(A, B, X), blue); draw(A--Z--B, lightblue); draw(X--Z, lightblue); draw(circumcircle(A, P, Q), heavycyan); draw(circumcircle(B, P, Q), heavycyan); draw(X--T--Y, lightblue); draw(T--B, lightblue); draw(A--Q, heavygreen); dot("$X$", X, dir(X)); dot("$A$", A, dir(240)); dot("$B$", B, dir(B)); dot("$Y$", Y, dir(240)); dot("$Z$", Z, dir(270)); dot("$T$", T, dir(200)); dot("$P$", P, dir(50)); dot("$Q$", Q, dir(10)); [/asy]$

From the harmonic condition we also know that $\overline{AB}$ concurs with the tangents at $X$ and $Y$ at some point $T$ . Thus $\overline{AP}$ is a symmedian in $\triangle AXY$ . Now we angle chase: first $\angle XAQ = \angle YAB = \angle YXB$ , so $\angle AQP = \angle XAQ+\angle AXQ = \angle YXB+\angle AXQ = \angle X = \angle ZAP.$ Therefore, $\overline{ZA}$ is tangent to both $(XAYB)$ and $(APQ)$ at $A$ , which proves the desired tangency.

Remark: The key fact here is that the midpoint of the symmedian chord $Q$ satisfies $\triangle QAX\sim \triangle QXB$ , after which the tangencies follow by angle chasing. This idea was also used in USAMO 2008/2.

This problem is also fairly guessable after some initial observations. By Power of a Point, we know $PX+PY = 11$ and $PX\cdot PY = AP\cdot PB = 15\implies \{PX, PY\} = \tfrac{11}{2}\pm\tfrac{\sqrt{61}}{2}.$ Then since $PQ^2$ is rational, it would be reasonable to expect that $Q$ is the midpoint, which gives $PQ = \tfrac{\sqrt{61}}{2}$ .

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spartacle

538 posts

spartacle

#9 h Mar 14, 2019, 5:09 PM • 1 Y

Y by Adventure10

Oof. I attempted to Cartesian coordinate bash this one... I could have done it if I hadn't misremembered the distance to a line formula as $\frac{|Ax + By + C|}{\sqrt{A^2 + B^2 + C^2}}$
although tbh even if I had remembered that there's a fair chance I would have forgotten to multiply by 2

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trumpeter

3332 posts

trumpeter

#10 h Mar 14, 2019, 5:19 PM • 1 Y

Y by Adventure10

Solution

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budu

1515 posts

budu

#11 h Mar 14, 2019, 5:25 PM • 2 Y

Y by Adventure10, Mango247

Solution

Shorter

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mira74

1010 posts

mira74

#12 h Mar 14, 2019, 5:26 PM • 2 Y

Y by Adventure10, Mango247

#13 was harder IMO
solution

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TheUltimate123

1740 posts

TheUltimate123

#13 h Mar 14, 2019, 5:54 PM • 3 Y

Y by Lol_man000, hurdler, Adventure10

symmedian not necessary
Simple solution

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tigershark22

559 posts

tigershark22

#14 h Mar 14, 2019, 6:54 PM • 2 Y

Y by Frestho, Adventure10

I was pressed for time so I assumed Q was the center of the circle, which wAs wrong, but I still got the right answer

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usaboer

10 posts

usaboer

#15 h Mar 14, 2019, 7:12 PM • 33 Y

Y by tapir1729, 62861, AforApple, mathisawesome2169, dual, popcorn1, Mudkipswims42, mira74, Ultroid999OCPN, pad, Makorn, brainiac1, hwl0304, Vfire, pretzel, eisirrational, wu2481632, blep, Lol_man000, Toinfinity, JacobGuo, fjm30, OlympusHero, IAmTheHazard, skyscraper, Kagebaka, ike.chen, rayfish, popop614, Adventure10, ESAOPS, mal7896, Sedro

what the heck guys what is with this symmedian stuff
keep it simple, stupid

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HamstPan38825

8857 posts

HamstPan38825

#39 h Oct 26, 2021, 1:16 AM • 1 Y

Y by centslordm

I think the canonical solution to this problem is definitely not elementary? Two approaches that I found relatively quickly.

Approach 1, Inversion. Invert about $P$ , say with radius 1. It's pretty straightforward why you would do so, because 1) it gets rid of two circles, and 2) all the lines pass through $P$ . In the inverted diagram, furthermore, one discovers that $(X^*Y;PQ^*)=-1$ because the picture is basically a symmedian picture. The rest is computation: one obtains $X^*P = \frac 1{\frac{11-\sqrt{61}}2}, Y^*P = \frac 1{\frac{11+\sqrt{61}}2}$ with some computation on the original diagram. Set $Y^*Q^* = x$ ; the equation is $\frac x{x+\frac 1{\frac{11-\sqrt{61}}2}+\frac 1{\frac{11+\sqrt{61}}2}} = \frac{11-\sqrt{61}}{11+\sqrt{61}}.$ In about one minute one can find $x = \frac{121\sqrt{61}-671}{1830},$ so $PQ^* = \frac{121\sqrt{61}-671}{1830}+\frac{11-\sqrt{61}}{60} = \frac{60\sqrt{61}}{1830} = \frac{2\sqrt{61}}{61},$ so $PQ^2 = \frac{61}4$ , giving the answer $\boxed{065}$ .

Approach 2, Symmedians. The inverted diagram being such a classic symmedian configuration hints at symmedians being involved somehow. Inspired by this solution to 2016 AIME I/15, radical axis on $\omega, \omega_1, \omega_2$ implies that $\overline{XY}, \overline{AA}, \overline{BB}$ concur. But this directly implies that $\overline{XY}$ is an $X$ -symmedian of $\triangle ABX$ — in turn, $\overline{AP}$ is an $A$ -symmedian of $\triangle AXY$ .

Now we claim that $\overline{AQ}$ is the $A$ -median. It suffices to show that if $T = \overline{AA} \cap \overline{BB}$ , then $AQBT$ is cyclic by the symmedian lemma. But $\measuredangle QAT = \measuredangle QPA = \measuredangle QPB = \measuredangle QBT$ by tangent-chrod angles, so this is obviously true. The rest is computation.

Remark. This configuration is eerily similar to 2016 AIME I/15...

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MatBoy-123

396 posts

MatBoy-123

#40 h Nov 2, 2021, 12:19 PM

Y by

Vfire wrote:

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 $m+n$ .

Nice !!
First note that $ABOQ$ is cyclic because of tangencies , So $\angle AQO = \angle ABO = 90 - \angle BYA = 90 - \angle PQA$ giving $OQ \perp XY$ . So $Q$ is the midpoint of $XY$ . Now from Power of point $PX.PY = PA.PB = 15$ , and $PY+ PX = XY =11$ , Solving we get $PX=\frac{11\pm\sqrt{61}}{2}$ . So $PQ^2=(XQ-XP)^2= (\frac{XY}{2} - XP)^{2} = 61/4$ . So we get $m + n = \boxed{065}$ . $\blacksquare$

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ike.chen

1162 posts

ike.chen

#41 h Nov 27, 2021, 7:22 AM

Y by

Walk-Through:

By Radical Axes, we know $AA, BB, XY$ concur, i.e. $AXBY$ is harmonic.
Because $(APQ)$ is tangent to $(AXY)$ , it's easy to see $\angle XAB = \angle XAP = \angle QAY$ so isogonality implies $Q$ is the midpoint of $XY$ .
We clearly know that $PX + PY = 11$ , and POP yields $PX \cdot PY = 15$ , so solving gives $PX, PY =\frac{11 \pm \sqrt{61}}{2}.$
Extract $PQ = \frac{11}{2} - \frac{11 - \sqrt{61}}{2} = \frac{\sqrt{61}}{2} \implies \boxed{065.}$

Remarks: $Q$ is actually the $X$ -Dumpty point of $ABX$ and the $Y$ -Dumpty point of $ABY$ .

This result can be shown without utilizing my previous isogonality argument. If we let $K = AA \cap BB \cap XY$ , then we have $\angle CAB = \angle CBA = \angle CBP = \angle BQP = \angle CQB$ which clearly suffices.

Note: Considering the second intersections between $\omega$ and $AQ, BQ$ leads to a nice solution involving trapezoids.

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Mogmog8

1080 posts

Mogmog8

#42 h Jan 13, 2022, 5:23 AM • 2 Y

Y by centslordm, megarnie

Let $O,O_1,$ and $O_2$ be the centers of $\omega,\omega_1,$ and $\omega_2,$ respectively. Let $D=\overline{BQ}\cap\omega.$ Consider the homothety $\omega\mapsto\omega_2$ centered at $B.$ Then, $D\mapsto Q$ and $A\mapsto P$ so $\overline{AD}\parallel\overline{PQ}.$ Also, letting $C=\overline{AA}\cap\overline{BB}$ (with respect to $\omega$ ), we know $\operatorname{pow}_{\omega_1}(P)=CA^2=CB^2=\operatorname{pow}_{\omega_2}(P)$ so $C$ lies on $\overline{XY}.$ Hence, by USAJMO 2011/5, $Q$ is the midpoint of $\overline{XY}.$ By PoP, $(PY)(11-PY)=15$ so $PY=\tfrac{1}{2}(11-\sqrt{61})$ and $PQ^2=\tfrac{61}{4}$ or $\boxed{65}.$

This post has been edited 1 time. Last edited by Mogmog8, Jul 16, 2022, 9:00 PM
Reason: grammar

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megarnie

5590 posts

megarnie

#43 h Jan 31, 2022, 4:00 AM

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The key claim is that $Q$ is midpoint of $XY$ .

First we will solve the problem assuming we have proven it. By Power of a Point. $PX\cdot PY=PY\cdot (11-PY)=3\cdot 5=15\implies PY=\frac{11+\sqrt{61}}{2}.$ So $PQ=\frac{11+\sqrt{61}}{2}-\frac{11}{2}=\frac{\sqrt{61}}{2}$ , which gives an answer of $\boxed{065}$ .

Now we will prove the claim: Let $D$ be the second intersection point of $\overline{BQ}$ and $\omega$ and $C$ be the radical center of the three circles. If $DA\parallel XY$ , then we are done by 2011 USAJMO #5.

Consider the homothety from $\omega$ to $\omega_2$ , centered at $B$ . Then $A$ maps to $P$ and $D$ maps to $Q$ . Thus, $AD\parallel PQ=XY$ .

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Bluesoul

894 posts

Bluesoul

#44 h Jul 15, 2022, 10:54 PM

Y by

I think it is an easier way to prove $Q$ is the midpoint

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 $XY$ , the rest is easy, answer is $\boxed{065}$

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franzliszt

23531 posts

franzliszt

#45 h Jul 16, 2022, 8:37 PM • 1 Y

Y by megarnie

How come 3 of the past 4 solutions (mine included) cite USAJMO 2011/5?

Note by Radical Axis that the tangents from $A$ and $B$ and the line $XY$ concur at a point we will call $Z$ . Let $AQ$ meet $\omega$ again at $D$ .

Lemma: In $\triangle ABC$ with circumcircle $\Omega$ , let $D$ be a point on $\Omega$ such that $AD$ is the $A$ -symmedian. Let $T$ be the midpoint of $AD$ and let $BT$ meet the line through $C$ parallel to $AD$ at $E$ . Show that $E$ lies on $\Omega$ . (USAJMO 2011/5)

Proof. We now employ Barycentric Coordinates. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ . Since $D$ is on the $A$ -symmedian, it can be parameterized by $D=(t:b^2:c^2)$ but it is also on $\Omega$ so $\text{Pow}_\Omega (D)=0$ . Of course, $\Omega$ has equation $a^2yz+b^2zx+c^2xy=0$ , so plugging in $D$ gives $a^2b^2c^2+2tb^2c^2=0 \Rightarrow t=-\frac{a^2}2$ . So $D=\left(-\frac{a^2}2,b^2,c^2\right)=(-a^2:2b^2:2c^2).$ Thus, the point at infinity along the line $A$ -symmedian has coordinates $P_\infty=(b^2+c^2:-b^2:-c^2).$
Normalize $D$ by multiplying each component by $\frac{1}{-a^2+2b^2+2c^2}$ . The midpoint formula on normalized $D$ and $A$ gives $T=(-a^2+b^2+c^2:b^2:c^2)$ . Note that $E$ is the intersection of cevians $BT$ and $CP_\infty$ . Points on $BT$ can be parameterized by $(-a^2+b^2+c^2:s:c^2)$ while points on $CP_\infty$ can be parameterized by $(b^2+c^2:-b^2:t)$ . Hence, their intersection must have coordinates $E=[(b^2+c^2)(-a^2+b^2+c^2):-b^2(-a^2+b^2+c^2):c^2(b^2+c^2)].$ It remains to check that $\text{Pow}_\Omega (E)=0$ . Plugging $E$ into the circumcircle equation yields $a^2[-b^2(-a^2+b^2+c^2)c^2(b^2+c^2)]+b^2[(b^2+c^2)(-a^2+b^2+c^2)c^2(b^2+c^2)]+c^2[(b^2+c^2)(-a^2+b^2+c^2)(-b^2)(-a^2+b^2+c^2)]$ which is clearly true after factoring. $\square$

Consider the homothety centered at $A$ that maps $\triangle AQP\mapsto ADB$ . By construction, $QP\parallel DB$ so we can conclude that $Q$ is the midpoint of $\overline{XY}$ by our lemma.

Unfortunately, we cannot do the answer extraction with bary since we are not given the side lengths of $\triangle XAB$ , but Power of a Point on $P$ with respect to $\omega$ and minimal computations give away an answer of $PQ^2=\frac{61}{4}$ or $\boxed{65}.$

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pinkpig

3761 posts

pinkpig

#46 h Aug 4, 2022, 3:30 PM • 1 Y

Y by hungrypig

Can someone tell me what is wrong with my solution?

Solution

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CoolJupiter

925 posts

CoolJupiter

#47 h Jan 23, 2023, 12:39 PM

Y by

Bro what are all of these solutions, just spam Evan Chen Lemmas

Suppose $T$ is the intersection of the tangents to circle $\omega$ at $A$ and $B$ respectively. Note that the arc formed between $AB$ and circles $\omega_1$ and $\omega_2$ must be the same, so $\angle BQP = \angle AQP$ . (Homotheties) We also have $BT = AT$ , implying that quadrilateral $ATBQ$ is cyclic. Since $\overline{XT}$ is the $X$ -symmedian of $\triangle ABX$ , it is well-known that $(ATB)$ passes through the midpoint of $XY$ , therefore, $Q$ is the midpoint of $XY$ . By PoP, $AP \cdot PB = XP \cdot PY \implies 15 = (11 - PY)(PY) \implies PY = \frac{11 - \sqrt{61}}{2}$ . Therefore, $PQ = \frac{11}{2} - \frac{11 - \sqrt{61}}{2} = \frac{\sqrt{61}}{2} \implies PQ^2 = \frac{61}{4}$ . The requested sum is $61 + 4 = \boxed{65}$ .

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Bluesoul

894 posts

Bluesoul

#48 h Jan 24, 2023, 6:47 PM • 1 Y

Y by bjump

Fun fact: this question appears the first time in 1997 china second round P1

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peace09

5417 posts

peace09

#49 h Jan 29, 2023, 7:27 PM • 2 Y

Y by Mango247, centslordm

pov: your geo level is literally elementary

Part I: Definitions. Let $M$ be the midpoint of $\overline{XY}$ , $N$ the midpoint of $\overline{PQ}$ , $O,O_1,O_2$ the centers of $\omega,\omega_1,\omega_2$ , and $O',O_1',O_2'$ the projections of $O,O_1,O_2$ onto $\overline{AB}$ . Additionally, extend $AB$ and $O_1O_2$ to meet at $C$ , and $AB$ and $OM$ to meet at $D$ .

Part II: $\triangle AO_1O_1'\sim\triangle AOO'\cong\triangle BOO'\sim\triangle BO_2BO_2'$ . We begin by computing the lengths along $\overline{AB}$ : as $AP=5$ and $BP=3$ , we have that $AO_1'=PO_1'=\tfrac{5}{2}$ and $BO_2'=PO_2'=\tfrac{3}{2}$ . The additional point $O'$ splits $\overline{PO_1'}$ into $PO'=1$ and $O_1'O'=\tfrac{3}{2}$ , because $AO'=BO'=4$ . Hence,
$\begin{align*} \tfrac{O_1O_1'}{OO'}&=\tfrac{AO_1'}{AO'}=\tfrac{5/2}{5/2+3/2}=\tfrac{5}{8}\\ \tfrac{O_2O_2'}{OO'}&=\tfrac{BO_2'}{BO'}=\tfrac{3/2}{3/2+3/2+1}=\tfrac{3}{8}, \end{align*}$ and so $OO':O_1O_1':O_2O_2'=8:5:3$ .

Part III: $\triangle CO_2O_2'\sim\triangle CO_1O_1'\sim\triangle DOO'$ . First, let $BC:=k$ . The said similar triangles yield that $\tfrac{CO_2'}{CO_1'}=\tfrac{O_2O_2'}{O_1O_1'}$ , and substituting known lengths gives us
$\tfrac{k+3/2}{k+11/2}=\tfrac{3}{5}\implies k=\tfrac{3\cdot11/2-5\cdot3/2}{5-3}=\tfrac{9}{2}.$ Next, let $CD:=\ell$ . The said similar triangles yield that $\tfrac{CO_2'}{DO'}=\tfrac{O_2O_2'}{OO'}$ , and substituting known lengths gives us
$\tfrac{6}{\ell+17/2}=\tfrac{3}{8}\implies\ell=6\cdot\tfrac{8}{3}-\tfrac{17}{2}=\tfrac{15}{2}.$
Part IV: $\triangle CNP\sim\triangle DMP$ . Now for the magic: in these similar triangles, $\tfrac{CP}{DP}=\tfrac{15/2}{15}=\tfrac{1}{2}$ , which implies $\tfrac{PN}{PM}=\tfrac{1}{2}$ ―that is, the midpoint $N$ of $\overline{PQ}$ is the midpoint of $\overline{PM}$ , and so $M=Q$ and $PQ=PM$ !

Part V: Power of a Point. We extract $PM$ by considering $\text{Pow}_{\omega}(P)$ , which produces $PX\cdot PY=PA\cdot PB$ or $xy=15$ , where $PX:=x$ and $PY:=y$ . But $x+y=XY=11$ , and so $x$ and $y$ are the roots of the quadratic $t^2-11t+15$ . Therefore, by the Quadratic Formula, $x,y=\tfrac{11\pm\sqrt{61}}{2}$ , and finally
$PM=XM-XP=\tfrac{11}{2}-\tfrac{11-\sqrt{61}}{2}=\tfrac{\sqrt{61}}{2}\implies PM^2=\tfrac{61}{4}\implies\boxed{065},$ as requested.

Remark. (The above solution doesn't even use $Q$ ...)

https://www.geogebra.org/calculator/fyz9pdem

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metricpaper

54 posts

metricpaper

#50 h Jan 15, 2024, 7:54 AM

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Immediately from the power of point $P$ we know that $PX\cdot PY=15$ . Then since $PX+PY=11$ , we get $PX=\tfrac{1}{2}(11-\sqrt{61})$ and $PY=\tfrac12 (11+\sqrt{61})$ .

To take advantage of the tangency conditions, we draw the tangent lines to the larger circle at $A$ and $B$ , and let those tangent lines intersect at $T$ . Then we know that $\angle PQB=\angle PBT$ , and similarly $\angle PQA=\angle PAT$ . But note that by equal tangents, we have $TA=TB$ so $\angle TAB=\angle TBA=\alpha$ , which means $\angle PQA=\angle PQB=\alpha$ . So $XY$ bisects $\angle AQB$ . We can also note that $TBQA$ is a cyclic quadrilateral.

Now we study point $O$ . Since $OA$ and $OB$ are perpendicular to $TA$ and $TB$ , respectively, we find that $TAOB$ is cyclic as well, so $T, A, O, Q, B$ are all concyclic. Then $AOQB$ is a cyclic quadrilateral. Hence from $\angle BAO=90-\alpha$ we know that $\angle BQO=90+\alpha$ . But from prior explorations we already know that $\angle BQA=2\alpha$ . This means that $\angle PQO=90^\circ$ . So $OQ$ is perpendicular to $XY$ . This implies that $XQ=QY$ , from which we find $PQ=\tfrac{\sqrt{61}}{2}$ .

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Jack_w

109 posts

Jack_w

#51 h Sep 15, 2024, 5:45 PM

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Solved at 3am last night.
Let $T$ be the radical center of $\omega$ , $\omega_1$ , $\omega_2$ , $O$ be the center of $\omega$ , and let $M$ be the midpoint of $AB$ .
Claim - $T$ , $A$ , $O$ , $Q$ , $B$ lie on the circle with diameter $TO$ .
Proof. Invert at $T$ around the circle $\Gamma$ with radius $TA = TB$ . By PoP, $TP \cdot TQ$ = $TA^2$ , and since $O$ is the pole of $AB$ wrt $\Gamma$ , $A$ , $M$ , $P$ , $B$ invert to $A$ , $O$ , $Q$ , $B$ respectively, so $AOQB$ is cyclic. Now $\angle{TAO} = \angle{TBO} = 90^{\circ}$ implies the result.

Now, we see $\angle{OQP} = 90^{\circ}$ as well, so $Q$ is the midpoint of $XY$ . (Alternatively, similar triangles to get $TM \cdot TO = TA^2 = TP \cdot TQ$ and use $MOQP$ cyclic.)
Then PoP at $P$ allows us to get $PX \cdot PY = 15$ , so we can simply solve for $PX$ and extract $PQ = \frac{11}{2} - PX = \frac{\sqrt{61}}{2} \implies \boxed{065}.$

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blueprimes

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blueprimes

#52 h Nov 21, 2024, 3:06 AM

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New solution I believe

WLOG $XP < YP$ , PoP easily yields $XP = \dfrac{11 + \sqrt{61}}{2}$ , $YP = \dfrac{11 - \sqrt{61}}{2}.$ By Radical Axis, note that $\overline{XY}$ , the tangents to $\omega$ at $A$ and $B$ , concur at some point $T$ . Due to harmonic quadrilaterals, we have that $(T, P ; X, Y)$ is a harmonic bundle so
$XT \cdot YP = XP \cdot YT = XP \cdot (XT + 11) \implies XT = \dfrac{11 \cdot XP}{YP - XP} = \dfrac{121 \sqrt{61} - 11 \cdot 61}{2 \cdot 61}$ so $PT = XT + XP = \dfrac{30 \sqrt{61}}{61}$ . Now $\angle QAT = 180^\circ - \angle QPA = \angle QPB = 180^\circ - \angle QBT$ so $QATB$ is cyclic, then PoP gives
$PQ \cdot PT = AP \cdot PB \implies PQ = 3 \cdot 5 \cdot \dfrac{61}{30 \sqrt{61}} = \dfrac{61}{2 \sqrt{61}} \implies PQ^2 = \dfrac{61}{4}.$ The requested sum is $61 + 4 = \boxed{65}$ .

This post has been edited 1 time. Last edited by blueprimes, Nov 21, 2024, 3:07 AM

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gladIasked

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gladIasked

#53 h Nov 30, 2024, 9:13 PM

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ok what how did I miss $Q$ midpoint of $XY$ .

Construct the tangent lines to $\omega$ at $A$ and $B$ . Let these lines intersect at $W$ . Note that the power of $W$ with respect to $\omega_1$ and $\omega_2$ is the same, so $W$ lies on the radical axis of $\omega_1$ and $\omega_2$ , or just $\overleftrightarrow {PQ}$ . We can now conclude a lot of things. First, $\angle WAP = \angle AQP=\angle AYB$ and $\angle WBP=\angle BQP$ . Since $\triangle WAB$ is isosceles, $\angle WAP = \angle WBP = \angle AQP = \angle BQP=\angle AYB$ , so in fact $PQ$ is the angle bisector of $\angle AQB$ . These angle equalities also tell us $\triangle WAP\sim \triangle WQA$ and $\triangle WBP\sim \triangle WQB$ . We know that $WAQB$ is cyclic, so if we set $PQ=x$ we have $WP=\frac{15}{x}$ by power of a point.

Extend $BQ$ to meet $\omega$ at $C\ne B$ . We have $\angle XBQ = \frac{\overset{\frown}{BX}+\overset{\frown}{CY}}{2} = \angle AYB = \frac{\overset{\frown}{BX}+\overset{\frown}{AX}}{2}\implies \overset{\frown}{CY} = \overset{\frown}{AX}$ . Thus, $\angle CBY=\angle AYX\implies \angle QBY=\angle QYA$ . Using the same logic we compute $\angle QAY=\angle QYB$ , so $\triangle QBY\sim \triangle QYA$ . We now have enough information to finish the problem. First, compute $PX=\frac{11-\sqrt{61}}{2}$ and $PY = \frac{11+\sqrt{61}}{2}$ by power of a point on $P$ with respect to $\omega$ . From our earlier two similarities, we have $\frac{AQ}{5}=\frac{WQ}{AW}\implies AQ=5\cdot \frac{WQ}{AW}$ and $\frac{BQ}{3}=\frac{WQ}{AW}\implies BQ = 3\cdot \frac{WQ}{AW}$ . Also, because $\triangle QBY\sim \triangle QYA$ , we have $\frac{QY}{AQ} = \frac{BQ}{QY}\implies QY^2=(AQ)(BQ)$ . Thus, $QY^2 = 15\cdot \frac{WQ^2}{AW^2}$ .

By power of a point on $W$ with respect to either $\omega_1$ or $\omega_2$ , we obtain $AW^2=WP\cdot WQ=15+\frac{225}{x^2}$ . Also, we can compute $QY=\frac{11+\sqrt{61}}{2} - x$ and $WQ=x+\frac{15}x$ . Finally, if we substitute all of these expressions into $QY^2 = 15\cdot \frac{WQ^2}{AW^2}$ , we can solve for $x=\frac{\sqrt{61}}{2}\implies PQ^2 = \boxed{\frac{61}{4}}$ , as desired.

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