Ratio conditions; prove angle XPA = angle AQY

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Ratio conditions; prove angle XPA = angle AQY

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Source: USA TSTST 2011/2012 P2

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5850 posts

#1 h Jul 26, 2011, 9:30 PM • 4 Y

Y by Arshia.esl, ImSh95, Adventure10, Rounak_iitr

Two circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $B$ . Line $\ell$ is tangent to $\omega_1$ at $P$ and to $\omega_2$ at $Q$ so that $A$ is closer to $\ell$ than $B$ . Let $X$ and $Y$ be points on major arcs $\overarc{PA}$ (on $\omega_1$ ) and $AQ$ (on $\omega_2$ ), respectively, such that $AX/PX = AY/QY = c$ . Extend segments $PA$ and $QA$ through $A$ to $R$ and $S$ , respectively, such that $AR = AS = c\cdot PQ$ . Given that the circumcenter of triangle $ARS$ lies on line $XY$ , prove that $\angle XPA = \angle AQY$ .

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2176 posts

#2 h Jul 27, 2011, 3:21 AM • 5 Y

Y by Dukejukem, Arshia.esl, ImSh95, Adventure10, Rounak_iitr

solution(slightly better than what i had on the test)

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#3 h Jul 30, 2011, 12:40 AM • 3 Y

Y by Muaaz.SY, ImSh95, Adventure10

Let $M$ be the point on the perpendicular bisector of $PQ$ such that $M$ and $B$ are on the same side of $PQ$ and $\angle{PMQ}=180^\circ - \angle{PAQ}$ . Since $AR=AS$ , $\angle{ROA}=2\angle{RSA}=180^\circ - \angle{RAS}=\angle{PMQ}$ . Since $OR=OA$ , triangles $ORA$ and $MPQ$ are similar. Hence $OA/MP=OR/MP=RA/PQ=c$ . Since $\angle{OAR}=\angle{PAQ}/2$ , it follows by tangent-angle theorem that

$\angle{OAX} = \angle{OAP}-\angle{XAP}\\ = (180^\circ - \angle{PAQ}/2)-(180^\circ -\angle{XPQ})\\ = (180^\circ - \angle{MPQ})-(180^\circ -\angle{XPQ})\\ = \angle{XPM}$
This combined with the fact that $OA/MP=c=XP/XA$ implies that triangles $XPM$ and $XAO$ are similar which in turn implies that triangles $XMO$ and $XPA$ are similar. By the same argument, triangles $YQA$ and $YMO$ are similar. Hence $c=XM/XO=YM/YO$ which by the angle bisector theorem and the fact that $X$ , $O$ and $Y$ are collinear implies that $\angle{XMO}=\angle{YMO}$ and hence that $\angle{XPA}=\angle{YQA}$ .

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#4 h Aug 15, 2011, 6:03 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Spiral similarity in the figure is obvious (could be proved by seg-angle-seg , on triangles)

centered at $X$ : $A \sim P,R \sim Q$ , and assume $O \sim O_{1}$ , then $\angle XPA=\angle XO_{1}O$
centered at $Y$ : $A \sim Q,S \sim P$ , and assume $O \sim O_{2}$ , then $\angle YQA=\angle YO_{2}O$

$\frac{dis<O,AR>}{dis<O_{1},PQ>}=\frac{dis<O,AS>}{dis<O_{2},QP>}=c$ , hence assume $O_{1}=O_{2}=Z$ , then $\angle XPA=\angle XZO$ , $\angle YQA=\angle YZO$ .

then $\frac{XO}{XZ}=\frac{YO}{YZ}=c$ , the given condition says that $X,O,Y$ are conlinear, so $ZO$ bisects $\angle XZY$ , so $\angle XZO=\angle YZO$ , hence $\angle XPA=\angle YQA$ .

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#5 h May 30, 2015, 3:58 AM • 2 Y

Y by ImSh95, Adventure10

From the length condition we immediately obtain that there are spiral similarities

$X : RA \mapsto QP$
$Y : SA \mapsto PQ$

We immediately obtain $PQRX$ and $PQSY$ are cyclic. Let $O \in XY$ be the circumcenter of $ASR$ , and let $QPP'Q'$ be an isosceles trapezoid such that $\angle PP'Q = \angle PQ'Q = \angle ASR = \angle ARS$ . Let $O'$ be the circumcenter of the trapezoid. Then by the spiral similarities,

$X : O \mapsto O' \Rightarrow \angle XOO' = \angle XRQ = 180 - \angle XPQ$
$Y : O \mapsto O' \Rightarrow \angle YOO' = \angle YSP = 180 - \angle YQP$

and we immediately get $PX || QY$ . The problem follows quickly by angle chasing.

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#6 h Mar 15, 2016, 5:50 PM • 2 Y

Y by ImSh95, Adventure10

By the given length conditions, $AX / AR = PX / PQ.$ Moreover, by inscribed arcs, we have $\angle XAR = 180^{\circ} - \angle XAP = \angle XPQ.$ Therefore, $\triangle XAR \sim \triangle XPQ.$ Similarly, $\triangle YAS \sim \triangle YQP.$

Let $O$ be the circumcenter of $\triangle ARS.$ Define points $O'$ and $O''$ so that $\triangle XAR \cup O \sim \triangle XPQ \cup O'$ and $\triangle YAS \cup O \sim \triangle YQP \cup O''.$

Because $O$ lies on the perpendicular bisectors of $\overline{AR}$ and $\overline{AS}$ , it follows that $O'$ and $O''$ lie on the perpendicular bisector of $\overline{PQ}.$ Therefore,
$\begin{align*} \angle O'QP = \angle O'PQ = \angle OAR = \angle OAS = \angle O''QP. \end{align*}$ It follows that $O' \equiv O''.$ Now, by spiral similarity, $\triangle XOO' \sim \triangle XAP.$ Therefore, $\angle XPA = \angle XO'O.$ Similarly, $\angle YQA = \angle YO'O.$ Thus, we need only show that $\angle XO'O = \angle YO'O$ , which, by the Angle-Bisector Theorem in $\triangle O'XY$ is equivalent to $XO / XO' = YO / YO'.$ Indeed,
$\begin{align*} \frac{XO}{XO'} = \frac{XA}{XP} = \frac{YA}{YQ} = \frac{YO}{YO'}. \end{align*}$

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993 posts

khina

#7 h Jan 1, 2020, 7:50 PM • 3 Y

Y by rashah76, ImSh95, Adventure10

What a... strange problem. Anyways, here's a solution.

solution

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#10 h Jan 15, 2021, 5:56 PM • 5 Y

Y by ImSh95, Mango247, Mango247, Mango247, Rounak_iitr

abacadaea wrote:

solution(slightly better than what i had on the test)

The only word which can describe this solution in this: WOW

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6871 posts

#11 h Mar 1, 2021, 3:04 PM • 2 Y

Y by ImSh95, Rounak_iitr

Solution from Twitch Solves ISL:

We begin as follows.
Claim: There is a spiral similarity centered at $X$ mapping $AR$ to $PQ$ . Similarly there is a spiral similarity centered at $Y$ mapping $SA$ to $PQ$ .
Proof. Since $\measuredangle XAR = \measuredangle XAP = \measuredangle XPQ$ , and $AR/AX = PQ/PX$ is given. $\blacksquare$
Now the composition of the two spiral similarities $AR \xmapsto{X} PQ \xmapsto{Y} SA$ must be a rotation, since $AR = AS$ . The center of this rotation must coincide with the circumcenter $O$ of $\triangle ARS$ , which is known to lie on line $XY$ .
$[asy] import graph; size(10cm); pen uququq = rgb(0.25098,0.25098,0.25098); pen zzttqq = rgb(0.6,0.2,0.); draw((13.97096,-7.13004)--(20.23013,-5.53182)--(27.19232,-12.23004)--cycle, linewidth(0.6) + zzttqq); draw(circle((25.,-3.), 9.48683), linewidth(0.6) + uququq); draw(circle((13.,-3.), 4.24264), linewidth(0.6) + uququq); draw((13.97096,-7.13004)--(16.,0.), linewidth(0.6) + red); draw((13.97096,-7.13004)--(11.14589,0.81605), linewidth(0.6) + red); draw((11.14589,0.81605)--(25.35640,-1.57297), linewidth(0.6) + blue); draw((27.19232,-12.23004)--(20.85410,5.53296), linewidth(0.6) + red); draw((27.19232,-12.23004)--(16.,0.), linewidth(0.6) + red); draw((20.85410,5.53296)--(9.74299,-7.13207), linewidth(0.6) + blue); draw((13.97096,-7.13004)--(20.23013,-5.53182), linewidth(0.6) + zzttqq); draw((20.23013,-5.53182)--(27.19232,-12.23004), linewidth(0.6) + zzttqq); draw((27.19232,-12.23004)--(13.97096,-7.13004), linewidth(0.6) + zzttqq); draw((20.23013,-5.53182)--(19.26755,-9.17733), linewidth(0.6)); draw((11.14589,0.81605)--(20.85410,5.53296), linewidth(0.6)); dot("$A$", (16.,0.), dir(100)); dot("$Q$", (20.85410,5.53296), dir(120)); dot("$P$", (11.14589,0.81605), dir(120)); dot("$X$", (13.97096,-7.13004), dir(280)); dot("$Y$", (27.19232,-12.23004), dir(280)); dot("$R$", (25.35640,-1.57297), dir((8.043, 16.110))); dot("$S$", (9.74299,-7.13207), dir(225)); dot("$O$", (19.26755,-9.17733), dir(45)); dot("$O'$", (20.23013,-5.53182), dir(135)); [/asy]$

Thus, we may let $O'$ be the image under the rotation at $X$ , so that $\triangle XPA \overset{+}{\sim} \triangle XO'O, \qquad \triangle YQA \overset{+}{\sim} \triangle YO'O.$ Because $\frac{XO}{XO'} = \frac{XA}{XP} = c \frac{YQ}{YA} = \frac{YO}{YO'}$ it follows $\overline{O'O}$ bisects $\angle XO'Y$ . Finally, we have $\measuredangle XPA = \measuredangle XO'O = \measuredangle OO'Y = \measuredangle AQY.$
Remark: Indeed, this also shows $\overline{XP} \parallel \overline{YQ}$ ; so the positive homothety from $\omega_1$ to $\omega_2$ maps $P$ to $Q$ and $X$ to $Y$ .

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#12 h May 16, 2022, 2:10 PM • 1 Y

Y by ImSh95

Claim $: XAR$ and $XPQ$ , $YAS$ and $YQP$ are similar.
Proof $:$ Note that $\frac{AX}{PX} = c = \frac{AR}{PQ} \implies \frac{AX}{AR} = \frac{PX}{PQ}$ . we prove the other one with same approach.
Let $T$ be point on perpendicular bisector of $PQ$ such that $\angle PTQ = \angle 180 - \angle PAQ$ so Now we have $\angle PTQ = \angle AOR$ so $\angle XPA = \angle XTO$ and with same approach $\angle YQA = \angle YTO$ also Note that $\frac{XO}{XT} = \frac{AX}{PX} = c = \frac{AR}{PQ} = \frac{YO}{YT}$ so $\angle YQA = \angle XPA$ .
we're Done.

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#13 h Jul 31, 2022, 11:11 AM

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hint
solution

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#14 h Oct 26, 2023, 12:54 PM • 2 Y

Y by mxlcv, Rounak_iitr

oly kid tries to use isogonal conjugation in a quadrilateral, lives to regret it
solution

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793 posts

#15 h Feb 4, 2024, 2:55 AM

Y by

Suppose the spiral similarity at $X$ sending $AR \mapsto PQ$ also sends $O \mapsto K$ and the spiral similarity at $Y$ sending $AS \mapsto QP$ also sends $O \mapsto K'$ . Then
$\triangle OAR \sim \triangle KPQ, \quad \triangle OSA \sim \triangle K'PQ.$ $\triangle XOK \sim \triangle XAP, \quad \triangle YOK' \sim \triangle YAQ.$
The first two similarities tells us $K = K'$ . The last two similarities then finishes, as
$\frac{XO}{XK} = \frac{XA}{XP} = c = \frac{YA}{YQ} = \frac{YO}{YK}$ $\implies \angle XOK = \angle YOK \implies \angle XPA = \angle AQY. \quad \blacksquare$

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1318 posts

#16 h Mar 14, 2024, 3:34 PM

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The center of the spiral similarity sending $AR$ to $PQ$ is centered at $X$ and the center sending $AS$ to $PQ$ is centered at $Y$ , since $\measuredangle{XAR} = \measuredangle XAP = \measuredangle XPQ$ and by our length conditions we have $\triangle XAR \sim \triangle XQP$ and analogously with $Y$ .
Let $O$ be the center of $(ASR)$ . Then let the spiral similarity at $X$ send $O \to O_x$ and the spiral similarity at $Y$ send $O \to O_y$ .
Then since $\triangle OAR \cong \triangle OAS$ and $\triangle OAR \sim \triangle O_xPQ$ and $\triangle OAS \sim O_yPQ$ we have $\triangle O_yPQ \cong \triangle O_xPQ$ so $O_y = O_x = O'$ .
And since $\frac{XO}{XO'} = \frac{XA}{XP} = \frac{YA}{YQ} = \frac{YO}{YO'}$ and by angle bisector theorem we get that $\angle XO'O = \angle YO'O$ from which we get by our similarity that $\angle XO'O = \angle YO'O \implies \angle XPA = \angle AQY$ as desired.

This post has been edited 4 times. Last edited by dolphinday, Mar 15, 2024, 2:19 PM

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OronSH

#17 h Mar 17, 2024, 8:26 PM • 1 Y

Y by dolphinday

Invert at $A.$ From now on, we take point names to mean the image of the point under inversion.

First, notice that the length condition becomes $\frac{AQ\cdot PX}{PQ}=AR=AS=\frac{AP\cdot QY}{PQ}.$ Then from $XP,YQ$ tangent to $(APQ)$ we have $\measuredangle QAR=\measuredangle QPX$ and $\measuredangle PAS=\measuredangle PQY.$ Thus we have $\triangle QRA\sim\triangle QXP$ and $\triangle PSA\sim\triangle PYQ,$ so $\measuredangle QRA=\measuredangle QXP$ and $\measuredangle PSA=\measuredangle PYQ$ and thus $PQRX,PQSY$ are cyclic.

Let $X'$ be the point on $QY$ with $XX'\parallel PQ$ and let $Y'$ be the point on $PX$ with $YY'\parallel PQ.$ We see $X',Y'$ lie on $(PQRX),(PQSY)$ respectively. We have $\measuredangle SY'P+\measuredangle Y'PA=\measuredangle SY'P+\measuredangle PQS=0,$ so $AP\parallel SY'.$ Similarly, $AQ\parallel RX'.$

Now let $A'$ be the reflection of $A$ over $RS.$ We see that $ARA'S$ is a rhombus. Thus we have $\measuredangle Y'A'X'=\measuredangle PAQ=\measuredangle PQY=\measuredangle Y'YX,$ so $A'$ lies on $(XYX'Y').$ However, it is given that $A'$ lies on $(AXY),$ so we have that $A$ lies on $(XYX'Y').$

To finish, we have $\measuredangle PXA=\measuredangle Y'A'A=\measuredangle SA'A=\measuredangle AA'R=\measuredangle AA'X=\measuredangle AYQ.$

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1880 posts

#18 h Apr 3, 2025, 1:45 AM • 1 Y

Y by OronSH

Solved with redbean.

Too lazy to type this up so you get the version I sent on Discord instead.

XPQ sim XAR
because XPQ=180-XAP=XAR
and PX/AX=PQ/AR
O' = image of O under spiral similarity at X sending A to P
then if Q' is the image of S
then O' is the circumcenter of Q'PQ
but now
consider the other spiral similarity at Y
then O'_2 is the circumcenter of PQP'
but clearly Q'PQP' cyclic
because equal angles and stuff
so therefore the two O' are the same point
great
so now note that
O'XY=AXP by definition
similarly O'YX=AYQ
now note that
PO'Q=2PQ'Q=180-Q'PQ=180-SAR=180-PAQ
but recall that O'XY=AXP=APQ, similarly O'YX=AYQ=AQP
so O'XY sim APQ
so finally, 180-PAQ=180-XO'Y
so finally
that means O' has an isogonal conjugate in PQYX
thus because of our two angle equivalences (at X and Y), A must exactly be that isogonal conjugate
so PXA=QPA=XPO'=XAO
so PX parallel AO
and now its not hard to finish
since XPA=180-PAO=180-QAO=YQA

This post has been edited 1 time. Last edited by cj13609517288, Apr 3, 2025, 1:57 AM

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