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Source: Iran team selection test third exam p1

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824 posts

andria

#1 h Jun 4, 2015, 3:30 PM • 6 Y

Y by Rmasters, Davi-8191, itslumi, Adventure10, Mango247, Rounak_iitr

Point $A$ is outside of a given circle $\omega$ . Let the tangents from $A$ to $\omega$ meet $\omega$ at $S, T$ points $X, Y$ are midpoints of $AT, AS$ let the tangent from $X$ to $\omega$ meet $\omega$ at $R\neq T$ . points $P, Q$ are midpoints of $XT, XR$ let $XY\cap PQ=K, SX\cap TK=L$ prove that quadrilateral $KRLQ$ is cyclic.

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tranquanghuy7198

253 posts

tranquanghuy7198

#2 h Jun 4, 2015, 4:58 PM • 4 Y

Y by aleksam, nguyendangkhoa17112003, Adventure10, Mango247

My solution:
$XS\cap{\omega} = \{S, I\}$
Notice that: $XR, XT$ is tangent to $\omega \Rightarrow (TRIS) = -1$
$\Rightarrow T(XRIS) = -1$ (1)
$TR\cap{XY} = W$
$XP = PT, XQ = QR \Rightarrow XK = KW$
$\Rightarrow T(XWKS) = -1$
$\Rightarrow T(XRKS) = -1$ (2)
(1), (2) $\Rightarrow TI \equiv TK$
$\Rightarrow I = TK\cap{SX} = L$
$\Rightarrow L\in{\omega}$
$\Rightarrow \angle{KLR} = \angle{RST} = \angle{XRT} = \angle{XQP} = \angle{KQR}$
$\Rightarrow K, L, R, Q$ are concyclic
Q.E.D

Attachments:

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824 posts

andria

#3 h Jun 4, 2015, 5:29 PM • 5 Y

Y by Tafhim, rashah76, Eliot, Adventure10, solidgreen

Thank you for your solution.
Here's my solution:
Let $KT\cap \omega=L'$ note that $XY$ is radical axis of point $A,\omega$ and $PQ$ is radical axis of $X,\omega$ so $K$ is radical center of $A,X,\omega$ so $KA^2=KX^2=KL'.KT$ so $KX$ is tangent to $\odot (\triangle XL'T)\longrightarrow \angle XL'K=\angle KXT=\angle 180-\angle STA$ so $X, L', S$ are collinear and $L\equiv L'$ now observe that $\angle KQR=\angle XQP=\angle XRT=\angle KLR$ so $KQLR$ is cyclic.
DONE

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2226 posts

#4 h Jun 4, 2015, 5:39 PM • 4 Y

Y by anantmudgal09, reveryu, Adventure10, Mango247

Let $TR\cap XY=M$ and let the midpoints of $XS, XY$ be $B, N$ . Note that $\angle YMR=\angle RTS=\angle YSR$ , so $MYRS$ is cyclic. Due to a homothety centered at $X$ , this implies $KNQB$ is cyclic. Because the radical axes of $A, X, \omega$ concur, we get $KA=KX\implies \triangle AXK\sim\triangle TAS$ . Therefore $XK=\frac{(AS)(AX)}{ST}\implies (XK)(XN)=\frac{XR^2}{2}=(XR)(XQ)$ , so $KNQR$ is cyclic. Finally, $(XL)(XB)=\left(\frac{XS}{1+\frac{ST}{XK}}\right)\left(\frac{XS}{2}\right)=(XK)(XN)$ , and so $KNBL$ is cyclic. These three combines give $KNQLBR$ is cyclic, as required.

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1000 posts

#5 h Nov 20, 2016, 2:00 PM • 1 Y

Y by Adventure10

What does "radical axes of point and circle " mean? ı think radical axes exist between circle and circle

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den_thewhitelion

262 posts

den_thewhitelion

#6 h Nov 20, 2016, 2:04 PM • 1 Y

Y by Adventure10

A point is a circle with radius 0

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924 posts

#7 h Oct 23, 2019, 6:04 PM • 3 Y

Y by amar_04, Eliot, Adventure10

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(30cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -25.983372532706174, xmax = 74.06437578769733, ymin = -40.10491409386843, ymax = 25.965579439801473; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((8.26,-3.26), 9), linewidth(2) + wrwrwr); draw((xmin, -0.34995352372502203*xmin-9.904574766277786)--(xmax, -0.34995352372502203*xmax-9.904574766277786), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.2690234650970968*xmin + 3.8378588610526587)--(xmax, 0.2690234650970968*xmax + 3.8378588610526587), linewidth(2) + wrwrwr); /* line */ draw((xmin, -1.1255014513011363*xmin-7.513525349498235)--(xmax, -1.1255014513011363*xmax-7.513525349498235), linewidth(2) + wrwrwr); /* line */ draw((xmin, 3.341464458272747*xmin + 7.2452672799432785)--(xmax, 3.341464458272747*xmax + 7.2452672799432785), linewidth(2) + wrwrwr); /* line */ draw((xmin, 27.076223935693086*xmin + 222.04753627358411)--(xmax, 27.076223935693086*xmax + 222.04753627358411), linewidth(2) + wrwrwr); /* line */ draw((-8.139965143017417,1.6480172325085285)--(5.28719790285137,-11.754848303012167), linewidth(2) + wrwrwr); draw((-9.05011357700883,-22.995365580963366)--(5.921918693006509,5.430993947868541), linewidth(2) + wrwrwr); draw((-0.22944998216551502,-6.248183227365741)--(1.5319932317881242,-9.237785955259287), linewidth(2) + wrwrwr); draw((-9.05011357700883,-22.995365580963366)--(5.28719790285137,-11.754848303012167), linewidth(2) + wrwrwr); draw((5.921918693006509,5.430993947868541)--(5.28719790285137,-11.754848303012167), linewidth(2) + wrwrwr); draw((-3.303985955614646,-3.7948843613753795)--(-0.22944998216551502,-6.248183227365741), linewidth(2) + wrwrwr); draw((1.5319932317881242,-9.237785955259287)--(-9.05011357700883,-22.995365580963366), linewidth(2) + wrwrwr); draw(circle((-8.010519033264748,-12.846422661194547), 10.202048774834777), linewidth(2) + wrwrwr); draw((1.5319932317881242,-9.237785955259287)--(5.921918693006509,5.430993947868541), linewidth(2) + wrwrwr); draw((xmin, -0.670286327440097*xmin-8.210911838260941)--(xmax, -0.670286327440097*xmax-8.210911838260941), linewidth(2) + wrwrwr); /* line */ /* dots and labels */ dot((8.26,-3.26),dotstyle); label("$O$", (8.513610816731658,-2.619491508885604), NE * labelscalefactor); dot((-22.201848979041344,-2.134959482851484),dotstyle); label("$A$", (-18.12247802181733,-1.255204031789175), NE * labelscalefactor); dot((5.921918693006509,5.430993947868541),linewidth(4pt) + dotstyle); label("T", (6.1748322845663814,5.956029775720519), NE * labelscalefactor); dot((5.28719790285137,-11.754848303012167),linewidth(4pt) + dotstyle); label("S", (5.525171581187138,-11.259978863829652), NE * labelscalefactor); dot((-8.139965143017417,1.6480172325085285),linewidth(4pt) + dotstyle); label("$X$", (-7.857838908425279,2.187997696120859), NE * labelscalefactor); dot((-8.457325538094988,-6.944903892931825),linewidth(4pt) + dotstyle); label("$Y$", (-8.182669260114901,-6.452489658823189), NE * labelscalefactor); dot((1.5319932317881242,-9.237785955259287),linewidth(4pt) + dotstyle); label("$R$", (1.8221055719254495,-8.72630212065057), NE * labelscalefactor); dot((-3.303985955614646,-3.7948843613753795),linewidth(4pt) + dotstyle); label("$Q$", (-3.050349703418877,-3.2691522122648555), NE * labelscalefactor); dot((-1.109023225005454,3.5395055901885346),linewidth(4pt) + dotstyle); label("$P$", (-0.8415033119294489,4.072013735920689), NE * labelscalefactor); dot((-9.05011357700883,-22.995365580963366),linewidth(4pt) + dotstyle); label("$K$", (-8.76736389315622,-22.499109032290708), NE * labelscalefactor); dot((-0.22944998216551502,-6.248183227365741),linewidth(4pt) + dotstyle); label("$L$", (0.0030556024635676683,-5.737862885106012), NE * labelscalefactor); dot((-8.298645340556202,-2.64844333021165),linewidth(4pt) + dotstyle); label("$B$", (-8.052737119439053,-2.0997629461822025), NE * labelscalefactor); dot((-1.4263836200830151,-5.053415535251828),linewidth(4pt) + dotstyle); label("$C$", (-1.1663336636190706,-4.503507548685434), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
Reims FTW!!!
Claim: $L\in \omega$
Proof:We'll use phantom points. Let $L'=XS\cap \omega$ and let $B=\odot(L'RK) \cap RS$ . Now let $KL' \cap \omega=T'$ We have by reims theorem that $BK\| ST' \implies T=T'$ as by thales we have $BK \| ST$ .Now we have that $L' \in KT$ as well as $L' \in XS$ and these together imply $L'=L$ and we are done. The rest of the proof follows smoothly from trivial angle chase.

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162 posts

#8 h Jul 27, 2020, 3:51 AM • 1 Y

Y by Eliot

Storage

This post has been edited 2 times. Last edited by Gaussian_cyber, Jul 27, 2020, 3:55 AM
Reason: latex

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1079 posts

#9 h Aug 1, 2021, 11:19 AM

Y by

$[asy]import olympiad;import geometry; size(10cm); defaultpen(fontsize(10pt));pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,S,T,A,X,Y,R,P,Q,K,L,F,M; O=(0,0);S=dir(250);T=dir(350);A=intersectionpoint(perpendicular(S,line(S,O)),perpendicular(T,line(T,O)));X=midpoint(A--T);path w=circle(O,1);Y=midpoint(A--S); R=intersectionpoints(circumcircle(O,T,X),w)[1];P=midpoint(X--T); Q=midpoint(X--R);K=extension(X,Y,P,Q);L=extension(T,K,S,X);M=midpoint(X--Y); F=extension(T,R,X,Y); draw(X--F,deep);draw(P--K,deep); draw(S--X,deep); draw(w,deep);draw(circumcircle(K,R,L),med+dashed); draw(S--Y--A--X,med, StickIntervalMarker(3,2,med)); draw(R--Q--X--P--T,light, StickIntervalMarker(4,1,light)); draw(T--F--A,deep);draw(circumcircle(R,M,A),deep); clip((1,0.2)--(1.1,-1.4)--(-0.4,-1.4)--(-0.4,0.2)--cycle); dot("$S$",S,dir(S)); dot("$T$",T,dir(T)); dot("$A$",A,dir(A)); dot("$O$",O,N); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$R$",R,dir(R)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$K$",K,dir(K)); dot("$L$",L,N); dot("$M$",M,M); dot("$F$",F,F); [/asy]$

Let $M$ be the midpoint of $XY$ . Note that $OTXMR$ is cyclic. Let $F=TR\cap XY$ , then $FRMA$ is cyclic as $\measuredangle FMA=90^\circ=\measuredangle TRA=\measuredangle FRA$ .

Now, $\begin{align*} \measuredangle XAM=90^\circ-\measuredangle MXA=90^\circ-\measuredangle MXT=90^\circ-\measuredangle MRT=90^\circ-\measuredangle MRF=90^\circ-\measuredangle MAF=\measuredangle AFM, \end{align*}$ therefore $XA^2=XM\cdot XF$ and as $K$ is the midpoint of $XF$ and $M$ the midpoint of $XY$ , we get that $XT^2=XA^2=XK\cdot XY$ . Now, we claim that $L$ lies on $\omega$ .
$\begin{align*} \measuredangle LTX=\measuredangle KTX=\measuredangle XYT=\measuredangle XST=\measuredangle LST, \end{align*}$ hence $L$ lies on $\omega$ .
Now, $\begin{align*} \measuredangle QRL=\measuredangle RTL=\measuredangle RTK=\measuredangle QKL, \end{align*}$ we are done.

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Mahdi_Mashayekhi

695 posts

Mahdi_Mashayekhi

#10 h Jan 12, 2022, 11:06 AM

Y by

Note that K lies on radical axis of A and circle $\omega$ and also lies on radical axis of X and $\omega$ . So if TK meets $\omega$ at L' we have :
KA^2 = KX^2 = KL'.KT. We will prove X,L',S are collinear so L' is L. we want to prove ∠XL'T = ∠L'ST + L'TS. ∠L'ST + ∠L'TS = ∠STX = ∠KXA and we have
KX^2 = KL'.KT so L'TX is tangent to KX so ∠XL'T = ∠KXA so L' is L. ∠RLK = ∠RST = ∠XRT = ∠XQP = ∠RQK so RLQK is cyclic.
we're Done.

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53 posts

mcmp

#12 h Nov 18, 2024, 10:06 AM

Y by

YES I FINALLY HAVE IT I CAN ACTUALLY IDENTIFY SIMILAR TRIANGLES AND POINT CIRCLES LETSSS GOOOOO

Let $Z$ be the midpoint of $XY$ and $U$ the midpoint of $ST$ ; I claim that $KZLR$ and $KZQR$ are concyclic which clearly finishes.

Let’s start with the first one. First I claim $L\in\omega$ , from which the conclusion that $KZLR$ will follow by Reims. Let $L’$ be the second intersection of $\overline{KT}\cap\omega\neq T$ . I claim that $X-L-S$ . However notice that $KYLS$ is cyclic because $\measuredangle YKL=\measuredangle STK=\measuredangle STL=\measuredangle YSL$ as desired.

We now show that $\overline{XT}$ tangent to $(TYK)$ before we continue. Notice by radax on point circles $(A)$ , $(X)$ and $\omega$ it’s clear $AK=KX$ . Hence $\measuredangle AXK=\measuredangle KAX=\measuredangle XYA$ and so $\triangle AKX\stackrel{-}{\sim}\triangle YAX$ , and so $TX^2=AX^2=XY\cdot XK$ as desired. Hence $\measuredangle XSY=\measuredangle XTY=\measuredangle TKX=\measuredangle LKY=\measuredangle LSY$ as desired. So we have that $L\in\omega$ .

For the final argument it suffices to show that $XZ\cdot XK=XQ\cdot XR$ however multiplying both sides by two it STP that $XR^2=XT^2=XY\cdot XK$ , which is what we just showed.

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