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Easy Geometry in Taiwan TST

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Source: 2020 Taiwan TST Round 3

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42 posts

Li4

#1 h May 22, 2020, 3:50 AM • 1 Y

Y by MS_asdfgzxcvb

Let $\Omega$ be the $A$ -excircle of triangle $ABC$ , and suppose that $\Omega$ is tangent to lines $BC$ , $CA$ , and $AB$ at points $D$ , $E$ , and $F$ , respectively. Let $M$ be the midpoint of segment $EF$ . Two more points $P$ and $Q$ are on $\Omega$ such that $EP$ and $FQ$ are both parallel to $DM$ . Let $BP$ meet $CQ$ at point $X$ . Prove that the line $AM$ is the angle bisector of $\angle XAD$ .

Proposed by Shuang-Yen Lee

This post has been edited 1 time. Last edited by Li4, May 22, 2020, 3:50 AM
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RaduAndreiLecoiu

59 posts

RaduAndreiLecoiu

#2 h May 22, 2020, 8:56 AM

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Let $DM \cap \Omega = \{R\}$ . Then the whole problem boils down to proving the following claim :
Claim: $A , X$ and $R$ are collinear.
The proof is just a trig bash, using only the sine rule and the trigonometric version of Ceva’s theorem : $\prod \frac{\sin{\angle BAR}}{\sin{\angle CAR}} = \prod \frac{\frac{FR \sin{\angle AFR}}{AR}} {\frac{ER \sin{\angle AFR}}{AR}} = \prod \frac{FR}{ER}\cdot \frac{\sin{\angle FDR}}{\sin{\angle EDR}} = \prod \frac{FR^2}{ER^2} = \frac{FR^2}{ER^2} \cdot \frac{DP^2}{FP^2} \cdot \frac{QE^2}{QD^2} = 1$ since $FR = QD , PD = ER$ and $PF = QE$ from the isosceles trapezoids inscribed in $\Omega$

With this claim proved we can reformulate the problem like this : Let $\bigtriangleup DEF$ and $\Omega$ its circumcenter . The median $DM$ intersects $\Omega$ again at $R$ . If $A$ is the intersecton of the tangents at $E$ and $F$ the prove that $\angle RAM = \angle DAM$

Let $DR \cap \Omega =\{G\}$ . Now if we reflect every point about the perpendicular bisector of $BC$ is easy to see that $R$ goes to $G$ since arcs $EG$ and $FR$ are equal because $DG$ is symmedian and $DM$ is median . Then the line $AR$ goes to line $AD$ . Since the refelction is about $AM$ , the conclusion follows.

This post has been edited 2 times. Last edited by RaduAndreiLecoiu, May 22, 2020, 8:58 AM
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520 posts

zuss77

#3 h May 22, 2020, 9:38 AM • 1 Y

Y by Mango247

Diagram: https://youtu.be/xNqNUgSpo5k

Attachments:

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2312 posts

#4 h May 22, 2020, 12:13 PM • 17 Y

Y by A-Thought-Of-God, MarkBcc168, Li4, jty123, AmirKhusrau, amar_04, yayups, mijail, Antara_Dey_temporaryacc, Wizard_32, Kobayashi, Late_in_autumn, Limerent, Cindy.tw, JTY26, centslordm, MS_asdfgzxcvb

Let $T$ be the second intersection of $AD$ with $\Omega.$ Note that $AD$ is the D-symmedian of $\triangle DEF,$ so $\measuredangle BDP = \measuredangle EDM = \measuredangle TDF = \measuredangle TFB.$ i.e. $BT, BX$ are isogonal conjugate WRT $\angle B.$ Similarly, we can prove that $CT, CX$ are isogonal conjugate WRT $\angle C,$ so $T, X$ are isogonal conjugate WRT $\triangle ABC.$ $\qquad \blacksquare$

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1566 posts

#5 h Jun 10, 2020, 5:35 AM • 2 Y

Y by GeoMetrix, amar_04

Let $DM \cap \Omega=\{D,R\}.$ As in #2, it suffices to show $A,X,R$ are collinear. We prove a more general result:

General 2020 Taiwan TST Round 3 wrote:

Let $ABC$ be a triangle with excircle $\gamma$ and tangent to $BC,CA,AB$ in $D,E,F$ respectively. Let $K,L,M \in \gamma$ such that $DK,EL,FM$ are concurrent at some point $T.$ Then $AK, BL, CM$ are concurrent.

(In the problem $T$ is a point at infinity and $DK$ bisects $EF).$

I believe there's a simpler proof, and will add it in when I find it. For now define $EF \cap KK=K'$ and $L',M'$ similarly. The projective dual of this result is to show $K',L',M'$ are collinear. Apply Pascal: on
$\begin{align*} &KKXYZM \implies \{K', T, XY \cap MK\} \text{ collinear} \\ &LLYZXK \implies \{L',T, YZ \cap KL\} \text{ collinear} \\ &MMZXYL \implies \{M', T, ZX \cap LM\} \text{ collinear.} \end{align*}$ By Desargues theorem, $\{XY \cap MK, YZ \cap KL, ZX \cap LM\}$ are collinear. Thus, $K',L',M'$ are collinear.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -36.61222179675041, xmax = 45.56755452972322, ymin = -31.861536355948605, ymax = 21.829250844014005; /* image dimensions */ pen qqzzff = rgb(0,0.6,1); pen ccwwff = rgb(0.8,0.4,1); pen ffqqff = rgb(1,0,1); /* draw figures */ draw(circle((8.476987806654387,-0.5678994570230728), 13.078301294072654), linewidth(0.6) + qqzzff); draw((-3.4980234120517437,4.689578174922191)--(-3.6612590742666473,-5.4366700817653415), linewidth(0.6) + red); draw((9.566058821463328,13.432816595696329)--(-3.6612590742666473,-5.4366700817653415), linewidth(0.6) + linetype("2 2") + ccwwff); draw((-3.4980234120517437,4.689578174922191)--(1.9563167609084644,17.11298524088268), linewidth(0.6) + green); draw((1.9563167609084644,17.11298524088268)--(14.170921758181409,11.205845949354217), linewidth(0.6) + green); draw((-1.0807697096160467,10.195381952239671)--(9.566058821463328,13.432816595696329), linewidth(0.6) + green); draw((-11.126028072288594,13.173614934861904)--(14.170921758181409,11.205845949354217), linewidth(0.6) + ffqqff); draw((-11.126028072288594,13.173614934861904)--(-3.6612590742666473,-5.4366700817653415), linewidth(0.6) + blue); draw((-3.4980234120517437,4.689578174922191)--(14.170921758181409,11.205845949354217), linewidth(0.6) + ffqqff); draw((-11.126028072288594,13.173614934861904)--(-30.52979683688744,-19.3145452708328), linewidth(0.6) + red); draw((-30.52979683688744,-19.3145452708328)--(10.315097195074383,-13.516386573854318), linewidth(0.6) + blue); draw((2.7344405149824063,-12.318010747621697)--(-19.779430535159815,-1.31496864643835), linewidth(0.6) + blue); draw((-3.4980234120517437,4.689578174922191)--(-19.779430535159815,-1.31496864643835), linewidth(0.6) + ffqqff); draw((4.672212678012367,11.944720211096373)--(-3.4980234120517437,4.689578174922191), linewidth(0.6) + ffqqff); draw((-3.4980234120517437,4.689578174922191)--(-30.52979683688744,-19.3145452708328), linewidth(0.6) + ffqqff); draw((-1.0807697096160467,10.195381952239671)--(10.315097195074383,-13.516386573854318), linewidth(0.6) + linetype("2 2") + ccwwff); draw((1.9563167609084644,17.11298524088268)--(2.7344405149824063,-12.318010747621697), linewidth(0.6) + linetype("2 2") + ccwwff); draw((14.170921758181409,11.205845949354217)--(10.315097195074383,-13.516386573854318), linewidth(0.6) + red); draw((4.672212678012367,11.944720211096373)--(2.7344405149824063,-12.318010747621697), linewidth(0.6) + red); /* dots and labels */ dot((-3.4980234120517437,4.689578174922191),linewidth(4pt) + dotstyle); label("$F$", (-4.6795087098920884,5.706361393296371), NE * labelscalefactor); dot((4.672212678012367,11.944720211096373),linewidth(4pt) + dotstyle); label("$D$", (4.7907321810634445,13.063408035856845), NE * labelscalefactor); dot((14.170921758181409,11.205845949354217),linewidth(4pt) + dotstyle); label("$E$", (14.495772432951759,11.811144777548678), NE * labelscalefactor); dot((-3.6612590742666473,-5.4366700817653415),linewidth(4pt) + dotstyle); label("$M$", (-5.383906792690434,-7.20760345800659), NE * labelscalefactor); dot((-1.0807697096160467,10.195381952239671),linewidth(4pt) + dotstyle); label("$B$", (-3.2707125442953977,9.85448343644217), NE * labelscalefactor); dot((9.566058821463328,13.432816595696329),linewidth(4pt) + dotstyle); label("$C$", (9.878051667940383,14.08087193323223), NE * labelscalefactor); dot((1.9563167609084644,17.11298524088268),linewidth(4pt) + dotstyle); label("$A$", (0.6426101379176327,18.228993976378028), NE * labelscalefactor); dot((-11.126028072288594,13.173614934861904),linewidth(4pt) + dotstyle); label("$M'$", (-12.897486342539452,14.31567129416501), NE * labelscalefactor); dot((2.7344405149824063,-12.318010747621697),linewidth(4pt) + dotstyle); label("$K$", (0.48607723062911146,-13.62545265683594), NE * labelscalefactor); dot((10.315097195074383,-13.516386573854318),linewidth(4pt) + dotstyle); label("$L$", (10.50418329709447,-16.364778534385053), NE * labelscalefactor); dot((-19.779430535159815,-1.31496864643835),linewidth(4pt) + dotstyle); label("$K'$", (-22.759059501716287,-0.7114878055329794), NE * labelscalefactor); dot((-30.52979683688744,-19.3145452708328),linewidth(4pt) + dotstyle); label("$L'$", (-33.24676429004721,-20.512900577530853), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
P.S. A homography or moving points can be also be used to trivialize this generalization.

This post has been edited 3 times. Last edited by Wizard_32, Jul 29, 2020, 12:43 PM
Reason: typo

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1915 posts

#6 h Jun 11, 2020, 3:26 PM • 5 Y

Y by GeoMetrix, Wizard_32, Bumblebee60, Mango247, Mango247

@above Ig your Generazilation is the Extraversion of the Steinbart Rabonitwz Theorem. Anyway here is another Generalization.

Jerabek's Theorem: Let $\Delta UVW$ and $\Delta XYZ$ be the circumcevian triangles of $P$ and $Q$ with respect to $\triangle ABC$ . Then the triangle formed by the lines $\{XU, YV, ZW\}$ is perspective with $\Delta ABC$ . $(\bigstar)$

Let $UX\cap BC=A^*$ ; $VY\cap AC=B^*$ ; $ZW\cap AB=C^*$ . Fix Point $Q$ , $\Delta ABC$ and a line $\ell$ passing through $A$ . Now Animate $\{P\}$ on $\ell$ . Now $P\mapsto V\mapsto B^*$ and $P\mapsto W\mapsto C^*$ are homographies. Let $A^*B^*\cap AB=C'$ . Then $P\mapsto C^*$ is a Homography. Hence, $C^*,C'$ coincide on three Input values of $\{P\}$ on $\ell$ . Now considering the cases when $P\equiv A,U,\ell\cap BC$ we are done. $\blacksquare$

Now the above generalization mentioned by Wizard_32 is the Extraversion of $(\bigstar)$ when $P=Q$ . $\blacksquare$

This post has been edited 6 times. Last edited by amar_04, Jun 11, 2020, 6:13 PM

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