A quick solution: Let R be the foot of the perpend. from X to BC. Let's assume Q and R are in the interior of the segms AC and BC (respectively) and P in the ext of AD. P, R, Q are colinear (Simson's thm). PQ tangent to circle XRD iff XRQ=XDR iff Pi-XCA=XDR iff XBA=XDR=XDC=ADB iff XBC+ABC=ADB=DAC+ACB iff XAC+ABC=DAC+ACD iff ABC=ACD=ACB iff AB=AC. It's the same for all the other cases.
Problem
Source:
Tags: geometry, circumcircle, cyclic quadrilateral, Triangle, IMO Shortlist
halfway28
19.10.2003 22:14
Let $ ABC$ be a triangle. $ D$ is a point on the side $ (BC)$. The line $ AD$ meets the circumcircle again at $ X$. $ P$ is the foot of the perpendicular from $ X$ to $ AB$, and $ Q$ is the foot of the perpendicular from $ X$ to $ AC$. Show that the line $ PQ$ is a tangent to the circle on diameter $ XD$ if and only if $ AB = AC$.
AnonymousBunny
14.09.2014 07:51
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.603627333234334, xmax = 8.114996713560203, ymin = -2.032092176769243, ymax = 5.503651436984822; /* image dimensions */ /* draw figures */ draw((1.440000000000002,4.780000000000006)--(-1.659228685335962,2.891407519873402)); draw((-1.659228685335962,2.891407519873402)--(3.260000000000003,1.640000000000002)); draw((3.260000000000003,1.640000000000002)--(1.440000000000002,4.780000000000006)); draw(circle((0.8141449736680563,2.319791035692952), 2.538567048955324)); draw((1.440000000000002,4.780000000000006)--(-1.092292214876374,0.6435392920170349)); draw((-2.244684856553164,2.534645165537919)--(-1.092292214876374,0.6435392920170349)); draw((-1.092292214876374,0.6435392920170349)--(2.597840944535727,2.782406282504303)); draw((3.260000000000003,1.640000000000002)--(2.597840944535727,2.782406282504303)); draw((-1.092292214876374,0.6435392920170349)--(-0.5896709572115705,2.619321656513676)); draw(circle((-0.5351462559143464,1.553628717842023), 1.067086867404043)); draw((-2.244684856553164,2.534645165537919)--(2.597840944535727,2.782406282504303)); draw((-1.659228685335962,2.891407519873402)--(-2.244684856553164,2.534645165537919)); /* dots and labels */ dot((1.440000000000002,4.780000000000006),dotstyle); label("$A$", (1.500654065679991,4.878415067284703), NE * labelscalefactor); dot((3.260000000000003,1.640000000000002),dotstyle); label("$C$", (3.327002408751392,1.735779630107790), NE * labelscalefactor); dot((-1.659228685335962,2.891407519873402),dotstyle); label("$B$", (-1.592620605467969,2.986252369508027), NE * labelscalefactor); dot((0.02199970304768386,2.463718143667010),dotstyle); label("$D$", (0.08564543951656240,2.558459063923735), NE * labelscalefactor); dot((1.440000000000002,4.780000000000006),dotstyle); dot((-1.092292214876374,0.6435392920170349),dotstyle); label("$X$", (-1.033198590473125,0.7485643095286542), NE * labelscalefactor); dot((-2.244684856553164,2.534645165537919),dotstyle); label("$Q$", (-2.184949797815450,2.640727007305330), NE * labelscalefactor); dot((2.597840944535727,2.782406282504303),dotstyle); label("$P$", (2.668858861698635,2.887530837450114), NE * labelscalefactor); dot((-0.5896709572115705,2.619321656513676),dotstyle); label("$R$", (-0.5231373415072381,2.722994950686925), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] WLOG assume the configuration shown in the diagram above; the proofs for other configurations follow analogously. Let $R$ be the foot of perpendicular from $X$ on $BC.$ First, note that $P, R, Q$ are collinear because they all lie on the Simson line of $X$. Also, note that since $XR \perp RD,$ $R$ lies on the circle with diameter $XD.$ Now, $QR$ is tangent to $(XRD)$ iff $\angle QRX = \angle RDX.$ However, note that $\angle RDX = \angle ADC.$ Also, since $XQ \perp AB$ and $XR \perp BR,$ quadrilateral $BQXR$ is cyclic, so $\angle QRX = \angle QBX.$ Again, from cyclic quadrilateral $(ABXC),$ we deduce that \[\angle QBX = 180^{\circ} - \angle ABX = \angle ACX = \angle BCA + \angle BCX = \angle BCA + \angle BAD.\] Also, note that from $\triangle ADB,$ $\angle ADC = \angle BAD + \angle ABC.$ Hence, we need $\angle BAD + \angle ABC = \angle BAD + \angle BCA \iff \angle ABC = \angle BCA \iff AB = AC,$ as desired. $\blacksquare$
Sardor
14.09.2014 08:32
It's not hard geometry ! I solve this problem by Simson line theorem and angle chasing.
babysama
22.10.2014 19:20
Great solution!!
geniusramanujan
29.01.2015 11:15
simsons line theorem is very useful
hayoola
20.02.2015 20:24
Let PQ and CB. meet each other at T. and. M is the mid point of DX.so beacuse of simson teorm we know that XT is perpendicular on CB. So DTX=90. and we find that TM=MD=MX. so our sircle must tangant at T on CP. and we must have the angles TXD=PTB=PXB so we must have TDM=PBX. So it proved
jayme
21.02.2015 10:26
Dear Mathlinkers, by introducing the second point of intersection R' of XR with (O), AR' // PQ... now a fine reasoning implies that ABC must be A-isoceles. Sincerely Jean-Louis
yojan_sushi
02.06.2015 04:55
Let $\angle XAC=\theta$, so that $\angle XAB=a-\theta$. Also, denote the second intersection of the circle with diameter $XB$ with $BC$ as $Y$. We first prove that if $PQ$ is tangent to $XYD$, then $AB=AC$. Clearly, $\angle XYD=90$, so $Y$ is the foot of the perpendicular from $X$ to $BC$. Since $P, Q, Y$ lie on the Simson Line, $PQ$ must be tangent to $(XYD)$ at $Y$. By Angle-by-Tangent, the equality $\angle PYB=\angle YXA$ must hold. Note that $c+\theta=180-(b+a-\theta)=\angle ADB=\angle YDX$, so that $\angle YXA=90-c-\theta$. From right triangle $APX$, we find that $\angle AXP=90+\theta-a$, and since $AQXP$ is cyclic, $90+\theta-a=\angle AXP=\angle AQP$. Since $\angle AQX=90$, $\angle PQX=90-(90+\theta-a)=a-\theta$. Again, since $PAQX$ is cyclic, $\angle PXQ=180-a$, so $180-a=\angle QXY+\angle YXA+\angle AXP=\angle QXY+(90-a+\theta)+(90-c-\theta)=\angle QXY+b$, and $\angle QXY=180-a-b=c$. In triangle $YXQ$, we can compute $\angle QYX=180-(a+c-\theta)=b+\theta$. Hence $\angle PYB+90+(b+\theta)=180$, and $\angle PYB=90-b-\theta$. Since $\angle PYB=\angle YXA$, $90-b-\theta=90-c-\theta$, and $b=c$, so $AB=AC$.
The other direction, showing that $AB=AC$ implies that $PQ$ is tangent to $(XYD)$, follows easily from the angle chasing above. We may conclude.