two tangent circles

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Source: FKMO 2021 Problem 5

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#1 h May 8, 2021, 1:19 PM • 3 Y

Y by chrono223, jhu08, Rounak_iitr

The incenter and $A$ -excenter of $\triangle{ABC}$ is $I$ and $O$ . The foot from $A,I$ to $BC$ is $D$ and $E$ . The intersection of $AD$ and $EO$ is $X$ . The circumcenter of $\triangle{BXC}$ is $P$ .
Show that the circumcircle of $\triangle{BPC}$ is tangent to the $A$ -excircle if $X$ is on the incircle of $\triangle{ABC}$ .

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#2 h May 8, 2021, 1:41 PM • 5 Y

Y by chrono223, SMSGodslayer, jhu08, Infinityfun, egxa

$[asy] import olympiad; import geometry; size(13cm); pair A= dir(104); pair B=dir(230); pair C=dir(310); pair I=incenter(A, B, C); pair O=excenter(B, C, A); pair D=foot(A, B, C); pair E=foot(I, B, C); pair X=intersectionpoint(line(E, O), line(A, D)); pair F=foot(O, B, C); pair P=circumcenter(X, B, C); pair J1[]=intersectionpoints(line(X, I), circle(O*2-F, F)); pair J=J1[0]; dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); filldraw(A--B--C--cycle, invisible, red); draw(circle(I*2-E, E), orange); draw(circle(O*2-F, F), orange); draw(A--D, red); draw(X--O, red); draw(A--(A+(B-A)*1.8), red); draw(A--(A+(C-A)*1.5), red); dot("$I$", I, NE); dot("$O$", O, S); dot("$D$", D, SW); dot("$E$", E, SW); dot("$X$", X, NW); dot("$F$",F, S); dot("$P$", P, NE); dot("$J$", J1[0], N); draw(X--J, magenta+dashed); draw(B--X--C, red); draw(circle(B, J, C), lightred+dashed); [/asy]$

Let $F$ be the touchpoint of the excircle at $BC$ , and let $J$ be the point on the $A$ -excircle so that $\odot(BJC)$ is tangent to the excircle. It suffices to prove that $PJ$ bisects $\angle BJC$ , since this would imply that $P$ is the midpoint of arc $BC$ of $\odot(BJC)$ .

We start with recalling ISL 2002 G7 from which we know that $OX$ bisects $\angle BXC$ . Since $X$ lies on the incircle, we have $\angle DXE=\angle XEI=\angle EXI$ ; thus $XI$ and $XD$ are isogonal WRT $\angle BXC$ . Hence $X, I, P$ are collinear. Note that it is well-known that $F$ lies on this line as well.

Now it suffices to prove that $F, I, J$ are collinear, and that $IJ$ bisects $\angle BJC$ . But this is the excenter version of ISL 2002 G7 (which can be proved analogously), so we're done.

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#3 h Aug 21, 2021, 1:54 AM • 10 Y

Y by Fermat_Theorem, jhu08, KPBY0507, Mathematicsislovely, math31415926535, HamstPan38825, johndooo_e, ike.chen, mathverse06, Kingsbane2139

It's known that $X$ coincides with the midpoints of $\overline{AD}$ and also lies on line $\overline{IF}$ .
$[asy] size(13cm); pen zzttqq = rgb(0.6,0.2,0); pen fuqqzz = rgb(0.95686,0,0.6); pen qqwuqq = rgb(0,0.39215,0); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw((-2.43454,2.03316)--(-2.6,0)--(-0.2,0)--cycle, linewidth(1) + zzttqq); draw((-2.43454,2.03316)--(-2.6,0), linewidth(1) + zzttqq); draw((-2.6,0)--(-0.2,0), linewidth(1) + zzttqq); draw((-0.2,0)--(-2.43454,2.03316), linewidth(1) + zzttqq); draw(circle((-1.4,0.92565), 1.51553), linewidth(1)); draw(circle((-1.89059,0.65401), 0.65401), linewidth(1) + fuqqzz); draw((-2.43454,2.03316)--(-2.43454,0), linewidth(1)); draw(circle((-0.90940,-1.83376), 1.83376), linewidth(1) + qqwuqq); draw((-2.43454,0)--(-1.4,-0.58987), linewidth(1)); draw((-2.43454,2.03316)--(-1.4,-0.58987), linewidth(1)); draw(circle((-1.4,0.32644), 1.24360), linewidth(1) + qqwuqq); draw((-2.6,0)--(-2.73712,-1.68502), linewidth(1)); draw((-2.43454,1.01658)--(-0.90940,-1.83376), linewidth(1)); draw((-0.90940,-1.83376)--(-0.90940,0), linewidth(1)); draw((-1.4,-0.91688)--(-0.90940,0), linewidth(1)); dot("$A$", (-2.43454,2.03316), dir((1.347, 3.366))); dot("$B$", (-2.6,0), dir(225)); dot("$C$", (-0.2,0), dir(-45)); dot("$I$", (-1.89059,0.65401), dir((1.479, 2.608))); dot("$D$", (-2.43454,0), dir((1.347, 2.711))); dot("$E$", (-1.89059,0), dir((1.479, 2.711))); dot("$M$", (-1.4,-0.58987), dir((1.224, 2.797))); dot("$O$", (-0.90940,-1.83376), dir((1.306, 2.649))); dot("$X$", (-2.43454,1.01658), dir(135)); dot("$P$", (-1.4,0.32644), dir((1.224, 2.716))); dot("$F$", (-0.90940,0), dir((1.306, 2.711))); dot("$N$", (-1.4,-0.91688), dir(225)); [/asy]$

Claim: $(BXC)$ is tangent to the incircle and passes through the midpoint $N$ of $\overline{EO}$ .
Proof. Follows by 2002 G7. $\blacksquare$
Hence by homothety at $X$ the line $\overline{XIF}$ passes thru $P$ .

Claim: $(BXNC)$ and the $A$ -excircle are orthogonal.
Proof. It suffices to show $BXCN$ is fixed under inversion around the $A$ -excircle. To this end, we prove that $ON \cdot OX = OF^2.$ Indeed, this follows from the similar isosceles triangles $\triangle ONF \sim \triangle OFX \sim \triangle EIX.$ $\blacksquare$
Hence it follows that inversion centered at $P$ with radius $PB = PC$ will fix the $A$ -excircle. Since $BC$ is tangent to the $A$ -excircle at $F$ , the inverse image of $F$ is the desired tangency point.

This post has been edited 1 time. Last edited by v_Enhance, Aug 21, 2021, 1:55 AM

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#4 h Apr 21, 2025, 9:32 PM

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Let $D'$ be the $A$ -extouch point. Let $E'$ be the antipode of $E$ in the incircle. Let $XE'\cap BC=E''$ . Let $PD'$ intersect the $A$ -excircle again at $Q$ . Let $AE$ intersect the incircle again at $R$ . Let $AI\cap BC=K$ . Let $OD'$ intersect $(BIC)$ again at $I'$ .

Considering the homothety centered at $E$ sending $AD$ to the vertical diameter of the $A$ -excircle, we get that $X$ must be the midpoint of $AD$ , therefore $X-I-D'$ collinear.

Claim: $EX$ bisects $\angle D'XD$ .
Proof: $\angle D'XE=\angle IXE=\angle IEX=\angle EXD$ .

Claim: $(XE'',XE;XB,XC)=-1$ .
Proof: $(R,E;X,E')\overset{E}{=}(A,K;O,I)=-1\implies RR,XE',BC$ concur at $E''$ which must be the harmonic conjugate of $E$ w.r.t. $B,C$ . This implies the claim.

Since $XE'\perp XE$ , $EX$ bisects $\angle BXC$ and therefore $\angle DXP$ (isogonal lines). But since it also bisects $\angle DXD'$ , $P$ must lie on $XID'$ .

Since $II'\parallel BC\implies II'D'E$ is a rectangle, $PE=PD',P\in ID'\implies P$ is the center $\implies PI'=PD$ . We also have $OD'=OQ\implies \angle PI'O=\angle PI'D'=\angle PD'I'=\angle QD'O=\angle D'QO=\angle PQO\implies PI'QO$ cyclic.

Therefore $D'P\cdot D'Q=D'I'\cdot D'O=D'B\cdot D'C\implies BPCQ$ cyclic. Since we have $BP=PC$ , by shooting lemma the circle through $D'$ and $Q$ tangent to $BC$ must be tangent to $(BPC)$ , but since this circle is unique it must be the $A$ -excircle, hence we're done.

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