# Show that $A-$excircle is tangent to $(AST)$

Elmo is now learning olympiad geometry. In triangle $$ABC$$ with $$AB\neq AC$$, let its incircle be tangent to sides $$BC$$, $$CA$$, and $$AB$$ at $$D$$, $$E$$, and $$F$$, respectively. The internal angle bisector of $$\angle BAC$$ intersects lines $$DE$$ and $$DF$$ at $$X$$ and $$Y$$, respectively. Let $$S$$ and $$T$$ be distinct points on side $$BC$$ such that $$\angle XSY=\angle XTY=90^\circ$$. Finally, let $$\gamma$$ be the circumcircle of $$\triangle AST$$.

(a) Help Elmo show that $$\gamma$$ is tangent to the circumcircle of $$\triangle ABC$$.

(b) Help Elmo show that $$\gamma$$ is tangent to the incircle of $$\triangle ABC$$.

Very Hard problem ..

My Progress: WLOG $$AB . Define $$N=BC\cap AY$$.

Claim: $$\angle ASX=\angle XST$$

Proof: Since $$\angle XSY=90$$ , by lemma 9.18 from EGMO it's enough to show that $$(A,N;X,Y)=-1$$ .

Now, note that $$AEFD$$ is a harmonic quad . So $$(A,D;E,F)=-1$$ . So projecting through $$D$$ on $$AI$$, we get $$(A,DD\cap AI=N;DE\cap AI=X,DF\cap AI=Y )=-1$$ . So $$\angle ASX=\angle XST$$.

Similarly , we can get $$\angle ATX=\angle XTN$$

So we get $$X$$ and $$Y$$ as the incentre and $$A$$-excentre of $$\Delta AST$$ respectively .

Also we get $$AS$$ and $$AT$$ isogonals [ since $$X$$ is the incentre and $$X\in AI$$ ]

Now, let $$S'=AS\cap (ABC)$$ , $$T'=AT\cap (ABC)$$ . By angle chase , we get $$ST||S'T'$$ and hence $$AST~AS'T'$$ and by homothety , we get $$(AST)$$ and $$(ABC)$$ tangent.

This proves part $$A$$.

For Part $$B$$: I applied inversion $$\psi$$ centered at $$A$$ with radius $$\sqrt{AX\cdot AY}$$ radius followed by reflection about the angle bisector of $$\angle AST$$.

Note that $$\psi:X\leftrightarrow Y$$, $$\psi:S\leftrightarrow T$$ and $$\psi:BC\leftrightarrow (AST)$$ .

Let us assume that $$\angle AXB=\angle AYC=90$$ [ I haven't got the proof of this , but it looks very true]

Now, let $$I-A$$ be the $$A$$-excentre of $$\Delta ABC$$ , $$F^*$$ be touch point of $$A$$-excircle to line $$AC$$, $$E^*$$ be touch point of $$A$$ -excircle to line $$AB$$, And $$K^*$$ be touch point of $$A$$-excircle to $$BC$$.

Note that $$K^*,C,F^*,I-A$$ is cyclic .

But it's known that $$\angle K^*CF^*=180-C\implies \angle CK^*F^*=\angle CF^*K^*=C/2$$ . So $$\angle XYF*=90+C/2$$ .

Also note that $$\Delta DXC$$ is isosceles , hence $$\angle XEC=90-C/2$$ .

Hence $$XEF^*Y$$ is cyclic. Similarly we can get $$XFE^*Y$$ is cyclic [$$BK*I-AE*$$ is cyclic]

Hence $$AX\cdot AY=AE\cdot AF^*=AF\cdot AE^*$$ .

So $$\psi:E\leftrightarrow E^*$$ and $$\psi:F\leftrightarrow F^*$$ .

Define $$D^*$$ as the inverse image of $$D$$ by $$\psi$$.

Note that $$(D^*E^*F^*)$$ will be a circle tangent to $$AB$$ at $$E^*$$, tangent to $$AC$$ at $$F^*$$ and $$(AST)$$ at $$D^*$$ . If I am able to show that $$(D^*E^*F^*)$$ is the $$A$$-excircle , then I am done . So basically, I need to prove that

Show that $$A-$$excircle is tangent to $$(AST)$$

After Proofing the above statement , we get that $$\psi:$$ incircle of $$ABC$$ $$\leftrightarrow$$ $$A$$-excircle of $$ABC$$. Now since $$A$$-excircle touches $$BC$$ at $$K^*$$ , and inversion preserves tangency , the inverted images will also be tangent to each other i.e incircle of $$ABC$$ is tangent to $$(AST)$$ and we will be done ..

If possible can someone post solution using the way I have proceeded ( using $$\sqrt{AX\cdot AY}$$) ? Thanks in advance!

Here's the ggb link for the diagram: https://www.geogebra.org/geometry/xzzqzmuh

• This issue looks connected with isogonal transform:with respect to an angle bissector users.math.uoc.gr/~pamfilos/eGallery/problems/Isogonal.html I suspect, without being sure that this property of circle $\gamma$ can be explained with results on this transform. Commented Sep 28, 2020 at 8:20
• A relationship that can be useful here Commented Sep 28, 2020 at 12:26
• Your username changed but it's easy to tell, same old beautiful questions are being asked ;)
– BLM
Commented Sep 30, 2020 at 7:12
• @BLM :smile: That's so sweet of you :) Commented Sep 30, 2020 at 9:46

Wonderful approach!

Filling the blanks you left...

Let us assume that $$\angle AXB=\angle AYC=90$$ [ I haven't got the proof of this , but it looks very true]

Proof. Note that $$\angle IED +\angle IXE = \angle AIE=\dfrac{\angle B+\angle C}{2}$$ and as $$\angle IED = \dfrac{\angle C}{2}\implies \angle IXE =\dfrac{\angle B}{2}=\angle DXY=\angle IBD\implies X\in\odot(BID)\implies \angle IXB=90^\circ$$ and similarly, we get $$\angle AYC=90$$ completing the proof.

If I am able to show that $$(D^∗E^∗F^∗)$$ is the $$A$$-excircle , then I am done.

So basically, all you wanna show is that the incircle maps to $$A$$-excircle under the $$\sqrt{AX\cdot AY}$$ inversion along with reflection about angle bisector $$\angle SAT$$. As you have already shown that $$\Psi: \{F, E\}\leftrightarrow \{F',E'\}$$ so its enough to show that there is at least one point $$P\in \odot(I), P\not\in\{E,F\}$$ such that $$\Psi(P)\in\odot(I_A)$$. As you have already shown that $$\{AS,AT\}$$ are isogonal, we know that angle bisector of $$\angle SAT$$ is nothing but $$AI$$. As reflection of $$A$$-excircle about $$AI$$ is $$A$$-excircle, its enough to show that there is a point $$P\in\odot(I)$$ (other than $$E,F$$) such that it maps to some point on $$A$$-excircle after inverting about the circle centered at $$A$$ with radius $$\sqrt{AX\cdot AY}$$. As you found that $$XYFE'$$ is cyclic, by power of point, we have $$AX\cdot AY = AF\cdot AE'$$. Let $$D$$-antipode in $$\odot(I)$$ be $$M$$. It is well known that $$A-M-K'$$ are collinear. Let $$W= AK'\cap \odot(I_A)$$, $$W\neq K'$$. Note that the homothety centered at $$A$$ that maps incircle to $$A$$-excircle tells us that $$MF\|K'E'$$ and thus, $$\angle AFM=\angle AE'K'=\angle E'WK'\implies MWE'F$$ is cyclic. Thus, $$AX\cdot AY = AE'\cdot AF= AM\cdot AW$$ and thus, $$M$$ maps to $$W$$ under the $$\sqrt{AX\cdot AY}$$ inversion. Thus, such a $$P$$ exsist and hence, under the $$\sqrt{AX\cdot AY}$$ inversion, the incircle gets maped to $$A$$-excircle. Thus, $$(D^∗E^∗F^∗)$$ is the $$A$$-excircle and we are done!$$\tag*{\blacksquare}$$

• umm...how did you get B/2=CXY ? Commented Sep 28, 2020 at 11:42
• @SunainaPati It was a typo... I meant angle DXY = B/2 Commented Sep 28, 2020 at 12:11
• WOW... I totally forgot about antipode of D wrt incircle .. Thank you so much Anand! Learnt a lot :) Commented Sep 28, 2020 at 12:58
• @SunainaPati Happy to help :) Commented Sep 28, 2020 at 13:59