A-yemyem? B-yemyem? Then C-yemyem as well!

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A-yemyem? B-yemyem? Then C-yemyem as well!

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Source: DeuX Mathematics Olympiad 2020 Shortlist G3 (Level II Problem 4)

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#1 h Jul 12, 2020, 6:02 PM

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Given an acute triangle $ABC$ . Define $D,E,F$ as the foot of $A,B,C$ altitude respectively. Let $X_A, X_B, X_C$ be the incenters of $\triangle AEF, \triangle BDF, \triangle CDE$ . We name a triangle $A$ -yemyem if the circle $(AX_B X_C)$ is tangent to $(ABC)$ . Define $B$ -yemyem and $C$ -yemyem similarly.
Prove that if a triangle is $A$ -yemyem and $B$ -yemyem, then it is $C$ -yemyem as well.

Proposed by Farrel Dwireswara Salim, Indonesia

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Reason: italic

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#2 h Jul 12, 2020, 7:01 PM • 1 Y

Y by Mango247

major spoil

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#3 h Jul 12, 2020, 7:58 PM • 1 Y

Y by AlastorMoody

Solved with AlastorMoody

Claim : $AX_AX_BB$ is cyclic.
Cyclic variations hold.

Proof : Let $I$ denote the incenter of $\triangle ABC$ .
Note that it suffices to show $IA \cdot IX_A=IB \cdot IX_B$ . Now, an easy angle chase yields that $\triangle AX_AE \sim \triangle DX_BB \sim \triangle AIB$ . Using $AF \cdot AB=AE \cdot AC$ , we get that $\frac{AX_A}{AI}=\frac{AE}{AB}=\frac{AF}{AC}=\cos A \Rightarrow AX_A=AI \cos A \Rightarrow IA \cdot IX_A=IA(IA-AX_A)=IA^2(1-\cos A)=2IA^2 \sin ^2 \frac{A}{2}=2r^2$ where $r$ is the inradius of $\triangle ABC$ . Similarly, we have $IB \cdot IX_B=2r^2$ , giving $IA \cdot IX_A=IB \cdot IX_B$ . $\blacksquare$ .

Define $P$ as the intersection of tangent at $A$ to $\odot (ABC)$ and $BC$ . Define $Q$ and $R$ similarly.
Note that by Pascal on $CCAABB$ , we have that $P,Q,R$ are collinear .

Next note that by radical axes on $\odot (ABC),\odot (AX_BX_C) ,\odot (BX_BX_CC)$ , we have that $P \in X_BX_C$ . Similarly $Q \in X_CX_A$ .

Note that since $AX_A \cap BX_B \cap CX_C \equiv I$ . Hence , triangles $\triangle ABC$ and $\triangle X_AX_BX_C$ are perspective from a point. By Desargues Theorem, they are perspective from a line as well . Hence $X_BX_A \cap AB \in PQ$ . So $R \in X_AX_B$ .

So we have $(RC)^2=(RA) \cdot (RB) = (RX_A) \cdot (RX_B)$ .

So $\odot (CX_AX_B)$ is tangent to $\odot (CAB)$ as desired . $\blacksquare$

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#4 h Oct 11, 2021, 10:48 PM

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Wow this is actually cute.

Let $\overline{AA}$ denote the tangent from $A$ to $(ABC)$ .

Claim. [Synthetic] $BX_BX_CC$ is cyclic and so on.
Proof. Let $I$ be the incenter of $\triangle ABC$ and let $I'$ be the reflection of $I$ over $BC$ .
Note that $\triangle BX_BF\sim\triangle BI'C \sim\triangle EX_CC$ . Thus by spiral sim, $\measuredangle I'X_BI=\measuredangle I'X_CI=90^\circ$ , which yields that $II'X_BX_C$ is cyclic quadrilateral. Hence, $\measuredangle X_BX_CC=90^\circ+\measuredangle X_BX_CI'=90^\circ+\measuredangle X_BII'=\measuredangle X_BBC$ , where the last equality holds as $\overline{II'}\perp BC$ , the claim follows.

As $(AX_B X_C)$ is tangent to $(ABC)$ and $(BX_C X_A)$ is tangent to $(ABC)$ , by radical axis theorem, we get that $\overline{BC},\overline{X_BX_C}$ and $\overline{AA}$ are concurrent, and, $\overline{AC},\overline{X_AX_C}$ and $\overline{BB}$ are concurrent.

By Pascal's theorem on $AABBCC$ , we get that $\overline{BC}\cap \overline{AA}$ , $\overline{AC}\cap \overline{BB}$ and $\overline{AB}\cap \overline{CC}$ are collinear.
As $\overline{AX_A},\overline{BX_B}$ and $\overline{CX_C}$ concur at the incenter of $\triangle ABC$ , by Desargue' theorem, we get that $\overline{X_BX_C}\cap \overline{BC}$ , $\overline{X_CX_A}\cap \overline{CA}$ and $\overline{X_AX_B}\cap \overline{AB}$ are collinear.

We conclude that $\overline{AB},\overline{X_AX_B}$ and $\overline{CC}$ are concurrent. By PoP, $\overline{CC}$ is also tangent to $(CX_AX_B)$ , we are done.

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#5 h Oct 12, 2021, 2:40 AM

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What's up with "yemyem"?

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