Amazing Problem!

Let $L,M = (AIC) \cap BC, (AIB) \cap BC$ evidently $L,M \in (BI_1H_1),(BI_2H_2)$ respectively. Let $O = (ALM) \cap (ABC)$, redefine $E = AO \cap BC$ Claim 1: External common tangents of circles $(BI_1H_1)$ and $(CI_2H_2)$ meet at $E$.

Proof

Let $\mathbb{l}$ denote the line through $A$ parallel to $BC$ Let $F = BI \cap \mathbb{l}$ and $J = CI \cap \mathbb{l}$, evidently $F,J \in (AI_1H_1),(AI_2H_2)$. $K = (ABC) \cap (AI_1H_1)$ then $K \in ZC,(AI_1H_1)$. Claim 2: $Z$ is exsimilicenter of $(AI_1H_1),(CI_2H_2)$

Proof

Now the tangents to $(BI_1H_1),(AI_1H_1)$ must be parallel as they have same circumradius, also it must be parallel to $AB$ as $I_1H_1 \perp AB$ Claim 3: $EZ \parallel AB$

Proof

For the last computation, $\frac{AC}{AB} = \frac{EC}{EL}= 1+\frac{CL}{EL}=1+\frac{AC}{EY}$ which simplifies to give $EY = \frac{bc}{|b-c|}$ which is symmetric in $b$ and $c$

@above: I think Claim 3 might not be true: If so, I_1H_1 would have bisected AB

navi_09220114 wrote: @above: I think Claim 3 might not be true: If so, I_1H_1 would have bisected AB Oops mb, we only need the tangents parallel to $AB$. (I initially thought it would be $AB$) Fixed now thanks

Official Solution: (A little longer than required, but reveals many hidden properties of the diagram) ............................................................... Let us start with a few preliminary observations. Claim 1: Suppose the line through $A$ parallel to $BC$ intersect $BI_1$ and $CI_2$ again at $P$ and $Q$ respectively. Then $P$ and $Q$ lies on the circles $(AI_1H_1)$ and $(AI_2H_2)$ respectively. Proof: Note that $\angle API_1=\angle I_1BD=\angle ABI_1=\angle AH_1I_1$, hence $P$ lies on $(AI_1H_1)$. Likewise $Q$ lies on $(AI_2H_2)$. $\square$ Claim 2: $AXYZ$ is cyclic, with its center $V$ lies on $PQ$. Proof: We angle chase that: $$\angle ZAP=\angle (AH_2, BC)=90^{\circ}-\angle BCI_2=90^{\circ}-\angle AQI_2=90^{\circ}-\angle AXZ$$which implies that the circumcenter of triangle $AXZ$ lies on $PQ$. Likewise, the circumcenter of triangle $AYZ$ lies on $PQ$. So if the perpendicular bisector of $AX$ meet $PQ$ at $V$, then $VA=VX=VZ=VY$ from the above, namely $AXYZ$ is cyclic and the center $V$ lies on $PQ$. $\square$ Claim 3: $V$ is the exsimilicenter of $(AI_1H_1)$ and $(AI_2H_2)$. Proof: Note that $\angle PXA=\angle PI_1A=180^{\circ}-\angle AI_1B=45^{\circ}$. Likewise $\angle QXA=\angle 45^{\circ}$ which implies $(AXYZ)$ is an Apollonius circle with $AX$ bisects the angle $\angle PXQ$. Thus $VP\cdot VQ=VA^2$. Moreover, as $AP$ and $AQ$ subtend the same angle in their respective circles, and are parallel (as they are the same line), then the exsimilicenter of the two circles must lie on $PQ$. As $V$ lies on $PQ$ and satisfies $$\frac{VA}{VP}=\frac{VQ}{VA}$$then the exsimilicenter must be $V$. $\square$ Now, we prove the most important claim of the entire problem. Claim 4: $Y$ is the exsimilicenter of the circles $(BI_1H_1)$ and $(AI_2H_2)$. Proof: Let $AH_1$ intersect $BC$ at $K$, then $\angle BKH_1=90^{\circ}+\angle I_1BK=90^{\circ}+\angle AI_1B=\angle BI_1H_1$, so $K$ lies on $(BI_1H_1)$. Moreover, $\angle BKA=180^{\circ}-\angle BKH_1=180^{\circ}-\angle BI_1H_1=\angle BAK$, so $BI_1$ is the perpendicular bisector of segment $AK$. Next, let $XI_1$ intersect $(AI_2H_2)$ again at $L$, we claim that the triangles $\triangle BI_1K$ and $\triangle QLA$ are homothetic. To prove this, note that $\angle BI_1K=\angle AI_1B=135^{\circ}=180^{\circ}-\angle QXA=\angle QLA$. Moreover, $\angle (I_1B, BK)=\angle (I_1P, PA)=\angle (LX, XA)=\angle (LQ, QA)$, so the two triangles $\triangle BI_1K$ and $\triangle QLA$ are similar. Finally, $BK$ and $QA$ being parallel implies the two triangles are homothetic in similar orientation. Hence the lines $AK, LI_1, QB$ are concurrent at the homothetic center, which is $Y$. This center of homothety is also the exsimilicenter of circles $(ALQ)\equiv(AI_2H_2)$ and $(BI_1K)\equiv(BI_1H_1)$, as desired. $\square$ By a similar argument like in Claim $4$, $Z$ is the exsimilicenter of circles $(CI_2H_2)$ and $(AI_1H_1)$. Till this end, let us bring $U$ back to the diagram, and the following claim will solve the problem immediately. Claim 5: $UYVZ$ is a rhombus. Proof: Recall that $V$ is the center of the circle $(AXYZ)$, so $VY=VZ$. Hence it suffice to prove that $UYVZ$ is a parallelogram. We will just prove that $UZ\parallel VY$, as $UY\parallel VZ$ will follow similarly. We claim that in fact the lines $UZ$ and $VY$ are both parallel to $AB$. The key observation is that the circles $(AI_1H_1)$ and $(BI_1H_1)$ have the same radius as $(AI_1B)$ and hence their exsimilicenter must be a point at infinity with direction perpendicular to $I_1H_1$. This implies that by Monge's Theorem applied to circles $\{(BI_1H_1), (CI_2H_2), (AI_1H_1)\}$, we get that $UZ\perp I_1H_1$ which implies $UZ\parallel AB$. In the same way by applying Monge's Theorem on circles $\{(BI_1H_1), (CI_2H_2), (AI_2H_2)\}$, we get $VY\parallel AB$ as well. This proves that $UZ\parallel VY\parallel AB$, likewise $UY\parallel VZ\parallel AC$, and together with $VY=VZ$ implies that $UYVZ$ is a rhombus. $\square$ Finally by Claim $5$, then we get $UY=UZ$ as desired. $\blacksquare$

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