Subfield: Geometry


Question: Let $A B C$ be a triangle with incenter $I$ and let $X, Y$ and $Z$ be the incenters of the triangles $B I C, C I A$ and $A I B$, respectively. Let the triangle $X Y Z$ be equilateral. Prove that $A B C$ is equilateral too.

Solution:
$A Z, A I$ and $A Y$ divide $\angle B A C$ into four equal angles; denote them by $\alpha$. In the same way we have four equal angles $\beta$ at $B$ and four equal angles $\gamma$ at $C$. Obviously $\alpha+\beta+\gamma=\frac{180^{\circ}}{4}=45^{\circ}$; and $0^{\circ}<\alpha, \beta, \gamma<45^{\circ}$.

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Easy calculations in various triangles yield $\angle B I C=180^{\circ}-2 \beta-2 \gamma=180^{\circ}-\left(90^{\circ}-2 \alpha\right)=$ $90^{\circ}+2 \alpha$, hence (for $X$ is the incenter of triangle $B C I$, so $I X$ bisects $\angle B I C$ ) we have $\angle X I C=$ $\angle B I X=\frac{1}{2} \angle B I C=45^{\circ}+\alpha$ and with similar aguments $\angle C I Y=\angle Y I A=45^{\circ}+\beta$ and $\angle A I Z=\angle Z I B=45^{\circ}+\gamma$. Furthermore, we have $\angle X I Y=\angle X I C+\angle C I Y=\left(45^{\circ}+\alpha\right)+$ $\left(45^{\circ}+\beta\right)=135^{\circ}-\gamma, \angle Y I Z=135^{\circ}-\alpha$, and $\angle Z I X=135^{\circ}-\beta$.

Now we calculate the lengths of $I X, I Y$ and $I Z$ in terms of $\alpha, \beta$ and $\gamma$. The perpendicular from $I$ on $C X$ has length $I X \cdot \sin \angle C X I=I X \cdot \sin \left(90^{\circ}+\beta\right)=I X \cdot \cos \beta$. But $C I$ bisects $\angle Y C X$, so the perpendicular from $I$ on $C Y$ has the same length, and we conclude

$$
I X \cdot \cos \beta=I Y \cdot \cos \alpha
$$

To make calculations easier we choose a length unit that makes $I X=\cos \alpha$. Then $I Y=\cos \beta$ and with similar arguments $I Z=\cos \gamma$.

Since $X Y Z$ is equilateral we have $Z X=Z Y$. The law of Cosines in triangles $X Y I, Y Z I$ yields

$$
\begin{aligned}
& Z X^{2}=Z Y^{2} \\
\Longrightarrow & I Z^{2}+I X^{2}-2 \cdot I Z \cdot I X \cdot \cos \angle Z I X=I Z^{2}+I Y^{2}-2 \cdot I Z \cdot I Y \cdot \cos \angle Y I Z \\
\Longrightarrow & I X^{2}-I Y^{2}=2 \cdot I Z \cdot(I X \cdot \cos \angle Z I X-I Y \cdot \cos \angle Y I Z) \\
\Longrightarrow & \underbrace{\cos ^{2} \alpha-\cos ^{2} \beta}_{\text {L.H.S. }}=\underbrace{2 \cdot \cos \gamma \cdot\left(\cos \alpha \cdot \cos \left(135^{\circ}-\beta\right)-\cos \beta \cdot \cos \left(135^{\circ}-\alpha\right)\right)}_{\text {R.H.S. }} .
\end{aligned}
$$

A transformation of the left-hand side (L.H.S.) yields

$$
\begin{aligned}
\text { L.H.S. } & =\cos ^{2} \alpha \cdot\left(\sin ^{2} \beta+\cos ^{2} \beta\right)-\cos ^{2} \beta \cdot\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right) \\
& =\cos ^{2} \alpha \cdot \sin ^{2} \beta-\cos ^{2} \beta \cdot \sin ^{2} \alpha
\end{aligned}
$$



$$
\begin{aligned}
& =(\cos \alpha \cdot \sin \beta+\cos \beta \cdot \sin \alpha) \cdot(\cos \alpha \cdot \sin \beta-\cos \beta \cdot \sin \alpha) \\
& =\sin (\beta+\alpha) \cdot \sin (\beta-\alpha)=\sin \left(45^{\circ}-\gamma\right) \cdot \sin (\beta-\alpha)
\end{aligned}
$$

whereas a transformation of the right-hand side (R.H.S.) leads to

$$
\begin{aligned}
\text { R.H.S. } & =2 \cdot \cos \gamma \cdot\left(\cos \alpha \cdot\left(-\cos \left(45^{\circ}+\beta\right)\right)-\cos \beta \cdot\left(-\cos \left(45^{\circ}+\alpha\right)\right)\right) \\
& =2 \cdot \frac{\sqrt{2}}{2} \cdot \cos \gamma \cdot(\cos \alpha \cdot(\sin \beta-\cos \beta)+\cos \beta \cdot(\cos \alpha-\sin \alpha)) \\
& =\sqrt{2} \cdot \cos \gamma \cdot(\cos \alpha \cdot \sin \beta-\cos \beta \cdot \sin \alpha) \\
& =\sqrt{2} \cdot \cos \gamma \cdot \sin (\beta-\alpha)
\end{aligned}
$$

Equating L.H.S. and R.H.S. we obtain

$$
\begin{aligned}
& \sin \left(45^{\circ}-\gamma\right) \cdot \sin (\beta-\alpha)=\sqrt{2} \cdot \cos \gamma \cdot \sin (\beta-\alpha) \\
\Longrightarrow & \sin (\beta-\alpha) \cdot\left(\sqrt{2} \cdot \cos \gamma-\sin \left(45^{\circ}-\gamma\right)\right)=0 \\
\Longrightarrow & \alpha=\beta \quad \text { or } \quad \sqrt{2} \cdot \cos \gamma=\sin \left(45^{\circ}-\gamma\right) .
\end{aligned}
$$

But $\gamma<45^{\circ}$; so $\sqrt{2} \cdot \cos \gamma>\cos \gamma>\cos 45^{\circ}=\sin 45^{\circ}>\sin \left(45^{\circ}-\gamma\right)$. This leaves $\alpha=\beta$. With similar reasoning we have $\alpha=\gamma$, which means triangle $A B C$ must be equilateral.