Subfield: Geometry


Question: The vertices $X, Y, Z$ of an equilateral triangle $X Y Z$ lie respectively on the sides $B C$, $C A, A B$ of an acute-angled triangle $A B C$. Prove that the incenter of triangle $A B C$ lies inside triangle $X Y Z$.

Solution:
We will prove a stronger fact; namely, we will show that the incenter $I$ of triangle $A B C$ lies inside the incircle of triangle $X Y Z$ (and hence surely inside triangle $X Y Z$ itself). We denote by $d(U, V W)$ the distance between point $U$ and line $V W$.

Denote by $O$ the incenter of $\triangle X Y Z$ and by $r, r^{\prime}$ and $R^{\prime}$ the inradii of triangles $A B C, X Y Z$ and the circumradius of $X Y Z$, respectively. Then we have $R^{\prime}=2 r^{\prime}$, and the desired inequality is $O I \leq r^{\prime}$. We assume that $O \neq I$; otherwise the claim is trivial.

Let the incircle of $\triangle A B C$ touch its sides $B C, A C, A B$ at points $A_{1}, B_{1}, C_{1}$ respectively. The lines $I A_{1}, I B_{1}, I C_{1}$ cut the plane into 6 acute angles, each one containing one of the points $A_{1}, B_{1}, C_{1}$ on its border. We may assume that $O$ lies in an angle defined by lines $I A_{1}$, $I C_{1}$ and containing point $C_{1}$ (see Fig. 1). Let $A^{\prime}$ and $C^{\prime}$ be the projections of $O$ onto lines $I A_{1}$ and $I C_{1}$, respectively.

Since $O X=R^{\prime}$, we have $d(O, B C) \leq R^{\prime}$. Since $O A^{\prime} \| B C$, it follows that $d\left(A^{\prime}, B C\right)=$ $A^{\prime} I+r \leq R^{\prime}$, or $A^{\prime} I \leq R^{\prime}-r$. On the other hand, the incircle of $\triangle X Y Z$ lies inside $\triangle A B C$, hence $d(O, A B) \geq r^{\prime}$, and analogously we get $d(O, A B)=C^{\prime} C_{1}=r-I C^{\prime} \geq r^{\prime}$, or $I C^{\prime} \leq r-r^{\prime}$.

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Fig. 1

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Fig. 2

Finally, the quadrilateral $I A^{\prime} O C^{\prime}$ is circumscribed due to the right angles at $A^{\prime}$ and $C^{\prime}$ (see Fig. 2). On its circumcircle, we have $\overparen{A^{\prime} O C^{\prime}}=2 \angle A^{\prime} I C^{\prime}<180^{\circ}=\overparen{O C^{\prime} I \text {, hence } 180^{\circ} \geq}$ $\overparen{I C^{\prime}}>\overparen{A^{\prime} O}$. This means that $I C^{\prime}>A^{\prime} O$. Finally, we have $O I \leq I A^{\prime}+A^{\prime} O<I A^{\prime}+I C^{\prime} \leq$ $\left(R^{\prime}-r\right)+\left(r-r^{\prime}\right)=R^{\prime}-r^{\prime}=r^{\prime}$, as desired.
Assume the contrary. Then the incenter $I$ should lie in one of triangles $A Y Z, B X Z, C X Y$ - assume that it lies in $\triangle A Y Z$. Let the incircle $\omega$ of $\triangle A B C$ touch sides $B C, A C$ at point $A_{1}, B_{1}$ respectively. Without loss of generality, assume that point $A_{1}$ lies on segment $C X$. In this case we will show that $\angle C>90^{\circ}$ thus leading to a contradiction.

Note that $\omega$ intersects each of the segments $X Y$ and $Y Z$ at two points; let $U, U^{\prime}$ and $V$, $V^{\prime}$ be the points of intersection of $\omega$ with $X Y$ and $Y Z$, respectively $\left(U Y>U^{\prime} Y, V Y>V^{\prime} Y\right.$; see Figs. 3 and 4). Note that $60^{\circ}=\angle X Y Z=\frac{1}{2}\left(\overparen{U V}-\overparen{U^{\prime} V^{\prime}}\right) \leq \frac{1}{2} \overparen{U V}$, hence $\overparen{U V} \geq 120^{\circ}$.



On the other hand, since $I$ lies in $\triangle A Y Z$, we get $\sqrt{U V^{\prime}}<180^{\circ}$, hence $\sqrt{U A_{1} U^{\prime}} \leq \sqrt{U A_{1} V^{\prime}}<$ $180^{\circ}-\overparen{U V} \leq 60^{\circ}$.

Now, two cases are possible due to the order of points $Y, B_{1}$ on segment $A C$.

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Fig. 3

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Fig. 4

Case 1. Let point $Y$ lie on the segment $A B_{1}$ (see Fig. 3). Then we have $\angle Y X C=$ $\frac{1}{2}\left(\overparen{A_{1} U^{\prime}}-\overparen{A_{1} U}\right) \leq \frac{1}{2} \overparen{U A_{1} U^{\prime}}<30^{\circ}$; analogously, we get $\angle X Y C \leq \frac{1}{2} \overparen{U A_{1} U^{\prime}}<30^{\circ}$. Therefore, $\angle Y C X=180^{\circ}-\angle Y X C-\angle X Y C>120^{\circ}$, as desired.

Case 2. Now let point $Y$ lie on the segment $C B_{1}$ (see Fig. 4). Analogously, we obtain $\angle Y X C<30^{\circ}$. Next, $\angle I Y X>\angle Z Y X=60^{\circ}$, but $\angle I Y X<\angle I Y B_{1}$, since $Y B_{1}$ is a tangent and $Y X$ is a secant line to circle $\omega$ from point $Y$. Hence, we get $120^{\circ}<\angle I Y B_{1}+\angle I Y X=$ $\angle B_{1} Y X=\angle Y X C+\angle Y C X<30^{\circ}+\angle Y C X$, hence $\angle Y C X>120^{\circ}-30^{\circ}=90^{\circ}$, as desired.