IMO Shortlist

We will prove a stronger fact; namely, we will show that the incenter $I$ of triangle $ABC$ lies inside the incircle of triangle $XYZ$ (and hence surely inside triangle $XYZ$ itself). We denote by $d(U, VW)$ the distance between point $U$ and line $VW$.

Denote by $O$ the incenter of $\triangle XYZ$ and by $r$, $r'$, and $R'$ the inradii of triangles $ABC$, $XYZ$ and the circumradius of $XYZ$, respectively. Then we have $R' = 2r'$, and the desired inequality is $OI \leq r'$. We assume that $O \neq I$; otherwise the claim is trivial.

Let the incircle of $\triangle ABC$ touch its sides $BC$, $AC$, $AB$ at points $A_1$, $B_1$, $C_1$ respectively. The lines $IA_1$, $IB_1$, $IC_1$ 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 $IA_1$, $IC_1$ and containing point $C_1$ (see Fig. 1). Let $A'$ and $C'$ be the projections of $O$ onto lines $IA_1$ and $IC_1$, respectively.

Since $OX = R'$, we have $d(O, BC) \leq R'$. Since $OA' \parallel BC$, it follows that $d(A', BC) = A'I + r \leq R'$, or $A'I \leq R' - r$. On the other hand, the incircle of $\triangle XYZ$ lies inside $\triangle ABC$, hence $d(O, AB) \geq r'$, and analogously we get $d(O, AB) = C'C_1 = r - IC' \geq r'$, or $IC' \leq r - r'$.

Fig. 1

Fig. 2

Finally, the quadrilateral $IA'O C'$ is circumscribed due to the right angles at $A'$ and $C'$ (see Fig. 2). On its circumcircle, we have $\widehat{A'O C'} = 2 \angle A'I C' < 180^{\circ} = \widehat{O C'I}$, hence $180^{\circ} \geq \widetilde{IC'} > \widetilde{A'O}$. This means that $IC' > A'O$. Finally, we have $OI \leq IA' + A'O \leq IA' + IC' \leq (R' - r) + (r - r') = R' - r' = r'$, as desired.

Assume the contrary. Then the incenter $I$ should lie in one of triangles $AYZ$, $BXZ$, $CXY$ - assume that it lies in $\triangle AYZ$. Let the incircle $\omega$ of $\triangle ABC$ touch sides $BC$, $AC$ at point $A_1$, $B_1$ respectively. Without loss of generality, assume that point $A_1$ lies on segment $CX$. In this case we will show that $\angle C > 90^{\circ}$ thus leading to a contradiction.

Note that $\omega$ intersects each of the segments $XY$ and $YZ$ at two points; let $U$, $U'$, and $V$, $V'$ be the points of intersection of $\omega$ with $XY$ and $YZ$, respectively ($UY > U'Y$, $VY > V'Y$; see Figs. 3 and 4). Note that $60^{\circ} = \angle XYZ = \frac{1}{2}(\widehat{UV} - \widehat{U'V'}) \leq \frac{1}{2} \widehat{UV}$, hence $\widehat{UV} \geq 120^{\circ}$.

On the other hand, since $I$ lies in $\triangle AYZ$, we get $\widehat{UU'} < 180^{\circ}$, hence $\widehat{UA_1U'} \leq \widehat{UA_1V'} < 180^{\circ} - \widehat{UV} \leq 60^{\circ}$.

Now, two cases are possible due to the order of points $Y$, $B_1$ on segment $AC$.

Fig. 3

Fig. 4

Case 1. Let point $Y$ lie on the segment $AB_1$ (see Fig. 3). Then we have $\angle YXC = \frac{1}{2}(\widehat{A_1U'} - \widehat{A_1U}) \leq \frac{1}{2} \widehat{A_1U'} < 30^{\circ}$; analogously, we get $\angle XYC \leq \frac{1}{2} \widehat{A_1U'} < 30^{\circ}$. Therefore, $\angle YCX = 180^{\circ} - \angle YXC - \angle XYC > 120^{\circ}$, as desired.

Case 2. Now let point $Y$ lie on the segment $CB_1$ (see Fig. 4). Analogously, we obtain $\angle YXC < 30^{\circ}$. Next, $\angle IYX > \angle ZYX = 60^{\circ}$, but $\angle IYX < \angle IYB_1$, since $YB_1$ is a tangent and $YX$ is a secant line to circle $\omega$ from point $Y$. Hence, we get $120^{\circ} < \angle IYB_1 + \angle IYX = \angle B_1YX = \angle YXC + \angle YCX < 30^{\circ} + \angle YCX$, hence $\angle YCX > 120^{\circ} - 30^{\circ} = 90^{\circ}$, as desired.

Assume the contrary. As in Solution 2, we assume that the incenter $I$ of $\triangle ABC$ lies in $\triangle AYZ$, and the tangency point $A_1$ of $\omega$ and $BC$ lies on segment $CX$. Surely, $\angle YZA \leq 180^{\circ} - \angle YZX = 120^{\circ}$, hence points $I$ and $Y$ lie on one side of the perpendicular bisector to $XY$; therefore $IX > IY$. Moreover, $\omega$ intersects segment $XY$ at two points, and therefore the projection $M$ of $I$ onto $XY$ lies on the segment $XY$. In this case, we will prove that $\angle C > 120^{\circ}$.