51st IMO Shortlisted Problems

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 $△XYZ$ and by $r$, $r^{\prime }$, and $R^{\prime }$ the inradii of triangles $ABC$, $XYZ$ and the circumradius of $XYZ$, respectively. Then we have $R^{\prime }=2r^{\prime }$, and the desired inequality is $OI\le r^{\prime }$. We assume that $O\ne I$; otherwise the claim is trivial.

Let the incircle of $△ABC$ touch its sides $BC$, $AC$, $AB$ 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 $OX=R^{\prime }$, we have $d(O,BC)\le R^{\prime }$. Since $O{A}^{\prime }\parallel BC$, it follows that $d(A^{\prime },BC)=A^{\prime }I+r\le R^{\prime }$, or $A^{\prime }I\le R^{\prime }-r$. On the other hand, the incircle of $△XYZ$ lies inside $△ABC$, hence $d(O,AB)\ge r^{\prime }$, and analogously we get $d(O,AB)=C^{\prime }{C}_1=r-I{C}^{\prime }\ge r^{\prime }$, or $I{C}^{\prime }\le r-r^{\prime }$.

Fig. 1 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 $\widehat{A^{\prime }O{C}^{\prime }}=2\angle A^{\prime }I{C}^{\prime }<180^{\circ }=\widehat{O{C}^{\prime }I}$, hence $180^{\circ }\ge \widetilde{I{C}^{\prime }}>\widetilde{A^{\prime }O}$. This means that $I{C}^{\prime }>A^{\prime }O$. Finally, we have $OI\le I{A}^{\prime }+A^{\prime }O\le I{A}^{\prime }+I{C}^{\prime }\le (R^{\prime }-r)+(r-r^{\prime })=R^{\prime }-r^{\prime }=r^{\prime }$, as desired.

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Alternative solution.

Assume the contrary. Then the incenter $I$ should lie in one of triangles $AYZ$, $BXZ$, $CXY$ - assume that it lies in $△AYZ$. Let the incircle $\omega$ of $△ABC$ 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 $U$, $U^{\prime }$, and $V$, $V^{\prime }$ be the points of intersection of $\omega$ with $XY$ and $YZ$, respectively ($UY>U^{\prime }Y$, $VY>V^{\prime }Y$; see Figs. 3 and 4). Note that $60^{\circ }=\angle XYZ=\frac{1}{2}(\widehat{UV}-\widehat{U^{\prime }{V}^{\prime }})\le \frac{1}{2}\widehat{UV}$, hence $\widehat{UV}\ge 120^{\circ }$.

On the other hand, since $I$ lies in $△AYZ$, we get $\widehat{U{U}^{\prime }}<180^{\circ }$, hence $\widehat{U{A}_1{U}^{\prime }}\le \widehat{U{A}_1{V}^{\prime }}<180^{\circ }-\widehat{UV}\le 60^{\circ }$.

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 $A{B}_1$ (see Fig. 3). Then we have $\angle YXC=\frac{1}{2}(\widehat{A_1{U}^{\prime }}-\widehat{A_{1}U})\le \frac{1}{2}\widehat{A_1{U}^{\prime }}<30^{\circ }$; analogously, we get $\angle XYC\le \frac{1}{2}\widehat{A_1{U}^{\prime }}<30^{\circ }$. Therefore, $\angle YCX=180^{\circ }-\angle YXC-\angle XYC>120^{\circ }$, as desired.

Case 2. Now let point $Y$ lie on the segment $C{B}_1$ (see Fig. 4). Analogously, we obtain $\angle YXC<30^{\circ }$. Next, $\angle IYX>\angle ZYX=60^{\circ }$, but $\angle IYX<\angle IY{B}_1$, since $Y{B}_1$ is a tangent and $YX$ is a secant line to circle $\omega$ from point $Y$. Hence, we get $120^{\circ }<\angle IY{B}_1+\angle IYX=\angle B_{1}YX=\angle YXC+\angle YCX<30^{\circ }+\angle YCX$, hence $\angle YCX>120^{\circ }-30^{\circ }=90^{\circ }$, as desired.

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Alternative solution.

Assume the contrary. As in Solution 2, we assume that the incenter $I$ of $△ABC$ lies in $△AYZ$, and the tangency point $A_1$ of $\omega$ and $BC$ lies on segment $CX$. Surely, $\angle YZA\le 180^{\circ }-\angle YZX=120^{\circ }$, 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 $\angle C>120^{\circ }$.