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Saudi Arabian Mathematical Competitions

Geometry Difficulty 5.0 AIME, harder Prove it Saudi Arabia

Let $I$ be the incenter of triangle $A B C$ and $J$ the excenter of the side $B C$ . Let $M$ be the midpoint of $C B$ and $N$ the midpoint of arc $B A C$ of circle $(A B C)$ . If $T$ is the symmetric of the point $N$ by the point $A$ , prove that the quadrilateral $J M I T$ is cyclic.

Solution

Suppose that $A I$ cuts $B C$ at $D$ and cuts $(A B C)$ again at $P$ . We have known that $B, C, I, J$ lie on circle $(P, \frac{I J}{2})$ .

Suppose that $A N$ meets $B C$ at $E$ , then $A E$ is the external angle bisector of $\angle B A C$ , so $(D E, B C) = -1$ . We have
$\overline{D E} \cdot \overline{D M} = \overline{D B} \cdot \overline{D C} = \overline{D I} \cdot \overline{D J}$
means that $I M J E$ is cyclic.

In the other hand, we have known that $(A D, I J) = -1$ . Combining with $D$ is orthocenter of $\triangle N P E$ , we get
$\overline{A T} \cdot \overline{A S} = -\overline{A N} \cdot \overline{A S} = \overline{A D} \cdot \overline{A P} = \overline{A I} \cdot \overline{A J},$
implies that $I J E T$ is cyclic.

From (1) and (2), then the points $J, I, M, T, E$ lie on a circle. $\square$

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