Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.4 Shortlist Prove it Romania

Let $ABCD$ be a cyclic quadrangle, and let its diagonals, $AC$ and $BD$ , cross at $X$ . Let $I$ be the incenter of the triangle $XBC$ , and let $J$ be the center of the circle tangent to the side $BC$ and the extensions of the sides $AB$ and $DC$ beyond $B$ and $C$ , respectively. Prove that the line $IJ$ bisects the arc $BC$ of the circle $ABCD$ , not containing the vertices $A$ and $D$ of the quadrangle.

Solution

Let $M$ be the midpoint of the arc $BC$ of the circle $ABCD$ , not containing the vertices $A$ and $D$ of the quadrangle.

An easy angle chase shows the vertices $B$ and $C$ , the incenter $K$ of the triangle $ABC$ , and its $A$ -excenter $K_A$ equally distanced from $M$ , so they all lie on a circle $\gamma$ centred at $M$ — the so-called trillium lemma for the incenter, respectively, excenter.

Similarly, the incenter $L$ and the $D$ -excenter $L_D$ of the triangle $DBC$ both lie on $\gamma$ .

Finally, apply Pascal's theorem to the hexagram $BK_AKCL_DL$ with vertices on $\gamma$ , to conclude that $J = BK_A \cap CL_D$ , $M = K_AK \cap LL_D$ and $I = KC \cap LB$ are collinear.

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