Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.3 National olympiad, round 2 Prove it Saudi Arabia

Let $ABC$ be an acute, non-isosceles triangle, $AX$ , $BY$ , $CZ$ are the altitudes with $X$ , $Y$ , $Z$ belonging to $BC$ , $CA$ , $AB$ respectively. Respectively denote $(O_{1})$ , $(O_{2})$ , $(O_{3})$ as the circumcircles of triangles $AYZ$ , $BZX$ , $CXY$ . Suppose that $(K)$ is a circle that is internally tangent to $\left(O_{1} ight) \right.$ , $\left(O_{2} ight) \right.$ , $\left(O_{3} ight) \right.$ . Prove that $(K)$ is tangent to the circumcircle of triangle $ABC$ .

Solution

Let $H$ be the orthocenter of triangle $ABC$ .

We can see that $HA \cdot HX = HB \cdot HY = HC \cdot HZ = k$ . We consider the inversion with center $H$ and ratio $-k$ as the function $f$ .

It is easy to see that
$f(A) = X, \quad f(B) = Y, \quad f(C) = Z$
so $f((O)) = (O')$ with $(O')$ being the 9-point circle (which also passes through $X$ , $Y$ , $Z$ ).

On the other hand, because $\angle HEA = \angle HFA = 90^{\circ}$ , hence $H \in (O_{1})$ . After the inversion, the circles $(O_{1})$ become a line passing through the images of $X$ , $Y$ ; indeed, this line is $BC$ or $f((O_{1})) = BC$ .

Similarly, we also have $f((O_{2})) = CA$ and $f((O_{3})) = AB$ . So the circle $K$ that is tangent to $(O_{1})$ , $(O_{2})$ , $(O_{3})$ will become the incircle $(I)$ of triangle $ABC$ . But based on Feuerbach's theorem, the two circles $(O')$ , $(I)$ are tangent to each other.

Therefore, the circles $(K)$ and $(O)$ are also tangent to each other. $\square$

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