# An acute triangle is inscribed in a circle. The resulting three minor arcs of the circle are reflected about the corresponding sides of the triangle.

An acute triangle is inscribed in a circle. The resulting three minor arcs of the circle are reflected about the corresponding sides of the triangle. Are the reflected arcs concurrent?

Source: Problem Solving Through Problems by Loren C. Larson • After experimenting with a dynamical geometry software, it seems that the three arcs are indeed concurrent at the orthocenter of $ABC$. Moreover, if we consider three circles rather than three arcs, then those three circles seems to persist being concurrent at the orthocenter, even when the inscribed triangle is obtuse. – Adren Feb 22 '17 at 7:52
• @Adren Thanks for confirming that. It does look like the concurrent point is the orthocenter. But I wasn't sure how to do that and can't really see how that will lead to a solution for this problem. – jxie20 Feb 22 '17 at 8:17

It is a known property that, given a triangle $T$, the symmetric of the orthocenter with respect to a side of $T$ belongs to the circumcircle.

This can be viewed as a consequence of the following properties ($H$ denotes the orthocenter of $T$) :

1) The feet of the altitudes belong to the Euler circle of $T$

2) The homothety with center $H$ and ratio $2$ transforms the Euler circle into the circumcircle.

Hence the three arcs you mentioned must go through the orthocenter.

• The Euler circle connection is a nice one. This answer to "Why would the reflections of the orthocentre lie on the circumcircle?" has a somewhat more mundane demonstration. – Blue Feb 23 '17 at 1:35

It is clear when reflections of the entire circum-circle are mirrored on each side they should concur at triangle orthocenter.. due to perpendicularity with each side.

EDIT1:

Among all perpendiculars to $AC$ choose one through B, repeat for 3 sides. Unique concurrency point must be the orthocenter. Other points of concurrency and new orthocenters would result by choice of other points for$B$ as $B_1,B_2$ &c on the same circumcircle. • Can you elaborate on the reasoning, especially the perpendicularity part? – jxie20 Feb 25 '17 at 6:01