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
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.
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$\begingroup$
geometry
contest-math
euclidean-geometry
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0After 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. – 2017-02-22
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0@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. – 2017-02-22
2 Answers
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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.
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0The Euler circle connection is a nice one. [This answer](http://math.stackexchange.com/a/590777/409) to "Why would the reflections of the orthocentre lie on the circumcircle?" has a somewhat more mundane demonstration. – 2017-02-23
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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.
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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.
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0Can you elaborate on the reasoning, especially the perpendicularity part? – 2017-02-25