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Let ABC be a triangle which it not right-angled. Define a sequence of triangles $A_iB_iC_i$,with $i \geq 0$, as follows: $A_0B_0C_0$ is the triangle $ABC$; and, for $i \geq 0$, $A_{i+1}$, $B_{i+1}$, and $C_{i+1}$ are the reflections of the orthocentre of triangle $A_iB_iC_i$ in the sides $B_iC_i$, $C_iA_i$, and $A_iB_i$ , respectively. Assume that $\angle A_m = \angle A_n$ for some distinct natural numbers $m, n$. Prove that $\angle A = 60$.

Why would the reflections of the orthocentre lie on the circumcircle?

this is the 5th question of crmo

Blue
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maths lover
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    This question seems to be asking for an explanation of the initial observation in the solution given here: http://olympiads.hbcse.tifr.res.in/uploads/crmo-2013-solutions-3 – Blue Dec 02 '13 at 18:29

1 Answers1

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Let $Q$ be the orthocenter of $\triangle ABC$, and let $Q^\prime$ be the reflection of $Q$ in side $\overline{BC}$.

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Note that $\angle QCA$ and $\angle A$ are complements (as acute angles of a right triangle); likewise, $\angle QAC$ and $\angle C$ are complements. (Here, $\angle C$ refers the interior angle at vertex C of $\triangle ABC$; in other words, $\angle C = \angle BCA.$)

Then with the exterior angle theorem, $$\angle Q^\prime = \angle Q^\prime QC = \angle QAC + \angle QCA = \left(\; 90^\circ - \angle C \;\right) + \left(\; 90^\circ - \angle A \;\right) = 180^\circ - \angle A - \angle C = \angle B$$

Since points $B$ and $Q^\prime$ subtend the same angle with segment $\overline{AC}$, they lie on the same circle through $A$ and $C$. (This is an aspect of the Inscribed Angle Theorem.) Thus, the reflection of the orthocenter in a side of a triangle lies on the circumcircle of that triangle.

BakedPotato66
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Blue
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  • In the given solution in your lnked page in the comment section of question can you please explain how they got $\angle QPA=\angle QCA$ and what is the intuition behind the induction step.i mean how they formed the inductive function. – user471651 Nov 21 '17 at 17:36
  • Shouldn't it be: $C$ and $Q'$...? – PinkyWay Mar 04 '20 at 10:15
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    @ms._VerkhovtsevaKatya: "Since $B$ and $Q'$ subtend the same angle ..." is correct (if that's what you mean). However, I should have written $C$ instead of $B$ in "with segment $\overline{AB}$" and "the same circle through $A$ and $B$". I've made the corrections. Thanks. – Blue Mar 04 '20 at 14:25