Let $ABC$ be a triangle which it not right-angled. Define a sequence of triangles $A_iB_iC_i$, with $i \ge 0$, as follows: $A_0B_0C_0$ is the triangle $ABC$ and, for $i \ge 0$, $A_{i+1},B_{i+1},C_{i+1}$ are the reflections of the orthocentre of triangle $A_iB_iC_i$ in the sides $B_iC_i$,$C_iA_i$,$A_iB_i$, respectively. Assume that $\angle A_m = \angle A_n$ for some distinct natural numbers $m,n$. Prove that $\angle A = 60^{\circ}$.
Problem
Source: RMO- Problem 5
Tags: geometry, geometric transformation, reflection, geometry unsolved
sayantanchakraborty
23.02.2014 20:42
Get a recursion to prove that $A_n=(-2)^n(\angle{A}-60)+60$.Then $\angle{A_m}=\angle{A_n} \Rightarrow \angle{A}=60^{\circ}$.By the way the problem is wrong since we don't know whether $A_n$ is positive for each n.
aayush-srivastava
20.11.2015 21:39
The statement of the problem as stated is not correct. I give below the reason, and I shall also give the condition under which the statement becomes true. Let $P, Q, R$ denote the $reflections$ of$ H$ with respect to $BC, CA, AB$, respectively. Then $P, Q, R$ lie on the $circumcircle$ of the triangle. If $ABC$ is an acute-angled triangle then ∠$QP R $= ∠$QP A$ + ∠$RP A$ = ∠$QCA $+ ∠$RBA$ = $180$◦ − 2∠$A.$ Similarly, if ∠$A$ is obtuse then we get ∠$QP R$ = 2∠$A$ − $180$◦ . Therefore, for example, if ∠$A$ = 180◦/7 and ∠$B$∠$C$ = $540$◦/$7$ then we get that ∠$A_3$ = 180◦/7 = ∠$A_0$. Therefore the statement of the problem is not correct. However, the statement is correct provided all the triangles $A_iB_iC_i$ are acute-angled. Under this assumption we give below a proof of the statement. Let α, β, γ denote the angles of $T_0$. Let$ f_k(x)$ =$(-2)^k.x$ − ($(-2)^k$ − 1)60◦ . We claim that the angles of $T_k $are $f_k$(α), $f_k$(β) and$ f_k$(γ). Note that this claim is true for k = 0 and k = 1. It is easy to check that$ f_{k+1}(x)$ = 180◦ − 2$f_k$(x), so the claim follows by induction. If $T_m$ = $T_n$, then $f_m$(α) = $f_n$(α), so $\alpha$($(-2)^m$ − $(-2)^n$) = 60◦. ($(-2)^m$ − $(-2)^n$). Therefore, since $m$ is not equal to $n$, it follows that α = 60◦ .
Wavefunction
04.10.2017 09:44
Can anyone attach a diagram, i wanna know whether they are just talking about $orthic$ triangles or not?