AoPS - Problem

Source: IMO LongList 1959-1966 Problem 32

Tags: inequalities, geometry, triangle inequality, similar triangles, IMO Shortlist, IMO Longlist

orl

02.09.2004 04:27

The side lengths $a,$ $b,$ $c$ of a triangle $ABC$ form an arithmetical progression (such that $b-a=c-b$). The side lengths $a_{1},$ $b_{1},$ $c_{1}$ of a triangle $A_{1}B_{1}C_{1}$ also form an arithmetical progression (with $b_{1}-a_{1}=c_{1}-b_{1}$). [Hereby, $a=BC,$ $b=CA,$ $c=AB, $ $a_{1}=B_{1}C_{1},$ $b_{1}=C_{1}A_{1},$ $c_{1}=A_{1}B_{1}.$] Moreover, we know that $\measuredangle CAB=\measuredangle C_{1}A_{1}B_{1}.$ Show that triangles $ABC$ and $A_{1}B_{1}C_{1}$ are similar.

We can enlarge $A_1B_1C_1$ s.t. it still keeps the same shape, but $A_1C_1=AC$. Now we can place $A_1B_1C_1$ in the plane s.t. $A_1=A,C_1=C$. Because of the angle congruence, the points $A,B,B_1$ are collinear. We can assume WLOG that $AB_1>AB$. In this case, we have $CB+BA=CB_1+B_1A\iff CB=CB_1+B_1B$, and this is an equality in the triangle inequality, so $B_1=B$, and this means that $ABC$ a homothety which stretches $b_1$ to match $b$ also stretches $c_1$ to match $c$, and this, together with $\angle CAB=\angle C_1A_1B_1$, means that $ABC,A_1B_1C_1$ are similar.