Saudi Arabia Mathematical Competitions

We will use the following relation: For any angle $\theta$, $$\cos 3\theta =\cos \theta -4\cos \theta {\sin }^2\theta$$ Let $a,b,c$ be the sides of the triangle and $\alpha ,\beta ,\gamma$ be the respective angles opposite these sides. Since the triangles $A_{1}BC$, $A{B}_{1}C$ and $AB{C}_1$ are all congruent to the triangle $ABC$, we have that $\widehat{C_{1}A{B}_1}=3\alpha$ or $|2\pi -3\alpha |$, $\widehat{A_{1}B{C}_1}=3\beta$ or $|2\pi -3\beta |$, and $\widehat{B_{1}C{A}_1}=3\gamma$ or $|2\pi -3\gamma |$.

Applying the Cosine Law to triangle $B_{1}C{A}_1$ yields that $$A_1{B}_1^2=a^2+b^2-2ab\cos 3\gamma$$ where $R$ is the circumradius of triangle $ABC$, the Cosine and the Sine Laws applied to that triangle yields that $$2ab\cos \gamma =a^2+b^2-c^2$$ and $$\sin \gamma =\frac{c}{2R}$$ Applying the above to the previous equation yields that $$\begin{array}{c}{A}_1{B}_1^2=a^2+b^2-2ab\cos 3\gamma =a^2+b^2-2ab\cos \gamma (1-4{\sin }^2\gamma )\\ =a^2+b^2-(a^2+b^2-c^2)(1-4{\sin }^2\gamma )\\ =(a^2+b^2)(1-1+4{\sin }^2\gamma )+c^2(1-4{\sin }^2\gamma )\\ =(a^2+b^2-c^2)(4{\sin }^2\gamma )+c^2=\frac{(a^2+b^2-c^2)c^2}{R^2}+c^2\\ =\frac{c^2}{R^2}(a^2+b^2-c^2+R^2)\end{array}$$ Similarly, $$B_1{C}_1^2=\frac{a^2}{R^2}(b^2+c^2-a^2+R^2)$$ and $$C_1{A}_1^2=\frac{b^2}{R^2}(c^2+a^2-b^2+R^2)$$ It follows that $A_1{B}_1=B_1{C}_1$ if and only if $$c^2(a^2+b^2-c^2+R^2)-a^2(b^2+c^2-a^2+R^2)=0$$ Factoring the left side yields that $$(c^2-a^2)(R^2+b^2-a^2-c^2)=0$$ The equality of other pairs of sides can be similarly handled. Thus, triangle $A_1{B}_1{C}_1$ is equilateral if and only if the following system of three equations is valid: $$\begin{array}{rl}& (c^2-a^2)(R^2+b^2-a^2-c^2)=0\\ & (a^2-b^2)(R^2+c^2-b^2-a^2)=0\\ & (b^2-c^2)(R^2+a^2-b^2-c^2)=0\end{array}$$ If $a$ is unequal to both $b$ and $c$, then $$R^2=a^2+c^2-b^2=b^2+a^2-c^2$$ so that $b=c$. Hence, the triangle is isosceles in any case.

Assume that $b=c$. Then from the middle equation, we obtain that $$(a^2-b^2)(R^2-a^2)=0$$ Therefore, either $a=b$, in which case the triangle is equilateral, or $R=a$ and $\alpha =a/2R=1/2$. Therefore, $\alpha =30^{\circ }$ or $\alpha =150^{\circ }$. Thus, there are three possible sets of angles for the triangle $ABC$: $(60^{\circ },60^{\circ },60^{\circ })$, $(30^{\circ },75^{\circ },75^{\circ })$ and $(150^{\circ },15^{\circ },15^{\circ })$.