Saudi Arabia Mathematical Competitions

Let $A_1$, $B_1$, $C_1$ be the tangency points of the incircle of triangle $ABC$ with the sides $BC$, $CA$, $AB$, respectively.



Since $I{A}^{\prime }=I{B}^{\prime }=I{C}^{\prime }$ it follows that triangles $I{A}_1{A}^{\prime }$, $I{B}_1{B}^{\prime }$, and $I{C}_1{C}^{\prime }$ are congruent. We get $A_1{A}^{\prime }=B_1{B}^{\prime }=C_1{C}^{\prime }$, hence $$\begin{array}{c}\left|(s-b)-\frac{a}{2}\right|=\left|(s-c)-\frac{b}{2}\right|=\left|(s-a)-\frac{c}{2}\right|,\end{array}\text{\hspace{1em}(1)}$$ where $a$, $b$, $c$ are the length sides of triangle $ABC$, and $s$ its semiperimeter. The relations (1) are equivalent to $$\begin{array}{c}|c-b|=|a-c|=|b-a|.\end{array}\text{\hspace{1em}(2)}$$ From (2), considering all 6 possible orders for $a$, $b$, $c$, it follows $a=b=c$.



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Alternative solution.

The relations $I{A}^{\prime }=I{B}^{\prime }=I{C}^{\prime }$ imply that $I=O_9$, the center of the Euler nine-point circle of triangle $ABC$. Hence $HI=IO$, where $H$ is the orthocenter and $O$ the circumcenter of triangle $ABC$.



But $\widehat{BAH}=\widehat{OAC}=\frac{\pi }{2}-\hat{B}$, that is $\widehat{HAI}=\widehat{OAI}$. That is triangle $HAO$ is isosceles, hence $AH=AO$.

In similar way, we get $BH=BO$ and $CH=CO$. Since $OA=OB=OC$, it follows $HA=HB=HC$, that is $H=O$, hence triangle $ABC$ is equilateral.



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Alternative solution.

We have $I{A}^2=r^2+(s-a{)}^2$, $I{B}^2=r^2+(s-b{)}^2$, $I{C}^2=r^2+(s-c{)}^2$, where $a$, $b$, $c$ are the length sides of triangle $ABC$, $s$ the semiperimeter, and $r$ the inradius.



Applying the Median Theorem in triangle $BIC$, we get $I{A}^{\prime 2}=\frac{1}{2}(I{B}^2+I{C}^2)-\frac{1}{4}B{C}^2=r^2+\frac{1}{2}[(s-b{)}^2+(s-c{)}^2]-\frac{a^2}{4}$, and similarly $$\begin{array}{rl}& I{B}^{\prime 2}=r^2+\frac{1}{2}[(s-a{)}^2+(s-c{)}^2]-\frac{b^2}{4}\\ & I{C}^{\prime 2}=r^2+\frac{1}{2}[(s-a{)}^2+(s-b{)}^2]-\frac{c^2}{4}\end{array}$$ It follows that $I{A}^{\prime }=I{B}^{\prime }$ if and only if $I{A}^{\prime 2}=I{B}^{\prime 2}$, that is $$\begin{array}{c}(a-b)\left[c-\frac{1}{2}(a+b)\right]=0\end{array}\text{\hspace{1em}(1)}$$ Also, $I{B}^{\prime }=I{C}^{\prime }$ if and only if $$\begin{array}{c}(b-c)\left[a-\frac{1}{2}(b+c)\right]=0\end{array}\text{\hspace{1em}(2)}$$ Relations (1) and (2) hold if and only if $a=b=c$.