Problem 4. Let $\triangle A^{\prime} B^{\prime} C$ be the image of $\triangle A B C$ under a rotation with center $C$. Denote by $M, E$ and $F$ the midpoints of the segments $B A^{\prime}, A C$ and $B^{\prime} C$, respectively. If $A C \neq B C$ and $E M=F M$, find $\Varangle E M F$.

Ivailo Kortezov
4. Set $B C=B^{\prime} C=a, A C=A^{\prime} C=b$, $A B=A^{\prime} B^{\prime}=c$ and $\Varangle C=\gamma$. Since the triangles $A A^{\prime} C$ and $B B^{\prime} C$ are similar we have

$$
\frac{A A^{\prime}}{A C}=\frac{B B^{\prime}}{B C}=k
$$

Therefore $A A^{\prime}=k b$ and $B B^{\prime}=k a$.

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-159.jpg?height=407&width=527&top_left_y=1050&top_left_x=1138)

Using the median formula several times we get

$$
\begin{aligned}
& 4 E M^{2}=k^{2} b^{2}+c^{2}+b^{2}+a^{2}-A^{\prime} B^{2}-b^{2} \\
& 4 F M^{2}=k^{2} a^{2}+c^{2}+b^{2}+a^{2}-A^{\prime} B^{2}-a^{2}
\end{aligned}
$$

Hence the condition $E M=F M$ is equivalent to $\left(k^{2}-1\right)\left(a^{2}-b^{2}\right)=0$, whence $k=1$ (since $a \neq b$ ). This is true exactly when $\triangle A A^{\prime} C$ is equilateral, i.e. for $\pm 60^{\circ}$ rotation. We consider here only the case of a positively oriented $\triangle A B C$ and rotation through $60^{\circ}$; the remaining cases are analogous. We have from $\triangle C B A^{\prime}$ that

$$
A^{\prime} B^{2}=a^{2}+b^{2}-2 a b \cos \left(60^{\circ}-\gamma\right)
$$

and we obtain from $\triangle A B^{\prime} C$ that

$$
4 E F^{2}=B^{\prime} A^{2}=a^{2}+b^{2}-2 a b \cos \left(60^{\circ}+\gamma\right)
$$

We shall prove that $E F=E M$. It follows from the above expressions for $E F$ and $E M$ that this is equivalent to the identity

$$
c^{2}+2 a b \cos \left(60^{\circ}-\gamma\right)=a^{2}+b^{2}-2 a b \cos \left(60^{\circ}+\gamma\right)
$$

Since $a^{2}+b^{2}-c^{2}=2 a b \cos \gamma$ we rewrite the above identity as

$$
2 a b \cos \left(60^{\circ}-\gamma\right)+2 a b \cos \left(60^{\circ}+\gamma\right)=2 a b \cos \gamma
$$

This is equivalent to $2 \cos 60^{\circ} \cos \gamma=\cos \gamma$, which is true. Therefore $E F=E M$ and $\Varangle E M F=60^{\circ}$.