Balkan Mathematical Olympiad Shortlist

Let the lines through $B$ and $C$ parallel to $XY$ meet $AM$ at $P$ and $Q$, respectively. Since $BM=MC$, we have $PM=MQ$ and $$\frac{BX}{XA}+\frac{CY}{YA}=\frac{PT}{TA}+\frac{QT}{TA}=\frac{PT+QT}{TA}=\frac{2MT}{TA}=2.$$ Let $I_a$, $I_b$, and $I_c$ be the excenters of $△ABC$ opposite $A$, $B$, and $C$, respectively. Since the figures $B{H}_{a}C{I}_a$, $C{H}_{b}A{I}_b$, and $A{H}_{c}B{I}_c$ are parallelograms, we have $$\begin{array}{rl}2& =\frac{BX}{XA}+\frac{CY}{YA}=\frac{B{H}_a}{H_{b}A}+\frac{C{H}_a}{H_{c}A}=\frac{I_{a}C}{C{I}_b}+\frac{I_{a}B}{B{I}_c}\\ & =\frac{r_a}{r_b}+\frac{r_a}{r_c}=\frac{S/(p-a)}{S/(p-b)}+\frac{S/(p-a)}{S/(p-c)}=\frac{p-b}{p-a}+\frac{p-c}{p-a},\end{array}$$ which is equivalent to $2a=b+c$.

On the other hand, $H_{a}T\perp BC\iff B{H}_a^2-H_a{C}^2=B{T}^2-T{C}^2$. Let $U$ be the tangency point of the ex-circle opposite $A$ and $BC$. Then $$B{H}_a^2-H_a{C}^2=C{I}_a^2-I_a{B}^2=C{U}_a^2-U{B}^2=(p-b{)}^2-(p-c{)}^2=a(c-b).$$ Also, since $BT$ and $CT$ are medians in $△ABM$ and $△ACM$, respectively, we have $$\begin{array}{rl}B{T}^2-T{C}^2& =\frac{1}{2}(B{A}^2+B{M}^2)-\frac{1}{4}A{M}^2-\frac{1}{2}(C{A}^2+C{M}^2)+\frac{1}{4}A{M}^2\\ & =\frac{1}{2}(B{A}^2-C{A}^2)=\frac{1}{2}(c-b)(c+b).\end{array}$$ If $b=c$, then $H_{a}T\perp BC$ by symmetry. If $b\ne c$, then the above implies that $$H_{a}T\perp BC\iff B{H}_a^2-H_a{C}^2=B{T}^2-T{C}^2\iff a=\frac{1}{2}(b+c),$$ as needed.