10.6. Let $ABC$ be an acute-angled triangle, where $m(\angle BAC) = 60^{\circ}$, and $BC = a$. Let $M$ be the midpoint of side $BC$, and $[BB_1]$ and $[CC_1]$ be the altitudes of the triangle. Calculate the sum of the distances from the orthocenter of triangle $ABC$ to the sides of triangle $MB_1C_1$.
## Solution.

1. In the right-angled triangle $B C B_{1}$, the segment $M B_{1}$ is a median, so $M B_{1}=\frac{B C}{2}=\frac{a}{2}$. Similarly, $M C_{1}=\frac{a}{2}$.
2. Consider triangles $A B_{1} C_{1}$ and $A B C$. We will show that they are similar.

These triangles have a common angle at $A$.

We will show that the sides forming angle $A$ are proportional.

$\frac{A C_{1}}{A C}=\cos 60^{\circ}=\frac{1}{2}=\frac{A B_{1}}{A B}$, thus $\triangle A B_{1} C_{1} \sim \triangle A B C$, then

![](https://cdn.mathpix.com/cropped/2024_06_06_f4f7cfcf3dfd8b0307cag-1.jpg?height=545&width=689&top_left_y=1529&top_left_x=1206)

$\frac{B_{1} C_{1}}{B C}=\frac{A B_{1}}{A B}=\frac{1}{2} \Rightarrow B_{1} C_{1}=\frac{a}{2}$. We observe that triangle $M B_{1} C_{1}$ is equilateral with side length $\frac{a}{2}$.

3. We calculate the area of triangle $M B_{1} C_{1}$:

$$
\mathcal{A}=\left(\frac{a}{2}\right)^{2} \cdot \frac{\sqrt{3}}{4}=\frac{a^{2} \sqrt{3}}{16} \text {. We can also write }
$$

$$
\begin{aligned}
& \mathcal{A}=\mathcal{A}_{M H B_{1}}+\mathcal{A}_{M H C_{1}}+\mathcal{A}_{B_{1} H C_{1}}=\frac{a}{4} \cdot d\left(H ; M B_{1}\right)+\frac{a}{4} \cdot d\left(H ; M C_{1}\right)+\frac{a}{4} \cdot d\left(H ; B_{1} C_{1}\right) \Leftrightarrow \\
& \Leftrightarrow \frac{a^{2} \sqrt{3}}{16}=\frac{a}{4} \cdot\left[d\left(H ; M B_{1}\right)+d\left(H ; M C_{1}\right)+d\left(H ; B_{1} C_{1}\right)\right] \Leftrightarrow d\left(H ; M B_{1}\right)+d\left(H ; M C_{1}\right)+d\left(H ; B_{1} C_{1}\right)=\frac{a \sqrt{3}}{4} .
\end{aligned}
$$

Answer: $\frac{a \sqrt{3}}{4}$