In an acute-angled triangle $A B C$, the altitudes intersect at point $H$, and the medians intersect at point $O$. The angle bisector of $\angle A$ passes through the midpoint of segment $O H$. Find the area of triangle $A B C$ if $B C=2$, and the difference between angles $B$ and $C$ is $30^{\circ}$.
The intersection point of the altitudes, the intersection point of the medians, and the center of the circumscribed circle of a triangle lie on one straight line (Euler's line). Next, apply the Law of Sines.

## Solution

It is known that the distance from the intersection point of the altitudes of a triangle to its vertex is twice the distance from the center of the circumscribed circle to the side opposite this vertex.

Let $Q$ be the center of the circle circumscribed around triangle $ABC$, and $M$ be the midpoint of side $BC$. Then $AH = 2QM$. If the line $QH$ intersects the median $AM$ at point $O'$, then from the similarity of triangles $AO'H$ and $MO'Q$, it follows that $AO':O'M = AH:QM = 2:1$, which means that point $O'$ coincides with point $O$, the intersection of the medians of triangle $ABC$. Thus, it is proven that points $H$, $O$, and $Q$ lie on one straight line (Euler's line), and point $O$ is located on the segment $QH$ with $OH = 2OQ$.

If $K$ is the midpoint of $OH$, then $HK = KO = OQ$.

Extend the angle bisector of the triangle, drawn from vertex $A$, to intersect the circumscribed circle of triangle $ABC$ at point $P$. Then $P$ is the midpoint of the arc $BC$ not containing point $A$. Therefore, $PM$ is the perpendicular bisector of side $BC$. Hence, the line $PM$ passes through point $Q$.

Let $R$ be the radius of the circle. Denote $QM = x$. Then $AH = 2x$. From the similarity of triangles $PKQ$ and $AKH$, we find that $R = QP = AH \cdot \frac{QK}{KH} = 2AH = 4x$. Then $QC = R = 4x$.

Applying the Pythagorean theorem to the right triangle $QMC$, we get the equation

$$
QC^2 = QM^2 + CM^2, \text{ or } 16x^2 = x^2 + 1,
$$

from which $x = \frac{1}{\sqrt{15}}$. Then $R = 4x = \frac{4}{\sqrt{15}}$.

Let $\angle BAC = \alpha$, $\angle ABC = \beta$, $\angle ACB = \gamma$. Suppose $\beta - \gamma = 30^\circ$. By the Law of Sines,

$$
\sin \alpha = \frac{BC}{2R} = \frac{2}{2 \cdot \frac{4}{\sqrt{15}}} = \frac{\sqrt{15}}{4}, \quad AC = 2R \sin \beta, \quad AB = 2R \sin \gamma
$$

Therefore,

$$
\begin{gathered}
S_{\Delta ABC} = \frac{1}{2} \cdot AC \cdot AB \cdot \sin \alpha = \frac{1}{2} \cdot 2R \sin \beta \cdot 2R \sin \gamma \cdot \sin \alpha = \\
= 2R^2 \sin \beta \sin \gamma \sin \alpha = 2 \cdot \frac{16}{15} \cdot \sin \alpha \sin \beta \sin \gamma = \\
= 2 \cdot \frac{16}{15} \cdot \sin \alpha \cdot \frac{1}{2}(\cos (\beta - \gamma) - \cos (\beta + \gamma)) = \frac{16}{15} \cdot \frac{\sqrt{15}}{4} \cdot \left(\cos 30^\circ - \cos (180^\circ - \alpha)\right) = \\
= \frac{4}{\sqrt{15}} \left(\frac{\sqrt{3}}{2} + \cos \alpha\right) = \frac{4}{\sqrt{15}} \left(\frac{\sqrt{3}}{2} + \sqrt{1 - \frac{15}{16}}\right) = \frac{4}{\sqrt{15}} \left(\frac{\sqrt{3}}{2} + \frac{1}{4}\right) = \frac{2\sqrt{3} + 1}{\sqrt{15}}
\end{gathered}
$$

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_4bea0374a35e8bb7d318g-39.jpg?height=794&width=811&top_left_y=1701&top_left_x=632)

$\frac{2\sqrt{3} + 1}{\sqrt{15}}$

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