In triangle $ABC$, the bisector of the angle at vertex $A$ intersects side $BC$ at point $M$, and the bisector of the angle at vertex $B$ intersects side $AC$ at point $P$. The bisectors $AM$ and $BP$ intersect at point $O$. It is known that triangle $BOM$ is similar to triangle $AOP$, $BO=(1+\sqrt{3})OP$, and $BC=1$. Find the area of triangle $ABC$.
Prove that triangle $ABC$ is isosceles.

## Solution

From the similarity of triangles $BOM$ and $AOP$, it follows that the angles $MBP$ and $MAP$ are equal. Therefore, $\angle B = 2 \angle MBP = 2 \angle MAP = \angle A$. Hence, triangle $ABC$ is isosceles. This means that $MP \parallel AB$.

Let $CK$ be the height of triangle $ABC$. By the property of the angle bisector, $AB: AC = BM: MC$. Since $\frac{CM}{CB} = \frac{PM}{AB} = \frac{OP}{OB} = \frac{1}{1+\sqrt{3}}$, then $\frac{BM}{MC} = \sqrt{3}$. Therefore, $AB = AC \sqrt{3} = BC \sqrt{3} = \sqrt{3}$, $CK^2 = AC^2 - AK^2 = 1/2$, $S_{ABC} = \frac{\sqrt{3}}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-10.jpg?height=302&width=1058&top_left_y=2536&top_left_x=500)

## Answer

\section*{[ Properties and criteria of isosceles triangles. ]  Problem 53236 Topics:

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-11.jpg?height=48&width=306&top_left_y=279&top_left_x=813)  Difficulty: $3+$  [ The ratio in which the angle bisector divides the side ]  Grades: 8,9  [}

In triangle $ABC$, the height $BH$ divides side $AC$ in the ratio $AH: HC = 4$, and angle $HBC$ is half of angle $A$. The angle bisector $AE$ of angle $A$ intersects $BH$ at point $M$. Find the ratio of the area of triangle $ABM$ to the area of the circle circumscribed around this triangle.

## Hint

Prove that triangle $ABM$ is isosceles and find $\cos \angle A$.

## Solution

Let $\angle HBC = \alpha$. Then $\angle BAM = \angle CAM = \alpha$. Since $\angle BME = \angle AMH$, then $\angle BEM = \angle AHM = 90^\circ$, which means the angle bisector $AE$ of triangle $ABC$ is also its height. Therefore, triangle $BAC$ is isosceles.

Let $CH = x$. Then $AH = 4x$, $AB = AC = 5x$, $\cos 2\alpha = \cos \angle A = AH / AB = 4/5$, $\cos \alpha = \frac{3}{\sqrt{10}}$.

Since $\angle BMA = 90^\circ + \alpha$, the radius $R$ of the circumscribed circle of triangle $ABM$ is

$\frac{AB}{2 \sin \angle BMA} = \frac{5x}{2 \cos \alpha} = \frac{5x \sqrt{10}}{6}$.

From triangle $ABH$, we find that $BH = 3x$. Therefore, $S_{ABH} = 1/2 \cdot 4x \cdot 3x = 6x^2$.

By the property of the angle bisector of a triangle, $\frac{BM}{MH} = 5/4$. Therefore, $S_{ABM} = 5/9 S_{ABH} = 10/3 x^2$, and the desired ratio is $\frac{\frac{10x^2}{3}}{\frac{25 \cdot 10x}{36}} = \frac{12}{25 \pi}$

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-11.jpg?height=366&width=594&top_left_y=1561&top_left_x=732)

## Answer

$\frac{12}{25 \pi}$