In triangle $A B C$, a circle of radius $R$ is inscribed, touching side $A C$ at point $D$, side $A B$ at point $E$, and side $B C$ at point $F$. It is known that $A D=R$,

$D C=a$. Find the area of triangle $B E F$.

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Triangle $ABC$ is a right triangle.

## Solution

Let $O$ be the center of the circle. Since $AE = AD = OD = OE = R$ and $OD \perp AD$, then $AEO D$ is a square. Therefore, $\angle A = 90^\circ$.

Let $BE = BF = x$. We can reason in different ways.

First method. $2R(a + R + x) = 2S_{ABC} = (a + R)(x + R)$, from which

$x = \frac{R(a + R)}{a - R}, \quad x^2 + R^2 = \frac{2R^2(a^2 + R^2)}{(a - R)^2}$.

Let $K$ be the midpoint of $EF$. Triangles $BKF$ and $BFO$ are similar with a ratio of $\frac{BF}{OF} = \frac{x}{\sqrt{x^2 + R^2}}$, so

$S_{BEF} = 2S_{BKF} = \frac{x^3 R}{x^2 + R^2} = \frac{R^2(R + a)^3}{2(a - R)(a^2 + R^2)}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_65daf5394c369f17193bg-19.jpg?height=371&width=506&top_left_y=1781&top_left_x=776)

Second method. $\operatorname{tg} \angle C / 2 = OD / DC = R / a$. Therefore,

$\operatorname{tg} \angle C = \frac{2 \frac{R}{a}}{1 + \left(\frac{R}{a}\right)^2} = \frac{2aR}{a^2 + R^2}, \quad \sin \angle B = \cos \angle C = \frac{1 - \left(\frac{R}{a}\right)^2}{1 + \left(\frac{R}{a}\right)^2} = \frac{a^2 - R^2}{a^2 + R^2}$,

$x = (a + R) \operatorname{tg} \angle C - R = \frac{R(a + R)}{a - R}$

$S_{BEF} = \frac{x^2 \sin \angle B}{2} = \frac{R^2(a + R)^2(a^2 - R^2)}{2(a - R)^2(a^2 + R^2)} = \frac{R^2(a + R)^3}{2(a - R)(a^2 + R^2)}$

## Answer

$\frac{R^2(R + a)^3}{2(a - R)(a^2 + R^2)}$.

Given triangle $ABC$. On side $BC$, point $P$ is taken, and on side $AC$, point $M$ is taken, such that $\angle APB = \angle BMA = 45^\circ$. Segments $AP$ and $BM$ intersect at point $O$. It is known that the areas of triangles $BOP$ and $AOM$ are equal, $BC = 1$, and $BO = \frac{\sqrt{2}}{2}$. Find the area of triangle $ABC$.

## Hint

Prove that triangle $ABC$ is isosceles.

## Solution

Triangles $BOP$ and $AOM$ are similar by two angles. Since their areas are equal, they are congruent. Therefore, $\angle MAO = \angle PBO$, $AO = OB$, $\angle OAB = \angle OBA$, $\angle A = \angle B$.

Thus, triangle $ABC$ is isosceles ($AC = CB$). Therefore, $MP \parallel AB$.

Let $PO = MO = x$. Then $\frac{MC}{AC} = \frac{MP}{AB} = \frac{MO}{OB} = \frac{x}{\frac{\sqrt{2}}{2}} = x \sqrt{2}$. Therefore, $MC = AC \cdot x \sqrt{2} = x \sqrt{2}$.

By the cosine rule, $BC^2 = MC^2 + MB^2 - 2MC \cdot MB \cos 135^\circ$,

$1 = 2x^2 + \left(x + \frac{\sqrt{2}}{2}\right)^2 + 2x\left(x + \frac{\sqrt{2}}{2}\right), \quad 10x^2 + 4x\sqrt{2} - 1 = 0$. From this,

$x = \frac{\sqrt{2}}{10}, \quad MB = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{10} = \frac{3\sqrt{2}}{5}, \quad MC = x \sqrt{2} = \frac{1}{5}$

Therefore, $S_{ABC} = \frac{5}{4} S_{AMB} = \frac{5}{4} \cdot \frac{1}{2} MB \cdot MA \sin 45^\circ = \frac{5}{4} \cdot \frac{1}{2} \cdot \frac{3\sqrt{2}}{5} \cdot \frac{4}{5} \cdot \frac{\sqrt{2}}{2} = \frac{3}{10}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_65daf5394c369f17193bg-20.jpg?height=389&width=504&top_left_y=1435&top_left_x=777)

## Answer

$0.3$.