In triangle $ABC$, the ratio between the sides is given by $\frac{AC - AB}{BC + AB} = \frac{AB - BC}{AC + AB}$. Find the radius of the circumscribed circle if the distance from its center to the point of intersection of the medians is $d$, and the length of side $AB$ is $c$.
Let $A C=b, A B=c, B C=a$. From the condition of the problem, it follows that $a^{2}+b^{2}=2 c^{2}$. Let $O$ be the center of the circle circumscribed around triangle $A B C$, and $M$ be the point of intersection of the medians. Compute the scalar square of the vector

$$
\bar{O} \bar{M}=\frac{1}{3}(\bar{O} \bar{A}+\bar{O} \bar{B}+\bar{O} \bar{C})
$$

## Solution

Denote $A C=b, A B=c, B C=a$.

From the condition of the problem, it follows that

$$
\frac{A C-A B}{B C+A B}=\frac{A B-B C}{A C+A B}=\frac{b-c}{a+c}=\frac{c-a}{b+c} \Rightarrow b^{2}-c^{2}=c^{2}-a^{2} \Rightarrow a^{2}+b^{2}=2 c^{2}
$$

Let $O$ be the center of the circle with radius $R$ circumscribed around triangle $A B C$, and $M$ be the point of intersection of the medians. Then

$$
\bar{O} \bar{M}=\frac{1}{3}(\bar{O} \bar{A}+\bar{O} \bar{B}+\bar{O} \bar{C})
$$

Denote

$$
\angle A O B=2 \gamma, \angle A O C=2 \beta, \angle B O C=2 \alpha
$$

(half of each of these angles is equal to the corresponding angle of triangle $A B C$ or complements it to $180^{\circ}$). Then

$$
\begin{gathered}
d^{2}=\bar{O} \bar{M}^{2}=\frac{1}{9}\left(\bar{O} \bar{A}^{2}+\bar{O} \bar{B}^{2}+\bar{O} \bar{C}^{2}+2 \bar{O} \bar{A} \cdot \bar{O} \bar{B}+2 \bar{O} \bar{A} \cdot \bar{O} \bar{C}+2 \bar{O} \bar{B} \cdot \bar{O} \bar{C}\right)= \\
=\frac{1}{9}\left(R^{2}+R^{2}+R^{2}+2 R \cdot R \cdot \cos 2 \gamma+2 R \cdot R \cdot \cos 2 \beta+2 R \cdot R \cdot \cos 2 \alpha\right)= \\
=\frac{1}{9} R^{2}(3+2(\cos 2 \gamma+\cos 2 \beta+\cos 2 \alpha))=\frac{1}{9} R^{2}\left(3+2\left(1-2 \sin ^{2} \gamma+1-2 \sin ^{2} \beta+1-2 \sin ^{2} \alpha\right)\right)= \\
=\frac{1}{9} R^{2}\left(9-4\left(\sin ^{2} \gamma+\sin ^{2} \beta+\sin ^{2} \alpha\right)\right)=\frac{1}{9} R^{2}\left(9-4\left(\left(\frac{c}{2 R}\right)^{2}+\left(\frac{b}{2 R}\right)^{2}+\left(\frac{a}{2 R}\right)^{2}\right)\right)= \\
=R^{2}-\frac{1}{9} \cdot 4 \cdot \frac{1}{4}\left(a^{2}+b^{2}+c^{2}\right)=R^{2}-\frac{1}{9} \cdot 3 c^{2}=R^{2}-\frac{c^{2}}{3}
\end{gathered}
$$

Therefore, $R^{2}=d^{2}+\frac{c^{2}}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-52.jpg?height=746&width=774&top_left_y=-1&top_left_x=642)

## Answer

$\sqrt{d^{2}-\frac{c^{2}}{3}}$

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