Problem 12.2. The excircle to the side $A B$ of a triangle $A B C$ is tangent to the circle with diameter $B C$. Find $\Varangle A C B$ if the lengths of the sides $B C, C A$ and $A B$ form (in this order) an arithmetic progression.

Oleg Mushkarov
12.2. Let $M$ be the midpoint of $B C$, let $I$ be the center of the excircle of $\triangle A B C$ tangent to $A B$ and let $T$ be its tangent point to the line $B C$. In the standard notation for $\triangle A B C$ we have $I M=\frac{a}{2}+r_{c}, I T=r_{c}$ and $M T=p-\frac{a}{2}$ since $B T=\stackrel{p}{p}-a$. It follows from the right $\triangle M I T$ that

$$
\left(\frac{a}{2}+r_{c}\right)^{2}=r_{c}^{2}+\left(p-\frac{a}{2}\right)^{2}
$$

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-143.jpg?height=584&width=598&top_left_y=1844&top_left_x=1114)
that

Then $a r_{c}=p(p-a)$. Since $r_{c}=\frac{S}{p-c}$, we obtain by using Heron's formula

$$
a S=p(p-a)(p-c)=\frac{S^{2}}{p-b}
$$

$$
a(p-b)=S
$$

Since $a, b$ and $c$ form (in this order) an arithmetic progression, we have $a=$ $b-x, c=b+x$ and

$$
p=\frac{3 b}{2}, p-a=\frac{b}{2}+x, p-b=\frac{b}{2}, p-c=\frac{b}{2}-x
$$

Now by (1) and the Heron formula we obtain the equation

$$
(b-x)^{2}=3\left(\frac{b^{2}}{4}-x^{2}\right)
$$

which has a unique solution $x=\frac{b}{4}$. Therefore $a=\frac{3 b}{4}, c=\frac{5 b}{4}$ and then $a^{2}+b^{2}=c^{2}$, i.e. $\Varangle A C B=90^{\circ}$.