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Problem 1500

National Olympiad, first round

Geometry Difficulty 6.0 Prove it Bulgarian Mathematical Competitions · Bulgaria

The excircle to the side $AB$ of a triangle $ABC$ is tangent to the circle with diameter $BC$ . Find $\text{ACB}$ if the lengths of the sides $BC$ , $CA$ and $AB$ form (in this order) an arithmetic progression.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

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

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
$p=\frac{3b}{2},\quad p-a=\frac{b}{2}+x,\quad p-b=\frac{b}{2},\quad p-c=\frac{b}{2}-x$
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 $a=\frac{3b}{4}$ , $c=\frac{5b}{4}$ and then $a^{2}+b^{2}=c^{2}$ , i.e. $\text{ACB=90}$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.