Find the angles of this triangle.

Question

Find the angles of a triangle if the distance between the centre of the circumcircle and the orthocenter is $\frac{1}{2}$ the length of the largest side and equals the smallest side.

Answer 1

It easy to find a solution by equilater and right triangle properties

but maybe it should be proved that it is unique.

Answer 2

In the standard notation let $a\leq b\leq c$.

Hence, $c=2a$ and since $a+b>c=2a$, we obtain $b>a$.

Thus, $$OH^2=R^2(1-8\cos\alpha\cos\beta\cos\gamma)=\frac{a^2b^2c^2}{16S^2}\left(1-\frac{\prod\limits_{cyc}(a^2+b^2-c^2)}{a^2b^2c^2}\right)=$$ $$=\frac{\sum\limits_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)}{\sum\limits_{cyc}(2a^2b^2-a^4)}=\frac{(3a^2-b^2)(18a^2+ab-b^2)}{(b^2-a^2)(9a^2-b^2)}.$$

If $3a^2=b^2$ then $c^2=a^2+b^2$ and we got a triangle with measured angles $90^{\circ},$ $30^{\circ}$ and $60^{\circ}$ and since $$18a^2+ab-b^2=14a^2+c^2+ab-b^2>0,$$ we are done!