Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 6.6 National olympiad Prove it North Macedonia

Let $A' \in (BC)$ , $B' \in (AC)$ , $C' \in (AB)$ be the points of tangency of the excribed circles of the triangle $ABC$ with the sides of $ABC$ . Let $R'$ the circumradius of $A'B'C'$ . Show that
$R' = \frac{1}{2r} \sqrt{2R(2R - h_a)(2R - h_b)(2R - h_c)},$
where, as usual, $R$ is the circumradius of $ABC$ , $r$ is the inradius of $ABC$ , and $h_a, h_b, h_c$ are the lengths of the altitudes of $ABC$ .

Solution

The triangle $A'B'C'$ is the pedal triangle of the symmetrical point of the incenter $I$ of $ABC$ with respect to the circumcenter of $ABC$ . So, the relation between the areas $S = [ABC]$ and $S' = [A'B'C']$ is given by
$S' = S \cdot \frac{R^2 - \overline{OI}^2}{4R^2} = \frac{r}{2R}.$
In the triangle $A'B'C'$ , as $AB' = s - c = r \operatorname{ctg} \frac{C}{2}$ , $AC' = s - b = r \cdot \operatorname{ctg} \frac{B}{2}$ and
$\begin{aligned} \overline{B'C'}^2 &= a'^2 = r^2 \left[ \left( \operatorname{ctg} \frac{B}{2} + \operatorname{ctg} \frac{A}{2} \right)^2 - 4 \operatorname{ctg} \frac{A}{2} \operatorname{ctg} \frac{B}{2} \operatorname{ctg} \frac{C}{2} \right] = \\ &= r^2 \frac{\cos^2 \frac{A}{2}}{\sin^2 \frac{B}{2} \sin^2 \frac{C}{2}} \left( 1 - \sin B \sin C \right) = a^2 \left( 1 - \frac{bc}{4R^2} \right) \end{aligned}$
but as $bc = 2Rh_a$ , we get
$a'^2 = \frac{a^2}{2R} (2R - h_a)$

and as we have
$S'^2 = \frac{a'^2 b'^2 c'^2}{16R'^2} = \frac{a^2 b^2 c^2}{16R^2} \frac{1}{8RR'^2} (2R - h_a)(2R - h_b)(2R - h_c),$
the relation searched holds.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.