Track / Stage 6 / 241 of 400 #1241 of 2000

Problem 1241

National olympiad, first round

Geometry Difficulty 6.3 Prove it

Prove that - if points $A, B, C$ lie on a circle with center $O$ , and moreover, $AB$ is a side of a regular hexagon inscribed in the circle and $BD$ is equal to the side $(AC)$ of a square inscribed in the circle - then $AD$ , which lies on the extension of $OA$ , is nothing other than the side length of a decagon inscribed in the circle.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Applying Carnot's theorem to triangle $ABD$ , I obtain the following equation:

$B D^{2}=A B^{2}+A D^{2}-2 A B \times A D \cos B A D$

But $A B=r, B D=r \sqrt{2}$ and $B A D=120^{\circ}, \mathrm{s} \quad \cos 120^{\circ}=-\frac{1}{2}$ , so

$2 r^{2}=r^{2}+A D^{2}+r A D$

$A D^{2}+r A D-r^{2}=0$

from which:

$A D=-\frac{r}{2} \pm \sqrt{\frac{r_{2}}{4}+r^{2}}=\frac{r}{2}(\sqrt{5}-1)$

which expression is indeed the side of a regular decagon.

(Béla Grünhut, Real Gymnasium VII. class, Pécsett.)

The problem was also solved by: Bernát Friedmann, S.-A.-Ujhely; János Galter, Sz.-Udvarhely; Miksa Mayer, Budapest; Aladár Visnya and Lipót Weisz, Pécsett.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.