Track / Stage 5 / 50 of 400 #650 of 1964

Problem 650

AIME late

Geometry Difficulty 5.1 Prove it Indian National Mathematical Olympiad · India · 2011

Let $ABCDEF$ be a convex hexagon in which the diagonals $AD$ , $BE$ , $CF$ are concurrent at $O$ . Suppose the area of triangle $OAF$ is the geometric mean of those of $OAB$ and $OEF$ ; and the area of triangle $OBC$ is the geometric mean of those of $OAB$ and $OCD$ . Prove that the area of triangle $OED$ is the geometric mean of those of $OCD$ and $OEF$ .

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

Let $OA = a$ , $OB = b$ , $OC = c$ , $OD = d$ , $OE = e$ , $OF = f$ , $[OAB] = x$ , $[OCD] = y$ , $[OEF] = z$ , $[ODE] = u$ , $[OFA] = v$ and $[OBC] = w$ . We are given that $v^2 = zx$ , $w^2 = xy$ and we have to prove that $u^2 = yz$ . Since $\angle AOB = \angle DOE$ , we have

$\frac{u}{x} = \frac{\frac{1}{2} de \sin \angle DOE}{\frac{1}{2} ab \sin \angle AOB} = \frac{de}{ab}.$

$\frac{v}{y} = \frac{fa}{cd}, \quad \frac{w}{z} = \frac{bc}{ef}.$

Multiplying these three equalities, we get $uvw = xyz$ . Hence

$x^2 y^2 z^2 = u^2 v^2 w^2 = u^2 (zx)(xy).$

This gives $u^2 = yz$ , as desired.

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