AoPS - Problem

Source: Baltic Way 2000 Problem 1

Tags: geometry, parallelogram, trigonometry, similar triangles, geometry unsolved

April

13.02.2009 03:45

Let $ K$ be a point inside the triangle $ ABC$. Let $ M$ and $ N$ be points such that $ M$ and $ K$ are on opposite sides of the line $ AB$, and $ N$ and $ K$ are on opposite sides of the line $ BC$. Assume that $ \angle MAB = \angle MBA = \angle NBC = \angle NCB = \angle KAC = \angle KCA$. Show that $ MBNK$ is a parallelogram.

The problem is very easy to solve with a little trig. First, notation. Denote$ AB,BC,CA$ by $ c,a,b$ respectively. Denote $ AK=r$. As the triangles erected on each side of $ ABC$ are all similar, we easy get $ AM=\frac{cr}{b}$ and $ BN=\frac{ar}{b}$. We will prove that $ MK=BN$ and $ MB=NK$, using the cosine law. $ MK^2=(\frac{cr}{b})^2+r^2-2\frac{cr^2}{b}\cos\hat{A}$ $ MK^2=r^2(\frac{c^2}{b^2}+1-\frac{2c}{b}\cos\hat{A})$ Now to prove that $ MK=BN$ we only need to verify that $ \frac{c^2}{b^2}+1-\frac{2c}{b}\cos\hat{A}=\frac{a^2}{b^2}$ Clearing the denominators yield the cosine law for $ a$. Similarly we can prove that $ MB=NK$ so we are done. Note that we don't need $ K$ to be in the interior of $ ABC$.