MathDB

1/AM + 1/ AN <= 2/ AB

Source: 2008 Cuba 2.6

August 27, 2024

geometrygeometric inequality

Problem Statement

We have an isosceles triangle

ABC

with base

BC

. Through vertex

A

draw a line

r

parallel to

BC

. The points

P, Q

are located on the perpendicular bisectors of

AB

and

AC

respectively, such that

PQ\perp BC

. They are points

M

and

N

on the line

r

such that

\angle APM = \angle AQN = 90^o

. Prove that

\frac{1}{AM} + \frac{1}{AN}\le \frac{2}{ AB}

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