Area Inequality in Trapezoid


Miguel Ochoa Sanchez has posted an engaging problem at the CutTheKnotMath facebook page and an absolutely beautiful solution.

Area Inequality in square - source


In square $ABCD\;$ $P\in AB,\;R\in CD.\;$ A quadrilateral $PQRS\;$ is the intersection of triangles $CDP\;$ and $ABR.$

Area Inequality in square - problem

Prove that $[PQRS]\le\frac{1}{4}[ABCD],$

where $[X]\;$ denotes the area of shape $X.$


The proof is based on a simple but unexpected lemma which makes the solution almost immediate.

Area Inequality in square - solution

From Lemma in $RPBC,\;$ $\displaystyle X\le\frac{M+N}{2};\;$ in $PRDA,\;$ $\displaystyle Y\le\frac{V+W}{2}.\;$ Adding up gives

$\displaystyle X+Y\le\frac{M+N+V+W}{2}.$

Now adding $X+Y$ to both sides we obtain

$\displaystyle 2[PQRS]=2(X+Y)\le\frac{M+N+V+W+2X+2Y}{2}=\frac{1}{2}[ABCD].$

And the conclusion follows.


It is worth noting that the Miguel's proof works equally well if $ABCD$ is a trapezoid, with $AB\parallel CD.$ Thus $ABCD\;$ being a square is a red herring.

In addition, Marcello Tarquini has found an exact condition to insure that $[PQRS]\;$ attains its maximum.


What Is Red Herring

|Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny


Search by google: