# Difference between revisions of "1969 Canadian MO Problems/Problem 10"

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== Solution == | == Solution == | ||

− | Let <math>\displaystyle AQ=x.</math> Because | + | Let <math>\displaystyle AQ=x.</math> Because [[triangle]]s <math>\displaystyle APQ</math> and <math>\displaystyle BPR</math> both contain a [[right angle]] and a <math>\displaystyle 45^\circ</math> angle, they are [[isosceles]] [[right triangle]]s. Hence, <math>\displaystyle PQ=RC=x</math> and <math>\displaystyle QC=PR=BR=1-x.</math> |

Now let's consider when <math>\displaystyle \frac13 <x<\frac23,</math> or else one of triangles <math>\displaystyle APQ</math> and <math>\displaystyle PBR</math> will automatically have area greater than <math>\displaystyle \frac29.</math> In this case, <math>\displaystyle [QCRP]>[ABC]-[APQ]-[PBR]>\frac29.</math> Therefore, one of these three figures will always have area greater than <math>\displaystyle \frac29,</math> regardless of where <math>\displaystyle P</math> is chosen. | Now let's consider when <math>\displaystyle \frac13 <x<\frac23,</math> or else one of triangles <math>\displaystyle APQ</math> and <math>\displaystyle PBR</math> will automatically have area greater than <math>\displaystyle \frac29.</math> In this case, <math>\displaystyle [QCRP]>[ABC]-[APQ]-[PBR]>\frac29.</math> Therefore, one of these three figures will always have area greater than <math>\displaystyle \frac29,</math> regardless of where <math>\displaystyle P</math> is chosen. |

## Revision as of 13:55, 28 July 2006

## Problem

Let be the right-angled isosceles triangle whose equal sides have length 1. is a point on the hypotenuse, and the feet of the perpendiculars from to the other sides are and . Consider the areas of the triangles and , and the area of the rectangle . Prove that regardless of how is chosen, the largest of these three areas is at least .

## Solution

Let Because triangles and both contain a right angle and a angle, they are isosceles right triangles. Hence, and

Now let's consider when or else one of triangles and will automatically have area greater than In this case, Therefore, one of these three figures will always have area greater than regardless of where is chosen.