AoPSWiki
Art of Problem Solving holds many free classes called Math Jams.
Click here for transcripts to past Math Jams.

1969 Canadian MO Problems/Problem 10

From AoPSWiki

Problem

Let ABC be the right-angled isosceles triangle whose equal sides have length 1. P is a point on the hypotenuse, and the feet of the perpendiculars from P to the other sides are Q and R. Consider the areas of the triangles APQ and PBR, and the area of the rectangle QCRP. Prove that regardless of how P is chosen, the largest of these three areas is at least 2/9.

Solution

Let AQ=x. Because triangles APQ and BPR both contain a right angle and a 45^\circ angle, they are isosceles right triangles. Hence, PQ=RC=x and QC=PR=BR=1-x.

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

1969 Canadian MO (Problems)
Preceded by
Problem 9
1 2 3 4 5 6 7 8 9 10 Followed by
Last question


Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us