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

 
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== Solution ==
 
== Solution ==
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Let <math>\displaystyle AQ=x.</math> Because triangles <math>\displaystyle APQ</math> and <math>\displaystyle BPR</math> both contain a right and a <math>\displaystyle 45^\circ</math> angle, they are isosceles-right. Hence, <math>\displaystyle PQ=RC=x</math> and <math>\displaystyle QC=PR=BR=1-x.</math>
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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 $\displaystyle ABC$ be the right-angled isosceles triangle whose equal sides have length 1. $\displaystyle P$ is a point on the hypotenuse, and the feet of the perpendiculars from $\displaystyle P$ to the other sides are $\displaystyle Q$ and $\displaystyle R$. Consider the areas of the triangles $\displaystyle APQ$ and $\displaystyle PBR$, and the area of the rectangle $\displaystyle QCRP$. Prove that regardless of how $\displaystyle P$ is chosen, the largest of these three areas is at least $\displaystyle 2/9$.

Solution

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

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

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