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

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== Problem ==
 
== Problem ==
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Let <math>\displaystyle ABC</math> be the right-angled isosceles triangle whose equal sides have length 1. <math>\displaystyle P</math> is a point on the [[hypotenuse]], and the feet of the [[perpendicular]]s from <math>\displaystyle P</math> to the other sides are <math>\displaystyle Q</math> and <math>\displaystyle R</math>. Consider the areas of the triangles <math>\displaystyle APQ</math> and <math>\displaystyle PBR</math>, and the area of the [[rectangle]] <math>\displaystyle QCRP</math>. Prove that regardless of how <math>\displaystyle P</math> is chosen, the largest of these three areas is at least <math>\displaystyle 2/9</math>.
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Let <math>ABC</math> be the right-angled isosceles triangle whose equal sides have length 1. <math>P</math> is a point on the [[hypotenuse]], and the feet of the [[perpendicular]]s from <math>P</math> to the other sides are <math>Q</math> and <math>R</math>. Consider the areas of the triangles <math>APQ</math> and <math>PBR</math>, and the area of the [[rectangle]] <math>QCRP</math>. Prove that regardless of how <math>P</math> is chosen, the largest of these three areas is at least <math>2/9</math>.
  
 
== Solution ==
 
== Solution ==
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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 triangle | isosceles]] [[right triangle]]s. Hence, <math>\displaystyle PQ=RC=x</math> and <math>\displaystyle QC=PR=BR=1-x.</math>
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Let <math>AQ=x.</math> Because [[triangle]]s <math>APQ</math> and <math>BPR</math> both contain a [[right angle]] and a <math>45^\circ</math> angle, they are [[isosceles triangle | isosceles]] [[right triangle]]s. Hence, <math>PQ=RC=x</math> and <math>QC=PR=BR=1-x.</math>
  
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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.
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Now let's consider when <math>\frac13 <x<\frac23,</math> or else one of triangles <math>APQ</math> and <math>PBR</math> will automatically have area greater than <math>\frac29.</math> In this case, <math>[QCRP]>[ABC]-[APQ]-[PBR]>\frac29.</math> Therefore, one of these three figures will always have area greater than <math>\frac29,</math> regardless of where <math>P</math> is chosen.
  
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{{Old CanadaMO box|num-b=9|after=Last question|year=1969}}
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* [[1969 Canadian MO Problems/Problem 9|Previous Problem]]
 
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* [[1969 Canadian MO Problems|Back to Exam]]
 

Revision as of 21:43, 17 November 2007

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 • Followed by
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