Difference between revisions of "1969 Canadian MO Problems/Problem 10"
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== Problem == | == Problem == | ||
− | Let <math> | + | 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 == | ||
− | Let <math> | + | 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> |
− | Now let's consider when <math> | + | 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. |
− | - | + | {{Old CanadaMO box|num-b=9|after=Last question|year=1969}} |
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Latest revision as of 13:13, 30 December 2015
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.
1969 Canadian MO (Problems) | ||
Preceded by Problem 9 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Last question |