Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(box)
(→‎Solution)
 
Line 5: Line 5:
 
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>
 
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>\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.
+
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}}
 
{{Old CanadaMO box|num-b=9|after=Last question|year=1969}}

Latest revision as of 13:13, 30 December 2015

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
Last question


Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_10&oldid=74105"