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 "1971 Canadian MO Problems/Problem 8"

(Created page with "== Problem == A regular pentagon is inscribed in a circle of radius <math>r</math>. <math>P</math> is any point inside the pentagon. Perpendiculars are dropped from <math>P</math...")
 
m (circ)
 
(One intermediate revision by one other user not shown)
Line 23: Line 23:
 
<cmath>p_1+p_2+p_3+p_4+p_5=(5r^2/s)\sin 72^{\circ}.</cmath>
 
<cmath>p_1+p_2+p_3+p_4+p_5=(5r^2/s)\sin 72^{\circ}.</cmath>
  
−
We now find <math>s</math> in terms of <math>r</math>. We know that <math>\angle OAB=54^{\cicr}</math>, so <math>[AOB]=(1/2)rs\sin 54^{\circ}</math>. Therefore <math>s=r\left(\sin 72^{\circ}/\sin 54^{\circ}\right)</math>. This shows that
+
We now find <math>s</math> in terms of <math>r</math>. We know that <math>\angle OAB=54^{\circ}</math>, so <math>[AOB]=(1/2)rs\sin 54^{\circ}</math>. Therefore <math>s=r\left(\sin 72^{\circ}/\sin 54^{\circ}\right)</math>. This shows that
  
 
<cmath>p_1+p_2+p_3+p_4+p_5=5r\sin 54^{\circ}.</cmath>
 
<cmath>p_1+p_2+p_3+p_4+p_5=5r\sin 54^{\circ}.</cmath>
Line 30: Line 30:
  
 
== See Also ==
 
== See Also ==
 +
{{Old CanadaMO box|num-b=7|num-a=9|year=1971}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Latest revision as of 21:38, 11 March 2016

Problem

A regular pentagon is inscribed in a circle of radius $r$. $P$ is any point inside the pentagon. Perpendiculars are dropped from $P$ to the sides, or the sides produced, of the pentagon.

a) Prove that the sum of the lengths of these perpendiculars is constant.

b) Express this constant in terms of the radius $r$.

Solution

Let the pentagon be $ABCDE$, and the perpendiculars from $P$ to sides $AB$, $BC$, $CD$, $DE$, and $EA$ have lengths $p_1$, $p_2$, $p_3$, $p_4$, and $p_5$, respectively.

Since $P$ is inside the pentagon, we have that the area of the pentagon is the sum of the areas of the triangles $ABP$, $BCP$, $CDP$, $DEP$, and $EAP$. This sum is simply equal to

\[[ABCDE]=\dfrac{1}{2}(AB\cdot p_1+BC\cdot p_2+CD\cdot p_3+DE\cdot p_4+EA\cdot p_5).\]

This is constant. However, since $ABCDE$ is a regular pentagon, we have that all of its sides are equal to some positive $s$, and

\[p_1+p_2+p_3+p_4+p_5=2[ABCDE]/s.\]

Therefore the sum of the lengths of the perpendiculars is constant.

Let $O$ be the center of the pentagon. We know that $\angle AOB=72^{\circ}$, so $[AOB]=(1/2)r^2\sin 72^{\circ}$. We were given that $ABCDE$ is a regular pentagon, so it follows that

\[p_1+p_2+p_3+p_4+p_5=(5r^2/s)\sin 72^{\circ}.\]

We now find $s$ in terms of $r$. We know that $\angle OAB=54^{\circ}$, so $[AOB]=(1/2)rs\sin 54^{\circ}$. Therefore $s=r\left(\sin 72^{\circ}/\sin 54^{\circ}\right)$. This shows that

\[p_1+p_2+p_3+p_4+p_5=5r\sin 54^{\circ}.\]

Note: The value of $\sin 54^{\circ}$ can be proven to be $\frac{\sqrt{5}+1}{4}$.

See Also

1971 Canadian MO (Problems)
Preceded by
Problem 7 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed by
Problem 9
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1971_Canadian_MO_Problems/Problem_8&oldid=77508"