Difference between revisions of "1971 Canadian MO Problems/Problem 8"
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<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^{\ | + | 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> | ||
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== 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 . is any point inside the pentagon. Perpendiculars are dropped from 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 .
Solution
Let the pentagon be , and the perpendiculars from to sides , , , , and have lengths , , , , and , respectively.
Since is inside the pentagon, we have that the area of the pentagon is the sum of the areas of the triangles , , , , and . This sum is simply equal to
This is constant. However, since is a regular pentagon, we have that all of its sides are equal to some positive , and
Therefore the sum of the lengths of the perpendiculars is constant.
Let be the center of the pentagon. We know that , so . We were given that is a regular pentagon, so it follows that
We now find in terms of . We know that , so . Therefore . This shows that
Note: The value of can be proven to be .
See Also
1971 Canadian MO (Problems) | ||
Preceded by Problem 7 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 9 |