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Difference between revisions of "1985 AHSME Problems/Problem 14"

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Revision as of 12:01, 5 July 2013

Problem

Exactly three of the interior angles of a convex polygon are obtuse. What is the maximum number of sides of such a polygon?

$\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }5 \qquad \mathrm{(C) \ } 6 \qquad \mathrm{(D) \ } 7 \qquad \mathrm{(E) \ }8$

Solution

All angle measures are in degrees.

The sum of the interior angle measures of an $n$-gon is $180(n-2)=180n-360$. Let the three obtuse angle measures be $o_1, o_2,$ and $o_3$, and the $n-3$ acute angle measures be $a_1, a_2, a_3, \cdots$.


Since $90<o_i<180$, $3(90)=270<o_1+o_2+o_3<3(180)=540$.


Similarly, since $0<a_i<90$, $0(n-3)=0<a_1+a_2+a_3+\cdots<90(n-3)=90n-270$.


Thus, $270+0<o_1+o_2+o_3+a_1+a_2+a_3+\cdots<540+90n-270$ $\implies 270<180n-360<90n+270$.


Thus, $180n-360<90n+270\implies n<7$, so the largest possible $n$ is $6, \boxed{\text{C}}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AHSME Problems and Solutions

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