1985 AHSME Problems/Problem 14

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

Suppose that such a polygon has $n$ sides. Let the three obtuse angle measures, in degrees, be $o_1$ , $o_2$ , and $o_3$ and the $(n-3)$ acute angle measures, again in degrees, be $a_1,a_2,a_3, \dotsc a_{n-3}$ .

Since $90 < o_i < 180$ for each $i$ , we have $3(90) = 270 < o_1+o_2+o_3 < 3(180) = 540,$ and similarly, since $0 < a_i < 90$ for each $i$ , $0 < a_1+a_2+a_3+\dotsb+a_{n-3} < 90(n-3) = 90n-270.$ It follows that $270+0 < o_1+o_2+o_3+a_1+a_2+a_3+\dotsb+a_{n-3} < 540+90n-270,$ and recalling that the sum of the interior angle measures of an $n$ -gon is $180(n-2) = 180n-360$ , this reduces to $270 < 180n-360 < 90n+270$ . Hence $\frac{540}{180} < n < \frac{270+360}{90} \iff 3 < n < 7,$ so an upper bound is $n \leq 6$ , and it is easy to check that this bound can be attained by e.g. a convex hexagon with a right angle, $2$ acute angles, and $3$ obtuse angles, as shown below:

Accordingly, the maximum possible number of sides of such a polygon is $\boxed{\text{(C)} \ 6}$ .

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1985 AHSME Problems/Problem 14

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