# 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 $\boxed{\text{(C)} \ 6}$.