# 1999 AHSME Problems/Problem 7

## Problem

What is the largest number of acute angles that a convex hexagon can have?

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$

## Solution

The sum of the interior angles of a hexagon is $720$ degrees. In a convex polygon, each angle must be strictly less than $180$ degrees.

Six acute angles can only sum to less than $90\cdot 6 = 540$ degrees, so six acute angles could not form a hexagon.

Five acute angles and one obtuse angle can only sum to less than $90\cdot 5 + 180 = 630$ degrees, so these angles could not form a hexagon.

Four acute angles and two obtuse angles can only sum to less than $90\cdot 4 + 180\cdot 2 = 720$ degrees. This is a strict inequality, so these angles could not form a hexagon. (The limiting figure would be four right angles and two straight angles, which would really be a square with two "extra" points on two sides to form the straight angles.)

Three acute angles and three obtuse angles work. For example, if you pick three acute angles of $80$ degrees, the three obtuse angles would be $160$ degrees and give a sum of $80\cdot 3 + 160\cdot 3 = 720$ degrees, which is a genuine hexagon. Thus, the answer is $\boxed{(B) 3}$

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