Difference between revisions of "1999 AHSME Problems/Problem 7"

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==Solution==
 
==Solution==
  
{{solution}}
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The sum of the interior angles of a hexagon is <math>720</math> degrees.  In a convex polygon, each angle must be strictly less than <math>180</math> degrees. 
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Six acute angles can only sum to less than <math>90\cdot 6 = 540</math> degrees, so six acute angles could not form a hexagon.
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Five acute angles and one obtuse angle can only sum to less than <math>90\cdot 5 + 180 = 630</math> degrees, so these angles could not form a hexagon.
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Four acute angles and two obtuse angles can only sum to less than <math>90\cdot 4 + 180\cdot 2 = 720</math> degrees.  This is a strict inequality, so these angles could not form a hexagon.  (The limiting figure would be four right angles and two striaght angles, which would really be a square with two "extra" points on two sides to form the straight angles..)
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Three acute angles and three obtuse angles work.  For example, if you pick three acute angles of <math>80</math> degrees, the three obtuse angles would be <math>160</math> degrees and give a sum of <math>80\cdot 3 + 160\cdot 3 = 720</math> degrees, which is a genuine hexagon.  Thus, the answer is <math>\boxed{C}</math>
  
 
==See Also==
 
==See Also==
  
 
{{AHSME box|year=1999|num-b=6|num-a=8}}
 
{{AHSME box|year=1999|num-b=6|num-a=8}}

Revision as of 13:08, 8 August 2011

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 striaght 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 $\boxed{C}$

See Also

1999 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions