Difference between revisions of "1999 AHSME Problems/Problem 7"
m (→Solution) |
|||
| Line 20: | Line 20: | ||
{{AHSME box|year=1999|num-b=6|num-a=8}} | {{AHSME box|year=1999|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
Revision as of 13:34, 5 July 2013
Problem
What is the largest number of acute angles that a convex hexagon can have?
Solution
The sum of the interior angles of a hexagon is 
degrees. In a convex polygon, each angle must be strictly less than 
degrees.
Six acute angles can only sum to less than 
degrees, so six acute angles could not form a hexagon.
Five acute angles and one obtuse angle can only sum to less than 
degrees, so these angles could not form a hexagon.
Four acute angles and two obtuse angles can only sum to less than 
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 
degrees, the three obtuse angles would be 
degrees and give a sum of 
degrees, which is a genuine hexagon. Thus, the answer is 
See Also
| 1999 AHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
