Difference between revisions of "1999 AHSME Problems/Problem 7"
Talkinaway (talk | contribs) (→Solution) |
m (→Solution) |
||
| (4 intermediate revisions by 3 users not shown) | |||
| Line 13: | Line 13: | ||
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. | 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. | ||
| − | 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 | + | 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 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 <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{ | + | 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{(B) 3}</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}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:18, 5 November 2019
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 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 
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. 
