Difference between revisions of "2011 AMC 12A Problems/Problem 24"
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\textbf{(E)}\ 2\sqrt{7} </math> | \textbf{(E)}\ 2\sqrt{7} </math> | ||
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Since Area = <math>r \times</math> semi-perimeter, and perimeter is fixed, we can maximize the area. Let the angle between the 14 and 12 be <math>\alpha</math> degree, and the one between the 9 and 7 be <math>\beta</math>. | Since Area = <math>r \times</math> semi-perimeter, and perimeter is fixed, we can maximize the area. Let the angle between the 14 and 12 be <math>\alpha</math> degree, and the one between the 9 and 7 be <math>\beta</math>. | ||
Revision as of 11:20, 19 February 2011
Problem
Consider all quadrilaterals 
such that 
, 
, 
, and 
. What is the radius of the largest possible circle that fits inside or on the boundary of such a quadrilateral?
Solution
Since Area = 
semi-perimeter, and perimeter is fixed, we can maximize the area. Let the angle between the 14 and 12 be 
degree, and the one between the 9 and 7 be 
.
2(Area) = 
(Area) = 
By law of cosine, 
(simple algebra left to the reader)
(Area)
= 
(Area)
= 
, which reaches maximum when 
.
(and since it is a quadrilateral, it is possible to have 
(hence cyclic quadrilateral, that would be the best guess and the Brahmagupta's formula would work for area and the work is simple).
(Area)
(Area)
(Area)
, Area = semi-perimeter.
Hence, 
, choice 
See also
| 2011 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 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 | |
| All AMC 12 Problems and Solutions | |
