Difference between revisions of "2011 AMC 12A Problems/Problem 24"
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== Solution == | == Solution == | ||
+ | Answer: <math>(C) 2 \sqrt{6}</math> | ||
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+ | Given, a 14-9-7-12 quadrilateral ( which has an in-circle). | ||
+ | |||
+ | Find the largest possible in-radius. | ||
+ | |||
+ | <br /> | ||
+ | '''Solution:''' | ||
+ | |||
+ | 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>. | ||
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+ | 2(Area) = <math>(14)(12) \sin \alpha + (9)(7) \sin \beta</math> | ||
+ | |||
+ | <math>\frac{2}{21}</math> (Area) = <math>8 \sin \alpha + 3 \sin \beta</math> | ||
+ | |||
+ | <br /> | ||
+ | By law of cosine, <math>14^2 + 12 ^2 - 2(14)(12) \cos \alpha = 9^2 + 7^2 - 2(9)(7) \cos \beta</math> | ||
+ | |||
+ | <math>8 \cos \alpha - 3 \cos \beta = 5</math> (simple algebra left to the reader) | ||
+ | |||
+ | <br /> | ||
+ | |||
+ | <math>\frac{4}{441}</math> (Area)<math>^2 + 25</math> = <math>64 \sin^2 \alpha + 9 \sin^2 \beta + 64 \cos^2 \alpha + 9 \cos^2 \beta - 48 \cos \alpha \cos \beta + 48 \sin \alpha \sin \beta = 73 - 48 \cos (\alpha + \beta)</math> | ||
+ | |||
+ | <math>\frac{4}{441}</math> (Area)<math>^2</math> = <math>48 ( 1- \cos (\alpha + \beta)</math>, which reach maximum when <math>( 1- \cos (\alpha + \beta) = 2</math>. | ||
+ | |||
+ | (and since it is a quadrilateral, it is possible to have <math>\alpha + \beta = \pi</math> (hence cyclic quadrilateral, that would be the best guess and the extended Heron's formula which I forgot the name for would work for area and the work is simple). | ||
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+ | <math>\frac{4}{441}</math> (Area)<math>^2 \ge 96</math> | ||
+ | |||
+ | (Area)<math>^2 \ge 24 (441)</math> | ||
+ | |||
+ | (Area)<math> \ge 42 \sqrt{6}</math>, Area = <math>r \times</math> semi-perimeter. | ||
+ | |||
+ | Hence, <math>r = 2 \sqrt{6}</math>, choice <math>(C)</math> | ||
+ | |||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=23|num-a=25|ab=A}} | {{AMC12 box|year=2011|num-b=23|num-a=25|ab=A}} |
Revision as of 00:22, 15 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
Answer:
Given, a 14-9-7-12 quadrilateral ( which has an in-circle).
Find the largest possible in-radius.
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 reach maximum when .
(and since it is a quadrilateral, it is possible to have (hence cyclic quadrilateral, that would be the best guess and the extended Heron's formula which I forgot the name for 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 |