Difference between revisions of "2011 AMC 12A Problems/Problem 24"

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<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>
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<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</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>.  
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<math>= 73 - 48 \cos (\alpha + \beta)</math>
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<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).
 
(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).

Revision as of 00:24, 15 February 2011

Problem

Consider all quadrilaterals $ABCD$ such that $AB=14$, $BC=9$, $CD=7$, and $DA=12$. What is the radius of the largest possible circle that fits inside or on the boundary of such a quadrilateral?

$\textbf{(A)}\ \sqrt{15} \qquad \textbf{(B)}\ \sqrt{21} \qquad \textbf{(C)}\ 2\sqrt{6} \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 2\sqrt{7}$

Solution

Answer: $(C) 2 \sqrt{6}$

Given, a 14-9-7-12 quadrilateral ( which has an in-circle).

Find the largest possible in-radius.


Solution:

Since Area = $r \times$ semi-perimeter, and perimeter is fixed, we can maximize the area. Let the angle between the 14 and 12 be $\alpha$ degree, and the one between the 9 and 7 be $\beta$.

2(Area) = $(14)(12) \sin \alpha + (9)(7) \sin \beta$

$\frac{2}{21}$ (Area) = $8 \sin \alpha + 3 \sin \beta$


By law of cosine, $14^2 + 12 ^2 - 2(14)(12) \cos \alpha = 9^2 + 7^2 - 2(9)(7) \cos \beta$

$8 \cos \alpha - 3 \cos \beta = 5$ (simple algebra left to the reader)


$\frac{4}{441}$ (Area)$^2 + 25$ = $64 \sin^2 \alpha + 9 \sin^2 \beta + 64 \cos^2 \alpha + 9 \cos^2 \beta - 48 \cos \alpha \cos \beta + 48 \sin \alpha \sin \beta$

$= 73 - 48 \cos (\alpha + \beta)$

$\frac{4}{441}$ (Area)$^2$ = $48 ( 1- \cos (\alpha + \beta))$, which reach maximum when $( 1- \cos (\alpha + \beta)) = 2$.

(and since it is a quadrilateral, it is possible to have $\alpha + \beta = \pi$ (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).

$\frac{4}{441}$ (Area)$^2 \ge 96$

(Area)$^2 \ge 24 (441)$

(Area)$\ge 42 \sqrt{6}$, Area = $r \times$ semi-perimeter.

Hence, $r = 2 \sqrt{6}$, choice $(C)$

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
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
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