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Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 7"

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[[Brahmagupta's Formula]] states that the area of a cyclic quadrilateral is <math>\sqrt{(s-a)(s-b)(s-c)(s-d)}</math>, where <math>s</math> is the semiperimeter and <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are the side lengths of the quadrilateral. Therefore the area of <math>ABCD</math> is <math>\sqrt{4\cdot 3\cdot 2\cdot 1}=\sqrt{24}</math>. It is also a well-known fact that the area of a circumscriptable quadrilateral is <math>sr</math>, where <math>r</math> is the inradius. Therefore <math>5r=\sqrt{24}\Rightarrow r=\frac{\sqrt{24}}{5}</math>. Therefore the area of the inscribed circle is <math>\frac{24\pi}{25}</math>, and <math>p+q=\boxed{049}</math>.
 
[[Brahmagupta's Formula]] states that the area of a cyclic quadrilateral is <math>\sqrt{(s-a)(s-b)(s-c)(s-d)}</math>, where <math>s</math> is the semiperimeter and <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are the side lengths of the quadrilateral. Therefore the area of <math>ABCD</math> is <math>\sqrt{4\cdot 3\cdot 2\cdot 1}=\sqrt{24}</math>. It is also a well-known fact that the area of a circumscriptable quadrilateral is <math>sr</math>, where <math>r</math> is the inradius. Therefore <math>5r=\sqrt{24}\Rightarrow r=\frac{\sqrt{24}}{5}</math>. Therefore the area of the inscribed circle is <math>\frac{24\pi}{25}</math>, and <math>p+q=\boxed{049}</math>.
  
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==See also==
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==See Also==
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{{Mock AIME box|year=Pre 2005|n=3|num-b=6|num-a=8}}

Latest revision as of 09:23, 4 April 2012

Problem

$ABCD$ is a cyclic quadrilateral that has an inscribed circle. The diagonals of $ABCD$ intersect at $P$. If $AB = 1, CD = 4,$ and $BP : DP = 3 : 8,$ then the area of the inscribed circle of $ABCD$ can be expressed as $\frac{p\pi}{q}$, where $p$ and $q$ are relatively prime positive integers. Determine $p + q$.

Solution

Let $BP=3x$ and $PD=8x$. Angle-chasing can be used to prove that $\triangle ABP \sim \triangle DCP$. Therefore $\frac{AB}{DC}=\frac{AP}{DP}=\frac{BP}{PC}=\frac{1}{4}$. This shows that $AP=2x$ and $CP=12x$. More angle-chasing can be used to prove that $\triangle APD \sim \triangle BPC$. This shows that $\frac{BC}{AD}=\frac{BP}{AP}=\frac{CP}{DP}=\frac{3}{2}$. It is a well-known fact that if $ABCD$ is circumscriptable around a circle then $AB+CD=AD+BC$. Therefore $BC+AD=5$. We also know that $\frac{BC}{AD}=\frac{3}{2}$, so we can solve (algebraically or by inspection) to get that $BC=3$ and $AD=2$.

Brahmagupta's Formula states that the area of a cyclic quadrilateral is $\sqrt{(s-a)(s-b)(s-c)(s-d)}$, where $s$ is the semiperimeter and $a$, $b$, $c$, and $d$ are the side lengths of the quadrilateral. Therefore the area of $ABCD$ is $\sqrt{4\cdot 3\cdot 2\cdot 1}=\sqrt{24}$. It is also a well-known fact that the area of a circumscriptable quadrilateral is $sr$, where $r$ is the inradius. Therefore $5r=\sqrt{24}\Rightarrow r=\frac{\sqrt{24}}{5}$. Therefore the area of the inscribed circle is $\frac{24\pi}{25}$, and $p+q=\boxed{049}$.

See Also

Mock AIME 3 Pre 2005 (Problems, Source)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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