Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 7"
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Let <math>BP=3x</math> and <math>PD=8x</math>. Angle-chasing can be used to prove that <math>\triangle ABP \sim \triangle DCP</math>. Therefore <math>\frac{AB}{DC}=\frac{AP}{DP}=\frac{BP}{PC}=\frac{1}{4}</math>. This shows that <math>AP=2x</math> and <math>CP=12x</math>. More angle-chasing can be used to prove that <math>\triangle APD \sim \triangle BPC</math>. This shows that <math>\frac{BC}{AD}=\frac{BP}{AP}=\frac{CP}{DP}=\frac{3}{2}</math>. It is a well-known fact that if <math>ABCD</math> is circumscriptable around a circle then <math>AB+CD=AD+BC</math>. Therefore <math>BC+AD=5</math>. We also know that <math>\frac{BC}{AD}=\frac{3}{2}</math>, so we can solve (algebraically or by inspection) to get that <math>BC=3</math> and <math>AD=2</math>. | Let <math>BP=3x</math> and <math>PD=8x</math>. Angle-chasing can be used to prove that <math>\triangle ABP \sim \triangle DCP</math>. Therefore <math>\frac{AB}{DC}=\frac{AP}{DP}=\frac{BP}{PC}=\frac{1}{4}</math>. This shows that <math>AP=2x</math> and <math>CP=12x</math>. More angle-chasing can be used to prove that <math>\triangle APD \sim \triangle BPC</math>. This shows that <math>\frac{BC}{AD}=\frac{BP}{AP}=\frac{CP}{DP}=\frac{3}{2}</math>. It is a well-known fact that if <math>ABCD</math> is circumscriptable around a circle then <math>AB+CD=AD+BC</math>. Therefore <math>BC+AD=5</math>. We also know that <math>\frac{BC}{AD}=\frac{3}{2}</math>, so we can solve (algebraically or by inspection) to get that <math>BC=3</math> and <math>AD=2</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>. |
==See also== | ==See also== |
Revision as of 14:13, 3 April 2012
Problem
is a cyclic quadrilateral that has an inscribed circle. The diagonals of
intersect at
. If
and
then the area of the inscribed circle of
can be expressed as
, where
and
are relatively prime positive integers. Determine
.
Solution
Let and
. Angle-chasing can be used to prove that
. Therefore
. This shows that
and
. More angle-chasing can be used to prove that
. This shows that
. It is a well-known fact that if
is circumscriptable around a circle then
. Therefore
. We also know that
, so we can solve (algebraically or by inspection) to get that
and
.
Brahmagupta's Formula states that the area of a cyclic quadrilateral is , where
is the semiperimeter and
,
,
, and
are the side lengths of the quadrilateral. Therefore the area of
is
. It is also a well-known fact that the area of a circumscriptable quadrilateral is
, where
is the inradius. Therefore
. Therefore the area of the inscribed circle is
, and
.