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 | + | ==See Also== |
+ | {{Mock AIME box|year=Pre 2005|n=3|num-b=6|num-a=8}} |
Latest revision as of 09:23, 4 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 .
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 |