Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 12"
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Also, from the Power of a Point Theorem, <math>DO \cdot BO=AO\cdot CO\Rightarrow CO=\frac{3a}{2}</math> | Also, from the Power of a Point Theorem, <math>DO \cdot BO=AO\cdot CO\Rightarrow CO=\frac{3a}{2}</math> | ||
− | Notice <math>\ | + | Notice <math>\frac{[AOB]}{[BOC]}=(\frac{2}{3})^2\Rightarrow [BOC]=\frac{9}{4}[AOD]</math> |
It is given <math>\frac{[AOD]+[AOB]}{[AOB]+[BOC]}=\frac{[ADB]}{[ABC]}=\frac{1}{2} \Rightarrow</math> | It is given <math>\frac{[AOD]+[AOB]}{[AOB]+[BOC]}=\frac{[ADB]}{[ABC]}=\frac{1}{2} \Rightarrow</math> | ||
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<math>[AOB]=\frac{[AOD]}{4}</math> | <math>[AOB]=\frac{[AOD]}{4}</math> | ||
− | <math> | + | Note that <math>\sin{\angle AOB}=\sin{(180-\angle AOD)}=\sin{\angle AOD}</math> |
− | <math>[ | + | Then <math>[AOB]=\frac{\frac{2}{3}\cdot 1\cdot\sin{\angle AOB}}{2}=\frac{\sin{\angle AOD}}{3}</math> |
+ | and <math>[AOD]=\frac{\frac{2}{3}\cdot a\cdot\sin{\angle AOD}}{2}=\frac{a\sin{\angle AOD}}{3}</math> | ||
<math>\Rightarrow a=4</math> | <math>\Rightarrow a=4</math> | ||
− | Thus we need to find <math>[ABCD]=\frac{ | + | |
+ | <math>[COD]=9[AOD]</math> | ||
+ | |||
+ | Thus we need to find <math>[ABCD]=\frac{25}{2}[AOD]</math> | ||
Note that <math>\triangle AOD</math> is isosceles with sides <math>4, 4, \frac{2}{3}</math> so we can draw the altitude from D to split it to two right triangles. | Note that <math>\triangle AOD</math> is isosceles with sides <math>4, 4, \frac{2}{3}</math> so we can draw the altitude from D to split it to two right triangles. | ||
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<math>[AOD]=\frac{\sqrt{143}}{9}</math> | <math>[AOD]=\frac{\sqrt{143}}{9}</math> | ||
− | Thus <math>[ABCD]=\frac{ | + | Thus <math>[ABCD]=\frac{25\sqrt{143}}{18}\rightarrow\boxed{186}</math> |
− | |||
− | |||
==See also== | ==See also== |
Revision as of 13:20, 24 February 2009
Contents
Problem
In quadrilateral and is defined to be the intersection of the diagonals of . If , and the area of is where are relatively prime positive integers, find
Note*: and refer to the areas of triangles and
Solution
is a cylic quadrilateral.
Let
~
Also, from the Power of a Point Theorem,
Notice
It is given
Note that
Then and
Thus we need to find
Note that is isosceles with sides so we can draw the altitude from D to split it to two right triangles.
Thus
See also
Problem Source
AoPS users 4everwise and Altheman collaborated to create this problem.