Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 12"
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== Problem == | == Problem == | ||
− | In quadrilateral <math> | + | In [[quadrilateral]] <math>ABCD,</math> <math>m \angle DAC= m\angle DBC </math> and <math>\frac{[ADB]}{[ABC]}=\frac12.</math> <math>O</math> is defined to be the intersection of the diagonals of <math>ABCD</math>. If <math>AD=4,</math> <math>BC=6</math>, <math>BO=1,</math> and the [[area]] of <math>ABCD</math> is <math>\frac{a\sqrt{b}}{c},</math> where <math>a,b,c</math> are [[relatively prime]] [[positive integer]]s, find <math>a+b+c.</math> |
+ | |||
+ | |||
+ | Note*: <math>[ABC]</math> and <math>[ADB]</math> refer to the areas of [[triangle]]s <math>ABC</math> and <math>ADB.</math> | ||
+ | |||
+ | ==Solution== | ||
+ | <math>m\angle DAC=m\angle DBC \Rightarrow ABCD</math> is a cylic quadrilateral. | ||
+ | |||
+ | Let <math>DO=a, AO=b</math> | ||
+ | |||
+ | <math>\triangle AOD</math> ~ <math>\triangle BOC \Rightarrow b=\frac{2}{3}</math> | ||
+ | |||
+ | Also, from the Power of a Point Theorem, <math>DO \cdot BO=AO\cdot CO\Rightarrow CO=\frac{3a}{2}</math> | ||
+ | |||
+ | Notice <math>\frac{[AOD]}{[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 [AOB]=\frac{[AOD]}{4}</math> | ||
+ | |||
+ | Note that <math>\sin{\angle AOB}=\sin{(180-\angle AOD)}=\sin{\angle AOD}</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>[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. | ||
+ | |||
+ | <math>[AOD]=\frac{\sqrt{143}}{9}</math> | ||
+ | |||
+ | Thus <math>[ABCD]=\frac{25\sqrt{143}}{18}\rightarrow\boxed{186}</math> | ||
+ | |||
+ | ==See Also== | ||
+ | {{Mock AIME box|year=2006-2007|n=2|num-b=11|num-a=13}} | ||
− | |||
== Problem Source == | == Problem Source == | ||
AoPS users 4everwise and Altheman collaborated to create this problem. | AoPS users 4everwise and Altheman collaborated to create this problem. |
Latest revision as of 10:53, 4 April 2012
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
Mock AIME 2 2006-2007 (Problems, Source) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |
Problem Source
AoPS users 4everwise and Altheman collaborated to create this problem.