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Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 12"
Line 14: Line 14:
 
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>\sin{\angle AOD}=\sin{AOB}</math> , <math>[AOD]=\sin{\angle AOD}\cdot\frac{1}{2}\cdot a\cdot\frac{2}{3}=\frac{a}{3}\cdot\sin{\angle AOD}</math> , <math>[AOB]=\sin{AOB}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot 1=\frac{[AOD]}{a}</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>
Line 20: Line 20:
 
<math>[AOB]=\frac{[AOD]}{4}</math>
 
<math>[AOB]=\frac{[AOD]}{4}</math>
  
−
<math>[BOC]=\frac{9}{4}[AOD]</math>
+
Note that <math>\sin{\angle AOB}=\sin{(180-\angle AOD)}=\sin{\angle AOD}</math>
  
−
<math>[COD]=36[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>\Rightarrow a=4</math>
  
−
Thus we need to find <math>[ABCD]=\frac{79}{2}[AOD]</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.
 
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.
Line 32: Line 36:
 
<math>[AOD]=\frac{\sqrt{143}}{9}</math>
 
<math>[AOD]=\frac{\sqrt{143}}{9}</math>
  
−
Thus <math>[ABCD]=\frac{79\sqrt{143}}{18}\rightarrow\boxed{240}</math>
+
Thus <math>[ABCD]=\frac{25\sqrt{143}}{18}\rightarrow\boxed{186}</math>
−
 
 
−
 
 
  
 
==See also==
 
==See also==

Revision as of 14:20, 24 February 2009

Problem

In quadrilateral $ABCD,$ $m \angle DAC= m\angle DBC$ and $\frac{[ADB]}{[ABC]}=\frac12.$ $O$ is defined to be the intersection of the diagonals of $ABCD$. If $AD=4,$ $BC=6$, $BO=1,$ and the area of $ABCD$ is $\frac{a\sqrt{b}}{c},$ where $a,b,c$ are relatively prime positive integers, find $a+b+c.$


Note*: $[ABC]$ and $[ADB]$ refer to the areas of triangles $ABC$ and $ADB.$

Solution

$m\angle DAC=m\angle DBC \Rightarrow ABCD$ is a cylic quadrilateral.

Let $DO=a, AO=b$

$\triangle AOD$ ~ $\triangle BOC \Rightarrow b=\frac{2}{3}$

Also, from the Power of a Point Theorem, $DO \cdot BO=AO\cdot CO\Rightarrow CO=\frac{3a}{2}$

Notice $\frac{[AOB]}{[BOC]}=(\frac{2}{3})^2\Rightarrow [BOC]=\frac{9}{4}[AOD]$

It is given $\frac{[AOD]+[AOB]}{[AOB]+[BOC]}=\frac{[ADB]}{[ABC]}=\frac{1}{2} \Rightarrow$

$[AOB]=\frac{[AOD]}{4}$

Note that $\sin{\angle AOB}=\sin{(180-\angle AOD)}=\sin{\angle AOD}$

Then $[AOB]=\frac{\frac{2}{3}\cdot 1\cdot\sin{\angle AOB}}{2}=\frac{\sin{\angle AOD}}{3}$ and $[AOD]=\frac{\frac{2}{3}\cdot a\cdot\sin{\angle AOD}}{2}=\frac{a\sin{\angle AOD}}{3}$

$\Rightarrow a=4$


$[COD]=9[AOD]$

Thus we need to find $[ABCD]=\frac{25}{2}[AOD]$

Note that $\triangle AOD$ is isosceles with sides $4, 4, \frac{2}{3}$ so we can draw the altitude from D to split it to two right triangles.

$[AOD]=\frac{\sqrt{143}}{9}$

Thus $[ABCD]=\frac{25\sqrt{143}}{18}\rightarrow\boxed{186}$

See also


Problem Source

AoPS users 4everwise and Altheman collaborated to create this problem.

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