Difference between revisions of "2007 AIME I Problems/Problem 12"

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The [[union]] of the area is equal to <math>2</math> times the area of <math>\triangle ABC</math>, minus the union of the area of the two triangles.
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The [[union]] of the area is equal to <math>2</math> times the area of <math>\triangle ABC</math>, minus the intersection of the area of the two triangles.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2007|n=I|num-b=11|num-a=13}}
 
{{AIME box|year=2007|n=I|num-b=11|num-a=13}}

Revision as of 12:40, 16 March 2007

Problem

In isosceles triangle $ABC$, $A$ is located at the origin and $B$ is located at (20,0). Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$. If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$-axis, the area of the region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$, where $p,q,r,s$ are integers. Find $(p-q+r-s)/2$.

Solution

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The union of the area is equal to $2$ times the area of $\triangle ABC$, minus the intersection of the area of the two triangles.

See also

2007 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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