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 ∠BAC=75∘. If △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 triangle and the rotated triangle is in the form p2+q3+r6+s where p,q,r,s are integers. Find (p−q+r−s)/2.
Solution:
Let [XYZ] represent the area of triangle XYZ.
For future use, sin75∘=cos15∘=(6+2)/4. Let B′ and C′ be the images of B and C respectively under the given rotation. Let D denote the point at which BC intersects AB′, let E denote the point at which BC intersects B′C′, and let F denote the point at which AC intersects B′C′. Then the region common to the two triangles (shaded in the figure on the right) is ADEF, and its area is [ADEF]=[AB′F]−[EB′D]. Note that ∠B+∠B′AB=75∘+15∘=90∘ implies AB′⊥BC. Because AB′=AB=20, the Law of Sines applied to △B′FA gives B′F=20sin60∘/sin45∘=203/2=106, and thus [AB′F]=21⋅20⋅106sin75∘=1006(46+2)=50(3+3).
Note that B′D=20(1−cos15∘),BD=20sin15∘, and △EB′D∼△ABD. Because [ABD]=21⋅20cos15∘⋅20sin15∘=100sin30∘=50, it follows that [EB′D]=50(sin15∘1−cos15∘)2.
Using cos215∘=21+cos30∘=42+3 and sin215∘=21−cos30∘=42−3 yields
