Problem:
In △ABC,AB=13,BC=14,AC=15, and point G is the intersection of the medians. Points A′,B′, and C′ are the images of A,B, and C, respectively, after a 180∘ rotation about G. What is the area of the union of the two regions enclosed by the triangles ABC and A′B′C′ ?
Solution:
Let M be the midpoint of BC, let M′ be the reflection of M in G, and let Q and R be the points where BC meets A′C′ and A′B′, respectively. Note that since M is on BC,M′ is on B′C′. Because a 180∘ rotation maps each line that does not contain the center of the rotation to a parallel line, BC is parallel to B′C′, and △A′RQ is similar to △A′B′C′. Recall that medians of a triangle trisect each other to obtain
M′G=MG=(1/3)AM, so A′M=AM′=(1/3)AM=(1/3)A′M′
Thus the similarity ratio between triangles A′RQ and A′B′C′ is 1/3, and
[A′RQ]=(1/9)[A′B′C′]=(1/9)[ABC]
Similarly, the area of each of the two small triangles with vertices at B′ and C′, respectively, is 1/9 that of △ABC. The desired area is therefore
[ABC]+3(1/9)[ABC]=(4/3)[ABC]
Use Heron's formula, K=s(s−a)(s−b)(s−c), to find [ABC]=21⋅7⋅6⋅8=84. The desired area is then (4/3)⋅84=112.
