Problem: ABC and A′B′C′ are equilateral triangles with parallel sides and the same center, as in the figure. The distance between side BC and side B′C′ is 61 the altitude of △ABC. The ratio of the area of △A′B′C′ to the area of △ABC is
Answer Choices:
A. 361
B. 61
C. 41
D. 43
E. 369+83
Solution:
Let △ABC and △A′B′C′ have heights h and h′. The required ratio thus equals (hh′)2. Let O be the common center of the triangles and let M and M′ be the intersections of BC and B′C′ with the common altitude from A. Since altitudes are also medians in equilateral triangles, OM=3h and OM′=3h′. By hypothesis, MM′=6h so 3h=3h′+6h. Thus hh′=21 and (hh′)2=41.
