Problem:
Let ABC be equilateral, and D,E, and F be the midpoints of BC,CA, and AB, respectively. There exist points P,Q, and R on DE, EF, and FD, respectively, with the property that P is on CQ,Q is on AR, and R is on BP. The ratio of the area of triangle ABC to the area of triangle PQR is a+bc, where a,b, and c are integers, and c is not divisible by the square of any prime. What is a2+b2+c2?
Solution:
To see first that there is at most one set of points with the given property, suppose that P′,Q′, and R′ also have the given property. Notice that P′ is on PE if and only if Q′ is on QE, which is true if and only if R′ is on RD, which is true if and only if P′ is on PD. Conclude that P′=P. It follows that Q′=Q, because P determines the position of Q on EF, and R′=R, because Q determines the position of R on DF. Without loss of generality, let DC=DE=1 and DP=x. Triangles ABC and DEF have 120-degree rotational symmetry, hence triangle PQR must also. (If this were not true, then a 120-degree rotation would produce another set of points with the given property.) It follows that EQ=x and PE=1−x. The similarity of triangles DCP and EQP implies that EPDP=EQDC, or 1−xx=x1. Thus x2=1−x, whose positive solution is x=21(5−1). Apply the Law of Cosines to triangle PQE to obtain
