Problem:
If a,b, and c are positive real numbers such that a(b+c)=152,b(c+a)=162, and c(a+b)=170, then abc is
Answer Choices:
A. 672
B. 688
C. 704
D. 720
E. 750
Solution:
Adding the given equations gives 2(ab+bc+ca)=484, so ab+bc+ca=242. Subtracting from this each of the given equations yields bc=90,ca=80, and ab=72. It follows that a2b2c2=90⋅80⋅72=7202. Since abc>0, we have abc=720​.
The problems on this page are the property of the MAA's American Mathematics Competitions