Suppose A, B, and C are points in the plane with AB=40 and AC=42, and let x be the length of the line segment from A to the midpoint of BC. Define a function f by letting f(x) be the area of △ABC. Then the domain of f is an open interval (p,q), and the maximum value r of f(x) occurs at x=s. What is p+q+r+s?
Let θ=∠BAC. Since f(x) is the area of △ABC,[ABC]=21⋅40⋅42sinθ=840sinθ. Since sinθ≤1, max{[ABC]}=840. Thus, r=840, with equality when θ=90∘.
Find s:
When θ=90∘, let M be the midpoint of BC.By the Pythagorean theorem, BC2=402+422=22(202+212)=4(400+441)=4(841). Therefore, BC=2(29)=58. Since the median to the hypotenuse of a right triangle is half the hypotenuse, AM=2BC=29. Hence, s=29.
Find p, q:
Let M be the midpoint of BC, and let BC=a and AM=ma=x.By Stewart's Theorem, ma2=x2=2b2+2c2−4a2. Substituting b=42 and c=40, x2=2422+2402−4a2.
By the Triangle Inequality, 42−40<a<40+42⟹2<a<82. By Hint 3, 1<x2=2422+2402−4a2<1681. Since x is a length, 1<x<41. Therefore, the domain of f is (1,41), so p=1,q=41. By Hints 1 and 2, p+q+r+s=1+41+840+29=911.
