Problem 3 - Entrance Test
A triangle ABC has side lengths AB = 7, BC = 8, CA = 9. Let M be the midpoint of BC. Find the length of AM.
Correct: C
We can use Apollonius's Theorem, which states that for a triangle ABC with median AM, AB^2 + AC^2 = 2(AM^2 + BM^2).
Given AB = 7, AC = 9, and BC = 8.
M is the midpoint of BC, so BM = MC = BC/2 = 8/2 = 4.
Substitute the values into Apollonius's Theorem:
7^2 + 9^2 = 2(AM^2 + 4^2)
49 + 81 = 2(AM^2 + 16)
130 = 2(AM^2 + 16)
Divide by 2:
65 = AM^2 + 16
Subtract 16 from both sides:
AM^2 = 65 - 16
AM^2 = 49
AM = sqrt(49)
AM = 7.
Alternatively, using the Law of Cosines:
First, find cos(B) in triangle ABC:
AC^2 = AB^2 + BC^2 - 2(AB)(BC)cos(B)
9^2 = 7^2 + 8^2 - 2(7)(8)cos(B)
81 = 49 + 64 - 112cos(B)
81 = 113 - 112cos(B)
112cos(B) = 113 - 81
112cos(B) = 32
cos(B) = 32/112 = 2/7.
Now, apply the Law of Cosines to triangle ABM to find AM:
AM^2 = AB^2 + BM^2 - 2(AB)(BM)cos(B)
AM^2 = 7^2 + 4^2 - 2(7)(4)(2/7)
AM^2 = 49 + 16 - 2(4)(2)
AM^2 = 65 - 16
AM^2 = 49
AM = 7.