Problem:
The angles in a particular triangle are in arithmetic progression, and the side lengths are 4,5 , and x. The sum of the possible values of x equals a+b+c, where a,b, and c are positive integers. What is a+b+c ?
Answer Choices:
A. 36
B. 38
C. 40
D. 42
E. 44 Solution:
Let the angles of the triangle be α−δ,α, and α+δ. Then 3α=α−δ+α+α+δ=180∘, so α=60∘. There are three cases depending on which side is opposite to the 60∘ angle. Suppose that the triangle is ABC with ∠BAC=60∘. Let D be the foot of the altitude from C. The triangle CAD is a 30-60-90 triangle, so AD=21AC and CD=23AC. There are three cases to consider. In each case the Pythagorean Theorem can be used to solve for the unknown side.
If AB=5,AC=4, and BC=x, then AD=2,CD=23, and BD=∣AB−AD∣=3. It follows that x2=BC2=CD2+BD2=21, so x=21.
If AB=x,AC=4, and BC=5, then AD=2,CD=23, and BD=∣AB−AD∣=∣x−2∣. It follows that 25=BC2=CD2+BD2=12+(x−2)2, and the positive solution is x=2+13.
If AB=x,AC=5, and BC=4, then AD=25,CD=253, and BD=∣AB−AD∣=∣∣∣∣∣x−25∣∣∣∣∣. It follows that 16=BC2=C2+BD2=2475+(x−25)2, which has no solution because 475>16.
The sum of all possible side lengths is 2+13+21. The requested sum is 2+13+21=36.
OR
As in the first solution, there are three cases depending on which side is opposite to the 60∘ angle. In each case, the Law of Cosines can be used to solve for the unknown side. If the unknown side is opposite to the 60∘ angle, then
x2=42+52−2⋅4⋅5⋅cos(60∘)=21,
so x=21.
If the side of length 5 is opposite to the 60∘ angle, then
52=x2+42−2⋅4⋅x⋅cos(60∘)=x2−4x+16
and the positive solution is 2+13.
If the side of length 4 is opposite to the 60∘ angle, then
42=x2+52−2⋅x⋅5⋅cos(60∘)=x2−5x+25
which has no real solutions.
The sum of all possible side lengths is 2+13+21. The requested sum is 2+13+21=36.