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+bβ+cβ, 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 Ξ±βΞ΄,Ξ±, and Ξ±+Ξ΄. Then 3Ξ±=Ξ±βΞ΄+Ξ±+Ξ±+Ξ΄=180β, so Ξ±=60β. There are three cases depending on which side is opposite to the 60β angle. Suppose that the triangle is ABC with β BAC=60β. Let D be the foot of the altitude from C. The triangle CAD is a 30-60-90 triangle, so AD=21βAC and CD=23ββAC. 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=23β, and BD=β£ABβADβ£=3. It follows that x2=BC2=CD2+BD2=21, so x=21β.
If AB=x,AC=4, and BC=5, then AD=2,CD=23β, and BD=β£ABβADβ£=β£xβ2β£. It follows that 25=BC2=CD2+BD2=12+(xβ2)2, and the positive solution is x=2+13β.
If AB=x,AC=5, and BC=4, then AD=25β,CD=253ββ, and BD=β£ABβADβ£=β£β£β£β£β£βxβ25ββ£β£β£β£β£β. It follows that 16=BC2=C2+BD2=2475β+(xβ25β)2, which has no solution because 475β>16.
The sum of all possible side lengths is 2+13β+21β. The requested sum is 2+13+21=36β.
OR
As in the first solution, there are three cases depending on which side is opposite to the 60β 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 60β angle, then
x2=42+52β2β 4β 5β cos(60β)=21,
so x=21β.
If the side of length 5 is opposite to the 60β angle, then
52=x2+42β2β 4β xβ cos(60β)=x2β4x+16
and the positive solution is 2+13β.
If the side of length 4 is opposite to the 60β angle, then
42=x2+52β2β xβ 5β cos(60β)=x2β5x+25
which has no real solutions.
The sum of all possible side lengths is 2+13β+21β. The requested sum is 2+13+21=36β.