Summer Problem Solving Marathon Solution #16

Because the sides of the triangle are consecutive integers, we can call them x, x+1, and x+2 for some integer x.

We start with the law of sines. Call the three angles of the triangle A, B, and C, and let a, b, and c be the lengths of the sides opposite angles A, B, and C respectively. Then the law of sines says:

^a⁄_{sin A} = ^b⁄_{sin B} = ^c⁄_{sin C}

In our triangle, the largest angle is twice the smallest angle. Call the smallest angle θ. Then the largest angle is 2θ. The smallest angle is opposite the shortest side, and the largest angle is opposite the longest side. So from the law of sines, we have:

^x⁄_{sin θ} = ^x+2⁄_{sin 2θ}

Next we use the fact that sin 2θ = 2sinθcosθ. Making this substitution, we get:

^x⁄_{sin θ} = ^x+2⁄ _2sinθcosθ

Then multiplying both sides by sinθ, we have:

x = ^x+2⁄ _2cosθ

So cosθ = ^x+2⁄_2x

Now we appeal to the law of cosines. The law of cosines says:

c² = a² + b² - (2ab)cos C

We'll take the instance in which C is the smallest angle (θ). Then c = x, and a and b are x+1 and x+2. So we have:

x² = (x+1)² + (x+2)² - 2(x+1)(x+2)cos θ

But we already know that cosθ = ^x+2⁄_2x. Making this substitution, we have:

x² = (x+1)² + (x+2)² - 2(x+1)(x+2)(^x+2⁄_2x)

Or:

x² = x² + 2x + 1 + x² + 4x + 4 - (x³ + 5x² + 8x + 4)/x

Or (multiplying both sides by x):

x³ = x³ + x² - 3x - 4

Or:

x² - 3x - 4 = 0

Or:

(x - 4)(x + 1) = 0

So x = 4 or x = -1. But since x is the length of the shortest side, it cannot be negative. So x = 4. Thus the cosine of the shortest side is (4 + 2)/8 = 3/4.