Problem 3 of the Canadian Mathematical Olympiad 1980

Among all triangles having (i) a fixed angle A and (ii) an inscribed circle of fixed radius r, determine which triangle has the least perimeter.

Solution by Steve Dinh, a.k.a. Vo Duc Dien

Let θ = ∠A, α = ∠B, γ = ∠C, r be the radius of the incircle, D, E and F be the points the incircle tangents with BC, CA and AB, respectively. Now let a = BC, b = AC, c = AB and d = AF = AE .

Note that BF = BD, CE = CD and we have

		a + b + c = 2d + 2BD + 2CD = 2d + 2(BD + CD) = 2d + 2a		(1)

Since ∠A and r are fixed, d is also fixed, and the minimum value of a + b + c is obtained when a is minimum.

From (1) a = b + c – 2d (2)

Now apply the law of the sine function, we obtain a/sinθ = b/sinα = c/sinγ or c = bsinγ/ inα

Substitute them into (2), we have a = b + bsinγ/sinα - 2d = b [(sinα + sinγ)/sinα] – 2d = a [(sinα + sinγ)/sinθ ] – 2d

or a = (2a /sinθ ) [ cos ½ (α – γ) sin ½ (α + γ) ] – 2d

or a = 2d sinθ / [ 2cos ½ (α – γ) sin ½ (α + γ) – sinθ ]

Since d, sinθ and sin ½ (α + γ) are all constants, a is minimum when cos ½ (α – γ) is maximum or when it’s equal to 1 or when α – γ = 0 or α = γ.

The triangle has the least perimeter when ∠B = ∠C as in triangle AIJ shown on the graph.