**Problem 1 of the Canadian Mathematical Olympiad 1986**

In the diagram line segments AB and CD are of length 1 while angles ABC and CBD are 90° and 30° respectively. Find AC.

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

We have AC² = 1 + BC² and BC/ sinD = CD /sin30° = 2 or sinD = ½ BC

We also have (AC + 1)/ sin120° = AB /sinD = 1/sinD = 2/BC

or (AC + 1)/ \sqrt{3} = 1/ \sqrt{AC² - 1} or (AC + 1)³ (AC – 1) = 3 or

(AC²)² + 2AC³ -2AC - 4 = 0 or AC(AC³ - 2) + 2(AC³ - 2) = 0 or (AC³ - 2)(AC + 2) = 0 but AC + 2 > 0;

therefore, AC³ - 2 = 0 and AC = \sqrt[3]{2}.