If a triangle has two vertices on the boundary of the unit circle and another vertex is the center of the circle, what is the biggest possible area of the triangle?
Solution 1: Use the formula \(S=\frac{1}{2}ab\sin\theta\). Where \(\theta\) is the angle in between edges with lengths \(a, b\).
Let \(\theta\) be the central angle formed by two sides of the triangle, then the area of the triangle is \(\frac{1}{2}\sin\theta\). So when \(\theta=\frac{\pi}{2}\), the triangle get the maximum possible area \(\frac{1}{2}\).
Solution 2: Use the inequality \(ab\leq\frac{a^2+b^2}{2}\), where the equal sign holds if and only if \(a=b\).
Let the chord has length \(2x\), then by Pythagorean theorem, the height of the triangle with base to be this chord is \(\sqrt{1-x^2}\). So the area of the triangle is \(x\sqrt{1-x^2}\leq\frac{x^2+(1-x^2)}{2}=\frac{1}{2}\). The equal sign holds if and only if \(x=\sqrt{1-x^2}\). Solve to get \(x=\frac{\sqrt 2}{2}\). So solution exist and the maximum value can be attained.
Solution 3: Same as solution 2, but use calculus to find the extreme value.