Multiple choice

The sides of a triangle inscribed in a given circle subtend angles $\alpha, \beta, \gamma$ at the centre. The minimum value of the A.M. of $\cos\left(\alpha +\dfrac{\pi}{2}\right), \cos \left(\beta +\dfrac{\pi}{2}\right)$ and $\cos\left(\gamma +\dfrac{\pi}{2}\right)$ is equal to?

  1. $-\dfrac{\sqrt{3}}{2}$
  2. $\dfrac{\sqrt{3}}{2}$
  3. $-\dfrac{2}{\sqrt{3}}$
  4. $\sqrt{2}$
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A Correct answer
Explanation

The angles subtended at the center by sides of an inscribed triangle satisfy alpha + beta + gamma = 2*pi. The expression is (1/3) * (cos(alpha + pi/2) + cos(beta + pi/2) + cos(gamma + pi/2)) = -(1/3) * (sin(alpha) + sin(beta) + sin(gamma)). For a triangle inscribed in a circle, the sum of sines of central angles is maximized for an equilateral triangle (alpha=beta=gamma=2*pi/3). sin(2*pi/3) = sqrt(3)/2. Sum = 3 * sqrt(3)/2. Average = -(1/3) * 3 * sqrt(3)/2 = -sqrt(3)/2.

AI explanation

Since the three vertices of the triangle lie on the circumference, the central angles must sum to 2pi, meaning alpha + beta + gamma = 2pi. Using the phase shift identity, cos(alpha + pi/2) equals -sin(alpha), so the arithmetic mean is (-sin(alpha) - sin(beta) - sin(gamma)) / 3. Letting the central angles be equal (each 2pi/3) minimizes this mean, yielding -sin(2pi/3), which is -sqrt(3)/2.

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