Problem

As shown in Figure 1, M is the middle point of AC. C is moving along the half unit circle in the first quadrant, \(BC = 1\). What is the max OM?

Solution

Assume \(OM=r\) and \(\angle MOA= \theta\). Then the coordinates of \(M\) is \((r\cos\theta, r\sin\theta)\). Assume the coordinates of \(C\) are \((x_1, y_1)\). Since \(M\) is the middle point of \(AC\),

\[ r\cos\theta = \frac{x_1 + 3}{2}, \; \; \; \; \; \; r\sin\theta = \frac{y_1}{2} \]

Since \(BC=1\), \((2r\cos\theta-3)^2 + (2r\sin\theta - 4)^2 = 1\). Thus we can solve for r (only + solution is kept, since we want to find max r),

\[ r = \frac{1}{2} \left((3\cos\theta + 4\sin\theta) + \sqrt{(3\cos\theta + 4\sin\theta)^2 -24}\right) \] Since \(3\cos\theta + 4\sin\theta=5\times(\frac{3}{5}\cos\theta + \frac{4}{5}\sin\theta)=5\sin(x+\theta)\), the max value of \(3\cos\theta + 4\sin\theta\) is 5. Thus the max of r is 3.