Eigenman's problem I-15

Follow the angles, using the isoceles triangles in this figure

Angle $\alpha$ is equal to $90-\dfrac{3}{4}\beta$ , where $\beta=64$ for this problem

Let the radii of the small circle and big circle be 1 and $r$ respectively. Since $AB$ is a diameter, $\angle ADB = 90^\circ$ . Let $BD=a$ and $AD=b$ . Then by cosine rule , we have:

$\begin{cases} a^2 = (r-1)^2 + 1 - 2(r-1)\cos 64^\circ & ...(1) \\ b^2 = (r+1)^2 + 1 {\color{#3D99F6}\ + 2(r+1)\cos 64^\circ} & ...(2) & \small \color{#3D99F6} \text{Note that }\cos (180^\circ - \theta) = - \cos \theta \end{cases}$

By Pythagorean theorem , $a^2 + b^2 = (2r)^2$ :

$\begin{aligned} (1)+(2): \quad 4r^2 & = 2r^2 + 2 + 2 + 4\cos 64^\circ \\ r^2 & = 2(1+\cos 64^\circ) = 2(1+2\cos^2 32^\circ -1) \\ \implies r & = 2\cos 32^\circ \end{aligned}$

Now we note that $\angle COP = 2\alpha$ and $\angle OCP = 180^\circ - 64^\circ - 2\alpha = 116^\circ - 2\alpha$ . By sine rule :

$\begin{aligned} \frac {\color{#3D99F6}\sin (116^\circ - 2\alpha)}1 & = \frac {\sin 64^\circ}{\color{#D61F06}r} & \small \color{#3D99F6} \text{Note that }\sin (180^\circ - \theta) = \sin \theta \\ \color{#3D99F6}\sin (64^\circ + 2\alpha) & = \frac {\sin 64^\circ}{\color{#D61F06}\cos 32^\circ} & \small \color{#3D99F6} \text{And }r = \cos 32^\circ \\ 2\sin (32^\circ + \alpha)\cos (32^\circ + \alpha) & = \frac {2\sin 32^\circ \cos 32^\circ}{\cos 32^\circ} \\ \cos(58^\circ - \alpha)\cos (32^\circ+\alpha) & = \cos 58^\circ \\ \cos (2\alpha - 26^\circ) + \cos 90^\circ & = \cos 58^\circ \\ \cos (2\alpha - 26^\circ) & = \cos 58^\circ \\ \implies 2\alpha - 26^\circ & = 58^\circ \\ \alpha & = \boxed{42}^\circ \end{aligned}$