Three Semicircles

Borrowing Vreken's labelling, in this figure the variable is the slope $s$ of the line $CB$ . The radius $OD$ of the semicircle is $1$ , and $r$ is the radius of the semicircles to be maximized. $\;\;$ $\Delta CBQ$ is a right triangle.

Letting $O$ be the origin with coordinates $(0,0)$ , then the coordinates of $C$ is

$(Cx, Cy) = (1-r-rCsc(\theta),\; 0)$

The coordinates of $P$ is then

$(Px, Py) = ( 1-r-rCsc(\theta) + rCos(\theta), \; rSin(\theta))$

so that

$\sqrt { {(1-r-rCsc(\theta)+rCos(\theta))}^2 + {(rSin(\theta))}^2 } = 1-r$

If $s = Tan(\theta)$ , then

$r=\dfrac { 2s\left( 1-s+{ s }^{ 2 } \right) }{ 2s\left( 1-s+{ s }^{ 2 } \right) +(1-2s+2{ s }^{ 2 })\sqrt { 1+{ s }^{ 2 } } }$

The maximum value for $r$ occurs when $s$ is the real solution to the following cubic equation

$3s^3-3s^2+2s-1=0$

or $s = \dfrac{1}{3} \left( 1- \left( \dfrac{2}{5+\sqrt{29}} \right)^{\frac{1}{3}} + \left( \dfrac{1}{2} \left(5 + \sqrt{29} \right) \right) ^{\frac{1}{3}} \right)$

Plugging this value for $s$ into the expression for $r$ delivers the approximate answer $0.6098902004274422$

Label the diagram as follows, draw $AO$ and $PQ$ , let $y$ be the radius of the small semi-circle (so that $CP = AP = QB = QD = y$ ), let $x = CO$ , let $z = PB$ , and let $\theta = \angle OQP$ . Since $AO = OD = 1$ and $AP = QD = y$ , $PO = OQ = 1 - y$ , and since $\triangle POQ$ is an isosceles triangle, $\angle OPQ = \angle OQP = \theta$ , and by the exterior angle theorem of $\triangle OPQ$ , $\angle POC = 2\theta$ . Since $B$ is a point of tangency, $\angle B$ is a right angle.

By Pythagorean's Theorem on $\triangle CQB$ , $(y + z)^2 + y^2 = (x + 1 - y)^2$ , which rearranges to $z = \sqrt{(x + 1) (x - 2 y + 1)} - y$ .

By Pythagorean's Theorem on $\triangle PQB$ , $PQ = \sqrt{y^2 + z^2}$ .

By the law of cosines on $\triangle OPQ$ , $\cos \theta = \frac{(\sqrt{y^2 + z^2})^2 + (1-y)^2 - (1-y)^2}{2(1-y)\sqrt{y^2 + z^2}} = \frac{\sqrt{y^2 + z^2}}{2(1 - y)}$ .

By the law of cosines on $\triangle OPC$ , $\cos 2\theta = \frac{(1-y)^2 + x^2 - y^2}{2x(1-y)}$ .

Using cosine double angle formula $\cos 2 \theta = 2 \cos^2 \theta - 1$ and combining the above equations gives $(1 -y)((1 -y)^2 + x^2 - y^2) = xy^2 + x(\sqrt{(x + 1)(x - 2y + 1)} - y)^2 - 2x(1 - y)^2$ , which can be found numerically to have a maximum $y$ -value of $y \approx \boxed{0.60989}$ .

$7r^3-12r^2+8r-2=0~has ~one~ real~solution. ~~~~r=0.60687......$