Four Rounders (corrected)

Let $r$ denote the radius of an inner circle. Sangaku tells us that the diameter of each of the two adjacent circles is twice the difference of $|\overline{BD}|$ and $|\overline{DE}|$ . For this case, if all circles have the common radii, then

$\begin{aligned} 2r &= 2\left(|\overline{BD}| - |\overline{DE}|\right) = 2\left(|\overline{BD}| - 2r\right)\\ r &= |\overline{BD}| - 2r\\ |\overline{BD}| &= 3r \end{aligned}$

Then, by Pythagorean Theorem, $\sqrt{|\overline{BD}|^2 - |\overline{DE}|^2} = \sqrt{5}r$ . Therefore, the intersecting chord theorem tells us

$\begin{aligned} |\overline{BE}| \cdot |\overline{EC}| &= |\overline{AE}| \cdot |\overline{ED}| = |\overline{AE}| \cdot \left(|\overline{AD}| - |\overline{ED}|\right)\\ \left(\sqrt{5}r\right)^2 &= 2r(2 - 2r)\\ 5r^2 &= 4r - 4r^2\\ 0 &= 4r - 9r^2\\ 0 &= r(4 - 9r)\\ r &= \dfrac{4}{9} \text{ as }r\text{ is nonzero} \end{aligned}$

So for this problem, $p = 4$ and $q = 9$ , giving us $p + q = \boxed{13}$ .

Let the radius of the three congruent circles be $r$ , the centers of the unit circle and the bottom right circle be $O$ and $P$ respectively, and the tangent point of the two circles be $Q$ . Note that the line perpendicular to the tangent line at $Q$ passes through $P$ and $O$ , therefore $O$ , $P$ , and $Q$ are colinear.

Let $OQ$ intersect the circle centered $P$ at $R$ and the tangent point of the bottom two circles be $S$ . By tangent-secant theorem , we have:

$\begin{aligned} OS^2 & = OR \cdot OQ \\ (3r-1)^2 & = (1-2r) \cdot 1 \\ 9r^2 - 6r + 1 & = 1 - 2r \\ 9r^2 - 4r & = 0 & \small \blue{\text{Since }r>0} \\ 9r - 4 & = 0 \\ \implies r & = \frac 49 \end{aligned}$

Therefore $p+q = 4+9 = \boxed{13}$ .

Referring to the figure, points $E$ , $M$ and $G$ are points of contact of the circles and the cords. Let $O$ and $K$ be the centers of the big circle and the lower-left circle respectively. Let $R$ be the radius of the small circles and let $L$ be the point of contact of the lower-left circle with the big circle. Line $AD$ through the center of the upper small circle and the point of contact of the other two small circles is an axis of symmetry of the compound figure of the four circles and cord $BC$ . Hence, $BC\bot AD$ . Also, $KG\bot AD$ .

Now, $OM=OA-AM=1-2R$

$OG=MG-OM=EK-OM=R-\left( 1-2R \right)\Rightarrow OG=3R-1$

$OK=OL-KL=1-R$ Using Pythagorean theorem on $\triangle OGK$ ,

$\begin{aligned} O{{K}^{2}}=O{{G}^{2}}+K{{G}^{2}} & \Leftrightarrow {{\left( 1-R \right)}^{2}}={{\left( 3R-1 \right)}^{2}}+{{R}^{2}} \\ & \Leftrightarrow 9{{R}^{2}}-4R=0 \\ & \overset{R\ne 0}{\mathop{\Leftrightarrow }}\,R=\frac{4}{9} \\ \end{aligned}$ For the answer, $p+q=4+9=\boxed{13}$ .

Let $r$ be the radius of an inner circle and label the diagram as follows:

Then $HI = r$ , $OI = OE - IE = 1 - r$ , and $OH = FH - OF = GI - (AO - AF) = r - (1 - 2r) = 3r - 1$ .

By the Pythagorean Theorem on $\triangle OHI$ , $r^2 + (3r - 1)^2 = (1 - r)^2$ , which solves to $r = \frac{4}{9}$ for $r > 0$ . Therefore, $p = 4$ , $q = 9$ , and $p + q = \boxed{13}$ .