Circles in a rectangle: question 2

Let the radius of the red circle be $r_r$ , the radius of the yellow circle be $r_y$ , and the radius of the brown circle be $r_b$ and label the diagram as follows:

From right $\triangle JBD$ with sides $JB = BD = 1$ , $JD = \sqrt{2}$ , and from right $\triangle LCD$ with sides $LC = CD = r_r$ , $LD = \sqrt{2}r_r$ . Since $JK + KL + LD = JD$ , $1 + r_r + \sqrt{2}r_r = \sqrt{2}$ , which solves to $r_r = (\sqrt{2} - 1)^2$ .

By the kissing circles theorem on the blue, yellow, and red circles, $\cfrac{1}{\sqrt{JB}} + \cfrac{1}{\sqrt{ME}} = \cfrac{1}{\sqrt{LC}}$ , or $\cfrac{1}{\sqrt{1}} + \cfrac{1}{\sqrt{r_y}} = \cfrac{1}{\sqrt{(\sqrt{2} - 1)^2}}$ , which solves to $r_y = \frac{1}{2}$ .

By the Pythagorean Theorem on $\triangle LMN$ , $LN = \sqrt{LM^2 - MN^2} = \sqrt{(r_y + r_r)^2 - (r_y - r_r)^2} = 2 \sqrt{r_yr_r} = 2 \sqrt{\frac{1}{2} \cdot (\sqrt{2} - 1)^2} = 2 - \sqrt{2} = CE$ .

That means $AF = AD - CD + CE + EF = 2 - (\sqrt{2} - 1)^2 + (2 - \sqrt{2}) + \frac{1}{2} = \frac{3}{2} + \sqrt{2}$ , $BF = AF - AB = \frac{3}{2} + \sqrt{2} - 1 = \frac{1}{2} + \sqrt{2}$ , and $BE = BF - EF = \frac{1}{2} + \sqrt{2} - \frac{1}{2} = \sqrt{2} = OM$ . Also, $JO = JB - OB = JB - ME = 1 - \frac{1}{2} = \frac{1}{2}$ .

$\angle JMO$ and $\angle JTB$ are corresponding angles to parallel lines, so $\angle JMO = \angle JTB$ . Since $TR$ and $TE$ are tangent lines to the red circle, and since $TM$ passes through the center of the red circle, $\angle RTM = \angle MTE = \angle JTB$ . Since $\angle WUQ$ and $\angle RTE$ are alternate interior angles to parallel lines, $\angle WUQ = \angle RTE$ . Since $UW$ and $UQ$ are tangent lines to the brown circle, and since $UV$ passes through the center of the brown circle, $\angle WUV = \angle QUV$ . Therefore, if $\theta = \angle JMO$ , then $\theta = \angle JMO = \angle JTB = \angle WUV = \angle QUV$ , and $2\theta = \angle RTE = \angle GUS$ .

From $\triangle JMO$ , $\tan \theta = \cfrac{JO}{OM} = \cfrac{1}{2\sqrt{2}}$ , which means $\tan 2\theta = \cfrac{2\tan \theta}{1 - \tan^2 \theta} = \cfrac{2 \cdot \frac{1}{2\sqrt{2}}}{1 - (\frac{1}{2\sqrt{2}})^2} = \cfrac{4\sqrt{2}}{7}$ .

Since $\triangle JMO \cong \triangle MTE$ by AAS congruence, $ET = OM = \sqrt{2}$ . Therefore, $FT = ET - EF = \sqrt{2} - \frac{1}{2}$ .

From $\triangle TFS$ , $SF = FT \cdot \tan 2\theta = (\sqrt{2} - \frac{1}{2}) \cdot \frac{4\sqrt{2}}{7} = \frac{1}{7}(8 - 2\sqrt{2})$ , so $GS = GF - SF = 2 - \frac{1}{7}(8 - 2\sqrt{2}) = \frac{1}{7}(2 \sqrt{2} + 6)$ .

From $\triangle UGS$ , $UG = \cfrac{GS}{\tan 2\theta} = \cfrac{\frac{1}{7}(2 \sqrt{2} + 6)}{\frac{4\sqrt{2}}{7}} = \frac{1}{4}(2 + 3\sqrt{2})$ .

From $\triangle UWV$ , $\tan \theta = \cfrac{WV}{UW}$ , or $\cfrac{1}{2\sqrt{2}} = \cfrac{r_b}{\frac{1}{4}(2 + 3\sqrt{2}) - r_b}$ , which solves to $r_b = \cfrac{10 + \sqrt{2}}{28}$ .

Therefore, $a = 10$ , $b = 2$ , $c = 28$ , and $a + b + c = \boxed{40}$ .

Let the center of the blue circle be the origin $O(0,0)$ of the $xy$ -plane. It is given that the radius of the blue circle $r_1 = 1$ . In the previous problem we found that the radius of the yellow triangle $r_2 = \dfrac 12$ and the horizontal distance between the centers of blue and yellow circles $O'Q = 2\sqrt{r_1r_2} = \sqrt 2$ . Then the coordinates of the right top vertex of the rectangle are $B \left(\sqrt 2+\dfrac 12, 1\right)$ .

Let the tangent point of the black line and the blue circle be $P$ . Then $\angle OPR$ is given by:

$\begin{aligned} \theta & = \tan^{-1} \frac {2\sqrt{r_1r_2}}{r_1-r_2} - \tan^{-1} \frac {r_2}{2\sqrt{r_1r_2}} & \small \red{\text{Note that the three red dash lines length is }2\sqrt{r_1r_2}} \\ & = \tan^{-1} 2\sqrt 2 - \tan^{-1} \frac 1{2\sqrt 2} \end{aligned}$

$\begin{aligned} \tan \theta & = \frac {2\sqrt 2 - \frac 1{2\sqrt 2}}{1+1} = \frac 7{4\sqrt 2} \implies \sin \theta = \frac 79 \implies \cos \theta = \frac {4\sqrt 2}9 \end{aligned}$

Then $P\left(\dfrac {4\sqrt 2}9, \dfrac 79 \right)$ and $DP = 1 - \dfrac 79 = \dfrac 29$ . Note that $\triangle ADP$ and $\triangle OPR$ are similar. Then $\dfrac {AD}{DP} = \dfrac 7{4\sqrt 2} \implies AD = \dfrac 7{4\sqrt 2} \times \dfrac 29 = \dfrac 7{18\sqrt 2} = \dfrac {7\sqrt 2}{36}$ . And $AB = \sqrt 2 + \dfrac 12 - \dfrac {4\sqrt 2}9 + \dfrac {7\sqrt 2}{36} = \dfrac {3\sqrt 2+2}4$ . Again $\triangle ABC$ and $\triangle OPR$ are similar and $BC = \dfrac {4\sqrt 2}7 \times AB = \dfrac {\sqrt 2(3\sqrt 2 + 2)}7$ . Then $CA$ is given by:

$\begin{aligned} CA & = \sqrt{AB^2+BC^2} = \sqrt{\frac {(3\sqrt 2+1)^2}{4^2}+\frac {2(3\sqrt 2+2)^2}{7^2}} = \frac {9(3\sqrt 2+2)}{28} \end{aligned}$

The semiperimeter is as follows:

$\begin{aligned} s & = \frac {AB+BC+CA}2 = \frac {3\sqrt 2+2}2\left(\frac 14 + \frac {\sqrt 2}7 + \frac 9{28} \right) = \frac {3\sqrt 2 + 2}2 \times \frac {4+\sqrt 2}7 = \sqrt 2 + 1 \end{aligned}$

The are of $\triangle ABC$ , $A = \dfrac 12 AB \times BC = \dfrac 12 \times \dfrac {3\sqrt 2+2}4 \times \dfrac {\sqrt 2(3\sqrt 2+2)}7 = \dfrac {11\sqrt 2+12}{28}$

The brown circle is an incircle and its radius is given by:

$\begin{aligned} r & = \frac As = \frac {11\sqrt 2+12}{28(\sqrt 2 + 1)} = \frac {(11\sqrt 2+12)(\sqrt 2 -1)}{28} = \frac {10+\sqrt 2}{28} \end{aligned}$

Therefore $a+b+c = 10 + 2 + 28 = \boxed{40}$ .