Not Even Sangaku!

Draw the diagram as follow:

Without loss of generality, since $3$ green circles are respectively tangent to both red and blue circles at the same points, we can assume that the green circles are smaller than the red circle, and that none of the region takes place outside of the red circle. In this case, $\overline{BO_2}$ is the chord of the green circle whose diameter overlaps the radius of the red circle.

Here are the following variables for the problem:

• $O_1$ , $O_2$ and $O_3$ are centers of the red, blue and green circles, respectively.
• $A$ , $B$ and $C$ are distinct intersection points.

The first step is to determine the green circle's radius. Then, we determine the area of the whole tomoe.

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I. Finding Green Circle's Radius

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Since $|AO_2| = 11$ , the radius of the blue circle, and $|O_1O_2| = 24$ , the radius of the red circle, $|O_1O_2| = |AO_1| - |AO_2| = 13$ . By law of cosines , noting that $\angle AO_1B = 120^{\circ}$ , $\begin{array}{rl} |BO_2|^2 &= |O_1O_2|^2 + |O_1B|^2 - 2|O_1O_2||O_1B|\cos\left(120^{\circ}\right)\\ |BO_2| &= \sqrt{1057} \end{array}$ Second, by law of sines , from $\dfrac{\sin\left(AO_1B\right)}{|BO_2|} = \dfrac{\sin\left(\angle O_1BO_2\right)}{|O_2O_1|}$ we find that $\angle O_1BO_2 = \arcsin\left(\dfrac{13\sqrt{3171}}{2114}\right)$ . Then, since $\Delta BO_2O_3$ is an isosceles triangle (so there are two congruent right triangles in that triangle), $|BO_3| = |O_2O_3| = \dfrac{1057}{61}$ , which is the radius of the green circle.

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II. Computing Areas of Tomoes

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Most of the work involves computing the areas of the regions. Because the circles' centers are not located at the same position, there are few regions to consider:

• Swirl region $ABO_2$ : The area is found by determining the arc sector of $ABO_2$ and removing it by the area of the curved shape $BCO_2O_1$ .
• Minor segment $CO_2$ : Because the swirl region does not include the minor segment, this component is considered.
• Arc sector $AO_2C$ with measurement greater than $180^{\circ}$ : With minor segment and swirl region areas both computed, the final step is to compute the remaining area of the sector region.

II.i. Swirl Region

For the swirl region, since large circular sector $AO_1B$ includes $\Delta O_1O_2O_3$ and the small circular sector $O_2O_3B$ , the area of the swirl region is $A_{\text{swirl}} = {\color{#D61F06}A_{\text{sec}_1}} - \left( A_{\Delta O_1O_2O_3} + {\color{#20A900}A_{\text{sec}_3}}\right)$ where ${\color{#D61F06}A_{\text{sec}_1}}$ is the area of the large sector and ${\color{#20A900}A_{\text{sec}_3}}$ is the area of the small sector. Instead of simplifying the whole expression (which is quite messy!), the sub-goal is to determine the areas of these three regions.

For ${\color{#D61F06}A_{\text{sec}_1}}$ , since the angle measurement of the sector is $120^{\circ}$ , ${\color{#D61F06}A_{\text{sec}_1}} = \dfrac{\pi}{3} \cdot 24^2 = 576\pi$ For ${\color{#20A900}A_{\text{sec}_3}}$ , $\angle BO_3O_2 = \left(180 - 2\arcsin\left(\dfrac{13\sqrt{3171}}{2114}\right)\right)^{\circ}$ . Then, in radians, ${\color{#20A900}A_{\text{sec}_3}} = \dfrac{180 - 2\arcsin\left(\frac{13\sqrt{3171}}{2114}\right)}{2} \cdot \left(\dfrac{1057}{61}\right)^2$

Since the side lengths of $\Delta O_1O_2O_3$ are $\sqrt{1057}$ , $13$ and $|O_1O_3| = 24 - |BO_3| = \dfrac{407}{61}$ , by Heron's formula , $A_{\Delta O_1O_2O_3} = \dfrac{5291\sqrt{3}}{244}$ .

II.ii. Minor Segment

Since $\Delta CO_2O_3$ is an isosceles triangle of base lengths $\dfrac{1057}{61}$ and side length $11$ , $\angle CO_3O_2 = 2\arcsin\left(\dfrac{671}{2114}\right)$ . The area of the minor segment $CO_2$ is $A_{\text{minor}} = \arcsin\left(\dfrac{671}{2114}\right)\left(\dfrac{1057}{61}\right)^2 - \dfrac{1}{2}\left(\dfrac{1057}{61}\right)^2\sin\left(2\arcsin\left(\dfrac{671}{2114}\right)\right)$

II.iii. Arc sector blue region

The final steps is to observe the angles in $\Delta CO_3O_2$ and $O_1O_2O_3$ , which are both used to determine the measurement of the sector. For $\Delta CO_3O_2$ , $\angle CO_2O_3 = \arccos\left(\dfrac{671}{2114}\right)$ , and for $\Delta O_1O_2O_3$ , $\angle O_1O_2O_3 = \arccos\left(\dfrac{1993}{2114}\right)$ . Therefore, in radians,

${\color{#3D99F6}{A_{\text{sec}_2}}} = \dfrac{\angle O_1O_2O_3 + \angle CO_2O_3 + \pi}{2}(11)^2$

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III. Final Results

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The total area of all three tomoes is approximately $1478.68$ . Thus, $\left\lfloor A \right\rfloor = \boxed{1478}$ .