Clockwise and counterclockwise triangles

This is a partial solution

I decided to solve for the area of $\Delta KHL$ only. The area of $\Delta ABC$ is obvious, for the area of $\Delta GHI$ I have a link to a youtube video that explains the solution and is also similar to this approach. Also if you have solved the last question of the AMC 12B, you will probably get this.

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Let $AK \cap BL = X$

$m\angle AKL = 90^{\circ}$ and $m\angle BLH = 90^{\circ}$ since $KL$ and $HL$ are tangents.

Also $m\angle KLH = 60^{\circ}$ thus $m\angle KLX = 30^{\circ}$ giving $m\angle LKX = 60^{\circ}$ and $m\angle AKB = 120^{\circ}$ .

Let $XK = a$ , thus $KL = a\sqrt{3}$ and $XL=2a.$ Also let $AK=BL=b$ then $AB=2b$ , $AX = b-a$ and $XB=2a+b$ .

By Law of Cosines on $\angle AXB$ , we get ${(2b)}^2 = {(b-a)}^2 + {(2a+b)}^2 -2(b-a)(2a+b)\cos120$ .

Solving for $a$ in terms of $b$ , we get $a= \dfrac{b\sqrt{21} - 3b}{6}.$ Since $KL=LH=KH=a\sqrt{3}= \dfrac{b\sqrt{63} - 3\sqrt{3}b}{6}$

we get the area of $\Delta KLH$ as $\dfrac{1}{2}({5\sqrt{3} - 3\sqrt{7}})b^2$ .

Also we get the area of $\Delta ABC$ and $\Delta GHI$ as $4b^2\sqrt{3}$ and $\dfrac{1}{2}({5\sqrt{3} + 3\sqrt{7}})b^2$ .

Performing the required division above gives $\dfrac{5}{4}$ .

$\therefore A+B = \boxed{9}.$

To find the area $S_{GJI}$ of an equilateral triangle $\triangle GJI$ , we will let $M$ be the intersection of $AB$ and $IJ$ and examine $\triangle AJM$ and $\triangle BIM$ .

Let $r$ be the radius of any of the three congruent circles. Then $AJ = BI = r$ and $AB = 2r$ . Since $IJ$ is tangent to circle $B$ , $\angle BIJ = 90°$ . Since $JG$ is tangent to circle $A$ , $\angle AJG = 90°$ , and since $\triangle IJG$ is an equilateral triangle, $\angle IJG = 60°$ , which means $\angle AJI$ $=$ $\angle AJG - \angle IJG$ $=$ $90° - 60°$ $=$ $30°$ .

By law of sines on $\triangle AJM$ , $\frac{\sin \angle AMJ}{AJ} = \frac{\sin \angle AJM}{AM}$ or $\frac{\sin \angle AMJ}{r} = \frac{\sin 30°}{AM}$ , and by law of sines on $\triangle BIM$ , $\frac{\sin \angle BMI}{BI} = \frac{\sin \angle BIM}{MB}$ or $\frac{\sin \angle BMI}{r} = \frac{\sin 90°}{MB}$ . Since $\angle AMJ = \angle BMI$ as vertical angles, we can combine the two ratios and arrive at $\frac{\sin 30°}{AM} = \frac{\sin 90°}{MB}$ , and simplifying this gives $MB = 2AM$ . Since $AM + MB = AB = 2r$ , $AM + 2AM = 2r$ , so $AM = \frac{2}{3}r$ , and $MB = 2AM = 2\frac{2}{3}r = \frac{4}{3}r$ .

Using $\frac{\sin \angle AMJ}{r} = \frac{\sin 30°}{AM}$ once again along with $AM = 2r$ , we find that $\angle AMJ = \sin^{-1}\frac{3}{4}$ . Since the sum of the angles of $\triangle AJM$ is $180°$ , $\angle MAJ$ $=$ $180° - \angle AJM - \angle AMJ$ $=$ $180° - 30° - \sin^{-1}\frac{3}{4}$ $=$ $150° - \sin^{-1}\frac{3}{4}$ . Therefore, $\sin \angle MAJ$ $=$ $\sin (150° - \sin^{-1}\frac{3}{4})$ $=$ $\sin 150° \cos(\sin^{-1}\frac{3}{4}) - \cos 150° \sin(\sin^{-1}\frac{3}{4})$ $=$ $\frac{\sqrt{7} + 3\sqrt{3}}{8}$ .

My law of sines on $\triangle AJM$ once again, $\frac{\sin \angle AJM}{AM} = \frac{\sin \angle MAJ}{MJ}$ or $\frac{\sin 30°}{\frac{2}{3}r} = \frac{\frac{\sqrt{7} + 3\sqrt{3}}{8}}{MJ}$ , which means $MJ = (\frac{\sqrt{7} + 3\sqrt{3}}{6})r$ .

My Pythagorean's Theorem, $MI = \sqrt{MB^2 - BI^2} = \sqrt{(\frac{4}{3}r)^2 - r^2} = \frac{\sqrt{7}}{3}r$ .

Since $IJ = MI + MJ$ , $IJ$ $=$ $\frac{\sqrt{7}}{3}r + (\frac{\sqrt{7} + 3\sqrt{3}}{6})r$ $=$ $\frac{\sqrt{7} + \sqrt{3}}{2}r$ .

The area $S_{GJI}$ of an equilateral triangle $\triangle GJI$ formed by side lengths $IJ$ would then be $A$ $=$ $\frac{\sqrt{3}}{4}(\frac{\sqrt{7} + \sqrt{3}}{2}r)^2$ $=$ $\frac{5\sqrt{3} + 3\sqrt{7}}{8}r^2$ .

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To find the area $S_{KLH}$ of an equilateral triangle $\triangle KLH$ , we will examine $\triangle AKF$ .

If $r$ is the radius of any of the three congruent circles, then $AK = r$ . Also, since $A$ is the vertex of equilateral triangle $\triangle ABC$ with sides $2r$ , and $F$ is the center of that equilateral triangle, $AF = \frac{2\sqrt{3}}{3}r$ . Since $KL$ is tangent to circle $A$ , $\angle AKL = 90°$ , and since $\triangle KLH$ is an equilateral triangle, $\angle LKH = 60°$ , and since $F$ is the center of equilateral triangle $\triangle KLH$ , $\angle LKF = 30°$ , which means $\angle AKF$ $=$ $\angle AKL + \angle LKF$ $=$ $90° + 30°$ $=$ $120°$ .

By the law of cosines on $\triangle AKF$ , we have $AF^2 = AK^2 + FK^2 - 2 \cdot AK \cdot FK \cdot \cos \angle AKF$ or $(\frac{2\sqrt{3}}{3}r)^2 = r^2 + FK^2 - 2 \cdot r \cdot FK \cdot \cos 120°$ , which solves to $FK = \frac{\sqrt{21} - 3}{6}r$ .

Since $K$ is the vertex of equilateral triangle $\triangle KLH$ , and $F$ is the center of that equilateral triangle, $KL = \sqrt{3}FK$ . Therefore, since $FK = \frac{\sqrt{21} - 3}{6}r$ , $KL = \frac{\sqrt{7} - \sqrt{3}}{2}r$ .

The area $S_{KLH}$ of an equilateral triangle $\triangle KLH$ formed by side lengths $KL$ would then be $A$ $=$ $\frac{\sqrt{3}}{4}(\frac{\sqrt{7} - \sqrt{3}}{2}r)^2$ $=$ $\frac{5\sqrt{3} - 3\sqrt{7}}{8}r^2$ .

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Finally, the area $S_{ABC}$ of an equilateral triangle $\triangle ABC$ formed by side lengths $2r$ would be $A$ $=$ $\frac{\sqrt{3}}{4}(2r)^2$ $=$ $\sqrt{3}r^2$ .

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Therefore, $\frac{S_{GJI} + S_{KLH}}{S_{ABC}}$ $=$ $\frac{\frac{5\sqrt{3} + 3\sqrt{7}}{8}r^2 + \frac{5\sqrt{3} - 3\sqrt{7}}{8}r^2}{\sqrt{3}r^2}$ $=$ $\frac{5}{4}$ , and $5 + 4 = \boxed{9}$ .