Quick and cute summer Sangaku #10

First we calculate the inradius $r$ of $\triangle ABC$ .
On $\triangle AIE$ , $\sin \dfrac{A}{2}=\dfrac{IE}{AI}=\dfrac{r}{66+r}$ . Similarly, $\sin \dfrac{B}{2}=\dfrac{r}{40+r}$ and $\sin \dfrac{C}{2}=\dfrac{r}{\frac{55}{2}+r}$ .

Now, we use the trigonometric identity ${{\sin }^{2}}\dfrac{A}{2}+{{\sin }^{2}}\dfrac{B}{2}+{{\sin }^{2}}\dfrac{C}{2}=1-2\sin \dfrac{A}{2}\cdot \sin \dfrac{B}{2}\cdot \sin \dfrac{C}{2}$ and we get the equation ${{\left( \dfrac{r}{66+r} \right)}^{2}}+{{\left( \dfrac{r}{40+r} \right)}^{2}}+{{\left( \dfrac{r}{\frac{55}{2}+r} \right)}^{2}}=1-2\cdot \dfrac{r}{66+r}\cdot \dfrac{r}{40+r}\cdot \dfrac{r}{\frac{55}{2}+r} \ \ \ \ \ (1)$

With the substitution $x=\dfrac{1}{r}$ , $\left( 1 \right)\Leftrightarrow {{\left( \frac{1}{66x+1} \right)}^{2}}+{{\left( \frac{1}{40x+1} \right)}^{2}}+{{\left( \frac{1}{27.5x+1} \right)}^{2}}=1-2\frac{1}{66x+1}\cdot \frac{1}{40x+1}\cdot \frac{1}{27.5x+1}$ $\Leftrightarrow \ldots$ $\Leftrightarrow \left( 127776000{{x}^{5}}+22651200{{x}^{4}}+1488960{{x}^{3}}+46656{{x}^{2}}+699x+4 \right)\left( 165x-4 \right)=0 \ \ \ \ \ (2)$

Since all the coefficients of the 5th degree polynomial in the first brackets are positive numbers, this polynomial does not have any positive real roots, thus, for $x>0$ , $\left( 2 \right)\Leftrightarrow 165x-4=0\Leftrightarrow x=\frac{4}{165}$ This gives $\boxed{r=\frac{165}{4}}.$

$\ \ \$

Now, we calculate some useful trigonometric numbers:
$\sin \dfrac{A}{2}=\dfrac{r}{66+r}=\dfrac{\frac{165}{4}}{66+\frac{165}{4}}\Rightarrow \sin \dfrac{A}{2}=\dfrac{5}{13}$ .
Hence, $\cos \dfrac{A}{2}=\dfrac{12}{13}$ , $\sin \dfrac{A}{4}=\sqrt{\dfrac{1-\cos \dfrac{A}{2}}{2}}=\dfrac{1}{\sqrt{26}}$ , $\cos \dfrac{A}{4}=\dfrac{5}{\sqrt{26}}$ , thus, $\tan \dfrac{A}{4}=\dfrac{1}{5}$ .

On $\triangle AEI$ , $AE=AI\cos \dfrac{A}{2}=\left( 66+\dfrac{165}{4} \right)\times \dfrac{12}{13}=99$ .

Then, for the common tangent line segment $FE$ of the incircle and the circle $\left( D,{{r}_{1}} \right)$ , we know that $FE=2\sqrt{r\cdot {{r}_{1}}}$ , hence $FE=\sqrt{165\cdot {{r}_{1}}}$ (Pythagoras’ theorem on $\triangle IND$ .)

$\ \ \$

On $\triangle AFD$ , $\begin{aligned} & FD=AF\cdot \tan \dfrac{A}{4}=\dfrac{1}{5}\left( AE-FE \right) \\ & \Rightarrow {{r}_{1}}=\dfrac{1}{5}\left( 99-\sqrt{165\cdot {{r}_{1}}} \right) \\ & \Rightarrow \ldots \\ & \Rightarrow {{r}_{1}}=\dfrac{231}{10}-\dfrac{33\sqrt{13}}{10} \\ \end{aligned}$

$\ \ \$

Working in the same way, we can calculate the radii of the other two small circles:
${{r}_{2}}=\dfrac{555}{22}-\dfrac{45\sqrt{65}}{22}$ and ${{r}_{3}}=\dfrac{55}{2}-\dfrac{55\sqrt{5}}{6}$ .
The sum of the radii of the six small circles is $2\left( {{r}_{1}}+{{r}_{2}}+{{r}_{3}} \right)=\dfrac{25023-3025\sqrt{5}-1089\sqrt{13}-675\sqrt{65}}{165}$ .

For the answer, $\dfrac{a-e}{c}-b-f-d+g-h=\dfrac{25023-13}{5}-3025-675-1089+65-165=\boxed{113}$ .

• First we need to know more about the triangle, its inradius and its sides.
• We can easily use a simple equation to solve for the inradius (a calculator is needed) $\boxed{\sin ^{-1}\left(\frac{r}{66+r}\right)+\sin ^{-1}\left(\frac{r}{40+r}\right)+\sin ^{-1}\left(\frac{r}{r+\frac{55}{2}}\right)=90°}$
• We get that the inradius is equal to $\boxed{r=\frac{165}{4}}$
• Now using the Pythagorean theorem we easily get that the sides of the triangle are: $\boxed{a=125}$ , $\boxed{b=154}$ , $\boxed{c=169}$

---

• Now we need to calculate the radius of the 6 small circles, although we only need to make 3 calculations. Let's try to get $\boxed{r_1}$ , the radius of the pink circle
• First we need to use the Pythagorean theorem : $\boxed{\left(\frac{165}{4}+r_1\right)^2=\left(\frac{165}{4}-r_1\right)^2+y^2}$ $<=>$ $\boxed{y=\sqrt{165\cdot r_1}}$
• After getting the value of $\boxed{y}$ in terms of $\boxed{r_1}$ , we need to get the value of $\boxed{z}$ in terms of $\boxed{r_1}$
• We know that the center of the pink circle is on the angle bisector of angle $\boxed{AKA_1}$ , then we use the double angle identity the calculate the tangent of angle $\boxed{PAB_1}$ since $\boxed{tan(AKA_1=\frac{33}{56}}$ - $\boxed{\tan \left(AKA_1\right)=\frac{2\cdot \tan \left(PAB_1\right)}{1-\tan ^2\left(PAB_1\right)}}$ $<=>$ $\boxed{\tan \left(PAB_1\right)=\frac{3}{11}}$
• We can conclude that $\boxed{z=\frac{11\cdot r_1}{3}}$
• We know that $\boxed{y+z=70}$ $<=>$ $\boxed{\sqrt{165\cdot r_1}+\frac{11\cdot r_1}{3}=70}$ $<=>$ $\boxed{r_1=\frac{555-45\sqrt{65}}{22}}$
• We can use a similar method to find $\boxed{r_2}$ and $\boxed{r_3}$ and the three radii are :
• $\boxed{r_1=\frac{555-45\sqrt{65}}{22}}$
• $\boxed{r_2=\frac{55}{2}-\frac{55\sqrt{5}}{6}}$
• $\boxed{r_2=\frac{231-33\sqrt{13}}{10}}$

---

• $\boxed{2\cdot \left(r_1+r_2+r_3\right)=\frac{25023-3025\sqrt{5}-1089\sqrt{13}-675\sqrt{65}}{165}}$
• $\boxed{\frac{25023-13}{5}-3025-675-1089+65-165=113}$