Descartes Sangaku : The trickster

Using Descartes' Theorem, if $r_n$ is the radius of the $n$ th black circle in the pattern (and we let $r_0=25$ ) then $\frac{1}{\sqrt{r_n}} \; = \; \frac{1}{\sqrt{r_{n-1}}} + \frac14 \hspace{2cm} n \ge 1$ so that $\frac{1}{\sqrt{r_n}} \: = \; \frac15 + \frac14n \; = \; \frac{5n + 4}{20}$ so that $r_n \; = \; \frac{400}{(5n + 4)^2} \hspace{2cm} n \ge 0$ It is easy to use simple trigonometry to determine the sines and cosines of the angles $\alpha_n,\beta_n,\gamma_n$ between the lines from the centre of the $n$ th black circle to the centres of its two "parent" circles (the $(n-1)$ st and the yellow circle), and between these lines and the vertical, and we calculate that the area of the $n$ th triangle is $A_n \; = \; \tfrac12r_n^2(\sin\alpha_n + \sin\beta_n + \sin\gamma_n) \; = \; \frac{r_{n-1}r_n^2}{8(r_{n-1} + r_n)(\frac1{16} + r_n)} \; = \; \frac{320000}{(17 + 30n + 50 n^2) (41 + 40n + 25 n^2)}$ which tells us that $A_{15} = \frac{160000}{36709361}$ , making the answer $36709361 - 160000 = \boxed{36549361}$ .

We can solve this problem using coordinate geometry :

• We give the green circle the equation : $\boxed{x^2+\left(y-25\right)^2-625=0}$
• We give the orange circle the equation : $\boxed{(x-40)^2+\left(y-16\right)^2-256=0}$
• Using Descartes' Circle Theorem we find that the radius of the third circle is equal to : $\boxed{\frac{400}{81}}$
• Now we determine the coordinates of the center of the third circle by solving this system of equation:
• $\begin{cases} x^2+\left(y-25\right)^2-(25+\frac{400}{81})^2=0 \\(x-40)^2+\left(y-16\right)^2-(16+\frac{400}{81})^2=0 \end{cases}$
• We find that the center of the third circle is $\boxed{P(\frac{200}{9};\frac{400}{81})}$
• Then its equation has to be : $\boxed{\left(x-\frac{200}{9}\right)^2+\left(y-\frac{400}{81}\right)^2-\frac{400^2}{81^2}=0}$

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• Now we need to determine the coordinates of the three intersection points. We can name them $A$ , $B$ and $C$
• $A$ is the intersection point between the green circle and the third circle, then we need to solve this system of equations:
• $\begin{cases} x_A^2+\left(y_A-25\right)^2-625=0 \\\left(x_A-\frac{200}{9}\right)^2+\left(y_A-\frac{400}{81}\right)^2-\frac{400^2}{81^2}=0\end{cases}$
• We get $\boxed{A(\frac{1800}{97};\frac{800}{97})}$
• $B$ is the intersection point between the orange circle and the third circle, then we need to solve this system of equations:
• $\begin{cases} (x_B-40)^2+\left(y_B-16\right)^2-256=0 \\\left(x_B-\frac{200}{9}\right)^2+\left(y_B-\frac{400}{81}\right)^2-\frac{400^2}{81^2}=0\end{cases}$
• We get $\boxed{B(\frac{1400}{53};\frac{400}{53})}$
• $C$ is the intersection point between the black line and the third circle, then we need to solve this system of equations:
• $\begin{cases} y_C=0 \\\left(x_C-\frac{200}{9}\right)^2+\left(y_C-\frac{400}{81}\right)^2-\frac{400^2}{81^2}=0\end{cases}$
• We get $\boxed{C(\frac{400}{81};0)}$

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• Now we use the distance formula to determine $AB$ , $AC$ and $BC$ :
• $\boxed{AB=400\sqrt{\frac{2}{5141}}}$
• $\boxed{AC=\frac{800\sqrt{97}}{873}}$
• $\boxed{BC=\frac{400\sqrt{\frac{2}{53}}}{9}}$

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• Now by Heron's formula the area of $\boxed{\Delta ABC=\frac{320000}{10282}}$

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• Now we need to find a formula allowing us to express the area of the $n^{th}$ triangle in terms of $n$ . Using a similar method we find the area of the five first triangles :
• $\boxed{A_1=\frac{320000}{10282}}$
• $\boxed{A_2=\frac{320000}{61217}}$
• $\boxed{A_3=\frac{320000}{215002}}$
• $\boxed{A_4=\frac{320000}{563137}}$
• $\boxed{A_5=\frac{320000}{1227122}}$
• We see that the numerator is the same, but the denominator is tricky. Actually it's impossible (for me) to identify any pattern. But we know that the denominator of the area of the sequence of circles in this kind of configuration can be written as a fourth degree polynomial expression : $\boxed{a\cdot n^4+b\cdot n^3+c\cdot n^2+d\cdot n+e}$ where $a,b,c,d,e$ are integers. See this problem for more details about this kind of function.
• In order to find the value of $a,b,c,d,e$ we may solve this system of equations:
• $\begin{cases} a\cdot 1^4+b\cdot 1^3+c\cdot 1^2+d\cdot 1+e=10282\\a\cdot 2^4+b\cdot 2^3+c\cdot 2^2+d\cdot 2+e=61217 \\a\cdot 3^4+b\cdot 3^3+c\cdot 3^2+d\cdot 3+e=215002\\a\cdot 4^4+b\cdot 4^3+c\cdot 4^2+d\cdot 4+e=563137\\a\cdot 5^4+b\cdot 5^3+c\cdot 5^2+d\cdot 5+e=1227122\end{cases}$
• We get $\boxed{a=1250}$ , $\boxed{b=2750}$ , $\boxed{c=3675}$ , $\boxed{d=1910}$ , $\boxed{e=697}$
• We can conclude that $\boxed{A_n=\frac{320000}{1250\cdot n^4+2750\cdot n^3+3675\cdot n^2+1910\cdot n+697}}$
• Then $\boxed{A_{15}=\frac{320000}{73418722}}$ which can be simplified as $\boxed{A_{15}=\frac{160000}{36709361}}$
• Finally $\boxed{36709361-160000=36549361}$