Pi Day Challenge 2017 - Mitosis

I use $c,g,p,r$ for the diameters of the cyan, green, purple and red circles, and $C,G,P,R$ for their centres.

Translate the pink line, so that one of its endpoints is in $P$ . call its other endpoint $Q$ We now have a rectangular triangle $PQR$ with sides $QR=(r-p)/2, PQ=L$ (the requested length) and hypothenuse $PR$ .

First we will find $PR$

Drop a line from $R$ onto $CG$ , meeting it at $M$ , its length is $a$ We now have formed two right angled triangles:

• $RGM$ with sides $RG=\frac{r-g}{2}, GM=d, RM=a$
• $RCM$ with sides $RC=\frac{r-c}{2}, CM=e, RM=a$
• we also have $GM+MC=GC$ , so $d+e=\frac{g+c}{2}$

We get equations $(\frac{r-g}{2})^2=a^2+d^2$ and $(\frac{r-c}{2})^2=a^2+e^2$ .

$d^2-e^2=(\frac{r-g}{2})^2-(\frac{r-c}{2})^2 =\frac{2r(c-g)+g^2-c^2}{4}\Rightarrow d-e=\frac{g^2-c^2+2r(c-g)}{2(c+g)}$ and since $e+d=\frac{c+g}{2}$ we know $e$ and $d$ :

$d=\frac{(d+e)+(d-e)}{2}=\frac{g^2+cg+r(c-g)}{2(c+g)}$ $e=\frac{(d+e)-(d-e)}{2}=\frac{c^2+cg-r(c-g)}{2(c+g)}$ $a=\frac{\sqrt{rcg(r-c-g)}}{c+g}$

Drop a line from P onto CG, meeting it at N, its length is $b$ We now have formed another two right angled triangles:

• $PGN$ with sides $PG=\frac{p-g}{2}, GN=f, PN=b$
• $PCN$ with sides $PC=\frac{p-c}{2}, CN=h, PN=b$
• we also have $GN+NC=GC$ , so $f+h=\frac{g+c}{2}$

Using similar techniques we find

$f=\frac{(f+h)+(f-h)}{2}=\frac{g^2+cg+p(c-g)}{2(c+g)}$ $h=\frac{(f+h)-(f-h)}{2}=\frac{c^2+cg-p(c-g)}{2(c+g)}$

$b=\frac{\sqrt{pcg(p-c-g)}}{c+g}$

We also find $(r-p)(c-g)=(g+c)(d-e-f+h)=2(g+c)MN$ so that

$MN=\frac{(r-p)(c-g)}{2(c+g)}$

Finally have an expression for RP: $RP^2=(a+b)^2+MN^2$

And the required length L can be found. Via

$L=\sqrt{RP^2-(\frac{r-p}{2})^2}=\sqrt{(a+b)^2+MN^2-(\frac{r-p}{2})^2}$ we reach the general formula in terms of circle diameters: $L=\sqrt{(\frac{\sqrt{rcg(r-c-g)}}{c+g}+\frac{\sqrt{pcg(p-c-g)}}{c+g})^2+(\frac{(r-p)(c-g)}{2(c+g)})^2-(\frac{r-p}{2})^2}$

it can be rewritten to meet the appropriate leading term:

$L=\frac{2\sqrt{cg/6}}{c+g}\sqrt{3rp-\frac{3}{2}(r+p)(c+g)+3\sqrt{rp(r-c-g)(p-c-g)}}$

This is not yet the requested form so we have to solve (note that these are not the same $a,b,c$ as earlier) $(a\sqrt{b}+\sqrt{c})^2=3rp-\frac{3}{2}(r+p)(c+g)+3\sqrt{rp(r-c-g)(p-c-g)}$ so that $a^2b+c=3rp-\frac{3}{2}(r+p)(c+g)=54100578693$ and $a^2bc=9rp(r-c-g)(p-c-g)=2×3^5×11×59×167×907×43963×314159$ we find $a=9, b=11×907×43963=438618851, c=2×3×59×167×314159=18572451762$ $a+b+c=\boxed{19011070622}$

Let $r_1$ , $r_2$ , $r_3$ , and $r_4$ denote the radii of the green, blue, purple, and red circles, respectively, and let the green and blue circles be centered at $(0,r_1)$ and $(0,-r_2)$ , respectively. The distance between the centers of the purple and green circles is $r_3 - r_1$ , while the distance between the centers of the purple and blue circles is $r_3 - r_2$ . This means the center of the purple circle is at the right intersection of the circles $x^2 + (y - r_1)^2 = (r_3 - r_1)^2 \quad \text{and} \quad x^2 + (y + r_2)^2 = (r_3 - r_2)^2,$ which is at $(x_1,y_1) = \left(\sqrt{(r_3 - r_1)^2 - (y_1 - r_1)^2} , \frac {(r_1 - r_2)r_3}{r_1 + r_2} \right).$ Similarly, the center of the red circle is at the left intersection of the circles $x^2 + (y - r_1)^2 = (r_4 - r_1)^2 \quad \text{and} \quad x^2 + (y + r_2)^2 = (r_4 - r_2)^2,$ which is at $(x_2,y_2) = \left(\sqrt{(r_4 - r_1)^2 - (y_2 - r_1)^2} , \frac {(r_1 - r_2)r_4}{r_1 + r_2} \right).$

Now consider the right triangle in the figure. The hypotenuse is the segment from $(x_1,y_1)$ to $(x_2,y_2)$ , and one of the legs has length $r_4 - r_3$ . The other leg has the same length as the segment of the tangent line. By the Pythagorean theorem, this length is $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 - (r_4 - r_3)^2},$ and with a little effort one can show that this simplifies to $\frac {2 \sqrt{105458935}}{50381} \left(9 \sqrt{438618851} + \sqrt{18572451762} \right).$ Therefore, the desired value is $9 + 438618851 + 18572451762 = \boxed{19011070622}$ .