Four Circles and an Ellipse

There are infinitely many possible values of the eccentricity.

If $r$ is the radius of the red circles, then the radius of the larger circle is $r +b$ . Then we must have $a = (2r+b)\sin\theta + r$ where $\cos\theta = \tfrac{r}{2r+b}$ , so we deduce that $\begin{aligned} (a-r)^2 & = \; (2r+b)^2\sin^2\theta \; = \; (2r+b)^2 - r^2 \\ 2r^2 + 2(a + 2b)r & = \; a^2 - b^2 \\ \left[r + \tfrac12(a+2b)\right]^2 & = \; \tfrac14(3a^2 + 4ab + 2b^2) \\ r & = \; \tfrac12\left[\sqrt{3a^2 + 4ab + 2b^2} - a - 2b\right] \end{aligned}$ We must have $r > 0$ , which requires $a > b$ and hence $e > 0$ . Here are a couple of examples, showing that different values of the eccentricity are indeed possible:

There is an upper limit on the possible values of $e$ . As shown here , there is a limit on the position of the centre of the two circles on the major axis that enables the circles to be tangent to the ellipse at the single point at the end of the semimajor axis, and also be inscribed inside the ellipse. This condition is that $a - r \ge ae^2$ , which means that $\xi = \sqrt{1-e^2}$ must satisfy the inequality $2\xi^4 + 4\xi^3 + 3\xi^2 - 1 \ge 0$ for $0 < \xi < 1$ , which means that $1 > \xi \ge 0.4406197005$ , which in turn implies that $0 < e \le 0.8976938674$ . Thus this construction is possible provided that $0 < e \le 0.8976938674$ . Here is another picture showing what goes wrong if $e$ is too big:

Thus $\hat{e} = 0.8976938674$ is the critical value of $e$ , and so the answer is $\lfloor 10^5 \hat{e}\rfloor = \boxed{89769}$ .

Without loss of generality, we can take the radius of the big black circle to be $1$ . We'll also take the radius of the red circles to be $r$ . If the ellipse has a semi-major axis $a$ , and semi-minor axis $b$ , then by summing the distances vertically, we get $2 = 2 r + 2 b$ , i.e. $r + b = 1$ . Taking the origin of the Cartesian reference frame to be at the center of the ellipse, then the center of the red circle on the right is $(a - r, 0)$ , and the center of the big black circle is at $(0, 1 - b)$ . Hence, by the distance formula applied to these two centers,

$(1 + r)^2 = (a-r)^2 + (1 - b)^2$

Substituing $r = 1 - b$ , this becomes,

$(2 - b)^2 = (a + b - 1)^2 + (1 - b)^2$

Expanding, and simplifying,

$a^2 + b^2 + 2 a b - 2 a - 2 = 0 \hspace{12pt} (1)$

This equation has many solutions for $a$ and $b$ . However, if we make an additional constraint on the small red circle, then we can find a unique solution. That constraint, which is not mentioned but implied in the supplied diagram, is that the red circles at the right and left have the same curvature as the ellipse at the points to tangency.

The radius of curvature of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ at the point $(a, 0)$ is given by $\rho = \dfrac{b^2}{a}$ . This can be deduced by using the parameteric equation of the ellipse, which is $\mathbf{p}(t) = (a \cos t , b \sin t )$ , and using the formula for the radius of curvature, at $t = 0$ ,

$\rho(t) = \dfrac{ | \mathbf{\dot{p}}(t) |^3 }{ \sqrt{ | \mathbf{\dot{p}}(t) |^2 | \mathbf{ \ddot{p} }(t) |^2 - ( \mathbf{\dot{p}}(t) \cdot \mathbf{\ddot{p}}(t) )^2 } }$

Thus, we'll set $r = \rho$ , so that $1 - b = \dfrac{b^2}{a}$ which re-arranges to,

$b^2 + a b - a = 0 \hspace{12pt} (2)$

Solving equations $(1) , (2)$ results in

$a = 1.575385122$

$b = 0.694145721$

Therefore, the eccentricity is $e = \displaystyle \sqrt{1 - \left(\dfrac{a}{b}\right)^2 } = 0.897693867$

And the answer is $\lfloor 10^5 (0.897693867) \rfloor = \boxed{89769}$