Mutually Common Round Circles

Let $a$ and $b$ be the major and minor axis respectively. Then the radius $R$ of the circles is the radius of curvature at the two vertices, i.e. $R=\dfrac{{{b}^{2}}}{a}$ Denote by $A$ , ${A}'$ the vertices of the ellipse, by $B$ , ${B}'$ its co-vertices. Let $O$ be the center of the ellipse and $K$ the center of the circle tangent at $A$ . Let $P$ be the common point of the three circles seen in the figure.

By Pythagorean theorem on $\triangle OPK$ ,

$\begin{aligned} O{{P}^{2}}+O{{K}^{2}}=P{{K}^{2}} & \Rightarrow {{\left( 2R-b \right)}^{2}}+{{\left( a-R \right)}^{2}}={{R}^{2}} \\ & \Leftrightarrow \left( 4{{R}^{2}}-4bR+{{b}^{2}} \right)+\left( {{a}^{2}}-2aR+{{R}^{2}} \right)={{R}^{2}} \\ & \Leftrightarrow 4{{R}^{2}}-4bR-2aR+{{a}^{2}}+{{b}^{2}}=0 \\ & \Leftrightarrow 4{{\left( \dfrac{{{b}^{2}}}{a} \right)}^{2}}-4b\dfrac{{{b}^{2}}}{a}-2a\dfrac{{{b}^{2}}}{a}+{{a}^{2}}+{{b}^{2}}=0 \\ & \overset{\div {{a}^{2}}}{\mathop{\Leftrightarrow }}\,4{{\left( \dfrac{b}{a} \right)}^{4}}-4{{\left( \dfrac{b}{a} \right)}^{3}}-{{\left( \dfrac{b}{a} \right)}^{2}}+1=0 \\ & \Leftrightarrow \left( \dfrac{b}{a}-1 \right)\left( 4{{\left( \dfrac{b}{a} \right)}^{3}}-\dfrac{b}{a}+1 \right)=0 \\ & \overset{\frac{b}{a}<1}{\mathop{\Leftrightarrow }}\,\dfrac{b}{a}\approx 0.7606898534022837848020404 \\ \end{aligned}$ The eccentricity is given by the formula $E=\sqrt{1-{{\left( \dfrac{b}{a} \right)}^{2}}}$ Substituting the value we found for $\dfrac{b}{a}$ , we get $E\approx \text{0}\text{.6491155112388}$ .

For the answer, $\left\lfloor {{10}^{6}}E \right\rfloor =\boxed{649115}$ .

Let the equation of the ellipse be,

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \ \mathrm{where}, \ a > b$

Equations of the two circles on whose center lie on the major axis will be,

$\left( x - (b-r) \right)^2+y^2 = r^2 \ \mathrm{and} \ \left( x - (-b+r) \right)^2+y^2 = r^2$

The value of $r$ for which the circles are tangent to the ellipse is $\frac{b^2}{a}$ . Hence the two equations become,

$\left( x - \frac{a^2-b^2}{a} \right)^2+y^2 = \frac{b^4}{a^2} \ \mathrm{and} \ \left( x - \frac{b^2-a^2}{a} \right)^2+y^2 = \frac{b^4}{a^2}$

Solving for the points of intersection, we get,

$X = 0$ and $Y = \pm \sqrt{2b^2-a^2}$

Consider the point $(0, -\sqrt{2b^2-a^2})$ , this point must lie on the circle whose center lies on the positive side of the $\mathrm{y-axis}$ (from intuition). The equation for such a circle is given as,

$x^2 + \left( y - \frac{ab-b^2}{a} \right)^2 = \frac{b^4}{a^2}$

Substituting,

$(0)^2 + \left( -\sqrt{2b^2-a^2} - \frac{ab-b^2}{a} \right)^2 = \frac{b^4}{a^2}$

Now put $b = a \sqrt{1-e^2}$ (where $e$ is the eccentricity of the ellipse) in the above equation.

$\left( -\sqrt{2a^2(1-e^2)-a^2} - a\sqrt{1-e^2} + \frac{a^2(1-e^2)}{a} \right)^2 = \frac{a^4(1-e^2)^2}{a^2}$

$\implies \left( -\sqrt{2(1-e^2)-1} - \sqrt{1-e^2} + (1-e^2) \right)^2 - (1-e^2)^2 = 0$

$\implies \left( -\sqrt{1-2e^2} - \sqrt{1-e^2}\right)\left( -\sqrt{1-2e^2} -\sqrt{1-e^2} +2(1-e^2) \right) = 0$

$\implies -\sqrt{1-2e^2} - \sqrt{1-e^2} = 0$ or $-\sqrt{1-2e^2} -\sqrt{1-e^2} +2(1-e^2) = 0$

Notice that,

$-\sqrt{1-2e^2} - \sqrt{1-e^2} \neq 0 \ \forall \ e \ \in (0,1)$

Hence, our only option is to find the solutions of,

$-\sqrt{1-2e^2} -\sqrt{1-e^2} +2(1-e^2) = 0$

Use MATLAB and ignore the trivial $e=0$ solution. We get $e =0.649115511238803$ . Now $\lfloor{e \times 1E+6}\rfloor = \textbf{649115}$ .

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sol = fzero(@(x) fun(x), 0.5);

function y = fun(x)
y = -sqrt(1-2*x^2)-sqrt(1-x^2)+2*(1-x^2);
end