Tricky Tripping Triangular Triplet

Quite an interesting problem. To solve this we first consider an ellipse of major axis of length $2a$ along the $\mathrm{x-axis}$ and minor axis of length $2b$ along the $\mathrm{y-axis}$ .

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

The circles outside the triangle are tangent to the ellipse at each of the end points of the major axis, the equation of one such circle which lies entirely in the first and fourth quadrant is,

$(x-(a-r))^2 + y^2 = r^2$

Where $r$ is the radius the of the circle. This equation gives of the feeling that the circle is tangent to the ellipse at one of the endpoint of the major axis; however, we check for the range of $r$ (in terms of $a$ and $b$ ) for which the tangency is valid. After doing the necessary calculations, we obtain that the circle is tangent to the ellipse at one of the endpoint of the major axis of the ellipse if and only if $r = \frac{b^2}{a}$ . Another condition is that $2r < a$ so that the circle does not crosses origin (mid-point of the ellipse) so that the both the maximum area circles do not intersect.

A point on the aforementioned circle is given as $\left(\frac{a^2-b^2}{a} + \frac{b^2\sin(\alpha)}{a},\frac{b^2\cos(\alpha)}{a}\right)$ for some $\alpha \in [0,2\pi]$ (Here, I have by mistake parameterized the coordinates of the point a point on the circle differently from the convention, however, this would not affect the end result). A line tangent to the circle at that point will be,

$\left(\frac{a^2-b^2}{a} + \frac{b^2\sin(\alpha)}{a} - \frac{a^2-b^2}{a}\right)\left(x - \frac{a^2-b^2}{a}\right) + \frac{b^2\cos(\alpha)}{a}y = \left(\frac{b^2}{a}\right)^2$ $\implies \frac{b^2\sin(\alpha)}{a} \left(x - \frac{a^2-b^2}{a}\right) + \frac{b^2\cos(\alpha)}{a}y = \frac{b^4}{a^2}$

This line passes through the point $(0,b)$ due to symmetry of the problem. Then,

$\frac{b^2\sin(\alpha)}{a} \left(0 - \frac{a^2-b^2}{a}\right) + \frac{b^2\cos(\alpha)}{a}b = \frac{b^4}{a^2}$

Hence, we obtain the following condition on $\alpha$ .

$-\frac{a^2-b^2}{a}\sin(\alpha) + b\cos(\alpha) = \frac{b^2}{a}$

Now, we will solve for the second point of intersection of the tangent line and ellipse $(X,Y)$ . Substituting from the equation of line to the equation of ellipse and after much simplification we obtain,

$\left( \frac{1}{a^{2}} + \frac{\tan^2(\alpha)}{b^2}\right)X^2 - \frac{2\tan^2(\alpha)}{b^2}\left( \frac{a^2-b^2}{a} + \frac{b^2}{a\sin(\alpha}\right)X + \frac{\tan^2(\alpha)}{b^2}\left( \frac{a^2-b^2}{a} + \frac{b^2}{a\sin(\alpha)}\right)^2 - 1 =0$

$AX^2+BX+C = 0$

Solving this equation would give us two values of $X$ ; however, we realize that since we have already obtained $\alpha$ such that one of the intersection points is $(0,b)$ , $C$ must be zero.

When $C =0$ ,

$X = 0 ,\ \mathrm{(trivial)} ,\ \mathrm{and},\ X = -B/A$

Consequently we can obtain the $Y$ coordinate for the non-trivial case.

$Y = \frac{b^2}{a\cos(\alpha)} - \tan(\alpha)\left( -\frac{B}{A} - \frac{a^2-b^2}{a} \right)$

Now we can easily obtain the inradius of the triangle ( $r_{in}$ ).

$r_{in} = \frac{|X|\sqrt{|X|^2+(|Y|+b)^2}}{(|Y|+b)^2}$

Notice that $r_{in}$ and $r$ , both, are functions of $a$ and $b$ only. The choice of $a$ and $b$ is inconsequential to our final result so we just assume $a = 1$ and choose $c = \frac{b}{a} ,\ \mathrm{such that} ,\ 0 < c < 1$ (the choice of $a$ will only scale up or down the problem). Now we have obtained $r_{in}$ and $r$ as a function of $c$ .

Now we'll solve for the value of $c$ for which $r_{in} = r$ where $0 < c < 1/\sqrt{2}$ . I used MATLAB for this purpose. Using the function $\mathtt{fzero}$ we obtain $c = 0.576127110516100$ . Now, we can obtain the corresponding $\theta$ from $\alpha$ .

$\theta = \alpha - \pi = 4.238967975210149 - \pi = 1.097375...$

Hence, $\lfloor\theta \times 1E+6\rfloor = \boldsymbol{1097375}$ .

Here is the graph for variation of $r_{in}$ and $r$ with $c$ .