I have the following problem where I need to solve for the inradius of a particular inscribed triangle in an ellipse:

∆ABC is situated within an ellipse whose major axis is of length 10 and whose minor axis is of length 8. Point A is a focus of the ellipse, point B is an endpoint of the minor axis and point C is on the ellipse such that the other focus lies on BC. Compute the inradius of ∆ABC. Hint: recall the area formula for a triangle involving the inradius.

At the end I'm given a hint to use the area formula for a triangle $K = rs$ in solving for the inradius. I divided both sides by $s$ to get $K/s = r$, and then tried to solve for both $K$ and $s$. It was easy to solve for $s$, because if we let $A'$ be the other focus then

$$ \begin{equation} \begin{split} \begin{gathered} P = AB + BC + AC = AB + BA' + A'C + CA\\ AB = a\\ BA' = a\\ A'C + CA = 2a\\ P/2 = 2a \end{gathered} \end{split} \end{equation} $$

However, I'm having trouble solving for $K$. I realize it's easy to just write up an analytic equation for the ellipse and find the equation of the line $BA'$ then solve for $C$, but I'm interested if there's a better way to do this. I noticed that since B connects line segments passing through both foci the question may have something to do with the reflective property of the ellipse, but I'm not sure. How do I find the area of the triangle non-analytically?

  • $\begingroup$ One approach is to squash the ellipse by a factor of $b/a$ along its major axis, turning it into a circle. The problem here is solvable using tools such as the Power of a Point to relate the lengths of the subsegments of $\overline{BC}$. Once done, converting the area of the squashed triangle to that of the original is simply a matter of multiplying by $a/b$. (Area is cool like that.) The process is a bit tedious, but there may be a way to streamline it. $\endgroup$ – Blue Mar 13 '17 at 7:35

This isn’t the non-analytic solution you’re looking for, but the polar equation of the ellipse with a focus at the origin $$r={a(1-e^2)\over1-e\cos\theta}\tag{*}$$ provides a fairly easy way to compute the area of the triangle.

Since $AC+BC=3a$, $BC=BA'+A'C$ and $BA'=a$, we can conclude that $AC=2a-A'C$, so we can use Heron’s formula to find the triangle’s area, which after simplification yields $K=a\sqrt{2A'C\cdot(a-A'C)}$.

From (*) we can easily find that $r=a$ when $\cos\theta=e$, so $$A'C={a(1-e^2)\over1-e\cos(\pi+\theta)}={a(1-e^2)\over1+e\cos\theta}={1-e^2\over1+e^2}a.$$ Now, $e^2=1-b^2/a^2$, the semi-axis lengths are $a=5$ and $b=4$, and we have everything we need.

Update: Taking an idea from Craig Hicks’ answer, we know that for $\angle B=2\phi$, $\cos\phi=b/a$ and $\sin\phi=f/a=e$, so $\cos2\phi=\cos^2\phi-\sin^2\phi=b^2/a^2-e^2=2b^2/a^2-1$. From $$AB^2+BC^2-2(AB)(BC)\cos(\angle B)=AC^2$$ we thus have $$a^2+(a+A'C)^2-2a(2b^2/a^2-1)(a+A'C)=(2a-A'C)^2$$ which simplifies to a linear equation in $A'C$. The solution to this equation is ${ab^2\over 2a^2-b^2}={ab^2\over a^2+f^2}$ which you can verify is equal to the previous expression for $A'C$ in terms of eccentricity.

Update 2: Blue’s comment above inspired me to find another expression for the triangle’s area besides Heron’s formula and $\frac12(AB)(BC)\sin(\angle B)$.

Observe that the altitude from $A$ is the same for $\triangle{ABC}$ and $\triangle{ABA'}$, which means that $\operatorname{Area}(\triangle{ABC}):\operatorname{Area}(\triangle{ABA'})::BC:BA'$. The ”stretch factor” from $BA'$ to $BC$ is $1+{1-e^2\over1+e^2}={2\over1+e^2}$, so the big trangle’s area is ${2bf\over1+e^2}$.

That’s about as geometric as I can make it.

  • You know the angle(ABC) is 2t, where cos(t) = b/a.
  • You know the length(BC) is a + x2a, and the length(AC) is (1-x)2a, where x is unknown between 0 and 1.
  • Define two equations for Area:
    • One equation is the side-angle-side equation: Area = length(AB)length(BC)sin(angle(ABC)/2
    • One equation is Heron's Formula: Area = sqrt(P(P-length(AB)(P-length(BC)(P-length(AC))
  • Now you have two simultaneous equations with two unknowns (x, Area), to plug into your favorite symbolic solver.
  • $\begingroup$ By the way, I think amd's answer is vastly more elegant. $\endgroup$ – Craig Hicks Mar 13 '17 at 6:46
  • 1
    $\begingroup$ Your approach is basically the same as mine, but uses the known angle $\angle{ABC}=2\theta$ to compute the unknown “extra” length $|A'C|$ instead. One could argue that’s more a “geometric” approach. BTW, you can use $|AB|^2+|BC|^2−2\cos2\theta=|AC|^2$ to find this length, expanding $\cos2\theta$ as $2b^2/a^2−1$ (double-angle formula). That ends up being a linear equation in |A′C|, so you don’t even need a symbolic solver. $\endgroup$ – amd Mar 14 '17 at 9:32

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