The question is as follows:

Let $E$ be an ellipse with major axis length $4$ and minor axis length $2$. Inscribed an equilateral triangle $ABC$ in $E$ such that $A$ lies on the minor axis and $BC$ is parallel to the major axis. Compute the area of $\triangle ABC$.

My Work:


Pardon my lack of talent with paint, but you get the idea. Points $a$ and $b$ have coordinates $(-x,y)$ and $(x,y)$ respectively. The equation for the ellipse in question is $x^2/4 + y^2 = 1$, which yields $y = \frac{\pm\sqrt{4-x^2}}{2}$ when rearranged, but the triangle is in the upper half so we can say that $y = \frac{\sqrt{4-x^2}}{2}$ for simplicity. Both $a$ and $b$ are at a distance of $\sqrt{x^2+y^2}$ from the origin, and a distance of $2x$ from each other. Since this is an equilateral triangle, we can say that $$2x = \sqrt{x^2+y^2} \hspace{5 mm} \text{and}$$

$\hspace{67 mm} \displaystyle{y = \frac{\sqrt{4-x^2}}{2}},$

a system of equations that yield readily to substitution. From these equations, we get that $x = \frac{2\sqrt{13}}{13}$ and $y = \frac{4\sqrt{3}}{26}$. From the dotted lines in the diagram, I hope it's clear that the square's area is $4xy$, and the triangle is half the square, so its area is $2xy$. Multiplying everything together yields the result $\frac{8\sqrt{39}}{169}$. This is a tiny number, and it seems the actual answer is $\frac{192\sqrt{39}}{169}$. I've double checked my work, and have no idea where I went wrong. Any ideas on where I went wrong, or on how to get the correct answer?

  • 1
    $\begingroup$ Inscribed usually means the vertices are on the figure. And saying that $B$ is parallel to the major axis does not mean anything. Did you intend to write $BC$? The picture does not go well with the wording of the problem. $\endgroup$ – André Nicolas Jul 30 '14 at 21:16

As has been noted, the triangle you drew is not the triangle that is described.

That said, the area of the triangle you drew can be found in a simpler fashion. Note that the ellipse is $$x^2 + 4y^2 = 4,$$ and for the triangle you drew, we must have $\sqrt{x^2+y^2} = 2x$, or equivalently, $3x^2 = y^2$. Substituting immediately gives $x^2 = 4/13$ and $y^2 = 12/13$, from which it immediately follows that the area of this triangle is $x^2 \sqrt{3} = 4 \sqrt{3}/13$.

But for the triangle that was described, each vertex is located on the boundary of the ellipse. One vertex is located at a boundary point on the minor axis; the other two are parallel to the major axis. So without loss of generality suppose the vertices are $(0,-1)$, $(x,y)$, and $(-x,y)$, where $x, y > 0$. Then the distance condition becomes $$x^2 + (y+1)^2 = (2x)^2,$$ hence $x^2 = \frac{1}{3}(y+1)^2$ and $$\frac{1}{3}(y+1)^2 + 4y^2 = 4.$$ I leave the rest to you.

  • $\begingroup$ @recursiverecursion: Alternately, take the line through $(0,-1)$ that makes a $30$ degree angle with the $y$-axis, see where it meets the ellipse. $\endgroup$ – André Nicolas Jul 30 '14 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.