# Minimal Ellipse Circumscribing A Right Triangle

Find the equation of the ellipse circumscribing a right triangle whose lengths of it's sides are $3,4,5$ and such that its area is the minimum possible one.

You may chose the origin and orientation of the $x,y$ axes as you want.

Motivation: It can be proved [Problem of the Week, Problem No. 8 (Fall 2008 Series), Department of Mathematics, Purdue University] that the area of this ellipse is $8\pi /\sqrt{3}$, without the need of using its equation, but I am also interested in finding it.

Edit: picture from this answer.

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The Steiner circumellipse: mathworld.wolfram.com/SteinerCircumellipse.html is what you seem to be interested in. –  Ｊ. Ｍ. Aug 28 '10 at 9:55
Also a minor grammatical note: it's "inscribed in", but "circumscribing a". :) –  Ｊ. Ｍ. Aug 28 '10 at 9:56
@J. Mangaldan: Thanks for your correction! English prepositions are very difficult to learn by foreigners. +1 in your comment! –  Américo Tavares Aug 28 '10 at 10:49
The proof you link to talks about an affine transformation that simultaneously maps the triangle to an equilateral triangle and the minimal circumellipse to a circle. You could trying finding this transformation explicitly using the triangle's coordinates, and applying its inverse to the circle to get the equation of the ellipse. –  Rahul Aug 28 '10 at 14:10
It's interesting that the tangent where the ellipse touches a vertex is parallel to the opposite side. This follows trivially from the same property of the circumcircle of an equilateral triangle and the fact that affine transformations preserve parallelism, but it's certainly non-obvious from looking at the original problem statement. –  Rahul Aug 28 '10 at 22:25
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If you look at the proof that you linked, the proof is using an Affine transformation.

Now you can also see that the centroid of the triangles are mapped to each other in the affine transformation, and so do the centers of the ellipses.

Thus the ellipse you need has the same center as your triangle! This property uniquely defines the ellipse with the maximum area.

Now consider the points $A = (-1,-4/3), B = (2, -4/3)$ and $C= (-1, 8/3)$. This is a 3-4-5 triangle whose center is origin which was gotten by starting with $(0,0), (3,0)$ and $(0,4)$ and translating so that the centroid is the origin.

Now the equation of an ellipse whose center is the origin is given by

$Px^2 + Qxy + Ry^2 = 1$.

Thus we must have that

(1) $P + 4Q/3 + 16R/9 = 1$
(2) $4P - 8Q/3 + 16R/9 = 1$
(3) $P -8Q/3 + 64R/9 = 1$

Solving these (see footnote) gives us the equation of the ellipse as

$x^{2}/3 + xy/4 + 3y^{2}/16 = 1$

In order to verify this, the area of $Px^2 + Qxy + Ry^2 = 1$ is given by $\displaystyle \frac{2\pi}{\sqrt{4PR - Q^2}}$ which comes out to $\displaystyle \frac{8\pi}{\sqrt 3}$

You can ignore the below if you like. This is just manually solving the equations

(1) $P + 4Q/3 + 16R/9 = 1$
(2) $4P - 8Q/3 + 16R/9 = 1$
(3) $P -8Q/3 + 64R/9 = 1$

Subtracting (1) and (2) gives $3P = 4Q$.

Subtracting (2) and (3) gives $3P = 48R/9$

Thus $Q = 3P/4$ and $R = 9P/16$

Thus using (3) we have that

$P - (8/3)*(3P/4) + 64/9 * (9P/16) = 1$

i.e $P - 2P + 4P = 1$ i.e $P = 1/3$.

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@Moron: +1 for your elementary solution. –  Américo Tavares Aug 28 '10 at 20:45
@Moron: Would you please inform how do you manage to change the background color of "the ellipse you need has the same center as your triangle!" text. Since this is not the correct place to ask you such a thing, I will delete this comment, if needed. –  Américo Tavares Aug 28 '10 at 20:49
@Américo Tavares: He surrounded that text with single backticks (this character: `), which produces the grey background with a monospaced font and ignores other markup inside it. –  Isaac Aug 28 '10 at 21:09
In other words: the centroid (the point where the three medians intersect) is the center of the ellipse. –  Ｊ. Ｍ. Aug 28 '10 at 23:46
@Americo: Thank you for asking this question. It is a nice problem. –  Aryabhata Aug 29 '10 at 7:25
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Given the proof to which you linked, as Rahul Narain said in a comment, there is an affine transformation that maps the unit circle and an inscribed equilateral triangle to the desired ellipse for a particular 3-4-5 right triangle. Represent this affine transformation by $$\begin{pmatrix}x'\\y'\\1\end{pmatrix}=\begin{pmatrix}a_{1,1} & a_{1,2} & b_1\\a_{2,1} & a_{2,2} & b_2\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}.$$ Let the 3-4-5 right triangle have vertices (0,0), (0,3), and (4,0), and let the equilateral triangle have vertices (1,0), $(-\frac{1}{2},\frac{\sqrt{3}}{2})$, and $(-\frac{1}{2},-\frac{\sqrt{3}}{2})$. Mapping these vertices in the order listed and solving for the parameters of the transformation gives the matrix $$\begin{pmatrix}-\frac{4}{3} & -\frac{4}{\sqrt{3}} & \frac{4}{3}\\-1 & \sqrt{3} & 1\\0 & 0 & 1\end{pmatrix}.$$

The unit circle has parametric representation $x=\cos t$, $y=\sin t$. Applying the affine transformation to this unit circle gives the parametric representation $x=\frac{4}{3}(1-\cos t-\sqrt{3}\sin t)$, $y=1-\cos t+\sqrt{3}\sin t$ for the desired ellipse.

edit: Looking at Moron's answer, his choice of x and y coordinates are swapped relative to mine, and as he said, his right triangle is translated to have its centroid at the origin, but accounting for these differences, my parametric equations satisfy his equation. That is, based on his equation, an equation in x and y alone for the ellipse with parametric equations as I specified is $9x^2+12xy+16y^2-36x-48y=0$.

edit 2: I don't know too much of the detailed mechanics of the different representations of affine transformations. The form that I used above is equivalent to $$\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}a_{1,1} & a_{1,2}\\a_{2,1} & a_{2,2}\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}b_1\\b_2\end{pmatrix}$$ and the 1s and 0s in the third row in the earlier version are just the glue that allows an affine transformation to be represented with a single matrix multiplication. My solution could have just as easily used this representation of the affine transformation.

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@Isaac: +1 for your elegant solution. For me it is relatively more difficult to follow, but the fault is mine. Thanks also for changing the tag name. –  Américo Tavares Aug 28 '10 at 20:53
@Américo Tavares: Thanks. It is a fairly terse answer--please let me know if there are parts where the explanation should be better. –  Isaac Aug 28 '10 at 21:07
@Isaac: I've read only a short text on the possibility of using a matrix to represent an affine transformation, but never done it by myself. So if you give more details on $b_1,b_2$ and the zeroes and ones it would help me. –  Américo Tavares Aug 28 '10 at 21:44
@Américo Tavares: Does it make better sense with my second edit? (It is much easier to get the LaTeX correct in the answer body than in comments.) –  Isaac Aug 28 '10 at 23:05
@Américo Tavares: If you're curious about the original 3×3 matrix in Isaac's answer, you can look up what are called homogeneous coordinates. In general, when the bottom row is not all zeroes and ones, it can represent not just affine but also projective transformations, which makes this very useful for rendering perspective projections in computer graphics. –  Rahul Aug 29 '10 at 1:33
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Not much to contribute at this point except this enhanced picture:

(dashed lines: axes; gray lines: medians)

and the Mathematica notebook used to produce it.

(Thanks Isaac!)

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@J. Mangaldan: +1 for the professional picture. I removed the figure I had in the question and replaced it by yours. –  Américo Tavares Aug 29 '10 at 8:28
Américo: I managed to produce a larger image; use the new one instead, please. :) –  Ｊ. Ｍ. Aug 29 '10 at 8:42
J. Mangaldan: done. –  Américo Tavares Aug 29 '10 at 9:00
It must be a version issue (as I got a warning when opening the notebook that it was created on an earlier version), but it seems like I'm getting a much crisper graphic from your notebook: imgftw.net/img/675511545.png --feel free to use it if you think it looks better. –  Isaac Aug 29 '10 at 9:02
Isaac: Yeah, I'm stuck on 5.2 (for a number of reasons too long to post here). Thanks for improving the pic (though since you're a mod, you could've just edited my post instead ;)). –  Ｊ. Ｍ. Aug 29 '10 at 9:32
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