Ratios of the major axes of ellipses inscribed in a triangle
I recently came across a question that went something like this:
In a triangle with vertices (0,0), (6,0), (2,4) an ellipse is inscribed such that it has the largest area. Now it is dilated, that is zoomed in, or enlarged while keeping it's center as well as orientation fixed, so that the new ellipse trisects the sides of the triangle. So what is the ratio of the lengths of the major axes of the new ellipse to the previous one?
And here is what I could gather from what I knew:
The inscribed ellipse or 'inellipse' of the largest area that can be inscribed in a triangle is known as the Steiner inellipse
and many of it's properties are known. The properties that might help us here are that it is tangent to the triangle sides at it's midpoints, the lengths of it's axes can be determined by the formula (g = semi major axis length, h = semi minor axis length)
$${g=\sqrt{a^2+b^2+c^2+2Z}/6}$$
$${h=\sqrt{a^2+b^2+c^2-2Z}/6}$$ where
$${Z=\sqrt{a^4+b^4+c^4-(ab)^2-(bc)^2-(ca)^2}}$$
and the centroid of the triangle is it's center.
So now we can find the two foci of the ellipse by the concept that the angle bisector of the angle formed by the normal and the tangent to the ellipse at point passes through the focus.
So we simply find the angle bisectors at the 3 points of tangency, because we know the equations of the tangent's and normal there, and then find the intersection points of these 3 angle bisectors.
They will thus determine 2 points which are the foci of the inellipse. So the line joining them must be the major axis of the ellipse.
Now when we extend the ellipse to make it trisect the 3 sides of the triangle it passes through 6 known points, and we know of a way to find the equation of a unique ellipse given 5 points of the ellipse. From here we can find out the equation of the new ellipse, and it's major axis will be the same line as the major axis of the previous ellipse because we have only dilated it, keeping orientation fixed.
But we already know the equation of the major axis of the previous ellipse. So we solve the equation of the second ellipse and the equation of the major axis to get the coordinates of the vertices of the ellipse. Thus we know the length of the major axis of the new ellipse as well.
Now we have to find the ratio of these 2 lengths.
However this is only what I have thought of doing, and I know that it is perfectly okay and will work, but could someone please tell me a better approach of trying this problem. Because although I know how to do it, implementing it manually would be no mean task, and certainly not doable in any examination hall. So a better and shorter method is highly solicited from you all, probably using any minute observation that may be useful and that I am missing.
Thanks in advance for the help!
geometry trigonometry
$\begingroup$ I suggest you change the title to something more likely to attract readers. $\endgroup$ – Peter Phipps Feb 10 '16 at 23:45
$\begingroup$ Ok @PeterPhipps I have changed it. However I am not too good with titles, and some of my previous questions got downvoted because of this. :'( Hope this is good. Else could you please edit it? :D $\endgroup$ – user260674 Feb 10 '16 at 23:49
$\begingroup$ I'm typing before thinking too much here, but ... Since the Steiner Ellipse of an arbitrary triangle can be affinely transformed into the incircle of an equilateral triangle, and since affine transforms preserve ratios of lengths of parallel segments, is it not enough to solve this problem in the equilateral case? $\endgroup$ – Blue Feb 11 '16 at 0:21
$\begingroup$ Actually it is, and for the equilateral case it can be done in a jiffy, and it is quite easy to see by projection as well, but I am not sure if the inverse transform is true as well, that is if the circle trisecting sides of the equilateral triangle can be tranformed into the ellipse trisecting the sides of the triangle? $\endgroup$ – user260674 Feb 11 '16 at 0:28
$\begingroup$ @Blue could you please post your solution? I have been looking forward to it. $\endgroup$ – user260674 Mar 28 '16 at 13:29
We take as given that the area-maximizing inellipse is the Steiner inellipse, which is tangent to the triangle's sides at their respective midpoints. We also take as given that any triangle even has a side-trisecting ellipse, and that this ellipse is a dilation of the Steiner inellipse with respect to its center (which coincides with the triangle's centroid). We note, therefore, that the latter indicates that the question's goal of the ratio of major axes is merely the dilation factor.
Now, I gave the short version of my argument as a comment to the question.
Since the Steiner Ellipse of an arbitrary triangle can be affinely transformed into the incircle of an equilateral triangle, and since affine transforms preserve ratios of lengths of parallel segments, [it is] enough to solve this problem in the equilateral case[.]
I'll explain in a bit more depth here.
An affine transformation moves points around in the plane (or space) under specific rules. In particular, (1) collinear points move to collinear points; (2) tangent curves move to tangent curves, and (3) ratios of lengths between collinear points match ratios of lengths between corresponding moved points.
We can see these rules in play when we observe that an affine transformation can move the vertices of any triangle onto the vertices of another. Consequently, under such a transformation, by (1), the sides of the original triangle move onto the sides of the target; and, by (3), the midpoints of the original sides (defined by the ratio $1/2$) move to midpoints of the targets sides, and side-trisecting points (ratio $1/3$ and/or $2/3$) move to side-trisecting points. Moreover, again by (1), original medians move to target medians, and thus also the original triangle's centroid moves to the target triangle's centroid.
We have more: The original triangle's Steiner inellipse moves to the target triangle's Steiner inellipse; this is because midpoints move to midpoints, tangent curves move to tangent curves, and we're told that the Steiner inellipse is the ellipse tangent to a triangle's sides at their respective midpoints. Further, the original triangle's side-trisecting ellipse moves to the target triangle's side-trisecting ellipse.
Now, here's the kicker: We're told that the side-trisecting ellipse is a dilation of the Steiner inellipse in the triangle's centroid $O$; that is, for any point $P$ on the inellipse, and $P^\prime$ is where $\overrightarrow{OP}$ meets the trisecting ellipse, then $|\overline{OP^\prime}|/|\overline{OP}|$ is some constant, the dilation factor. But ... affine transformations preserve centroids (such as $O$); and they preserve ratios of lengths of between collinear points (such as, say, $O$, $P$, $P^\prime$); therefore, affine transformations preserve the dilation factor between Steiner inellipses and side-trisecting ellipses! That is to say:
There's one dilation factor that works for all triangles.
We take advantage of this by computing the factor in the most-convenient of cases: the equilateral triangle.
In an equilateral triangle (of side-length, say, $6s$), the Steiner inellipse is the incircle, with radius $s \sqrt{3}$. The side-trisecting ellipse is a circle of radius $2 s$. Therefore,
The dilation factor is $2/\sqrt{3} = 1.154\dots$. $\square$
$\begingroup$ That's is really a nice solution. Affine transformations seem interesting. :) $\endgroup$ – user260674 Mar 30 '16 at 19:56
$\begingroup$ FYI: affine transformations are just linear transformations (origin-fixing rotations, reflections, dilations, shears), combined with a translation. When you learn to think of these kinds of problems in terms of vectors, the linear/affine transformation properties will become "obvious". $\endgroup$ – Blue Mar 30 '16 at 20:19
$\begingroup$ Yeah, I noticed, I think, homothety is also a part of affine transformations. It has the same properties. :) $\endgroup$ – user260674 Mar 31 '16 at 23:33
As I have already outlined in the question (and then I modified the solution to shorten it by a great extent), we can have the following steps leading to the solution:
We can find the perpendicular to the sides of the triangle at the midpoint of the sides, so that these are the normal to the ellipse at these points, because, being an inellipse, the side of the triangle is a tangent to it.
Now we find the equations of the angle bisectors of the angles between the tangent (side of triangle) and normal at the sides of the triangle, and their intersection points (two in number) will give the foci of the ellipse.
Now that we have the foci we can find the equation of the major axis. Note that this line is the major axis of the larger ellipse too.
We can find the equation of the minor axis by making it perpendicular to the major axis at the centre of the ellipse which is the centroid of the triangle.
Now we find the points of trisection of the sides of the triangle. And since these points lie on the bigger ellipse we use the last step which is step 6.
Given a point $P$ on the ellipse, if the perpendicular distance of $P$ from the major axis be $q$ and the perpendicular distance of $P$ from the minor axis be $p$, and the lengths of the semi major and semi minor axes be $a'$ and $b'$ respectively, then the equation of the ellipse can be written as, $${p^2/a'^2}+{q^2/b'^2}=1$$ and since the point P with it's coordinates as well as the equations of the major and minor axes are known, the only unknown quantities here are $a'$ and $b'$ which can be found by satisfying this equation for 2 of the given points on the larger ellipse. Now when we know $a'$ and $g$ where $g$ is the length of the major axis of the inellipse, $a'/g$ will be our answer.
This method will not be too long except the part where we have to find the angle bisector equations. But as some of you people pointed out, an affine transform kills the problem, I would like to see why it works.
While it is quite clear to me why the inellipse of the triangle will map to the incircle of an equilateral triangle, it is not immediately clear to me why the side-trisecting circle of the equilateral triangle maps to the outer ellipse of the original triangle. I will be eagerly waiting for that solution. :)
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