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Topic: Steiner Ellipses (Read 2398 times) |
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Icarus
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Steiner Ellipses
« on: Mar 20th, 2003, 5:28pm » |
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This one comes from Jakob Steiner, a Swiss geometer in the 19th century:
What is the area of the smallest ellipse that can be circumscribed around a 3-4-5 triangle?
What is the area of the largest ellipse that can be inscribed in a 3-4-5 triangle?
Exact answers only.
Small Hint: The dimensions of the triangle are not important - they were only given to offer some perspective to the problem.
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Chronos
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Re: Steiner Ellipses
« Reply #1 on: Mar 20th, 2003, 7:51pm » |
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Ah, but the dimensions do matter, to the extent that it's a right triangle.
Unless I'm mistaken, the answer for the circumscription is 4pi/25 (extra space to make answer look longer). I suspect that the inscription answer uses the same gimmick, but I can't convince myself of that, and I'm also not sure I'm remembering the right formula for the relevant dimensions. But if I'm remembering both of those correctly, then the area of the inscribed ellipse is approximately 3.14159265358979323846.
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Icarus
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Re: Steiner Ellipses
« Reply #2 on: Mar 20th, 2003, 8:05pm » |
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Even without knowing the answers, I could tell that you have a problem - your circumscribed ellipse has ~1/6 the area of your inscribed ellipse! A good trick! I assume you actually wanted 25PI/4, which is certainly the area of a circumscribed ellipse. However the smallest circumscribed ellipse is ~75% smaller by area.
Actually both answers are off. The problem is a bit more complex than that.
My hint could have been better stated, since of course the area of the ellipses depends on the size of the triangle. However the solution method does not require triangles with special properties.
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| « Last Edit: Mar 21st, 2003, 3:49pm by Icarus » |
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I wonder: Which is larger
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SWF
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Re: Steiner Ellipses
« Reply #3 on: Mar 21st, 2003, 5:40pm » |
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My first impression is that the answer should be: (hidden text)
Big ellipse area= (Triangle area)*4*pi/3/sqrt(3)
Small ellipse area= (Triangle area)*pi*sqrt(3)/9
If that is right, then this one is very easy with the right approach. Otherwise, I am glad to have gotten the wrong answer an easy way rather than after lots of work.
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Icarus
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Re: Steiner Ellipses
« Reply #4 on: Mar 22nd, 2003, 11:52am » |
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Those are the correct answers, though they might be better stated as:
Inscribed ellipse area = (pi/(3sqrt(3))) * Triangle area
Circumscribed ellipse area = (4pi/(3sqrt(3))) * Triangle area.
As Martin Gardner (whom I got this from) said "...it is one of the best examples I know of a problem that is difficult to solve by calculus or analytic geometry but is ridiculously easy if approached with the right turn of mind and some knowledge of elementary plane and projective geometry."
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Chronos
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Re: Steiner Ellipses
« Reply #5 on: Mar 25th, 2003, 11:08am » |
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Quote:| ...and projective geometry |
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OK, I've got it now, and in retrospect, it seems obvious. But then again, my original solution seemed obvious, too, at the time. 
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James Fingas
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Re: Steiner Ellipses
« Reply #6 on: Mar 26th, 2003, 12:34pm » |
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I think I have achieved an "aha" moment in this question (either that or I'm wrong). Consider the equilateral triangle. Obviously, the smallest ellipse that circumscribes it is a circle. The largest ellipse that inscribes it is also a circle. Both these circles have centers at the triangle's centroid ...
We know that any triangle can be transformed to an equilateral triangle through rotation and scaling in the x and y directions. These operations preserve the sizes of two inscribing or circumscribing ellipses relative to each other and relative to the triangle, and also preserve the inscribing or circumscribing property and elliptical shape of those ellipses.
Therefore, the ratio of the area of the triangle to the areas of the smallest circumscribed and largest inscribed ellipses does not depend on which triangle you are talking about.
An equilateral triangle of side 1 has an area of sqrt(3)/4, the circumscribed circle has an area of pi/3, and the inscribed circle has an area of pi/12.
Because the area of the 3-4-5 triangle is 6, then the area of the smallest circumscribed ellipse is pi/3*6/sqrt(3)*4 = pi*8/sqrt(3), and the area of the largest inscribed ellipse is one quarter that, or pi*2/sqrt(3).
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Earendil
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Re: Steiner Ellipses
« Reply #7 on: Mar 7th, 2004, 6:32pm » |
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I tried to solve the smallest elipse problem using a little bit of what was learned in Calculus classes  Not sure if its alright though:
The formula of an elipse is: (x-x')²/a² + (y-y')²/b² = 1 where (x',y') is its center.
And its area is given by: [pi]*a*b.
Now using Analytic Geometry, we can suppose that the triangles edges are at the points A(0,3/2) , B(0,-3/2) and C(4,3/2).
The elipse passes the three points so:
{ (-x')²/a² + (-3/2 - y')²/b² = 1 (e1)
{ (-x')²/a² + ( 3/2 - y')²/b² = 1 (e2)
{ (4 - x')²/a² + (3/2 - y')² = 1 (e3)
I) By (e2) - (e1) we obtain y' = 0
II) By (e3) - (e2) we obtain x' = 2
Substituting x' = 2 and y' = 0 in (e1) we obtain:
4/a² + 9/4b² = 1 -> 16b² + 9a² = 4b²a² ->
16b² = a²(4b²-9) -> 4b*sqrt(1/(4b²-9)) = a
Now we have:
Area elipse = f(a,b) = [pi]*a*b
And the restriction: 4b*sqrt(1/(4b²-9)) = a
Substituting...
f(b) = [pi]*4b²*sqrt(1/(4b²-9))
f'(b) =
[pi]*((16x³-72x)*sqrt(4x²-9))/(16x^4 - 72x²+ 81))
f'(b) = 0, Implies:
b = 0 (Not useful)
b = 3*sqrt(2)/2 , a = 2*sqrt(2)
b = 3/2 (Not useful)
So it's area is: a*b*[pi] = 6*[pi]
And its equation is: (x-2)²/8 + 2y²/9 = 1
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| « Last Edit: Mar 7th, 2004, 6:39pm by Earendil » |
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Barukh
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Re: Steiner Ellipses
« Reply #8 on: Mar 8th, 2004, 9:07am » |
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It's not a surprise I missed this woderful problem at this site: at the time the post next to last was written, I wasn't even aware about the existence of the site...  Thanks to Earendil for popping it up!
Let me propose a follow up: detertmine the center and both axes of the sought-for ellipses.
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Earendil
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Re: Steiner Ellipses
« Reply #9 on: Mar 8th, 2004, 12:46pm » |
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Barukh it might be easier to...
put the triangle in the center since there is no equation equivalent to it
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Barukh
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Re: Steiner Ellipses
« Reply #10 on: Mar 9th, 2004, 1:55am » |
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on Mar 8th, 2004, 12:46pm, Earendil wrote:| Barukh it might be easier to...<hide> |
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I am not sure I understand what you mean. My question was: given the triangle, where are the centers of the minimal (maximal) ellipse and what are their axes?
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Earendil
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Re: Steiner Ellipses
« Reply #11 on: Mar 9th, 2004, 8:04am » |
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Ops... sorry, I misunderstood what you had said. I did the same thing
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Barukh
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Re: Steiner Ellipses
« Reply #12 on: Mar 10th, 2004, 1:03am » |
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on Mar 7th, 2004, 6:32pm, Earendil wrote:| I tried to solve the smallest elipse problem using a little bit of what was learned in Calculus classes Not sure if its alright though: |
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I think I know what's the problem with your solution. You place the triangle in a certain position in the coordinate system (which is OK), and assume that the minimal ellipse will be also in a certain position. The latter needs to be justified, and I doubt it’s correct.
More specifically:
Quote:| Now using Analytic Geometry, we can suppose that the triangles edges are at the points A(0,3/2) , B(0,-3/2) and C(4,3/2). |
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So, the (right) triangle is situated so that its legs are parallel to the coordinate system axes.
Quote:| The formula of an elipse is: (x-x')²/a² + (y-y')²/b² = 1 where (x',y') is its center. |
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That’s not the formula of the ellipse in the general position: this one is situated so that its axes are also parallel to the coordinate system axes.
To make it right, you need to consider the general equation of an ellipse Ax2 + By2 + Cxy + Dx + Ey + F = 0.
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Earendil
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Re: Steiner Ellipses
« Reply #13 on: Mar 10th, 2004, 1:03pm » |
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It might be easier just to use
{ (x0 - x1)² + (y0 - y1)² = 9
{ (x1 - x2)² + (y1 - y2)² = 16
{ (x2 - x0)² + (y2 = y0)² = 25
{ x0²/a² + y0²/b² = 1
{ x1²/a² + y1²/b² = 1
{ x2²/a² + y2²/b² = 1
And try to isolate a and b (didn't try yet though)
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