[bonanova]

What's the area of the smallest ellipse that contains the 3-4-5 triangle?

What's the area of the largest ellipse that fits inside the 3-4-5 triangle?

This will test your calculus or your logic.

Enjoy.

[Guest]

I think the minor radius of the largest ellipse would be 3 and the major one would be 4. Therefore the area would be 12(pi) or 37.6991 (I'm not sure if I did it right)

I'm not sure how to find the area of the smallest ellipse.

[plainglazed]

6.25pi -- from the radius R of a circle circumscribing a triangle = the product of the lengths of the triangles sides / 4R and thinking the circumscribing circle would be the same area as the smallest circumscribing ellipses for a 3 4 5 triangle but cant formalize that logic.

[Guest]

What's the area of the smallest ellipse that contains the 3-4-5 triangle?

What's the area of the largest ellipse that fits inside the 3-4-5 triangle?

Smallest: 6 pi

Largest: 12pi/(2+sqrt(2))

[plainglazed]

Smallest: 6 pi

Largest: 12pi/(2+sqrt(2))

yes, I can see that now for the smallest. and tho you can see my math's not great, I can tell that number's smaller than mine.

[bonanova]

No one has it yet.

All answers so far are on the high side.

The problem simplifies with a particular point of view.

[Guest]

Smallest: 6 pi

Largest: 12pi/(2+sqrt(2))

Oops.

6pi/(3+2sqrt(2)) for the largest ellipse that fits inside the triangle. But if 6pi is incorrect for the other one, then that's probably strike two...

[bonanova]

Oops.

6pi/(3+2sqrt(2)) for the largest ellipse that fits inside the triangle. But if 6pi is incorrect for the other one, then that's probably strike two...

Right - 0 and 2.

Choke up on the bat a bit ... ... or just look at the puzzle from a different angle.

[plainglazed]

1.75pi(1.5)^.5. But that seems pretty small relative to the area of the 345 triangle.

[Guest]

A typical Sangaku problem. From what I understand you have to use affine transformation to solve the first part of the question. I don't know how to do the second one though.

For the first one: We have an ABC triangle that is a 3-4-5 triangle. We're looking for an ellipse whose area ratio over the triangle area is minimal. So for every ellipse which encloses the triangle, we have an affine transformation that sends the ellipse into the unit circle and the triangle ABC into a triangle A1B1C1 enclosed by the unit circle. After which you do some analytical geometry and use some common geometry sense to figure out that the triangle

A1B1C1 must be equilateral. So the ratio between the area of the unit circle and the area of the triangle A1B1C1 is

pi/((3/4)sqrt(3)) = (4pi)/(3sqrt(3))

So the minimal area of an ellipse which encloses a 3-4-5 triangle ABC is

(4pi)/(3sqrt(3)) * 6 = (8pi)/(sqrt(3))

PS: both situation require a circle not an ellipse. I think that's the angle we're all supposed to look at it from.

Edited July 13, 2009 by Dunpeal

[Guest]

I think I may have figured out the second one too. You just have to use some equations that are already provided for us in geometry. I'm assuming that the 3-4-5 triangle is a right triangle, since I don't think it'll be possible any other way. The radius of a circle inside a right triangle is found by the equation

r = ab/(a+b+c) = s - c

where

s = (a+b+c)/2

and c is the hypotenuse. So if I'm right and we solve we should get s = 6, r = 1. Therefore, the area of the circle is

A = 2pi

[Guest]

A typical Sangaku problem. From what I understand you have to use affine transformation to solve the first part of the question. I don't know how to do the second one though.

For the first one: We have an ABC triangle that is a 3-4-5 triangle. We're looking for an ellipse whose area ratio over the triangle area is minimal. So for every ellipse which encloses the triangle, we have an affine transformation that sends the ellipse into the unit circle and the triangle ABC into a triangle A1B1C1 enclosed by the unit circle. After which you do some analytical geometry and use some common geometry sense to figure out that the triangle

A1B1C1 must be equilateral. So the ratio between the area of the unit circle and the area of the triangle A1B1C1 is

pi/((3/4)sqrt(3)) = (4pi)/(3sqrt(3))

So the minimal area of an ellipse which encloses a 3-4-5 triangle ABC is

(4pi)/(3sqrt(3)) * 6 = (8pi)/(sqrt(3))

PS: both situation require a circle not an ellipse. I think that's the angle we're all supposed to look at it from.

That was very helpful. I haven't proved that the solutions yielded by this technique are optimal, but am not sure I can without getting into way too much calculus.

Applying a shear to a circle gives you an ellipse, and applying a stretch to an ellipse also gives you an ellipse. Shearing preserves area, and stretching multiplies area by the stretch factor. All we need is the area of the ellipse, so we don't have to bother with its other parameters.

So we start with a circle, and consider its largest inscribed triangle, and smallest circumscribed triangle. In each case the triangle in question is equilateral. We can then stretch and shear that triangle into the 3-4-5 right triangle to get our best-fit ellipse.

Considering a 3-3-3 equilateral triangle and using some basic geometry (30-degree right triangles in particular), we can figure out that the triangle's height is (3/2)*sqrt(3), and that the center is at 1/3 that height, or (1/2)*sqrt(3). The distance from the center to the apex is then sqrt(3).

To transform this equilateral triangle into the 3-4-5 right triangle we need to first shear the equilateral triangle to make it right. Shearing parallel to the base, we preserve both the base of 3 and the height of (3/2)*sqrt(3). Then we stretch the triangle vertically to make its height 4, which requires a scale factor 4/[(3/2)*sqrt(3)] = 8/[3*sqrt(3)]. That gives us our 3-4-5 triangle. We only scaled once, so we must transform the circle's area by the same factor of 8/[3*sqrt(3)] to find the area of the ellipse in question.

The inscribed circle to a 3-3-3 triangle has a radius equal to the distance from the triangle's center to its base, (1/2)*sqrt(3). So the circle's area is (3/4)*pi.

Shearing and stretching the 3-3-3 into the 3-4-5, we transform the circle to an ellipse with area 8/[3*sqrt(3)]*(3/4)*pi = 2*pi/sqrt(3).

The circumscribed circle to a 3-3-3 triangle has a radius equal to the distance from the triangle's center to its apex, sqrt(3). So the circle's area is 3*pi.

Applying the shear and stretch, we transform the circle to an ellipse with area 8/[3*sqrt(3)]*3*pi = 8*pi/sqrt(3). This is the same as Dunpeal's result above.

[bonanova]

gurn has it. with groundwork by Dunpeal

in the plane change all areas by the same factor.

Take a unit equilateral triangle and its circumscribed and inscribed circles

Find the area ratios of the circles to the triangle.

Multiply these ratios by 6, the area of the 345 triangle.