[Monte314]

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[ScurvyRose]

1:1

[muguly]

2:1

[ScurvyRose]

OK, how about this....It is all red! The yellow is a result of a green additive!

[Mozzes]

( Show Spoiler )

I think I see a strategy. You can use one side of the triangle, arbitrarily define the length (easiest would probably be the put it equal to 1) and use it as a chord to figure out the arc length of the portion of the circle bound by the two points. From there you can figure out the circumference, from the circumference you can figure out the radius and with the radius you can use the formula for a circular segment to figure out the area of the yellow parts. You can then subtract the area of the 3 circular segments from the total area of the circle to figure out the area of the triangle and from there figure out your ratio.

That's assuming of course that the ellipse that the triangle is inscribed inside of is a circle. I suspect it is but I don't have a proof for that off the top of my head.

Going old school aren't you Monte? I'm sure some Greek was drawing this one up in the sand 3000 years ago.
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[thod]

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The figure on the right is the same as the one on the left just tilted at an angle and rotated. Thus all we are doing is working out the ratio for the one on the left.

The sides of the triangle are 2rCos(30) long. So the red triangle has an area of 3r^2/2.
The yellow chords must therefore be r^2( Pi - 3/2). The ratio is therefore 1.5/(Pi-1.5) = 0.91

[ScurvyRose]

OK, so I rounded!

[Monte314]

*trembling*

Uh, Ms. Rose. Uh... you win the prize for FIRST RESPONSE. OK?

(Please don't trout slap the nice mathematician!)

As usual, Thod's method is right on target. However, I get a different value for the area of the triangle, so our answers do not agree. Thod, could you check your area computation?

[Wufnu]

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The length of half of a side of the equilateral triangle is going to be found by dividing the triangle up into some right triangles. You will find that using the radius of the circle, r, as the hypotenuse of one of these right triangles, r*cos(30) will be half the length of each arm length so each arm is 2*r*cos(30).

If side length L = 2*r*cos(30) = base then the height of the triangle is 2*r*cos(30)*sin(60), which would make the area = .5 * 4 * r^2 * (cos(30))^2 * sin(60) which simplifies down to: 1.2845452525 * r^2

To find the area of the yellow portions, it's simply pi*r^2 - 1.2845452525 * r^2 which = r^2 * (pi - 1.2845452525). This will give a ratio (red over yellow) of 1.2845452525 / (pi - 1.2845452525) = 0.69

So, the ratio of the red area to the yellow area is 0.69

[redbaren]

dang I just started Geometry so i think this over my head.

[Monte314]

Yay! Wufnu's answer is correct.

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As several of you have pointed out, the ratio for the right hand figure is *the same* as the ratio on the left hand figure. (That's why I included both!)

To create the right hand figure, the left hand figure is compressed. This compression affects the area inside the triangle in exactly the same way it affects area outside the triangle, so the problem for the left and right hand sides has the same answer. The left hand figure is easy to do using simple geometry.

The correct answer is 3*sqr(3)/((4*pi)-(3*sqr(3)) ~ 0.7050201618985662

(Here "sqr" is the positive square root).

[ScurvyRose]

	Originally Posted by Monte314 To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	*trembling* Uh, Ms. Rose. Uh... you win the prize for FIRST RESPONSE. OK? (Please don't trout slap the nice mathematician!) As usual, Thod's method is right on target. However, I get a different value for the area of the triangle, so our answers do not agree. Thod, could you check your area computation?

I guess alot when it comes to formulas! I am impressed with some of the thought processes that go into solving these!

(I struggled with statistics and algebra in college, obviously!)

[muguly]

Damn it! I suck at math :-(

[Wufnu]

These are fun, more more!

[ScurvyRose]

	Originally Posted by Wufnu To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	These are fun, more more!

Monte needs very little encouragement to post math!! I am sure he is already formulating the next one!

[enWTFp]

Oh!

( Show Spoiler )

Volume proportions are preserved under projections. Follows duality in the optimization problems.

As 1-dimensional volume, the lengths of the arcs on the ellipse preserve also the same proportion - 1/3 of its circumference. (a side note)

Red : Yellow = 1/2 sin120 : Pi/3 - (1/2 sin120) =~ 0.7

[Monte314]

enWTFp: nice treatment.

SR and Wufnu: many more are on the way!

[Silence]

Aargh. It's been too long between math classes, and my brain has clearly rusted shut.

That being said, the gears are trying to turn- and that's always a good thing.

[Monte314]

These problems are like WD40 for the mind....

[Lagawrd]

One who adopts math correctly from the beginning should automatically have a life time supply of WD-40 and would not get rusty except a few things that require memory.

I always arrive late to your problems monte. Even though I have never got the problems completely, I have always had a good approach, these will train me further.

[Monte314]

	Originally Posted by Lagawrd To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	One who adopts math correctly from the beginning should automatically have a life time supply of WD-40 and would not get rusty except a few things that require memory. I always arrive late to your problems monte. Even though I have never got the problems completely, I have always had a good approach, these will train me further.

This is why I post them (aside from the fun of cooking them up). A good problem helps people put familiar ideas together in unfamiliar but useful ways... and that is the essence of creativity.

[Avid]

Monte keep doing these so you brainwash me into having an appreciation of doing math for maths sake. In any instance where the math is directly useful to some goal I have I learn it...otherwise I ignore it. Math has always been my weakest subject due to pure apathy.

[reb]

Monte,
trying not to cheat and look at answers lol!

re: equilateral in circle, intuitively 'half'. ellipse with triangle...? the same? woont that be funny!
reb

[Monte314]

	Originally Posted by reb To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Monte, trying not to cheat and look at answers lol! re: equilateral in circle, intuitively 'half'. ellipse with triangle...? the same? woont that be funny! reb

Very insightful. It is a fact that the ratio is the same in BOTH pictures. That's why I put them both up there. The ratio is not 1:2, however.

[reb]

'half' as in 'triangle takes half the area...' bad communication on my part, and i undergradded in english...duh.

i've had 'blueprint reading', so the visual/spatial is pretty easy lots of times; i frequently don't need to look for a size on wrenches (or wenches)...i know which one to pick up, as i 'see' the nut/bolt size, and can match it to the wrench gap....down in the .001, though, i want my mic...lol! in .0001, i want my starrett....
reb