Another area challenge. Find the area of the largest triangle between these curves?

I think the maximum area is around 2.30674. This is a pure brute force problem, so I'll get back to this later. Not real sure about this figure.

No chance on any exact answers.

Edit: This is what happens when I have to leave for work: Everybody else jumps in with their answers. Ah well, first of all, I have no idea where I got that 2.30674. I have 1.34081, which is closer to Vikram P's answer, but I already know that 1.34081 is wrong and needs refinement, because the bottom chord intersects the Cosine curve. The real figure has to be slightly less than this, so, back to the drawing board.

Edit 2:.. 1.3408107. ...Still working. Actually, it's under 1.340812, not under 1.34081.

Edit 3:.. 1.34081127

Edit 4:.. 1.34081149

Edit 5:.. 1.34081160

Edit 6:.. 1.34081171

Edit 7:.. 1.340811769

Edit 8:.. 1.340811791

Edit 9:.. 1.340811802

Edit 10: 1.340811808

Edit 11: 1.340811810

Edit 12: 1.3408118106

Edit 13: 1.34081181065

Edit 14: 1.340811810657

Well, I think there's a convergence to a value here. But the bottom chord, while avoiding contact with the Cosine curve, intersects the Sine curve by an incredibly small margin beyond π/4, just by 0.00023901267. It is very difficult to distinguish between Cosine curve and the bottom chord in the vicnity of π/4.

Edit 15: Morning. Zeta, leave this open just a bit longer while I double check my equations.

Edit 16: Okay, forget everything above, I did make a small and really annoying mistake in my equations. The new revised best approximation, using a different approach, is:

1.340849434

With this mass of equations I'm having to deal with, it's easy to make a mistake. I might have to recheck everything all over again, and maybe improve accuracy, but later on.

Edit 17: 1.34084943628

This value has all the same digits as the 2 adjacent values which are less. For verification, the 3 points are at, along with area:

0.785573733397 on Sine

2.383114935032 on Sine

3.137496016263 on Cosine

1.340849436288

Method: Pick point A near π/4 on the Sine curve. Pick point B on the Cosine curve such that line AB clears the Cosine curve by a margin as small as practically possible. Use slope of line AB to determine point C on the Sine curve. Use Heron's formula to find area. Repeat. With the latest approximation, point A is 0.00017557 beyond π/4. Line AB misses the Cosine curve by 2.22 x 10^(-16). I think I quit now.

Editorial note: There are no other "sweet spots" in the vicinity of this solution. I checked the range of values for A from 0.0 beyond π/4 to 0.0003 beyond. It drops off quickly after 0.0003. There is a 2nd order extremal at about 0.00017557. Were it not for the fact that the chord cannot intersect the Cosine curve, the maximum occurs when A is 0.0 beyond π/4.

Edit 18: I tried to find another "sweet spot" elsewhere in the enclosed area between the curves, but any inscribed triangle eventually "snaps" to the maximal everybody's already been coming up with. The default case is where A is at π/4, and when intersections are not allowed, then it's what we have now.

Edit 19: The location of the 3 points for maximum area, if AB is allowed to intersect the Cosine curve, are:

0.7853981633974 on Sine & Cosine

2.3873614874707 on Sine

3.1276229438014 on Cosine

1.3410756431327

This agrees well with Zeta's first figures.

Edit 20: About that behavior near π/4, yes, that is odd. I checked it, and, sure enough, while keeping AB tangent to the Cosine curve, there is actually a minimum BEFORE π/4 is reached, right around π/4 + 0.00001 (very roughly), before it starts to go up again. So, you are not imagining things.

Edit 21: Dr D, actually, the largest possible area is 1.3410756431327, as also per Zeta's figures. But the triangle ABC isn't an inscribed one, one of the chords cuts across one of the curves. This is what makes the problem particularly messy.

Edit 22: Zeta, doing it by software only is possible, if you have a sufficiently sophisicated math package. Method by computer: Pick point X on the Cosine curve (close to π/4). Using equation of line tangent to it, find points A and B on the Sine and Cosine curve respectively. Using the slope of the tangent, find point C on the Sine curve. Use Heron's formula to find area. Reiterate to find max area to desired degree of precision. Problem is, I don't have that kind of math package.

Well, on the 2nd thought, maybe a workaround is possible for me to do this by computer with what I have anyway, but doing it "by hand" was much quicker. And I'm lazy.

Case 1: y = x^2 - a and y = -x^2 + a Both are parabola with y-axis as their axis. y = x^2 - a has vertex at (0, -a) and opens upwards y = -x^2 + a has vertex at (0, a) and opens downwards They both intersect at (-√a, 0) and (√a, 0). The area enclse between the curves is symmetrical about both axes. => Required area = 4 ∫(x=0 to x=√a) (-x^2 + a) dx = 4 [-x^3/3 +ax] (x=0 to √a) = 4 [-a√a/3 + a√a] = (8/3)a√a Any two of the four curves can be selected in 4C2 = 6 ways. Thus, there are 6 problems in 1. I shall be back to examine and try other cases within a few hours. Case 2: x = y^2 - a and x = -y^2 + a By symmetry, this area will be same as in case 1, i.e,, (8/3)a√a. Case 3: y = x^2 - a and x = y^2 - a Subtracting one eqn. from the other, x^2 - y^2 + x - y = 0 => (x - y)(x + y + 1) = 0 => x = y or y = - x - 1 x=y => x^2 - x - a = 0 => x = (1/2)[1 ± √(1 + 4a)] => (i) For a < - 0.25, the curves do not intersect, (ii) For a = - 0.25, the curves touch each other (iii) For a > - 0.25, the curves intersect in two points (iv) For a = 0, the intersection points are (0, 0) and (1, 1) y = - x - 1 => x^2 + x + 1 - a = 0 => x = (1/2)[ -1 ± √(4a - 3)] => (i) For - 0.25 < a < 3/4, the curves intersect in two points given by x = y = (1/2)[1 ± √(1 + 4a)] (ii) For a = 3/4, Two points of intersection are (-1/2, -1/2) and (3/2, 3/2) (iii) For a > 3/4, the curves intersect in fourpoints creating four areas between them and each area has to be worked out separately. I shall work out here the special case of a = 0 => y = x^2 and x = y^2 The points of intersection are (0, 0) and (1, 1) Area enclosed between the curves = ∫(x=0 to 1) (√x - x^2)dx = [ (2/3) x^(3/2) - x^3 / 3 ] (x=0 to 1) = 2/3 - 1/3 = 1/3. Case 4: y = -x^2 + a and x = -y^2 + a This case is similar to Case 3 with the corresponding curves being mirror images w.r.t. the coordinate axes.

The maximum area is 1.341075643. It is achieved when the points are

(2.38736148747, 0.684728546336)

(0.785398163397, 0.707106781186)

(3.12762294380, -0.999902425191)

Mathematica produces an answer instantly, with the simple code below:

pq = {q1 - p1, q2 - p2, 0};

pr = {r1 - p1, r2 - p2, 0};

Delta = 1/2 Norm[Cross[pq, pr]];

NMaximize[{Delta,

Pi/4 <= p1 <= 5 Pi/4 && Pi/4 <= q1 <= 5 Pi/4 && Pi/4 <= r1 <= 5 Pi/4

&& Cos[p1] <= p2 <= Sin[p1] && Cos[q1] <= q2 <= Sin[q1] &&

Cos[r1] <= r2 <= Sin[r1]}, {p1, p2, q1, q2, r1, r2}, WorkingPrecision -> whatever precision you want]

Okay, I see that it's picking up the intersection and the line joining the intersection and the lower point is actually intersecting the cosine past 0.785. (maybe this was the trap?) So we can tweak the answer slightly. Just move the point slightly forward on the sine so the segment doesn't touch the cosine.

I tweaked it a bit and moved the second point to (0.7856354716 , 0.7072745635). This makes the area come out to 1.340813653 which is still pretty good.

********

Sorry I haven't had a chance to get back to this problem. Although it's a very interesting one because of the kind of potential applications. For example, how would one approach solving this for arbitrary closed curve? Or in higher dimensions, for instance, let's say in a closed arbitrary surface, what is the maximum volume of a regular object (cube or tetrahedral, etc). This is what I've been thinking about the past few days.

Solving this by hand seems a little odd to me. I understand the method of Dr D. as a quite logical idea. But I was looking for a way to make the computer understand to avoid having the sides go out of bounds. This could be accomplished, for instance, by choosing a point where the line will be tangent to the curve and construct the triangle from there. But since, in our problem, the curve is cosine, it's very hard to find the intersection of cosine and a line. This is what made this job difficult.

I'm not worried about getting the "cake" lol. I would be happy with whatever choice of best answer you go with. However, it would be interesting if others can find a systematic way for a computer to understand avoiding intersecting the sides.

Thanks, falzoon, for a thought provoking question.

First let's find the area of the grey shaded part itself.

INT [pi/4, 5pi/4] sinx - cosx dx = -cosx - sinx [pi/4, 5pi/4]

= (1/sqrt2 + 1/sqrt2) - (-1/sqrt2 - 1/sqrt2)

= 4/sqrt2 = 2.828 approx.

I suggest that a triangle in this area is not going to fill much more than half of it (if that) so I am very skeptical of Scythian's answer.

My first choice for vertices of the triangle was the left intersection of the two curves, the minimum of the cosine curve, and a point on the sine curve. However, I found that the line between the first two just crosses the cosine curve so had to reject this.

(I suspect that this was the trap.)

I then decided to choose the tangent to the cosine curve at A and find where it cuts the curve again. This involved solving the equation.

x = 1 + pi/4 - (sqrt2)*cosx

Of course this has to be solved numerically and it gave (3.1974, -0.99844) approx as the point of intersection.

I then chose (3pi/4, 1/sqrt2) as the other vertex to make calculation easy.

This gives an area of (1/2)*(pi/2)*(0.99844 + 1/sqrt2) = 1.33953 approx.

This is very slightly larger than the previous answer but in good agreement with it.

EDIT. If the tangent to the cosine curve is to be used as the base of the triangle (and of course that's a big if) then the question arises as to what point on the sine curve would be furthest perpendicularly from it.

I have found that (3pi/4, 1/sqrt2) is the best point to use. I won't put all the working here but outline the method. Rotate the diagram by arcsin(1/sqrt3) so that the tangent line is horizontal. Work out the new equation of the sine curve and differentiate it to find maximum which is at x = pi*(sqrt6)/4 - pi/(4*sqrt3). This converts to x = 3pi/4 when the reverse rotation is applied..

This means that, if my first answer to the maximum triangle area can be exceeded, then a different baseline for the triangle must be used. Perhaps it is the line from the cosine minimum to a point on the sine curve just above (pi/4, 1/sqrt2) so that it is just tangent to the cosine curve near the same point.

I will return if I get anywhere with this.

I'm posting this because it is a simple method to find it numerically.

Choose two points on the cosine curve and one point on the sine curve. Because of symmetry this can always be reversed.

Write a program that selects points A, B and C sequentially between the two limits π/4 and 5π/4 in steps of say π/400. Find the coordinates which maximize that area.

Then zoom into these points, and select a smaller range just surrounding them and reduce the step size. Repeating the procedure will allow for increased accuracy.

The only problem is if you zoom into the wrong "hill".

I got 1.3409654 with points 5π/4 and 2.3406406452 on the cosine curve and 1.58666294 on the sine curve. When you reverse the direction, this is close to what Scythian has, but apparantly I zoomed into a different hill.

*EDIT*

I thought the fact that the triangle must be in the grey colored region means that it MUST be completely inscribed.

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What I get is 1.3364396+ units , which is way below what scythian has to offer but this is what my written program tells me, I may be wrong but have a look at following pic.

http://sites.google.com/site/vcpandya/TriangleArea...

lnnmk

very cool question - will add to my brainteasers after I get comfortable with a solution