/sci/ - Science & Math

>>14505169
2

1. solve for the hypotnuse which is root 2
2. root 2 x 2piR; R is half of root 2
3. divide again for half circle

>>14505175
the hypotenuse isnt root 2 because the angle opposite the hypotenuse isnt 90 degrees meaning a^2 = b^2 + c^2 doesnt work here

>>14505169
OP here, I forgot to mention that the angle between the 2 straight lines can be of any value.

>>14505169
Look up the brachistochrone problem
>Triangle area = .5ABsin(theta)
>Opposite length, L, of triangle = sqrt(A^2+b^2-2ABcos(theta)
>Area under curve x = integral x from 0 to L
>Sum of areas = 1
>Length of curve x = integral sqrt(dx^2+dy^2)

>>14505187
or maybe lagrange multipliers first

>>14505169
The shortest distance between 2 points is a stright line, so optimizing for the length of x it must have a stright shape because any curve between 2 points has greater length. Then the figure is an equilateral triangle.

x/2 = t

t**2 + h**2 = 1**2
h * t = 2 * 1
h = 2/t
t**2 + 4/t**2 = 1
t**2 = q
q**2 - 1 q + 4 =0

>>14505222
>equilateral
I've mistaken, English is not my native language.

*isosceles

If you assume a triangle and use herons formula, you get that x isn't real. Therefore it's not a triangle.

>>14505222
But an equilateral triangle with side length 1 still has an area less than 1, so the line must be a curve bulging outward

>>14505291
>>14505308
Yes, I have already figured that out (a square 1x1 has area of 1 but the maximum of area of the triangle is half of it). One has to resort to variational calculus to solve this, but I'm not good at it.

Calculus of variations

>>14505169
You could try variational Calculus. You could model the problem as follows with a certain transformation: you are looking for a function from 0 to sqrt(2) with minimal length (make sure to use arclength) under some area constraints. Total area in the original problem is the sum of the triangle area + the area of the curve in the transformed problem (here the area of the curve can be negative). You have am isoperimetric problem with am integral constraint, use lagrange multipliers. Once you get the curve, you atleast know the shape, then you have to transform it back into your picture somehow. Of course you could try to solve it directly as a planar curve with some parametric equation, might even be simpler

>>14505169
i guess it would be shaped like an ice cream, since a circle has the greatest ratio of area to circumference. I set up a triangle with area based on the angle, and a circle with area based on the diameter, which is the same as the third side of the triangle. Then i solved numerically for sum of the areas and got x=1.379. The angle for this is 52 degrees.

>>14505169
apparently it's a circular arc
https://youtu.be/IknAKDSDOw0

also https://math.stackexchange.com/questions/4455800/find-curve-that-minimizes-lenght-with-integral-constraint

>>14505523
Circle - yes, but, but I'm nkt sure it is the same for an arc between 2 points.

>>14505523
i made a mistake here. also solved it exactly, at least if the ice cream shape is true:
x = (pi/2)*sqrt(8*((1+pi-sqrt(2*pi-3))/(4+pi^2)))

>>14505576
that does not matter since you can transform your problem for your curve to lie on the real line (just translate the triangle so that one end is at the origin and the other end somewhere on the x-axis). Qualitatively nothing should change and I am sure the solutions are semicircular arcs either facing up (adding area), facing down (taking area away from the triangle) or the the trivial ''constant' semicircle with height 0, lying entirely on the x-axis (in this case the constraint area matches the area you would get completing the triangle by a straight line).

>>14505169
x=1

>>14505169
i guess its thales theorem

you elongate the arms and give them proportional lenght

>>14505169
Pi minus 1

>>14505724
Sorry, meant Pi minus 01.0

>>14505739
Sorry, pi 0.00

>>14505749
fuck

>>14505724
>>14505739
>>14505749
>>14505751
What I meant to say is that a unit equilateral triangle is approximately a cotangent quarter sine wave minus 1
Hope that makes sense

>>14505754
Also if the coordinate is a t or a z or a w or something it can be a Fourier penis ejaculating radiosity semen into your mother’s uterus

>>14505764
And if it’s a T it could be an even bigger and harder dick

>>14505291
>it's not a triangle.
no shit...

>>14505782
With enough very primitive bump maps a T could be a uterus as well I guess but the real time cock tracing will probably have to wait for more computational resources

>>14505816
You could also make an eggbryo but it might take that Java Minecraft engine or that URE thing or something

>>14505169
First observation: your shape cannot be concave. Making it concave results in longer x while reducing area.

So, the shape is for sure convex. Then it can be divided into two parts:

1. The triangle that is formed if you assume x to be just the straight line.

2. The rest (a funny bubble created by x)

Using trigonometry, you can express as a function of the angle:

1. The distance between ends of angle's rays. Let's call it D(angle).

2. The area that your "funny bubble created by x" should take. Let's call it H(angle) = (1 - triangle_area).

Observation: if you're able to solve the task for any given angle, then you can just use calculus to find the angle that gives you the smallest solution.

So, basically, your task boils down to:

Given a line segment of length D(angle), what is the shape that contains it on the border, has area equal to H(angle), and smallest length possible.

If you think about it for a second, it should indeed be some ellipse.

1. It cannot be concave, so we rule out all funny shapes.

2. It cannot have sharp corners (because round corners give you more area per unit of length).

At this point I would plug in ellipse equations, use calculus, and look for ones that satisfy our criteria.

>>14505885
> can't be concave
under the assumption that the Area is 1 you are right, but in general this isn't true. If the Area constraint is less than what you'd get connecting the triangle straight, then the curve has to bend to the inside

the triangle does not matter to the problem, other than adding a constant to the area constraint (which in itself is an Integral constraint).You can translate the triangle, so that the whole problem will resolve around finding a function y(x) on the real axis. Using Calculus of Variations with multipliers, the minimizers become circular arcs, with the trivial edge case of a direct line segment when the triangle would satisfy the area constraint directly (this solution corresponds to a circular arc of a circlewith an infinitely large radius).

>>14505169
Let's say the angle between the two straight lines is [math]\alpha[/math]. The problem can be reformulated into finding the [math]\alpha[/math] and a function [math]u : [0,\alpha] \to \mathbb{R}^+ [/math] that minmizes
[eqn]\int_0^\alpha \sqrt{(u(t))^2 + (\dot u(t))^2} dt [/eqn]
under the constraints
[eqn] \frac{1}{2} \int_0^\alpha (u(t))^2 dt = 1\\
u(0) = u(\alpha) = 1
[/eqn]

bunch of pseuds in this thread, this serves as a prove that /sci/ doesn't even have the necessary reading comprehension to tackle basic problems.

Why don't you solve it if you're so smart?

>>14505964
Here I did your homework, happy? I used the parameteic representation for the curve, using a single function y(x) should be equally solvable I think, you'll have to try it (after transforming the problem and triangle to the real axis). Lambda, c1 and c2 need to be found satisfying endpoint conditions and area constraints. x0 and y0 I assume to be given as one of the ends of the triangle. Notice that c1 and c2 contain derivatives of the curve, so one has to be careful when choosing those. Also I wouldn't be surprised if that problem can't be solved for arbitrary area constraints, imagine requiring A = 1000, the resulting curve will most likely not be smooth, also small areas will likely cut the triangle and fail too. I assume in those cases the minimizer will be made of two lines extending outwards and at some time connected by a circular arc.

>>14505945
minimizing [math] \int_0^\alpha \sqrt{(u(t))^2 + (\dot u(t))^2} dt[/math] is equivalent to
minimising [math] \int_0^\alpha \left\{(u(t))^2 + (\dot u(t))^2 \right\} dt[/math] which is
[math] \int_0^\alpha (u(t))^2 dt + \int_0^\alpha (\dot u(t))^2 dt[/math]
but [math] \int_0^\alpha (u(t))^2 dt = 2[/math] by constraint,
so we want to minimise [math] \int_0^\alpha (\dot u(t))^2 dt[/math] which can be written
[math] \int_1^1 \dot u(t) du[/math] which is zero;
but this means [math]u(t)[/math] is constant, so [math]u(t)=1[/math]
and the diagram is a sector of a circle of radius one, with area one,
so [math]\alpha =1[/math] radian = 57.3° (approximately)

>>14506129
...looks correct, except the last line. Should be:
minimizing [math] \displaystyle \int_0^\alpha \sqrt{(u(t))^2 + (\dot u(t))^2} dt[/math] is equivalent to
minimising [math] \displaystyle \int_0^\alpha \left\{(u(t))^2 + (\dot u(t))^2 \right\} dt[/math] which is
[math] \int_0^\alpha (u(t))^2 dt + \int_0^\alpha (\dot u(t))^2 dt[/math]
but [math] \int_0^\alpha (u(t))^2 dt = 2[/math] by constraint,
so we want to minimise [math] \int_0^\alpha (\dot u(t))^2 dt[/math] which can be written
[math] \int_1^1 \dot u(t) du[/math] which is zero;
but this means [math]u(t)[/math] is constant, so [math]u(t)=1[/math]
and the diagram is a sector of a circle of radius one, with area one,
so [math]\alpha =2[/math] radianS = 115° (approximately) and [math]x=2[/math]

There are two steps for solving this problem. First you need to find a way to minimize the length of the x-line between the straight lines. Then you need to find an angle between the straight lines for which the minimized x-line is the shortest.

>>14505945
how did you write the mathematical symbols?

I dk how to solve this, but bump anyways cuz i wanna know what's the true ans

>>14508511
true answer is >>14506028 if the angle is not a decision variable (meaning you can choose it one but for the problem it is fixed). OPs problem is not well formulated, you don't have certain coordinates from the start and end points and you don't have the angle given. The result is however an arc of a circle. This circle is determined by having the two end points of the segments on it's circumference and having radius such that the total area of the inscribed figure is 1. This shouldn't be too hard to solve, the main point however is that the minimizer is some sort of circular arc. This only works when you can find such a circular arc, if not I suspect the solution becomes a broken irregular extremal curve, where you extend both segments linearly and put a semicircle on top of the extensions at some point, such that the area constraint is satisfied. If the area should be really small, say 0.01, you can also imagine the solution will be two lines facing directly into the the triangle and completed with a circular arc somewhere inside the triangle. If the angle is fixed in the problem, then the triangle isn't even part of the problem in it's essence, you are just looking for a curve with minimal length satisfying some area (area of curve + constant area of triangle = 1).