Average Area of a Triangle Created from 3 points on $f(x)= \sin(x)$

Calculus Level 3

$x_{1}$ , $x_{2}$ , and $x_{3}$ are chosen randomly and independently on the interval $[0,2\pi]$ and let $f(x)=\sin(x)$ .

What is the average area of the triangle created with the three vertices below (to two significant figures)?

$P_{1}=\big( x_{1} , f( x_{1})\big),\quad P_{2}=\big( x_{2} , f( x_{2})\big),\quad P_{3}=( x_{3} , f( x_{3})\big)$

The answer is 0.67.

5 solutions

Chris Lewis
Dec 3, 2018

By the shoelace theorem , the area of the triangle is given by

$\begin{aligned} \Delta=\frac{\left| (x_1 \sin{x_2}+x_2 \sin{x_3}+x_3 \sin{x_1})-(x_2 \sin{x_1}+x_3 \sin{x_2}+x_1 \sin{x_3}) \right|}{2} \end{aligned}$

The average area can be evaluated numerically either by scanning over the parameter space (ie letting $x_1,x_2,x_3$ each step from $0$ to $2\pi$ in small intervals, calculating the area for each of these, then averaging) or with a Monte-Carlo implementation (ie picking a large number of values for $x_1,x_2,x_3$ uniformly at random and again averaging).

The result comes out at $\boxed{0.67}$ , and seems to be a little higher than $\frac{2}{3}$ . Did anyone manage to find an exact value without resorting to numerical methods?

Did the Monte-Carlo way an got 0.668537 with 10,000,000,000 samples.

Tom Stein - 2 years, 6 months ago

Agreed - My Monte Carlo ( GCC (and libraries) - double floating points - 7.3.0 x86 architecture) gives 0.66853x on about 10^9 samples. Rounding errors would probably go the other way, not adding very small areas to a very large total accumulator.

Ed Sirett - 2 years, 6 months ago

numerical integration using mpmath library gives the result of 0.668481 with an error of 1e-7

Ilya Lyubarsky - 2 years, 4 months ago

Vinod Kumar
Dec 5, 2018

Wrote a Python program generating pair of three random vertices (x, sin(x)) and calculated the average of area of 10000 random triangles using:

Area=abs[0.5{x1 (y2-y3)+x2 (y3-y1)+x3*(y1-y2)}]

Answer=0.668

Python3 Program on Ubuntu 18.04

import math

import random

n = int(input("Maximal Number? "))

TArea=0

counter = 0

while counter <= n-1:

x1=(2*math.pi)*random.random()

y1=math.sin(x1)

x2=(2*math.pi)*random.random()

y2=math.sin(x2)

x3=(2*math.pi)*random.random()

y3=math.sin(x3)

Area=abs(0.5*(x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2)))

TArea +=Area

counter += 1

print("Answer",TArea/counter)

Same here, and got the value 0.669. My code is almost identical to yours!

import sys
import math
import random

# Expected area of a random triangle determined by three points on a sine curve in [0,2 pi]

n=10**6         # Number of triplets simulated
i=1             # Counter
Area=0          # Area of the simulated triangle
Sum=0           # Cumulative area
EV=0            # Expected value of the area 

while i <= n:
    Ax=2*math.pi*random.random()                # x coordinate of the first random point on the sine curve (A).
    Ay=math.sin(Ax)                             # x coordinate of the first random point on the sine curve (A).
    Bx=2*math.pi*random.random()                # idem for points B and C.
    By=math.sin(Bx)
    Cx=2*math.pi*random.random()
    Cy=math.sin(Cx)

    Area=0.5*abs((Cx-Ax)*(By-Ay)-(Bx-Ax)*(Cy-Ay))    # The area this random triangle is
                                                     # one half of the abs. value of the cross product of vectors AB and AC.
    Sum+=Area                                        # We add all areas...
    i+=1

EV=Sum/n        # ... and calculate the average.

print ('The mean value of the areas simulated is', EV)

Gabriel Chacón - 2 years, 5 months ago

The program below is equivalent to calculating 800^3=512 million triangles, evenly distributed over the range. It is somewhat optimised for speed. I trust the answer upto 5 decimal places: 0.66852....

Some playing with parameter count and considering differences of the calculated values made me realize there a some small, but systematic underestimation due to discrete values being the same more likely than in a true random case. The error in these calculations is an underestimate that diminishes by a factor 4 when count doubles:

count	result	delta
25	0.661540
		0.0053
50	0.666807
		0.0013
100	0.668096
		0.00032
200	0.668419
		0.000081
400	0.668500
		0.000020
800	0.6685203

Considering this, the true value must be close to $\boxed{0.668527}$ .

import math;

def MeanTriangleArea(count):
    ThirdArea=0; MaxValue=2.0*math.pi
    x=[];y=[];
    for t in range(0,count):
        xvalue=(t+.5)*MaxValue/count
        yvalue=math.sin(xvalue)
        x.append(xvalue)
        y.append(yvalue)
    for t1 in range(0,count):
        for t2 in range (t1,count):
            x2=x[t2]-x[t1]; y2=y[t2]-y[t1]
            for t3 in range (t2,count):
                x3=x[t3]-x[t1]; y3=y[t3]-y[t1]
                ThirdArea += abs((x2)*(y3)-(y2)*(x3))

    # The area of each triangle is |a×b|/2, so only half the calculated value
    # We only considered t1<t2<t3, but there are 3!=6 permutations of (t1,t2,t3)
    # Combined we need to multiply with 3
    return 3.0*ThirdArea/(count*count*count)

print(MeanTriangleArea(800));

K T - 9 months, 3 weeks ago

Kevin Tong
Dec 4, 2018

Using the Shoelace Theorem, we have the area is $\frac{1}{2}\left| x_1\sin x_2+x_2\sin x_3+x_3\sin x_1 - x_2\sin x_1 - x_3\sin x_2 - x_1\sin x_2\right|$ We then use a program to find the average area to get a value of approximately $\boxed{0.67}$ .

Max Brodsky
Dec 5, 2018

One can find the exact average by executing a triple integral with integrand as the shoelace area mentioned in other answers and parameters of each x value going from 0 to 2pi and dividing by the input volume. This gives exactly two-thirds.

That integral is unsolvable and computing that integral numerically gives about 0.66 or 0.67 .

Hari Krishna - 2 years, 6 months ago

Unless you show how, this answer must be considered as bluff. And the answer is not exactly 2/3.

K T - 9 months, 3 weeks ago

A Former Brilliant Member
Dec 2, 2018

area = Function[{a, b, c}, Block[{p1 = {a, Sin[a]}, p2 = {b, Sin[b]}, p3 = {c, Sin[c]}}, Area[ SSSTriangle[ EuclideanDistance[p1, p2], EuclideanDistance[p2, p3], EuclideanDistance[p3, p1]] ] ] ];

z = Table[area @@ t, {t, RandomReal[{0, 2 [Pi]}, {1000000, 3}]}]

z1 = Select[z, # [Element] Reals &]

(* One generated triangle was so degenerate that it failed sum of side test for a triangle. I substituted 0 as its area. *)

Length[z1] is 999 999

Mean[Join[z1, {0.}]] is 0.669701546195

(* Mathematica 11.3 used *)

what are you sying

Nahom Assefa - 2 years, 6 months ago

I have been a programmer for more than a half century. I gave the source code for the Monte Carlo simulation I used. Line for line not including comments in (* *): 1. Define a function that computes the area of a triangle whose three sides are specified by three points whose x-coordinates are arguments and their y-coordinates by Sin[x-coordinate].

Compute a one million entry table of areas of such triangles.
Select those areas which give real results. There was one triangle that was degenerate.
Add in a zero area for the degenerate triangle and compute the mena of the table.

Q.E.D.

A Former Brilliant Member - 2 years, 6 months ago

Average Area of a Triangle Created from 3 points on f ( x ) = sin ⁡ ( x ) f(x)= \sin(x) f ( x ) = sin ( x )

The answer is 0.67.

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Average Area of a Triangle Created from 3 points on $f(x)= \sin(x)$