Mean perimeter for random triangle inscribed in the unit circle [closed] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Determine if circle is covered by some set of other circlesAverage distance between two points in a circular diskGiven the circumcircle, the 9-point circle, and the angular measures for a triangle, construct the triangle?Does the perimeter of a 2-D object “count” toward its area?Maximum perimeter for triangle inscribed in circleFind the biggest area of an isosceles triangle in an unit circleFinding two points B and C such that the perimeter of triangle ABC is minimal (an extension)What is the likelihood of two line segments crossing?Intersection of random line segments on the sphereIntersection of random line segments in the plane
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Mean perimeter for random triangle inscribed in the unit circle [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Determine if circle is covered by some set of other circlesAverage distance between two points in a circular diskGiven the circumcircle, the 9-point circle, and the angular measures for a triangle, construct the triangle?Does the perimeter of a 2-D object “count” toward its area?Maximum perimeter for triangle inscribed in circleFind the biggest area of an isosceles triangle in an unit circleFinding two points B and C such that the perimeter of triangle ABC is minimal (an extension)What is the likelihood of two line segments crossing?Intersection of random line segments on the sphereIntersection of random line segments in the plane
$begingroup$
I calculated an approximate value of 2.2 by using a Monte-Carlo-Simulation (random points on circumference). Can anybody confirm that or offer an analytical solution?
geometry analysis stochastic-analysis
asked Mar 26 at 16:43
user3451767user3451767
63
$endgroup$
closed as unclear what you're asking by callculus, Cesareo, clathratus, Shailesh, Santana Afton Mar 27 at 2:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I calculated an approximate value of 2.2 by using a Monte-Carlo-Simulation (random points on circumference). Can anybody confirm that or offer an analytical solution?
geometry analysis stochastic-analysis
asked Mar 26 at 16:43
user3451767user3451767
63
$endgroup$
closed as unclear what you're asking by callculus, Cesareo, clathratus, Shailesh, Santana Afton Mar 27 at 2:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
You need to describe in detail the distributions used to define the random triangle. Points on circumference? Angles of triangle?
$endgroup$
– herb steinberg
Mar 26 at 17:26
add a comment |
$begingroup$
I calculated an approximate value of 2.2 by using a Monte-Carlo-Simulation (random points on circumference). Can anybody confirm that or offer an analytical solution?
geometry analysis stochastic-analysis
asked Mar 26 at 16:43
user3451767user3451767
63
$endgroup$
I calculated an approximate value of 2.2 by using a Monte-Carlo-Simulation (random points on circumference). Can anybody confirm that or offer an analytical solution?
geometry analysis stochastic-analysis
geometry analysis stochastic-analysis
asked Mar 26 at 16:43
user3451767user3451767
63
asked Mar 26 at 16:43
user3451767user3451767
63
edited Mar 26 at 17:31
user3451767
asked Mar 26 at 16:43
user3451767user3451767
63
asked Mar 26 at 16:43
user3451767user3451767
63
asked Mar 26 at 16:43
user3451767user3451767
63
63
closed as unclear what you're asking by callculus, Cesareo, clathratus, Shailesh, Santana Afton Mar 27 at 2:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by callculus, Cesareo, clathratus, Shailesh, Santana Afton Mar 27 at 2:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
You need to describe in detail the distributions used to define the random triangle. Points on circumference? Angles of triangle?
$endgroup$
– herb steinberg
Mar 26 at 17:26
add a comment |
3
$begingroup$
You need to describe in detail the distributions used to define the random triangle. Points on circumference? Angles of triangle?
$endgroup$
– herb steinberg
Mar 26 at 17:26
3
3
$begingroup$
You need to describe in detail the distributions used to define the random triangle. Points on circumference? Angles of triangle?
$endgroup$
– herb steinberg
Mar 26 at 17:26
$begingroup$
You need to describe in detail the distributions used to define the random triangle. Points on circumference? Angles of triangle?
$endgroup$
– herb steinberg
Mar 26 at 17:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Nice question!
An inscribed triangle in the unit circle can be represented by three points representing the vertices of the triangle. Let's call them $P_0$, $P_1$ and $P_2$. WLOG, we can fix one point in $P_0 = (1,0)$, and work with two angles $theta_1$ and $theta_2$ such that $P_1 = (cos(theta_1), sin(theta_1))$ and $P_2 = (cos(theta_2), sin(theta_2))$. Both angles are random variables uniformly distributed in the set $[0, 2pi]$.
The perimeter of the triangle is:
$$P(theta_1, theta_2) = L_0,1 + L_0,2 + L_1,2, $$
where $L_i,j$ is the length of the side connecting vertices $i$ and $j$.
Therefore:
$$begincases
L_0,1 = sqrt(1-cos(theta_1))^2 + sin^2 (theta_1) = sqrt2 - 2 cos(theta_1)\
L_0,2 = sqrt(1-cos(theta_2))^2 + sin^2 (theta_2) = sqrt2 - 2 cos(theta_2)\
L_1,2 = sqrt(cos(theta_2)-cos(theta_1))^2 + (sin(theta_2)-cos(theta_1))^2 = sqrt2 - 2 cos(theta_1 - theta_2)\
endcases.$$
The average perimeter is:
$$beginalign*mathbbE[P] & = int_mathbbRint_mathbbRP(theta_1, theta_2) f_theta_1(theta_1)f_theta_2(theta_2) dtheta_1 dtheta_2\
& = int_0^2piint_0^2pi(L_0,1 + L_0,2 + L_1,2)frac12pifrac12pidtheta_1 dtheta_2 = frac12pi simeq 3.82,endalign*$$
where $f_theta_1(theta_1)$ and $f_theta_2(theta_2)$ are the pdf of the angles, and they are equal to $frac12pi$ in the set $[0, 2pi]$, $0$ outside.
I have used Wolfram to calculate the integral, even though I'm quite sure there is a closed form to solve it.
The result is quite different from yours. I've also tried using Matlab, which confirm my thoughts. I post the code.
%% Trials
N = 10000;
%% Angles
t0 = zeros(N, 1); % t0 = 2*pi*rand(N,1);
t1 = 2*pi*rand(N,1);
t2 = 2*pi*rand(N,1);
%% Perimeter of each trial
P = zeros(N,1);
%% Montecarlo loop
for i=1:N
L01 = sqrt((cos(t0(i))-cos(t1(i)))^2 + (sin(t0(i))-sin(t1(i)))^2);
L12 = sqrt((cos(t1(i))-cos(t2(i)))^2 + (sin(t1(i))-sin(t2(i)))^2);
L20 = sqrt((cos(t2(i))-cos(t0(i)))^2 + (sin(t2(i))-sin(t0(i)))^2);
P(i) = L01+L12+L20;
end
%% Results
fprintf('MonteCarlo result: %fn', mean(P))
fprintf('Theoretical result: %fn', 12/pi)
As a final remark, notice that degenerate triangles (i.e., $theta_1 = theta_2$, or $theta_1= 0$ or $theta_2 = 0$) represent a "zero-measure" subset, and thus they do not contribute to the calculation of the expected value of the perimeter.
Bonus
As a consequence, the average length of one side of these triangles is $frac4pi simeq 1.27.$
answered Mar 26 at 20:21
the_candymanthe_candyman
9,14032147
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Nice question!
An inscribed triangle in the unit circle can be represented by three points representing the vertices of the triangle. Let's call them $P_0$, $P_1$ and $P_2$. WLOG, we can fix one point in $P_0 = (1,0)$, and work with two angles $theta_1$ and $theta_2$ such that $P_1 = (cos(theta_1), sin(theta_1))$ and $P_2 = (cos(theta_2), sin(theta_2))$. Both angles are random variables uniformly distributed in the set $[0, 2pi]$.
The perimeter of the triangle is:
$$P(theta_1, theta_2) = L_0,1 + L_0,2 + L_1,2, $$
where $L_i,j$ is the length of the side connecting vertices $i$ and $j$.
Therefore:
$$begincases
L_0,1 = sqrt(1-cos(theta_1))^2 + sin^2 (theta_1) = sqrt2 - 2 cos(theta_1)\
L_0,2 = sqrt(1-cos(theta_2))^2 + sin^2 (theta_2) = sqrt2 - 2 cos(theta_2)\
L_1,2 = sqrt(cos(theta_2)-cos(theta_1))^2 + (sin(theta_2)-cos(theta_1))^2 = sqrt2 - 2 cos(theta_1 - theta_2)\
endcases.$$
The average perimeter is:
$$beginalign*mathbbE[P] & = int_mathbbRint_mathbbRP(theta_1, theta_2) f_theta_1(theta_1)f_theta_2(theta_2) dtheta_1 dtheta_2\
& = int_0^2piint_0^2pi(L_0,1 + L_0,2 + L_1,2)frac12pifrac12pidtheta_1 dtheta_2 = frac12pi simeq 3.82,endalign*$$
where $f_theta_1(theta_1)$ and $f_theta_2(theta_2)$ are the pdf of the angles, and they are equal to $frac12pi$ in the set $[0, 2pi]$, $0$ outside.
I have used Wolfram to calculate the integral, even though I'm quite sure there is a closed form to solve it.
The result is quite different from yours. I've also tried using Matlab, which confirm my thoughts. I post the code.
%% Trials
N = 10000;
%% Angles
t0 = zeros(N, 1); % t0 = 2*pi*rand(N,1);
t1 = 2*pi*rand(N,1);
t2 = 2*pi*rand(N,1);
%% Perimeter of each trial
P = zeros(N,1);
%% Montecarlo loop
for i=1:N
L01 = sqrt((cos(t0(i))-cos(t1(i)))^2 + (sin(t0(i))-sin(t1(i)))^2);
L12 = sqrt((cos(t1(i))-cos(t2(i)))^2 + (sin(t1(i))-sin(t2(i)))^2);
L20 = sqrt((cos(t2(i))-cos(t0(i)))^2 + (sin(t2(i))-sin(t0(i)))^2);
P(i) = L01+L12+L20;
end
%% Results
fprintf('MonteCarlo result: %fn', mean(P))
fprintf('Theoretical result: %fn', 12/pi)
As a final remark, notice that degenerate triangles (i.e., $theta_1 = theta_2$, or $theta_1= 0$ or $theta_2 = 0$) represent a "zero-measure" subset, and thus they do not contribute to the calculation of the expected value of the perimeter.
Bonus
As a consequence, the average length of one side of these triangles is $frac4pi simeq 1.27.$
answered Mar 26 at 20:21
the_candymanthe_candyman
9,14032147
$endgroup$
add a comment |
$begingroup$
Nice question!
An inscribed triangle in the unit circle can be represented by three points representing the vertices of the triangle. Let's call them $P_0$, $P_1$ and $P_2$. WLOG, we can fix one point in $P_0 = (1,0)$, and work with two angles $theta_1$ and $theta_2$ such that $P_1 = (cos(theta_1), sin(theta_1))$ and $P_2 = (cos(theta_2), sin(theta_2))$. Both angles are random variables uniformly distributed in the set $[0, 2pi]$.
The perimeter of the triangle is:
$$P(theta_1, theta_2) = L_0,1 + L_0,2 + L_1,2, $$
where $L_i,j$ is the length of the side connecting vertices $i$ and $j$.
Therefore:
$$begincases
L_0,1 = sqrt(1-cos(theta_1))^2 + sin^2 (theta_1) = sqrt2 - 2 cos(theta_1)\
L_0,2 = sqrt(1-cos(theta_2))^2 + sin^2 (theta_2) = sqrt2 - 2 cos(theta_2)\
L_1,2 = sqrt(cos(theta_2)-cos(theta_1))^2 + (sin(theta_2)-cos(theta_1))^2 = sqrt2 - 2 cos(theta_1 - theta_2)\
endcases.$$
The average perimeter is:
$$beginalign*mathbbE[P] & = int_mathbbRint_mathbbRP(theta_1, theta_2) f_theta_1(theta_1)f_theta_2(theta_2) dtheta_1 dtheta_2\
& = int_0^2piint_0^2pi(L_0,1 + L_0,2 + L_1,2)frac12pifrac12pidtheta_1 dtheta_2 = frac12pi simeq 3.82,endalign*$$
where $f_theta_1(theta_1)$ and $f_theta_2(theta_2)$ are the pdf of the angles, and they are equal to $frac12pi$ in the set $[0, 2pi]$, $0$ outside.
I have used Wolfram to calculate the integral, even though I'm quite sure there is a closed form to solve it.
The result is quite different from yours. I've also tried using Matlab, which confirm my thoughts. I post the code.
%% Trials
N = 10000;
%% Angles
t0 = zeros(N, 1); % t0 = 2*pi*rand(N,1);
t1 = 2*pi*rand(N,1);
t2 = 2*pi*rand(N,1);
%% Perimeter of each trial
P = zeros(N,1);
%% Montecarlo loop
for i=1:N
L01 = sqrt((cos(t0(i))-cos(t1(i)))^2 + (sin(t0(i))-sin(t1(i)))^2);
L12 = sqrt((cos(t1(i))-cos(t2(i)))^2 + (sin(t1(i))-sin(t2(i)))^2);
L20 = sqrt((cos(t2(i))-cos(t0(i)))^2 + (sin(t2(i))-sin(t0(i)))^2);
P(i) = L01+L12+L20;
end
%% Results
fprintf('MonteCarlo result: %fn', mean(P))
fprintf('Theoretical result: %fn', 12/pi)
As a final remark, notice that degenerate triangles (i.e., $theta_1 = theta_2$, or $theta_1= 0$ or $theta_2 = 0$) represent a "zero-measure" subset, and thus they do not contribute to the calculation of the expected value of the perimeter.
Bonus
As a consequence, the average length of one side of these triangles is $frac4pi simeq 1.27.$
answered Mar 26 at 20:21
the_candymanthe_candyman
9,14032147
$endgroup$
add a comment |
$begingroup$
Nice question!
An inscribed triangle in the unit circle can be represented by three points representing the vertices of the triangle. Let's call them $P_0$, $P_1$ and $P_2$. WLOG, we can fix one point in $P_0 = (1,0)$, and work with two angles $theta_1$ and $theta_2$ such that $P_1 = (cos(theta_1), sin(theta_1))$ and $P_2 = (cos(theta_2), sin(theta_2))$. Both angles are random variables uniformly distributed in the set $[0, 2pi]$.
The perimeter of the triangle is:
$$P(theta_1, theta_2) = L_0,1 + L_0,2 + L_1,2, $$
where $L_i,j$ is the length of the side connecting vertices $i$ and $j$.
Therefore:
$$begincases
L_0,1 = sqrt(1-cos(theta_1))^2 + sin^2 (theta_1) = sqrt2 - 2 cos(theta_1)\
L_0,2 = sqrt(1-cos(theta_2))^2 + sin^2 (theta_2) = sqrt2 - 2 cos(theta_2)\
L_1,2 = sqrt(cos(theta_2)-cos(theta_1))^2 + (sin(theta_2)-cos(theta_1))^2 = sqrt2 - 2 cos(theta_1 - theta_2)\
endcases.$$
The average perimeter is:
$$beginalign*mathbbE[P] & = int_mathbbRint_mathbbRP(theta_1, theta_2) f_theta_1(theta_1)f_theta_2(theta_2) dtheta_1 dtheta_2\
& = int_0^2piint_0^2pi(L_0,1 + L_0,2 + L_1,2)frac12pifrac12pidtheta_1 dtheta_2 = frac12pi simeq 3.82,endalign*$$
where $f_theta_1(theta_1)$ and $f_theta_2(theta_2)$ are the pdf of the angles, and they are equal to $frac12pi$ in the set $[0, 2pi]$, $0$ outside.
I have used Wolfram to calculate the integral, even though I'm quite sure there is a closed form to solve it.
The result is quite different from yours. I've also tried using Matlab, which confirm my thoughts. I post the code.
%% Trials
N = 10000;
%% Angles
t0 = zeros(N, 1); % t0 = 2*pi*rand(N,1);
t1 = 2*pi*rand(N,1);
t2 = 2*pi*rand(N,1);
%% Perimeter of each trial
P = zeros(N,1);
%% Montecarlo loop
for i=1:N
L01 = sqrt((cos(t0(i))-cos(t1(i)))^2 + (sin(t0(i))-sin(t1(i)))^2);
L12 = sqrt((cos(t1(i))-cos(t2(i)))^2 + (sin(t1(i))-sin(t2(i)))^2);
L20 = sqrt((cos(t2(i))-cos(t0(i)))^2 + (sin(t2(i))-sin(t0(i)))^2);
P(i) = L01+L12+L20;
end
%% Results
fprintf('MonteCarlo result: %fn', mean(P))
fprintf('Theoretical result: %fn', 12/pi)
As a final remark, notice that degenerate triangles (i.e., $theta_1 = theta_2$, or $theta_1= 0$ or $theta_2 = 0$) represent a "zero-measure" subset, and thus they do not contribute to the calculation of the expected value of the perimeter.
Bonus
As a consequence, the average length of one side of these triangles is $frac4pi simeq 1.27.$
answered Mar 26 at 20:21
the_candymanthe_candyman
9,14032147
$endgroup$
Nice question!
An inscribed triangle in the unit circle can be represented by three points representing the vertices of the triangle. Let's call them $P_0$, $P_1$ and $P_2$. WLOG, we can fix one point in $P_0 = (1,0)$, and work with two angles $theta_1$ and $theta_2$ such that $P_1 = (cos(theta_1), sin(theta_1))$ and $P_2 = (cos(theta_2), sin(theta_2))$. Both angles are random variables uniformly distributed in the set $[0, 2pi]$.
The perimeter of the triangle is:
$$P(theta_1, theta_2) = L_0,1 + L_0,2 + L_1,2, $$
where $L_i,j$ is the length of the side connecting vertices $i$ and $j$.
Therefore:
$$begincases
L_0,1 = sqrt(1-cos(theta_1))^2 + sin^2 (theta_1) = sqrt2 - 2 cos(theta_1)\
L_0,2 = sqrt(1-cos(theta_2))^2 + sin^2 (theta_2) = sqrt2 - 2 cos(theta_2)\
L_1,2 = sqrt(cos(theta_2)-cos(theta_1))^2 + (sin(theta_2)-cos(theta_1))^2 = sqrt2 - 2 cos(theta_1 - theta_2)\
endcases.$$
The average perimeter is:
$$beginalign*mathbbE[P] & = int_mathbbRint_mathbbRP(theta_1, theta_2) f_theta_1(theta_1)f_theta_2(theta_2) dtheta_1 dtheta_2\
& = int_0^2piint_0^2pi(L_0,1 + L_0,2 + L_1,2)frac12pifrac12pidtheta_1 dtheta_2 = frac12pi simeq 3.82,endalign*$$
where $f_theta_1(theta_1)$ and $f_theta_2(theta_2)$ are the pdf of the angles, and they are equal to $frac12pi$ in the set $[0, 2pi]$, $0$ outside.
I have used Wolfram to calculate the integral, even though I'm quite sure there is a closed form to solve it.
The result is quite different from yours. I've also tried using Matlab, which confirm my thoughts. I post the code.
%% Trials
N = 10000;
%% Angles
t0 = zeros(N, 1); % t0 = 2*pi*rand(N,1);
t1 = 2*pi*rand(N,1);
t2 = 2*pi*rand(N,1);
%% Perimeter of each trial
P = zeros(N,1);
%% Montecarlo loop
for i=1:N
L01 = sqrt((cos(t0(i))-cos(t1(i)))^2 + (sin(t0(i))-sin(t1(i)))^2);
L12 = sqrt((cos(t1(i))-cos(t2(i)))^2 + (sin(t1(i))-sin(t2(i)))^2);
L20 = sqrt((cos(t2(i))-cos(t0(i)))^2 + (sin(t2(i))-sin(t0(i)))^2);
P(i) = L01+L12+L20;
end
%% Results
fprintf('MonteCarlo result: %fn', mean(P))
fprintf('Theoretical result: %fn', 12/pi)
As a final remark, notice that degenerate triangles (i.e., $theta_1 = theta_2$, or $theta_1= 0$ or $theta_2 = 0$) represent a "zero-measure" subset, and thus they do not contribute to the calculation of the expected value of the perimeter.
Bonus
As a consequence, the average length of one side of these triangles is $frac4pi simeq 1.27.$
answered Mar 26 at 20:21
the_candymanthe_candyman
9,14032147
edited Mar 26 at 20:38
answered Mar 26 at 20:21
the_candymanthe_candyman
9,14032147
answered Mar 26 at 20:21
the_candymanthe_candyman
9,14032147
answered Mar 26 at 20:21
the_candymanthe_candyman
9,14032147
9,14032147
add a comment |
add a comment |
3
$begingroup$
You need to describe in detail the distributions used to define the random triangle. Points on circumference? Angles of triangle?
$endgroup$
– herb steinberg
Mar 26 at 17:26