Plot only equilateral triangles

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Mayur Deogade
Mayur Deogade am 4 Sep. 2020
Beantwortet: Alan Stevens am 5 Sep. 2020
Plot random 10 triangles and plot only almost equilateral triangles. Check whether the three edges differ by less than a chosen small constant, for example e = 0.01. I have drawn the random triangles but I need to plot only equilateral triangles. So far, I get any random triangles including obtuse, acute and equilateral triangles.
  3 Kommentare
Mayur Deogade
Mayur Deogade am 4 Sep. 2020
Bearbeitet: Mayur Deogade am 4 Sep. 2020

figure, hold on

for i=1:10 Fill(rand(3,1), rand(3,1), rand(1,3))

End

Axis([0 1 0 1])

Axis square off

Mayur Deogade
Mayur Deogade am 4 Sep. 2020
Here, I am able to plot 10 random triangles but I need to plot only triangles which are nearly equilateral

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Akzeptierte Antwort

Asad (Mehrzad) Khoddam
Asad (Mehrzad) Khoddam am 5 Sep. 2020
figure, hold on
e = 0.01;
i=0;
while i<= 10
x = rand(3,1);
y = rand(3,1);
c = rand(3,1);
L1 = sqrt((x(2)-x(1)).^2 + (y(2) - y(1)).^2);
L2 = sqrt((x(3)-x(1)).^2 + (y(3) - y(1)).^2);
L3 = sqrt((x(3)-x(2)).^2 + (y(3) - y(2)).^2);
L = sort([L1, L2, L3]);
if L(2)-L(1) < e && L(3) - L(2) <e && L(3)-L(1) < e
i = i + 1;
fill(x,y,c)
end
end
axis([0 1 0 1])
axis square off

Weitere Antworten (2)

Alan Stevens
Alan Stevens am 5 Sep. 2020
Here's an alternative approach:
e = 0.01;
for i = 1:10
xc = rand(1); yc = rand(1);
r = rand(1);
while r < e/sqrt(3)
r = rand(1);
end
theta1 = 2*pi*rand(1);
theta = [theta1;
theta1 + 2*pi/3;
theta1 + 4*pi/3];
x = xc + r*cos(theta);
y = yc + r*sin(theta);
plot([x; x(1)],[y; y(1)])
hold on
end
axis equal off

Asad (Mehrzad) Khoddam
Asad (Mehrzad) Khoddam am 5 Sep. 2020
figure, hold on
e = 0.0001;
i=0;
while i<= 10
x = rand(3,1);
y = rand(3,1);
c =rand(3,1);
L1 = sqrt((x(2)-x(1)).^2 + (y(2) - y(1)).^2);
L2 = sqrt((x(3)-x(1)).^2 + (y(3) - y(1)).^2);
L3 = sqrt((x(3)-x(2)).^2 + (y(3) - y(2)).^2);
L = sort([L1, L2, L3]);
if L(2)-L(1) < e || L(3) - L(2) <e
i = i + 1;
fill(rand(3,1), rand(3,1), rand(1,3))
end
end
axis([0 1 0 1])
axis square off
  1 Kommentar
Mayur Deogade
Mayur Deogade am 5 Sep. 2020
It's plotting all the triangles
Acute obtuse

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