circle inside triangle tangent points

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pauli relanderi
pauli relanderi 2021 年 11 月 22 日
コメント済み: pauli relanderi 2021 年 11 月 24 日
Hello !
I have a problem and have not gotten anywhere.
I have to find the centerpoint of the circle and tangent points of the circle
Looking for any sugestions. Prefer not to use solve, but its acceptable.
Points of the triangle are :
ax=1,ay=1;
bx=5,by=2;
cx=4,cy=4;
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
Only calculations ive gotten so far are these below. BUT they are very wrong or atleast i cant figure out how to solve this problem.
alfa=atand((ay-by)/(ax-bx))
beta=atand((ay-cy)/(ax-cx))
AC = sqrt((ax-cx)^2+(ay-cy)^2); %AC
AB = sqrt((ax-bx)^2+(ay-by)^2); %CB
CB = sqrt((bx-cx)^2+(by-cy)^2); %CB
s = (AB+AC+CB)/2
A = sqrt(s*(s-AB)*(s-AC)*(s-CB))
R = 2*(A)/(AB+AC+CB)
Fx = (bx+ax)/2; ,Fy = (by +ay)/2;
Ex = (cx+ax)/2; ,Ey = (cy +ay)/2;
Tx = (bx+cx)/2; ,Ty = (by +cy)/2;
theta=atand((ay-Ty)/(ax-Tx))
delta = ((ay-Ty)/(ax-Tx))
a=cosd(theta);
b=-cosd(delta);
c=sind(theta);
d=-sind(delta);
e=Tx-ax;
f=Ty-ay;
r=(d*e-b*f)/(a*d-b*c);
t=(a*f-c*e)/(a*d-b*c);
Px=ax+R*cosd(theta);
Py=ay+R*sind(theta);
Thank you for any help !

採用された回答

Yongjian Feng
Yongjian Feng 2021 年 11 月 22 日
This is more like a math problem, is it?
Can you compute two of the bisectors? They intercept right at the center of the circle, right?
  4 件のコメント
pauli relanderi
pauli relanderi 2021 年 11 月 24 日
Thank you man !
Helped me alot !
Glad to see that matlab commonity is this active !
Have a great rest of the week !

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その他の回答 (1 件)

Matt J
Matt J 2021 年 11 月 23 日
You can find the circle using incircle in this FEX submission
Once you've done that, the computation of the tangent points is easy.

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