circle inside triangle tangent points

조회 수: 8 (최근 30일)
pauli relanderi
pauli relanderi 2021년 11월 22일
댓글: pauli relanderi 2021년 11월 24일
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 :
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.
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;
delta = ((ay-Ty)/(ax-Tx))
Thank you for any help !

채택된 답변

Yongjian Feng
Yongjian Feng 2021년 11월 22일
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?
  댓글 수: 4
Yongjian Feng
Yongjian Feng 2021년 11월 23일
편집: Yongjian Feng 2021년 11월 23일
Just basic math concepts:
% plot the triabgle
plot([ax bx cx ax], [ay by cy ay]);
hold on
ab = distance([ax ay], [bx by]);
ac = distance([ax ay], [cx cy]);
bc = distance([bx by], [cx cy]);
% d on bc. bisector theorem
dx = cx + (bx-cx)*ac/(ac + ab);
dy = cy + (by-cy)*ac/(ac + ab);
% e on ab. bisector theorem
ex = ax + (bx-ax)*ac/(ac+bc);
ey = ay + (by-ay)*ac/(ac+bc);
% ad and ce intercept at the center of the circle
[ox, oy] = polyxpoly([ax dx], [ay dy], [cx ex], [cy ey]);
% radius is the shortest distance from o to bc
radius = point_to_line([ox oy 0], [bx by 0], [cx cy 0]);
% G is O's projection on bc, and is the tangent point
[gx, gy] = point_project([ox oy], [cx cy], [bx by]);
% H is O's projection on ab, and is the tangent point
[hx, hy] = point_project([ox oy], [ax ay], [bx by]);
% I is O's projection on ac, and is the tangent point
[ix, iy] = point_project([ox oy], [ax ay], [cx cy]);
% now draw everything
plot([ax dx], [ay dy]);
plot([cx ex], [cy ey]);
% draw the circle
p = nsidedpoly(300, 'center', [ox oy], 'Radius', radius);
plot(p, 'FaceColor', 'r');
% three vertices of the triangle
text(ax, ay, 'A');
text(bx, by, 'B');
text(cx, cy, 'C');
% Note D and E are NOT the tangent points. They are the intercept of
% bisector
text(dx, dy, 'D');
text(ex, ey, 'E');
% O is the center of the circle
text(ox, oy, 'O');
% G, H, I are the tangent points.
text(gx, gy, 'G');
text(hx, hy, 'H');
text(ix, iy, 'I')
function dist = distance(p1, p2)
dist = sqrt((p1(1) - p2(1))^2+(p1(2) - p2(2))^2);
% compute the shortest distance from a point to a line
function d = point_to_line(pt, v1, v2)
a = v1 - v2;
b = pt - v2;
d = norm(cross(a,b)) / norm(a);
% project point c on to line ab. If d is the projection point on ab, then
% the distance cd must be the min.
function [dx, dy] =point_project(c, a, b)
ab_sq = (a(1) - b(1))^2 + (a(2) - b(2))^2;
alpha = -((a(2)-c(2))*(b(2)-a(2))+(a(1)-c(1))*(b(1)-a(1)))/ab_sq;
dx = a(1) + alpha*(b(1)-a(1));
dy = a(2) + alpha*(b(2)-a(2));
pauli relanderi
pauli relanderi 2021년 11월 24일
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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추가 답변 (1개)

Matt J
Matt J 2021년 11월 23일
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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