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Math Forum / Mathematics / Recreational Math / October 2008Inscribed circle
Find the equation of the inscribed circle of the triangle formed by the lines 4x-3y+9 = 0 12x+5y-43 = 0 and 7x+24y+19 = 0 > Find the equation of the inscribed circle of the triangle formed by the > lines 4x-3y+9 = 0 > 12x+5y-43 = 0 and > 7x+24y+19 = 0 Let the lines be labeled: (a) 4x-3y+9 = 0 (b) 12x+5y-43 = 0 (c) 7x+24y+19 = 0 Using Gaussian solution of MX=B for matrices/vectors, we find the vertices of the triangle to be: A = [49/11, -23/11] (intersection of lines b,c) B = [-7/3, -1/9] (intersection of lines a,c) C = [3/2, 5] (intersection of lines a,b) And the opposite side lengths are a = 115/18 b = 169/22 c = 700/99 which satisfies the triangle inequalities on all 3 sides. The area of this triangle, using Heron's formula is area = 2093/99 showing that we do have a Heron triangle. The radius of the incircle = 2*area/(a+b+c) = 2 The centre of the incircle is at [ (a*A[1]+b*B[1]+c*C[1])/(a+b+c), (a*A[2]+b*B[2]+c*C[2])/(a+b+c) ] which is [1,1] For the incircle with centre at [1,1] and radius 2, we have the equation (1) (x - 1)^2 + (y - 1)^2 = 2^2 = 4 which is what we seek. Things get much more interesting if we have a triangle in 3d space. > > Find the equation of the inscribed circle of the > triangle formed by the [quoted text clipped - 50 lines] > Things get much more interesting if we have a > triangle in 3d space. To which the OP responded "not quite". But the required circle is indeed centered at (1,1) and has radius 2. Perhaps the OP is objecting to something about the method? Anyway here's another approach. First rewrite the equations so the y coeffs are positive. L1 -4x + 3y - 9 = 0 L2 12x + 5y - 43 = 0 L3 7x + 24y + 19 = 0. A graph shows the center of the inscribed circle is above line L3 and below lines L1 and L2. So the center is at the place where -dist(P,L1) = -dist(P,L2) = dist(P,L3). Here for any point X and line L dist(X,L) is the signed distance from X to line L which is positive for points above line L. In general for a line with equation ax+by+c=0 and b>0 this signed distance for P(x,y) is .. [ax+by+c)/sqrt(a^2+b^2). Applying this to the present problem, the following are to be equal: -(-4x+3y-9)/5 = -(12x+5y-43)/13 = +(7x+24y+19)/25. This has unique solution (x,y) = (1,1) at which a.s three signed distances are 2, so the required circle is as in previous response centered at (1,1) and has radius 2. That the equation is .. (x-1)^2 + (y-1)^2 = 4 may be seen after 3 pages of careful computation. > > > Find the equation of the inscribed circle of the > > triangle formed by the [quoted text clipped - 87 lines] > ... (x-1)^2 + (y-1)^2 = 4 > may be seen after 3 pages of careful computation. I should have said, "at which all three", rather than "at which a.s three"... my bad. Not quite. Not a big deal ! x^2+y^2-2x-2y-2 = 0 Do you have a general equation then for an inscribed sphere, given the equations of the four faces of an irregular tetrahedron ? That would be interesting. A1*x + B1*y + C1 = 0, A1^2 + B1^2 = 1 A2*x + B2*y + C2 = 0, A2^2 + B2^2 = 1 A3*x + B3*y + C3 = 0, A3^2 + B3^2 = 1 The following method works for any dimension First, find X and Y from (a,a)X + (a,b)Y +(a,c) = 0 (b,a)X + (b,b)Y +(b,c) = 0 where (a,a) = A1*A1 + A2*A2 + A3*A3 (a,b) = A1*B1 + A2*B2 + A3*B3 .. Then the center (x, y), and the radius (R) find from A1*x + B1*y + C1 = R*sign(A1*X + B1*Y + C1) A2*x + B2*y + C2 = R*sign(A2*X + B2*Y + C2) A3*x + B3*y + C3 = R*sign(A3*X + B3*Y + C3) > Do you have a general equation then for an inscribed > sphere, given the equations of the four faces of an > irregular tetrahedron ? That would be interesting. Hi Bill, We talked about a triangle in 3D, and in this case we proceed exactly as in 2D. Barycentric coordinates work similarly on tetrahedra. Let A, B, C, D, be the vertices of the tetrahedron, and Sa, Sb, Sc, Sd, be the areas of their respective faces. The incenter It of tetrahedron is determined as follows: It = (Sa*A + Sb*B + Sc*C + Sd*D)/(Sa+Sb+Sc+Sd) . The radius R of the inscribed sphere is R = 3*V/(Sa+Sb+Sc+Sd) , where V is the volume of the tetrahedron. Best regards, Avni |
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