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In triangle ABC the vertices are A(0,0), B(14,0), C(5,12). Consider…

Question

In triangle ABC the vertices are A(0,0), B(14,0), C(5,12). Consider the triangle whose three vertices are the incentre, the circumcentre, and the centroid of triangle ABC. If the area of this new triangle equals m/n in lowest terms, find m+n.

✓ Verified answer: 49checked by our engine — not a guess

Step-by-step solution

Side lengths: a = BC = sqrt((14-5)^2+12^2) = sqrt(81+144)=15; b = CA = sqrt(25+144)=13; c = AB = 14.
Incentre I = (a A + b B + c C)/(a+b+c) = (15(0,0)+13(14,0)+14(5,12))/42 = ((182+70)/42, (168)/42) = (252/42, 168/42) = (6, 4).
Centroid G = ((0+14+5)/3,(0+0+12)/3) = (19/3, 4).
Circumcentre O: on perpendicular bisector of AB it has x = 7.
Equate distances to A and C: 49+y^2 = (7-5)^2+(y-12)^2 => 49+y^2 = 4 + y^2 -24y +144 => 49 = 148 -24y => 24y = 99 => y = 33/8.
So O = (7, 33/8).
Area of triangle I O G with I=(6,4), O=(7,33/8), G=(19/3,4):
Since I and G share y = 4, base IG = 19/3 - 6 = 1/3 (horizontal). Height = |33/8 - 4| = 1/8.
Area = (1/2)(1/3)(1/8) = 1/48.
Thus m/n = 1/48, m+n = 49.

Final answer49

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