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A triangle has side lengths 4, 6, 8. A tangent is drawn to the incircle parallel to side 4 cutting other two sides at M and N, then the length of MN is written as a/b where a & b are coprime numbers . find a+b
The answer is 29.
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1 solution
Let D be the area of the triangle and sides 4,6,8 be a,b and c respectively ,vertices opposite to them be A,B and C resp. and s be the semiperimeter where 2s=a+b+c Radius of incircle r=D/s Since MN is parallel to BC hence BM/AB=CN/AC=ratio of altitude of trapezium MNCB(2r) and altitude from A w.r.t triangle ABC=(2D/s)/(2D/a)=a/s. Let P,Q and R be the points where AB,ACand MN touch the incircle. Since tangents from an external point are equal in length hence MR=MP , NR=NQ Hence MN=MR+NR=MP+NQ=BM-BP+CN-CQ=(BM+CN)-(BP+CQ) =a(b+c)/s-(s-b+s-c)=a(b+c)/s-a=a(b+c-a)/(a+b+c) put a=4 b=6 and c=8 to get MN=20/9