The incircle of the triangle PQR touches its sides QR, RP and PQ respectively at the points S, T and U. It is known that the respective lengths of QS, RT and PU (in that order) are consecutive integers and the length of the inradius of of the triangle PQR is 4 units.

Determine the respective lengths of each of the sides PQ, QR and RP of the triangle.

Let QS=QU=x, RT=RS=x+1, and PU=PT=x+2. So the perimeter (P) of triangle PQR is 6x+6 and the semiperimeter (S) is 3x+3.

Now the area of triangle PQR is half the inradius times the perimeter so A=2P=12(x+1).

Using Heron's formula, A^2 = S(S-a)(S-b)(S-c) -->

A^2 = (3x+3)(x+1)(x+2)(x) = 3x(x+2)(x+1)^2

So 144(x+1)^2 = 3x(x+2)(x+1)^2 --> x=6 -->

PQ=14, QR=13, and RP=15.

Posted by Dennis on 2007-12-06 11:20:22