Triangle PQR is equilateral with PQ = QR = RP = 2. The line QP is extended to meet at point S such that P lies between S and Q.
Y is length of the inradius
of Triangle SPR while Z is the length of the exradius
of Triangle SQR with respect to the side QR.
In the general case for a starting equilateral triangle of side length x, it appears that the sum of the lengths of the inradius and exradius of the requested circles is x*(¡î3/2), thus in this specific case, the total length would be exactly ¡î3.
This ties with the answer given in the initial post by Charlie.
Several geometric facts, including that ¥ÄSPR is isosceles with congruent sides x and apex angle 120¨¬ give rise to an equation for the inradius in terms of tan 15¨¬, the inradius Y, and leg length x*(¡î3/2), which is 1/2 the length of hypotenuse RS in ¥ÄQRS. Here, Inradius=x(¡î3 - 3/2)
For the exradius, we are presented with a triangle of angles 60¨¬, 45¨¬, and 75¨¬, the last angle of which is opposite the known side QR of length x. Use of the formula for sin(5¥è) in conjunction with the half angle formula for sin(30¨¬) to get sin 75¨¬ for a Law of Sines application. This provides length RE as ¡î3*¡î(2-¡î3) in ¥ÄRET, where E is the desired excenter, and ET is the desired exradius Z, T lying along line RS.
RET is also a 45¨¬ right triangle due to the circumstances of the construction. Using sin 45¨¬, RE, and our exradius Z, we get Z=x*[¡î3*¡î(4-2¡î3)]2. However, since ¡î(4-2¡î3)=¡î3-1, we finally get Z=x(3/2 - ¡î3/2)
So: Y+Z=x*(¡î3/2), with x=¡î3 for the starting triangle with side length 2.
EDIT: Not sure why the symbols got funky in this post. Sorry if it makes it illegible.
Edited on October 8, 2007, 3:40 pm
Posted by Mike C
on 2007-10-08 15:38:45