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Seeking Sums With Inradius And Exradius (Posted on 2007-10-08) Difficulty: 3 of 5
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.

Determine Y+Z.

See The Solution Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Idea of a solution (no proof listed just yet) | Comment 3 of 4 |

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*(3/2), thus in this specific case, the total length would be exactly 3.

This ties with the answer given in the initial post by Charlie.

Several geometric facts, including that SPR 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*(3/2), which is 1/2 the length of hypotenuse RS in QRS. Here, Inradius=x(3 - 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) 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 3*(2-3) in RET, 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*[3*(4-23)]2. However, since (4-23)=3-1, we finally get Z=x(3/2 - 3/2)

So: Y+Z=x*(3/2), with x=3 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

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