 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Seeking Sums With Inradius And Exradius (Posted on 2007-10-08) 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) Comments: ( Back to comment list | You must be logged in to post comments.) 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*��(4-2��3)]2. However, since ��(4-2��3)=��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 Please log in:
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