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.
Constructing the figure in Geometer's Sketchpad shows the sum to be constant at 1.73205, which seems to be the square root of 3. There's no guarantee it doesn't deviate from sqrt(3)
beyond the accuracy shown.
If we take one limiting case, where P and S coincide, angle RSQ is the same as RPQ; that is, 60 degrees, so the bisector, angle FSQ is 30 degree. (Point F being the center of the excircle.) Angle RQF is 60 degrees as it is half an angle of 120 degrees (the supplement of RQP). But angle FRQ is also 60 degrees, as half the supplement of QRP. So triangle FRQ is also equilateral in this case, and has the same altitude as triangle RQP; that is, square root of 3; and that is the radius of the excircle. Meanwhile the incircle is the mere point P (or S), having zero radius, so the total of the two radii is indeed sqrt(3).
To take the opposite limiting case, where S gets exceedingly far away without limit (goes to infinity). Line RS approaches being parallel to QP, and the centers of the two circles approach being half the height of the original triangle away from QP, thus adding up to the full height of the equilateral triangle, or, again, sqrt(3).
So if indeed there is one answer to this puzzle, as the sum of the two radii is constant, that constant is sqrt(3). We have not proved that it does not deviate ever so slightly from that value between the limiting cases, but I'd place my bets on its being constant, especially under the usual assumption that a puzzle has a unique answer.
Edited on October 8, 2007, 11:29 am
Posted by Charlie
on 2007-10-08 11:25:46