1. a) This is a simple case. The circumscribing triangle should be equilateral, so the floating tangent should be tangent at C.
b) The circumscribing triangle can be considered to be made up of three kites. Each kite is formed from two radii to points of tangency and segments of the tangent lines to their point of intersection. As each kite is made up of congruent right triangles adjacent at their common hypotenuse with the leg that is a radius of the circle being 1 unit, the area of the whole kite is equal to the segment length from the point of tangency (of either one) and the point of intersection. The fixed kite, from the center of the circle to A to S to B therefore has area sqrt(3).
So for part 1b, we want RT to be 8  sqrt(3), as the two segments that make it up correspond to kites with the same area as those two lengths. Also, mRT = mRA + mBT.
Since the angles at R and T add up to 120°, half those angles add up to 60°. The areas accounted for by the kites formed at those vertices add up to the sum of the reciprocals of the tangents of those halfangles, as those are the lengths of RA and BT.
So it is desired that 1/tan(T/2) + 1/yan(R/2) = 8  sqrt(3)
and T + R = 120°.
If the angle at T is 20.92618301 degrees and T is 7.14703277 units from point S, the correct size circumscribing triangle will be formed.
R T off by ST SR RT
99.07381699 20.92618301 0.000000000 7.14703277 2.58501804 6.26794919
DEFDBL AZ
dr = ATN(1) / 45
PRINT " r t discrepancy"
FOR t = 20.926183# TO 20.92618303# STEP .000000001#
r = 120  t
PRINT USING "###.######## ###.######## ###.#########"; r; t; 1 / TAN(t * dr / 2) + 1 / TAN(r * dr / 2)  8 + SQR(3);
PRINT USING "####.########"; 1 / TAN(t * dr / 2) + SQR(3); 1 / TAN(r * dr / 2) + SQR(3); 1 / TAN(t * dr / 2) + 1 / TAN(r * dr / 2)
NEXT
c) The area of the equilataral triangle is 3*sqrt(3)/4, so 8 times it is 6*sqrt(3), or about 10.3923048.
105.57299600 14.42700400 0.000000000 9.63289204 2.49146361 8.66025404
So the angle at T needs to be 14.4200400 degrees, 9.63289204 units from S. The sides are the ones shown, and Heron's formula verifies the area as 10.3923048.
DEFDBL AZ
dr = ATN(1) / 45
PRINT " r t discrepancy"
FOR t = 14.427# TO 14.42701# STEP .000001#
r = 120  t
PRINT USING "###.######## ###.######## ###.#########"; r; t; 1 / TAN(t * dr / 2) + 1 / TAN(r * dr / 2)  5 * SQR(3);
PRINT USING "####.########"; 1 / TAN(t * dr / 2) + SQR(3); 1 / TAN(r * dr / 2) + SQR(3); 1 / TAN(t * dr / 2) + 1 / TAN(r * dr / 2)
NEXT
2. The base of the right triangle is 1+sqrt(3) and the height is 3+sqrt(3), so the area is (1+sqrt(3))*(3+sqrt(3))/2 = (6 + 4*sqrt(3)) / 2 = 3 + 2*sqrt(3).

Posted by Charlie
on 20091002 19:44:51 