Source: Original.
To simplify typing, let b=2c and let m be the side length of the of the non-base sides of ABC. Then: 1) area(ABC) = bh/2 = ch 2) Let C1 be the inscribed unit circle with center pt P. Draw lines from P to the three vertices. This gives three triangles APB, BPC, APC with respective areas: b*1/2 (= c), m*1/2, m*1/2 (since each has altitude 1). So: area(ABC) = c + m = c + sqrt(c^2 + h^2) Thus: ch = c + sqrt(c^2 + h^2) => c(h-1) = sqrt(c^2 + h^2) => c^2(h-1)^2 = c^2 + h^2 => c^2 = h/(h-2) => A) b = 2sqrt(h/(h-2)) (b in terms of h) Let Q be the center of circle CR that inscribes the triangle. Then the altitude of ABC passes thru Q. Let x = length of QK, where K is the point where the altitude bisects the base. Then h = R+x R^2 = c^2 + x^2 = c^2 + (h-R)^2 = c^2 + h^2 - 2hR + R^2 => 2hR = h/(h-2) + h^2 => 2R = 1/(h-2) + h = (h-1)^2/(h-2) => B) R = (h-1)^2/2(h-2) When h=4, R=9/4 If b=2R, then ABC is a right triangle and its altitude = R. Thus (from A) 2R = 2sqrt(R/(R-2)) => R^2 = R/R-2 => R^2-2R-1 = 0 => C) h = R = 1+sqrt(2) By B, 2hR - 4R = h^2 - 2h + 1 => h^2 - (2+2R)h + (1+4R) = 0 => h = [(2+2R) +/- sqrt((2+2R)^2 - 4(1+4R))]/2 => D) h = 1+R +/- sqrt(R^2-2R) When R=4, h = 5+/-sqrt(8) = 7.828427125 or 2.171572875