 Geometric problem - Printable Version +- HP Forums (https://archived.hpcalc.org/museumforum) +-- Forum: HP Museum Forums (https://archived.hpcalc.org/museumforum/forum-1.html) +--- Forum: Old HP Forum Archives (https://archived.hpcalc.org/museumforum/forum-2.html) +--- Thread: Geometric problem (/thread-132642.html) Geometric problem - Hal Bitton in Boise - 02-10-2008 Hi folks, Came upon an interesting problem the other day which has been giving me fits ever since. It goes like this: Consider a circle inscribed in a right triangle such that it just touches all three sides. Describe the ratio of the area of the circle to the area of the triangle solely in terms of r (radius of the circle), and h (hypotenuse of the triangle). I would think there exists a fixed relationship between these two figures, but thus far it eludes me. I'm trying to relate the radius to any one of the sides of the triangle, which is certainly doable given carte blanche to use all the parameters, divide the triangle, etc, but keeping it constrained to just r and h is proving difficult. Does anybody have any words of wisdom as to how I should attack this thing? Thanks and best regards, Hal Re: Geometric problem - Monte Dalrymple - 02-10-2008 Well, it's been more years than I care to admit, but... for right triangle with sides a, b and h; and inscribed circle of radius r: area of triangle = ab/2 = r^2 + r(a-r) + r(b-r) = r(a + b - r) but we also have h = (a - r) + (b - r) so (a + b - r) = (h + r) substituting gives area of triangle = r(h + r) ... unless it's been too many years Re: Geometric problem - Hal Bitton in Boise - 02-11-2008 Many thanks Monte... Your memory serves you well! I had forgotten that the center point of the inscribed circle bisects all three angles of the triangle (as your solution implied). That was the key. Best regards, Hal