Proposed Problem
See complete Problem 275 at:
gogeometry.com/problem/p275_right_triangle_sagitta_inradius.htm
Level: High School, SAT Prep, College geometry
Sunday, March 29, 2009
Problem 275: Right Triangle, Circumcircle, Sagitta, Inradius
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Since ABC is a rt. triangle, we've: AB+BC=2r+AC ---(1)and 2b=AC-BC while 2a=AC-AB --(2) & (3)
ReplyDeleteNow, AB+BC=2r+AC or (AB + BC - AC)^2 = (2r)^2 or
4r^2=(AB+BC)^2+AC^2 -2AC(AB+BC)
= AB^2+BC^2+2AB*BC+AC^2 -2AC*AB-2AC*BC
= 2AC^2 + 2AB*BC - 2AC*AB - 2AC*BC
or 2r^2 = AC^2 - AC*BC - AC*AB + AB*BC
= AC(AC-AB) -AB(AC-AB)
= (AC-AB)(AC-BC)
= 2a*2b using (2) & (3)
or r^2 = 2ab :QED
Ajit: ajitathle@gmail.com
Solution is uploaded to the following link
ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJNUlWQ0V3ZnlUeDJfZ0tKcHZMM212UQ
Let AC = 2R, X = midpoint of BC, Y = midpoint of AB
ReplyDeleteNote segments a, b extended pass through O. Thus
AB = 2 OX = 2(R - a),
BC = 2 OY = 2(R - b),
AC = 2R
AB^2 + BC^2 = AC^2 implies
(R - a)^2 + (R - b)^2 = R^2,
R^2 - 2R(a + b) + (a^2 + b^2)= 0,
[R - (a + b)]^2 = 2ab
R = (a + b) ± √(2ab)
r = in-radius = (AB + BC - AC)/2 = R - a - b = + √(2ab)