Sunday, March 29, 2009

Problem 275: Right Triangle, Circumcircle, Sagitta, Inradius

Proposed Problem

Problem 275: Right Triangle, Circumcircle, Sagitta, Inradius.

See complete Problem 275 at:
gogeometry.com/problem/p275_right_triangle_sagitta_inradius.htm

Level: High School, SAT Prep, College geometry

3 comments:

  1. Since ABC is a rt. triangle, we've: AB+BC=2r+AC ---(1)and 2b=AC-BC while 2a=AC-AB --(2) & (3)
    Now, 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

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  2. Solution is uploaded to the following link

    https://docs.google.com/open?id=0B6XXCq92fLJJNUlWQ0V3ZnlUeDJfZ0tKcHZMM212UQ

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  3. Let AC = 2R, X = midpoint of BC, Y = midpoint of AB
    Note 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)

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