Tuesday, April 7, 2009

Problem 24: Right triangle, Altitude, Incircles

Proposed Problem

Problem 24: Right triangle, Altitude, Incircles.

See complete Problem 24 at:

Level: High School, SAT Prep, College geometry


  1. With the usual nomenclature, we've BD=ac/b and AD = c^2/b. Thus, r1 = (ca/b * c^2/b)/(ac/b+c^2/b+c) = ac^2/(b(a+b+c))
    Likewise, r2 = ca^2/(b(a+b+c))
    Therefore, r1^2 + r2^2 =
    a^2c^4/(b(a+b+c))^2 + c^2a^4/(b(a+b+c))^2
    = a^2c^2(a^2+c^2)/(b(a+b+c))^2
    = a^2c^2b^2/(b(a+b+c))^2
    = [ac/(a+b+c)]^2 = r^2
    Ajit: ajitathle@gmail.com

  2. Triangle ABD similar to tri. ABC and we have r1/r = AB/AC
    Triangle BDC similar to tri. ABC and we have r2/r = BC/AC
    ABC is a right triangle so we have AC^2=AB^2+BC^2
    And r^2=r1^2+r2^2

    Peter Tran

  3. Reference my proof for Problem 23

    r1 = rc/b and r2 = rc /b

    So r1^2 + r2^2 = r^2(a^2 + c^2)/b^2 = r^2 from Pythagoras

    Sumith Peiris
    Sri Lanka

  4. https://www.facebook.com/photo.php?fbid=10206230208662809&set=a.10205987640598759.1073741831.1492805539&type=3&theater

  5. Let the point M be the center of the circle with radius r1.
    Let the point N be the center of the circle with radius r2.
    Let the point O be the center of the circle with radius r.

    Then draw a line from M perpendicular onto BD. The line from M meets BD at the point P.

    Also draw a line from the point N perpendicular onto the radius r. This line meets
    the radius at point Q.

    We have to use the results from problem 23 BD=r1+r2+r and the result from problem 28 DE =r2-r1 .

    PB = BD-r1 = r+r2
    QN = r2-DE = r1


    OQ = r-r2
    MP = r1

    Triangles BPM and NQO are similar we have :

    MP/BP = OQ/NQ
    r1/(r+r2) = (r-r2)/r1
    r1^2 = (r+r2)(r-r2)
    r1^2 = r^2-r2^2