Tuesday, April 7, 2009

Problem 23: Right triangle, Altitude, Inradius

Proposed Problem

Problem 23: Right triangle, Altitude, Inradius.

See complete Problem 23 at:
gogeometry.com/problem/p023_right_triangle_incircles.htm

Level: High School, SAT Prep, College geometry

8 comments:

  1. Under the standard nomenclature, AD = c*sin(C)= c*c/b=c^2/b while BD=c*cos(C)= ac/b ----(1)
    So r1 = AD*BD/(AD+BD+AB)
    = (c^2/b*ac/b)/(c^2/b + ac/b + c)
    = ac^2/b(a+b+c)
    Similarly, r2 = ca^2/b(a+b+c)
    and r = ac/(a+b+c)
    So,r1+r2+r=ac^2/b(a+b+c)+ca^2/b(a+b+c)+ac/(a+b+c)
    = ac/b = BD by equation (1)
    Ajit: ajitathle@gmail.com

    ReplyDelete
  2. The statement that r=ac/(a+b+c) is incorrect; the numerator should be 2ac.

    However, it is easier to show that r=a+b-c and use that to obtain the desired result.

    ReplyDelete
  3. Prof Elie Achkar (Lebanon)

    BD + DC = BC + 2r1 in the triangle BDC
    AD + BD = AB + 2r2 in the triangle BDA
    AB + BC = AC + 2r in the triangle ABC

    then :
    BD+DC+AD+BD+AB+BC = 2r1 + 2r2 + 2r
    THEN :
    2BD = 2(r1 + r2 +r )
    follow : BD = r1 + r2 + r

    ReplyDelete
  4. Inradius = Tr. Area/Semisum = (ac/2)/[(a+b+c)/2]Hence r = ac/(a+b+c)
    So then why is the statement incorrect?

    ReplyDelete
  5. Dear Joe = Ajit:
    I agree, your statement r = ac/(a+b+c) is correct.

    ReplyDelete
  6. How is it possible that
    BD + DC =BC + 2r1
    .................
    .................

    Please tell me the condition no

    ReplyDelete
  7. In a right triangle ABC (right angled at B)
    r=s–b,is well-known.

    Easily remembered as :
    in-radius = semi perimeter–hypotenuse
    (or)
    in-diameter=perimeter–twice the hypotenuse

    So 2(r + r1 + r₂)
    =(AB+BC–AC)+(BD+AD–AB)+(BD+DC–BC)
    =2BD+(AD+DC)–AC
    =2BD+AC–AC
    =2BD

    Hence r+r1+r₂=BD

    ReplyDelete
  8. If O1, O2 and O are the centers of circles with radius r1,r2 and r then

    r1 = rc/b and r2 = ra/b since Tr.s AO1C, CO2B and AOC are similar

    So r+r1+r2 = r (a+b+c)/b ....(1)

    By writing the area of the right Tr. in 2 ways

    BD. b = r(a+b+c)

    So from (1) r+r1+r2 = BD

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete