Sunday, March 22, 2015

Problem 1102: Right Triangle, Incircle, Tangency Point, 90 Degree, Cathetus, Metric Relations

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1102: Right Triangle, Incircle, Tangency Point, 90 Degree, Cathetus, Metric Relations.

4 comments:

  1. Let BC=a, AC=b, AB=c, and s=(a+b+c)/2

    AD=s-c, DC=s-a
    DC/BC = DG/AB
    (s-a)/a = DG/c
    DG = (s-a)c/a

    AD×DG
    = (s-c)(s-a)c/a
    = 1/4 (b² - (a-c)²) c/a
    = 1/4 (2ac) c/a
    = 1/2 c²
    = 1/2 AB²

    Hence, AB² = 2 AD×DG

    ReplyDelete
  2. slight error on lines #2 to #5: with defs in line #1, AD=s-a and DC=s-c
    remaining lines are ok

    ReplyDelete
  3. (s - a)(s - c) / ac = cos² (B/2) = 1/2
    DG = (s - c) tan C = (s - c).c/a = c.(s - c)/a
    2.AD. DG =2.(s - a). c(s - c) / a
    = 2. [(s - a)(s - c) /ac].c²
    = 2. (1/2).c² = c² = AB²

    ReplyDelete
  4. AD = c - r where r is the inradius. From similar Tr.s DG = c(a-r)/a

    So AD. DG = c(a-r)(a-r)/a

    Using r = 1/2 (a+c -b) and simplifying we get that AD AG = c^2 /2 which is the required result

    In the simplification we have used the fact that b^2 = a^2 + c^2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete