Saturday, January 16, 2010

Problem 419: Triangle, Cevian, Incircle, Excircle, Inradius, Exradius, Congruence

Proposed Problem
Click the figure below to see the complete problem 418 about Triangle, Cevian, Incircle, Excircle, Inradius, Exradius, Congruence.

Problem 419: Triangle, Cevian, Incircle, Excircle, Inradius, Exradius, Congruence.
See also:
Complete Problem 419
Level: High School, SAT Prep, College geometry

3 comments:

  1. draw AH altitude on BC.
    name S point D meet BC, P point D meet AC

    from P 325 => AH = 4r = 4∙ 17 = 68

    AH∙BC = AC∙BP ( the same area of ABC, express two time )

    68∙(5/7)AC = AC∙BP =>

    BP = 340/7 (1)

    ▲DSB ~ ▲PBC => x/BD = (1/2)AC/(5/7)AC

    x/BD = 7/10 => BP/x = 17/7 => x/BP = 7/17

    x = (7/17)∙BP substitute (1)

    x = (7/17)∙(340/7)

    x = 20

    ReplyDelete
  2. Once we've AH=4*17=68, we can also say that x = Inradius of Tr. ABC= Area(Tr. ABC)/semiperimeter =(68/2)*(5b/7)/((5b/7+5b/7+b)/2) which directly gives x=20
    Ajit

    ReplyDelete
  3. bjhvash44@sbcglobal.net

    Iwas solving number 420.Why no comment or solution.Too easy?

    ReplyDelete