Monday, August 5, 2019

Geometry Problem 1443: Triangle Area, Incenter, Circumcenter, 90 Degree, Perpendicular, One-third, Measurement

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

Details: Click on the figure below.

Geometry Problem 1443: Triangle Area, Incenter, Circumcenter, 90 Degree, Perpendicular, One-third Measurement, iPad apps, Tutoring.

2 comments:

  1. Let BI extended meet AC at D and circle O at E

    < EAI = EIA = (A+B)/2 so AE = EI = IB since OI is perpendicular to BE

    Further, since < EAD = < ABE = B/2, Tr.s ADE & ABE are similar and so
    AD/c = AE/BE = 1/2 so that AD = c/2

    But AD = bc/(a+c), hence c/2 = bc/(a+c) from which, a+c= 2b

    Now S1 = rb/2 = (S/s)*(b/2) = Sb/(a+b+c) = Sb/(3b) = S/3

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  2. https://photos.app.goo.gl/oBvDsrGZ2wwrWe9V7

    let BI meet AC at N and arc AC at D
    Let CI meet arc AB at M
    We have I is the midpoint of BD
    and Triangle DIC is isoceles
    Triangle DNC simillar to DCB ( case AA)
    so DC^2=DN.DB=2DN.DI=2DN.DC
    so DN/DC=DN/DI=1/2 => N is the midpoint of ID
    and NI/NB=1/3
    Area(AIC)/Area(ABC)= NI/NB= 1/3

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