Wednesday, September 10, 2014

Geometry Problem 1043: Right Triangle, Incenter, Excenter, Congruence, Metric Relations

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

Click the figure below to view the complete problem 1043.

Online Math: Geometry Problem 1043: Right Triangle, Incenter, Excenter, Congruence, Metric Relations.

2 comments:

  1. C, I, E are collinear

    Connect I with E and A with B

    AE = AI = 3√2
    in the triangle AIC we have IC = 3√3 - 3
    triangle EBC similar to triangle AIC, EB/AI = EC/AC
    x = 3(1 + √3)/2
    x= 4.098

    Erina, NJ

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  2. Radius of incircle=r. Drop I onto ED to obtain point P.
    AB=x+r=PI, so triangle PIE is congruent to triangle BAC.
    BC=PE=x-r. DC=2x-r=6+r. Therefore
    1) x-r=3.
    Substitute this into (x-r)^2+(x+r)^2=36 in triangle PIE to get
    2) x+r=3*sqrt(3).
    Adding 1) and 2) yields 2x=3(1+sqrt(3))==>x=3(1+sqrt(3))/2

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