Friday, February 27, 2009

Problem 260: Equilateral Triangle, Incircle, Tangency Points, Vertices, Distances, Square

Proposed Problem
Problem 260. Equilateral Triangle, Incircle, Tangency Points, Vertices, Distances, Square.

See complete Problem 260 at:
gogeometry.com/problem/p260_equilateral_triangle_incircle_distance_square.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

2 comments:

  1. WLOG,let G be (0,0) and A:(-1,0), C;(1,0). Therefore, B is (0,V3) where V = square root.
    In-radius of Tr. ABC = 1/V3. Equation of the in-circle, x^2 + (y-1/V3)^2 = 1/3 or x^2+y^2-2y/V3+1/3=1/3 or x^2+y^2-2y/V3 =0
    or 3x^2+3y^2=2V3y ---(1)
    Now m^2+n^2+q^2=(x+1)^2+y^2+(x-1)^2+y^2+x^2+y-V3)^2 = 3x^2++3y^2-2V3y+5 = 5 using (1)
    Now E is (-1/2,V3/2) while F is (1/2,V3/2)
    Thus, d^2+e^2+f^2 = (x+1/2)^2+(y-V3/2)^2+x^+y^2+(x-1/2)^2+(y-V3/2)^2 = 3x^2+3y^2-2V3y+8/4=2 using(1)as before.
    Thus, (m^2+n^2+q^2)/(d^2+e^2+f^2) = 5/2
    QED
    Ajit: ajitathle@gmail.com

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  2. If the sides of the equilateral triangle ABC is

    AB=BC=AC= a

    then from problem 258 we get :

    n^2+m^2+q^2=(5/4)a^2

    (1.)
    4(n^2+m^2+q^2)=5a^2

    and from problem 259 we get :

    (d^2+e^2+f^2)=(1/2)a^2

    (2.)
    10(d^2+e^2+f^2)=5a^2

    From equation 1 and 2 we get

    4(n^2+m^2+q^2)=10(d^2+e^2+f^2)

    (n^2+m^2+q^2)/(d^2+e^2+f^2)=5/2

    ReplyDelete