Thursday, February 26, 2009

Problem 258: Equilateral Triangle, Incircle, Point, Vertices, Side, Distances, Squares

Proposed Problem
Problem 258. Equilateral Triangle, Incircle, Point, Vertices, Side, Distances, Squares.

See complete Problem 258 at:
gogeometry.com/problem/p258_equilateral_triangle_incircle_distance_square.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

3 comments:

  1. The sum of the distances between D and the sides of the triangle ABC are constant and could be expressed depending on a.

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  2. Let the midpoint of AC be(0,0). So we've:
    A:(-a/2,0), C: (a/2,0) and B:((0,aV3/2). The inradius =a/2V3 while the incentre:(0,a/2V3)
    Hence, incircle is: x^2+(y-a/2V3)^2 =a^2/12
    or x^2+y^2-ay/V3 = 0 or 3x^2+3y^2-ayV3 = 0----(1)
    d^2+e^2+f^2=(x-a/2)^2+(x+a/2)^2+2y^2+x^2+(y-aV3/2)^2=3x^2+3y^2-ayV3+5a^2/4 = 0 + 5a^2/4 using (1). Hence, d^2+e^2+f^2=5a^2/4 QED
    Ajit: ajitathle@gmail.com

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  3. Let P be the point of tangency with the circle on the line AB.
    Let Q be the point of tangency with the circle on the line BC.
    Let R be the point of tangency with the circle on the line AC.

    If we apply apollonius theorem on triangle ADB :

    e^2+d^2=2PD^2+a^2/2
    (1.)e^2+d^2-a^2/2=2PD^2

    If we apply apollonius theorem on triangle BDC :

    e^2+f^2=2DQ^2+a^2/2
    (2)e^2+f^2-a^2/2=2BQ^2

    If we apply apollonius theorem on triangle ADC :

    f^2+d^2=2DR^2+a^2/2
    (3.)f^2+d^2-a^2/2=2DR^2

    If we add equations 1 ,2 and 3 :

    (4.)2(PD^2+DQ^2+DR^2)=2e^2+2d^2+2f^2-(3/2)a^2

    The triangle PQR is equilateral with side a/2 therefore from problem 257 we have :

    PD^2+DQ^2+DR^2=2(a^2/4)
    (5.)PD^2+DQ^2+DR^2=a^2/2

    From equation 4 and 5 we get :


    2e^2+2d^2+2f^2=a^2
    2e^2+2d^2+2f^2=a+(3/2)a^2
    2e^2+2d^2+2f^2=5a^2/2

    d^2+e^2+f^2=5a^2/4

    ReplyDelete