Tuesday, February 24, 2009

Problem 257: Equilateral Triangle, Circumcircle, Point, Vertices, Side, Distances, Squares

Proposed Problem
Problem 257. Equilateral Triangle, Circumcircle, Point, Vertices, Side, Distances, Squares.

See complete Problem 257 at:
gogeometry.com/problem/p257_equilateral_triangle_circumcircle_distance_sides.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

2 comments:

  1. By an earlier problem, we know that d = e + f
    hence, LHS =(e+f)^2+e^2+f^2 = 2(e^2+f^2+ef)
    Using cosine rule in Tr. BDC, we've: a^2 =e^2+f^2-2efcos(BDC)=e^2+f^2-2ef(-1/2)=e^2+f^2+ef
    Hence, LHS = 2(e^2+f^2+ef) = 2a^2
    Ajit: ajitathle@gmail.com

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  2. We see that ∠ADB=∠ADC =60

    If we apply the cosine rule in triangl ABD with cos(60)=1/2:

    d^2+e^2-2decos(∠ADB)=a^2
    d^2+e^2-2decos(60)=a^2
    d^2+e^2-de=a^2

    And then we apply the cosine rule in triangle ADC :

    d^2+f^2-2dfcos(∠ADC)=a^2
    d^2+f^2-2dfcos(60)=a^2
    d^2+f^2-df=a^2

    Adding these two equations we get

    2d^2+e^2+f^2-de-df=2a^2
    2d^2+e^2+f^2-d(e+f)=2a^2

    In problem 256 we see that d=e+f :

    2d^2+e^2+f^2-d^2=2a^2
    d^2+e^2+f^2=2a^2

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