Friday, June 19, 2009

Problem 305: Square, Triangles, Angle, Sides

Proposed Problem
Click the figure below to see the complete problem 305.

 Problem 305: Square, Triangles, Angle, Sides.
See also:
Complete Problem 305
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

2 comments:

  1. Triangles BCE, FDE and FAB are similar to each other, therefore, we can say that --> AB/BF = ED/FE. Since AB=BC, 1/BF = ED/(FE*BC). Further, BC/BE=FD/FE or 1/BE=FD/(FE*BC). Thus, (1/BE)^2 + (1/BF)^2 = [ED/(FE*BC)]^2+[FD/(FE*BC)]^2 =(1/BC)^2*{FE^2+ED^2)/FE^2 =(FE/FE)^2*(1/BC)^2= 1/BC^2.
    Hence the proof.

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  2. Let BC = a, BE = p, BF = q and DF = r

    q/p = (a+r)/a

    q^2 = a^2 + (a+r)^2

    Substitute for a+r

    q^2 = a^2 + a*2 q^2/p^2

    Divide both sides by a^2 q^2

    1/a^2 = 1/ p^2 + 1/q^2

    Sumith Peiris
    Moratuwa
    Sri Lanka

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