Tuesday, January 19, 2010

Problem 420. Triangle, Angles, Altitude, Sides, Measurement

Proposed Problem
Click the figure below to see the complete problem 420 about Triangle, Angles, Altitude, Sides, Measurement.

 Problem 420. Triangle, Angles, Altitude, Sides, Measurement.
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Complete Problem 420
Level: High School, SAT Prep, College geometry

4 comments:

  1. Another way would be to extend AB to E such that BE=3c. Now it's easy to see that Tr. EAC is isosceles with BD // to altitude from E to AC which gives us: c/4c= 1/((x+1)/2) or x=7
    Ajit

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  2. Let AD = AE, E being on DC.

    BE bisects < ABC, hence c/3c = 2/(x-1) from which we have x = 7

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  3. Problem 420
    Suppose that the point E is symmetric of A with respect to the D. Then BE is the bisector of the angle ABC.Is c/3c=2/(x-1) or x=7.

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  4. [For easy typing, I use a instead of alpha]
    cosa=BD/c
    cos3a=BD/3c
    cos3a/cosa=1/3
    4(cosa)^2-3=1/3
    (cosa)^2=5/6
    (sina)^2=1/6
    (1/c)^2=1/6
    c=sqrt6

    In triangle BDC
    sin3a=x/3c
    3sina-4(sina)^3=x/3c
    3/c-4(1/c)^3=x/3c
    3c^2-4=(x*c^2)/3
    18-4=2x
    x=7

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