Monday, May 19, 2008

Elearn Geometry Problem 46



See complete Problem 46
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

7 comments:

  1. x=30 degrees is also a solution to the problem

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  2. Solucion:
    http://img8.imageshack.us/img8/9236/rsolmarzo7.jpg

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  3. In Tr. DBC, DC/DB = sin(3x+30)/sin(30) = 2sin(3x+30) While Tr. ADC gives,DC/AD =sin(2x)/sin(x)
    Now AD=DB gives, 2sin(3x+30)=sin(2x)/sin(x) or
    2sin(3x+30)=2sin(x)cos(x)/sin(x) or
    sin(3x+30)=sin(90-x) or 3x+30=90-x or x=15
    Alternatively, sin(3x+30)=sin(90-x) may mean
    sin(180 -(3x+30))=sin(90-x) giving us,
    180 -(3x+30) = 90- x giving us x=30
    Svvv is very correct in saying that 15 deg. and 30 deg. are both acceptable solutions to this problem.
    Ajit: ajitathle@gmail.com

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  4. Thanks for the comments.
    New condition: angle B is an obtuse angle.

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  5. Dear Antonio

    This problem is proving very difficult to crack thro Euclidean geometry alone.

    I tried to drop altitudes from B to DC and AC and work through the cyclic quad
    Secondly tried to find point M on DC such that D is the centre of ABM Tr.
    Thirdly found that if u find N on DC such that Tr. ANC is isoceles I found that < DBN is 30

    But I cannot clinch it

    Are any of my approaches right?

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    Replies
    1. Dear Sumith,
      Some ideas: equilateral triangle, isosceles triangle, congruence ....

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    2. Let BE = h be an altitude of triangle ABC and let CD = d

      So BD = DA = DE = CE = c/2

      CD is a median of triangle ABC, so
      d^2 = a^2/2 +b^2/2 -c^2/4 ....(1)

      Since triangles BDC and ADC are equal in area, easily
      1/2 da/2 = 1/2 bh/2 hence d = bh/a ....(2)

      Using Pythagoras h^2 = c^2 - (b-c/2)^2 = a^2 - (c/2)^2....(3)

      Eliminating d and h between these 3 equations,
      a^2/2 + b^2/2 - c^2/4 = (b^2/a^2)(a^2 - c^2/4).

      Simplifying
      a^/2 - c^2/4 = b^2/2 - b^2c^2/(4a^2) which gives,
      2(a^2 - b^2) = (c^2/a^2)(a^2 -b^2)

      Since a NOT = b, a^2 = (c/2)c
      So BC^2 = BD. BA and BC is therefore a tangent to circle ADC.

      Hence 2x = 30 and x = 15

      Sumith Peiris
      Moratuwa
      Sri Lanka

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