Monday, May 19, 2008

Geometry Problem 105



See complete Problem 105
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

4 comments:

  1. This Problem has two solutions for x
    x=Alfa
    x=90-Alfa-Beta

    ReplyDelete
  2. Angle DBC=180-90+α+β-x-2β-α =90-x-β. Now using the trigonometric version of Ceva's Theorem, we can say that: sin(90-α-β)/sin(x)*sin(β)/sin(β)*sin(α)/sin(90-x-β)=1 or sin(90-α-β)/sin(90-x-β) = sin(x)/sin(α). This equation is clearly true when x=α which makes both sides=1 or equally when x=90-α-β
    Ajit: ajitathle@gmail.com
    Ajit:

    ReplyDelete
  3. In particular, if AC=AB, then x=alpha.
    Then without loss of generality,
    we may assume that AC < AB.
    Let E be a point on the extended line of AC with AE=AB.
    Since the angle DCB=90-x-beta = the angle of DCB, the quadrilateral CDBE is a cyclic
    and so alpha =90-beta-x,
    that is, x=90-alpha-beta.


    September 21, 2009
    Bae deok rak
    bdr@korea.com

    From daechidong,
    Topmath academy(02-567-5114)
    Seoul, South Korea.

    ReplyDelete
  4. To Antonio: according to the conclusions reached by Joe and Anonymous, the problem 105 has two possible answers for the angle x, meaning that some condition is missing in enunciate. Is that true? What should be changed?

    ReplyDelete