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Just find two! (Posted on 2005-04-27) Difficulty: 4 of 5
Given a triangle ABC, how can you find a point P on AC and a point Q on BC, such that AP=PQ=QB?

N.B. A construction method is sought, and only compass and straightedge are allowed.

See The Solution Submitted by Federico Kereki    
Rating: 4.3333 (3 votes)

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Solution Solution | Comment 24 of 25 |
 
WOLOG let |AC| be less than or equal to |BC|.
Construct a circle with C as the center and |AC| as the radius.
Construct point D on line segment BC such that |DB| = |AC|.
Construct a line through D and parallel to AB intersecting the circle at point E (on the same side of AC as point B).
Point Q is the intersection of AE with BC.
Construct a line through Q and parallel to CE intersecting line AC at point P.
From similar triangles CAE and PAQ,
    |AE|   |AC|   |CE|
    ---- = ---- = ----.
    |AQ|   |AP|   |PQ|
From similar triangles AQB and EQD,
    |AE|   |DB|
    ---- = ----.
    |AQ|   |QB|
Therefore,
    |AC|   |CE|   |DB|
    ---- = ---- = ----.
    |AP|   |PQ|   |QB|
Since |AC| = |CE| = |DB|, we have
      |AP| = |PQ| = |QB|.
If you let E be the other intersection and perform
the rest of the construction, then you get a second
solution unless angle C is 60 degrees (AE and BC
would be parallel).
 

Edited on May 17, 2005, 12:03 am

Edited on May 17, 2005, 11:51 pm
  Posted by Bractals on 2005-05-14 00:12:51

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