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Square from four points (Posted on 2013-01-08) Difficulty: 3 of 5

Given four points A, B, C, and D.

Construct a square PQRS such that the points A, B, C, and D lie on lines
PQ, QR, RS, and SP respectively
or
describe the configuration of the four points where the construction is not possible.

Note: Don't confuse lines (of infinite extent) with sides (of finite extent).

  Submitted by Bractals    
Rating: 4.0000 (1 votes)
Solution: (Hide)
ANALYSIS:

Let a, b, c, and d denote lines that are constructed 
through points A, B, C, and D respectively such that

   c||a and  a⊥b and b||d.

The intersections of these lines 

   P = d∩a
   Q = a∩b
   R = b∩c
   S = c∩d

are the vertices of the rectangle PQRS. 
 
Clearly, the decription for the construction of only
one of these four lines, say d, is needed as the 
other three follow per definition.

CONSTRUCTION:

If BD⊥AC then

   if |BD|=|AC| then

      Construct d||AC.

   else

      Construction not possible.

else

      Construct line g through B⊥AC. The circle 
      with center B and radius |AC| intersects line g 
      at two points. Label one of the points D' such 
      that DD' and AC are not parallel. Construct line
      d through D'.

PROOF:

If BD⊥AC then

   if |BD|=|AC| then

      clearly, PQRS is a square since
      |PS| = |AC| = |BD| = |PQ|.

   else

      clearly, the construction when |BD| = |AC| will
      not work. If d is constructed perpendicular to
      AC, then P = Q = R = S = AC∩BD. But, what is
      the meaning of point X lies on line YZ if Y = Z?

      Assume line d is any line through point D (that
      is not parallel or perpendicular to AC) which
      forms a square PQRS. Construct a line through 
      S||AC intersecting line a at A' and a line 
      through S⊥AC intersecting line b at B'.

      ΔSPA' ~ ΔSRB' since corresponding sides are
      perpendicular.

      |SB'| = |BD| ≠ |AC| = |SA'| ⇒ |SP| ≠ |SR|. 
              
      This contradicts the assumption that 
      PQRS is a square.

else

   Construct a line through S||AC intersecting line 
   a at A' and a line through S⊥AC intersecting line 
   b at B'.

   ΔSPA' ~ ΔSRB' since corresponding sides are 
   perpendicular.

   |SB'| = |BD'| = |AC| = |SA'|. ⇒ |SP| = |SR|.

   Therefore, PQRS is a square.

QED

Comments: ( You must be logged in to post comments.)
  Subject Author Date
When it's not possible.Jer2013-01-10 11:27:52
Some ThoughtsAnalytic beginning.Jer2013-01-09 13:48:50
Some Thoughtsre(2): It's all relative (partial spoiler?)Steve Herman2013-01-08 22:32:50
re: It's all relative (partial spoiler?)Bractals2013-01-08 18:47:14
Some ThoughtsIt's all relative (partial spoiler?)Steve Herman2013-01-08 16:00:12
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