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# The "aha!" problems 5. Four-point square

11 replies to this topic

### #1 bonanova

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Posted 28 April 2014 - 08:05 AM

Take any four points A, B, C, D in the plane, no three of which are collinear.

They describe a unique quadrilateral, if we take the points as being its vertices.

But they also describe a square, if we require only that the points lie on its sides.

Using a compass and straightedge, construct a square such that the four points lie, one each, on its sides

Edit: or the extensions of its sides.

. . . . . . . . . . . . . . .

Hint: The points are not special.

Draw four similar points and do the construction on a sheet of paper if that helps.

The answer would then be to describe the process.

Edited by bonanova, 07 June 2014 - 12:25 AM.
Relax the constraints

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### #2 plasmid

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Posted 28 April 2014 - 03:07 PM

Are you sure the question is phrased correctly, particularly the part about the points lying one each on the sides?
Spoiler for

Edited by plasmid, 28 April 2014 - 03:14 PM.

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### #3 bonanova

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Posted 28 April 2014 - 05:33 PM

You may be right that not every set of four points will work as the OP asks.

So can you construct a square on the points roughly as shown?

Perhaps then derive conditions for the four points, that it can be done?

This may be a richer problem that I originally thought.

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### #4 plasmid

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Posted 01 May 2014 - 03:38 PM

For now, I'll point out some things which would render squaring the points impossible, placing some limits on the answer.
Spoiler for

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### #5 bonanova

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Posted 03 May 2014 - 12:31 PM

Yes, clearly not every set of four points (no three of which are collinear) lie on a square.

Let's limit the puzzle to a set of points similar to the ones in the OP.

Using standard constructions (compass, ruler) can you draw a square that has these four points (one each) on its sides?

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### #6 plasmid

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Posted 03 May 2014 - 03:02 PM

Yes, I believe you can.

Spoiler for hint

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### #7 bonanova

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Posted 04 May 2014 - 08:30 AM

By modifying the OP slightly I believe the four points may be completely arbitrary:

Take any four points A, B, C, D in the plane, no three of which are collinear.

They describe a unique quadrilateral, if we take the points as being its vertices.

But they also describe a square, if we require only that the points lie on its sides or their extensions.

The OP did not state, but it implied, that A B C D are taken in sequence.

That is, the sides assiciated with A and C are opposite sides, (and so also with B and D) for the construction to be unique.

That is, any four points, no three collinear, define two unique pairs of parallel lines,

one point on each line, that intersect to form a unique square.

Find a compass - ruler construction for those four lines.

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### #8 plasmid

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Posted 04 May 2014 - 03:52 PM

There's one more exception of points that can't be squared, proof of which would require giving my answer to the problem, so here goes.

Spoiler for solution

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### #9 bonanova

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Posted 19 May 2014 - 02:46 PM

Spoiler for Clue

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### #10 bonanova

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Posted 06 June 2014 - 05:18 PM

Spoiler for My solution

Plasmid, does this cover the problem cases you noted?

Edited by bonanova, 09 June 2014 - 03:35 AM.
Added the final figure to the solution

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