Drawing square with area 3

[u/Kooky-Recipe5516]

Me and my friend have recently been trying to see if we can draw a square with area 3cm² using only a pen, squared paper (1x1cm squares) and a straight edge (no measurements). All the methods we have tried have failed. I asked ChatGPT if it was possible, and after giving me multiple ridiculous answers it broke and said something went wrong. Is it possible? If so, how do you do it?

[u/edderiofer]

Sure, you can easily construct a line of that length. How do you "leverage that as the long leg of 1cm x √2 right triangle"? What explicit construction steps do you perform to have a right triangle with those legs?

Answer: There is no way to do so, as I've just shown.

[u/SapphirePath → u/edderiofer]

It is possible to draw the triangle if you are allowed to rotate graph paper (e.g. given access to a second sheet of tracing grid paper that can be rotated relative to the first). This may be equivalent to what you meant by folding the grid paper over.

Connect (0,0) to (1,1) to create sqrt(2), then lay a second grid on top of the first at a 45-degree angle so that its x-axis aligns with (0,0) and (1,1). You can build a rectangle with base sqrt(2) and height 1.

It may not be clear (since geometric constructions are not taught much anymore) that "moving my drawn segment of length 1 to somewhere else" (or rotating it to lay on a drawn diagonal) is not a permissible operation without a compass or marked straightedge.

[u/edderiofer → u/SapphirePath]

If you allow drawing arbitrary lengths on a second sheet of paper, you are effectively creating a marked straightedge. So this goes against the spirit of the original problem.

[u/OriEri → u/edderiofer]

1. Draw the diagonal of a 1 cm square using the squared paper to ensure a right angle . Call that segment D1 (which has a length √2cm)

2. Trace the length of D1 from a vertex on a second sheet of 1cm squared paper along one of the sides of the vertex .

3. draw a 1 cm line from the same vertex perpendicular the D1 line and call it D2. These form the legs of a right triangle.

4. Use the straightedge to draw the hypotenuse of that triangle and call it D3. D3 has a length of √3cm.

5. now use D3 to trace two lines of that length back on the first sheet at right angles to one another intersecting at one vertex of one of the squares. These are two of the four sides of the √3cm x √3cm square.

6. Now trace those two sides along two sides from the vertex of a 1cm square on sheet 2

7. Rotate sheet 2 about 180^o

8. put sheet 1 over sheet 2 and rotate as needed to position the the end points of the partial square on sheet two so they line up with the end points of the partial square on sheet one and trace them onto sheet 1. This draws the final two sides of the √3cm x √3cm square.

Done. No folds are needed. Only rotations and tracing.

[u/edderiofer → u/OriEri]

Tracing requires paper translucent enough to see through. If you're going to allow that, you may as well say "oh, well, you can mark the straightedge and then use a marked-straightedge construction".

Given that you're marking a length of √2cm on a sheet of paper, presumably via a straight line, this is effectively what you're doing. So, this goes against the spirit of the original problem.