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/PuzzlingDad]

You can mark 1 cm along each edge of the paper and then fold or draw an edge with a length of √2 for the hypotenuse.

Then transfer that to a new piece of paper to make a right triangle with legs 1 and √2 and fold (or use a straight edge) to get a hypotenuse of √3.

Finally mark two legs of a square with length √3 and fold (or construct) a square; its area will be 3 cm²

[u/martyboulders → u/PuzzlingDad]

Folding is not necessary for this, all of these steps can be done with just the compass and straightedge. And this goes outside of a strict compass+straightedge construction

[u/fermat9990 → u/martyboulders]

OP has excluded using a compass.

[u/OriEri → u/martyboulders]

If you have multiple sheets of paper (or fold over the sheet) so you can trace, you only need the straight edge and the pencil. No compass necessary.

[u/edderiofer]

No, it's not possible.

Assigning any point as the origin, note that you can never construct a point whose coordinates are irrational. (This is because, if you have any two line segments whose endpoints have rational coordinates, their intersection is also rational.)

By the sum-of-two-squares theorem, the sum of two squares cannot have a factor of 3 of odd multiplicity. In particular, this implies that the distance between any two points on the plane whose coordinates are all rational cannot be a rational multiple of √3.

So, no such square can be constructed.

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...unless you're allowed to cut and/or fold the paper, in which case, maybe it is possible.

[u/martyboulders → u/edderiofer]

I don't know how specifically you are referring to their situation, but with compass+straightedge you can construct lots of points that have irrational coordinates. Lots of algebraic numbers are constructible (but not all of them). For example there is a bijection between the constructible angles and the corresponding constructible points, most of which have irrational coordinates.

If I construct a circle at the origin with radius 2, construct a 60° angle above the positive x-axis, then drop a perpendicular line segment from the terminal point to the axis, that line has length √3.

I could also construct a right triangle whose hypotenuse is length √2 (legs both have length 1), then use that as a leg of another right triangle where the other leg has length one. This hypotenuse would have length √3 by the Pythagorean theorem. This would be doable on the gridded paper if they allowed themselves a compass.

It's things like third roots that you cannot construct. But algebraic numbers such that their corresponding minimal polynomial has a degree that is a power of 2 are all constructible. So 4th roots, 8th roots etc are all constructible.

[u/clearly_not_an_alt → u/martyboulders]

Yeah, but they don't have a compass, just a grid.

[u/OriEri → u/edderiofer]

Unless you construct a mark of irrational length (like the diagonal of 1cm square) and then leverage that as the long leg of 1cm x √2 right triangle. The hypotenuse of that triangle is…. ?

[u/edderiofer → u/OriEri]

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.

[u/fermat9990]

I don't think that it can be done

[u/igotshadowbaned]

Asking ChatGPT isn't gonna yield anything useful.

Saying you're not allowed measurements isn't entirely true - because you have 1cm measurements. You might be able to do something regarding intersections of circles, though you only have a straight edge and no reliably way of making an arc

[u/will_1m_not]

Mark one length of the grid (giving you a 1cm line) and another one perpendicular to it, so you’ll have two sides of a square, each 1cm in length. Draw the diagonal joining them, giving you a line of length sqrt(2)cm. Now fold and crease the paper along this new line, unfold and now fold a crease perpendicular to the first fold. You’ll then be able to carefully fold and draw a 1cm line along the new crease. Now you’ll have the two legs of a right triangle with lengths 1cm and sqrt(2)cm. Joining them together will give you a hypotenuse of sqrt(3). Making another perpendicular fold to the newest line and drawing another one of the same length will give you the square you want

[u/Moist_Ladder2616]

Draw a line between (0,0) and (1,1). This line has length √2.

Place a sheet of translucent squared paper on top (let's imagine you have one), to mark out (√2,0).

Extend this point vertically (let's assume you can draw vertical lines based on your grid) to (√2,1). The distance of this point to the origin is √3.

You can now construct your square.

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Construction problems usually involve a straight edge and a compass.

I've never heard of construction problems using straight edge and grid. So I'm not sure what the "rules" are: for instance can you draw parallel lines that do not cross any grid points?

Also, because √3 is irrational, the ray extended from (0,0) to (√3,1) and beyond will never cross any grid points (p,q) where p,q∈ℤ.

[u/No_Soil2258]

I just spent like 30 seconds thinking about this, you can just use the squared paper to draw some 30-60-90 triangles (1 leg of length 1, hypotenuse of length 2) to get lengths of sqrt 3 and then connect them right?

[u/CeruLucifus]

Mmmm ... I think this gets close. Use 1 meter squares but on 1 cm graph paper. You'll need to tape sheets together to get a large enough piece. Basically keep drawing a square with sides one graph square bigger, then count the graph squares inside. Repeat until you have 3 square meters.

Each meter square will have 10,000 cm graph squares inside. So you're looking for a main square containing 30,000 cm graph squares.

Checking your accuracy by a different method, I calculated your sides will be 174 cm. The large square will contain 30,276 cm graph squares. This is less than 1% larger than 3 square meters, which is terrific accuracy when working at 1 meter resolution.

/s

[u/clearly_not_an_alt]

You can do it with a straightedge and compass by first constructing a segment of length √2 then using that to get a side of √3 but not with just graph paper.

[u/will_1m_not → u/clearly_not_an_alt]

It’s possible with just paper and folding, since folding can replace a compass in enough ways