So, I previously asked the question Draw a Square Without a Compass, Only a Straightedge. From the comments and answers, it appears that that question is not solvable. Given that the question I originally saw was on an actual exam (entrance exam for Cambridge undergraduate from the 90s, or maybe 80s), this got me thinking: odds are, I've misremembered the question!

I think I have remembered it correctly now, and so pose the following question.

Is it possible to, given a square drawn on a plane, using only an unmarked straightedge, construct another square with twice the area? If so, how is this done?


From an arbitrary point A on the top half of the vertical side of the square, construct the sequence of points to finish with a square twice the area of the original square. That is, the diagonal of a smaller unit square being $\sqrt2$ and forming the side of the larger square with double the area.

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  • $\begingroup$ Perfect. That's exactly what I was looking for. I knew that one starts with an arbitrary point on the side (A, or B), but I couldn't remember (or work out) what to do with it. $\endgroup$ – Sam T Oct 3 '18 at 21:27

I assume that, although the straightedge starts unmarked, we can mark lengths on it. Thus, if the side length is 1, we can mark any length in $\mathbb N[\sqrt 2]$.

Suppose we have square ABCD, with A =(0,0) B = (1,0), C = (1,1), D = (0,1). By the line AC and taking the point $\frac {3\sqrt 2}{2}$ from the origin, we can construct the point E = (1.5,1.5). By taking the ray DB and taking the point $\frac {3\sqrt 2}{2}$ along it, we can construct F = (1.5,-.5). The intersection of the diagonals of the square gives us G = (.5,.5). We can extend the sides of the square to give us the $x$ and $y$ axes, and then taking the distance from F to the $x$-axis gives us .5 (note that we can create the line segment between F and the $x$-axis by first taking the line segment EF). We can then take H = (1,.5), and draw the ray GH. Taking the point that is distance 2 away from G gives us I = (2.5,.5). GFIE is then a square with area 2.

  • $\begingroup$ If you can mark lengths on a straightedge don't you almost have a compass? You can construct any point which lies at the intersection of a circle and a line (though not at the intersection of two circles). $\endgroup$ – Rahul Oct 3 '18 at 19:06
  • $\begingroup$ And anyway, if you can mark lengths you can just construct the points $E=(-1,0)$ and $F=(0,-1)$ and then you have the square $BDEF$. $\endgroup$ – Rahul Oct 3 '18 at 19:08
  • $\begingroup$ Nope, you can do it with no marks on the straightedge at all, as has been demonstrated in another answer. $\endgroup$ – Oscar Lanzi Oct 3 '18 at 21:16
  • $\begingroup$ While it does seem strange to add marks (and this wasn't mentioned in the OP) -- and, as Rahul points out, this makes it trivial -- this answer was posted before the one to which Oscar refers $\endgroup$ – Sam T Oct 3 '18 at 21:25

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