You’re locked in a room with nothing but a pencil, a math compass, and paper. You do not have a straightedge. Your captors have informed you that you cannot leave until you construct (the endpoints of) a line segment of length $\sqrt{7}$.

What do you do?

  • 3
    $\begingroup$ Is this a puzzle you created, and do you know of a solution? $\endgroup$
    – user46002
    Mar 8, 2019 at 3:25
  • 12
    $\begingroup$ Presumably you have access to something of length 1? $\endgroup$
    – Dr Xorile
    Mar 8, 2019 at 3:55
  • 4
    $\begingroup$ I see three VTCs saying this looks like a math textbook problem rather than a geometry puzzle. Since anti-VTCs (or would they be VTNCs?) don't exist, I'm logging my differing opinion here: Constructing numbers by limited means is the fundamental form of all geometry puzzles, and this question definitely should not be closed on those grounds. $\endgroup$
    – Bass
    Mar 8, 2019 at 9:32
  • 2
    $\begingroup$ Seeing as I define the coordinate system, I'll draw two dots as endpoints and declare it's length to be root(7). Perhaps OP would like to define a base unit? $\endgroup$ Mar 8, 2019 at 13:08

2 Answers 2


Suppose you can set your pair of compasses to length 1. Then

draw a unit circle. Mark a point on the circumference. Step off 6 points with the compasses around the circumference. Take 1 of those points as the center for another circle and repeat. In this way you can mark off straight lines from the center of the original circle of any integer length. These can radiate out at $k.60°$ from each other. Suppose the lengths are $a$ and $b$.

Then we know that the length of the line that would join those two points is

$\sqrt{a^2+b^2-ab}$. It could also be $+ab$ by taking 120°. Anyway, there's a bunch of solutions. Such as $a=3$ and $b=2$.

The simplest construction would use:

Just two circles. Here we use that $a=2$, $b=1$ and 120° gives us $\sqrt{7}$. construction

  • $\begingroup$ Simpler than my own solution. Nice! $\endgroup$ Mar 27, 2019 at 14:31
  • $\begingroup$ Neat puzzle. Quite surprising $\endgroup$
    – Dr Xorile
    Mar 29, 2019 at 1:00

Edited: added a drawing for the first step.
Edited again: Dr Xorile very clever solution eliminates the second step.

You can solve the problem in two steps. First construct points on a line that correspond to integer length 1, 2, 3, 4, .... This part is easily done by repeatedly constructing vertices of a regular hexagon. On the drawing below AB is a unit segment. Then A is 0, B is 1, E is 2, G is 3, etc

enter image description here
The second step is constructing the solution - a segment X which is a geometric mean of 1 and 7. This is done by a well known elegant method described in https://www.cut-the-knot.org/pythagoras/fastGM.shtml

In fact, as Dr Xorile pointed out in his very clever solution, no second step is necessary - segment HG is already a solution. Its length is $\sqrt{7}$

  • $\begingroup$ Incorporated @Dr Xorile very clever observation into the drawing. All the credit belongs to his solution! $\endgroup$
    – ppgdev
    Mar 8, 2019 at 5:46

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