Here is a grid graph with $7$ horizontal and $7$ vertical lines which are $1$ unit apart. It is trivial to mark two points which have a distance of $\sqrt{36}$.

Drawing at most two extra lines as helpers, could you mark two points which have a distance of $\sqrt{3.6}$?

enter image description here

  • $\begingroup$ By points, do you mean the intersection of these lines? Or anywhere on the lines? $\endgroup$ – Ébe Isaac Feb 20 at 6:58
  • $\begingroup$ @ÉbeIsaac arbitrary point can be made only if that helps.. but the exact position can't be explicitly specified unless it's in an intersection. $\endgroup$ – athin Feb 20 at 7:00

Of course we need to use Pythagoras.

So we need to write $3.6=\frac{18}{5}$ as the sum of two squares.
$\frac{18}{5} = \frac{90}{25}= \frac{81+9}{25}= (\frac{9}{5})^2+(\frac{3}{5})^2$

This leads to the following solution:

enter image description here

Here is another more compact solution.

enter image description here
The two lines have the equations $$y=4-2x\\2y=x-1$$ Their intersection point is $(\frac{9}{5},\frac{2}{5})$. From there to point $(0,1)$ we have $\Delta x=\frac{9}{5}$ and $\Delta y=\frac{3}{5}$ giving the same distance as the first solution.

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  • $\begingroup$ So fast, yep this is correct, well done! If you want more challenge, could you solve it in a grid graph of only $4$ horizontal and $4$ vertical lines? :) $\endgroup$ – athin Feb 20 at 7:18
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    $\begingroup$ @athin Ébe Isaac's solution would fit. $\endgroup$ – Jaap Scherphuis Feb 20 at 7:28
  • $\begingroup$ Unfortunately, their solution uses $5$ horizontal lines, so the challenge is still running :) $\endgroup$ – athin Feb 20 at 7:42
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    $\begingroup$ @athin I have now added a solution to your bonus challenge. $\endgroup$ – Jaap Scherphuis Feb 20 at 10:12
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    $\begingroup$ But Ebe's solution doesn't actually produce a distance of sqrt(3.6). $\endgroup$ – Gareth McCaughan Feb 20 at 11:10

A viable solution:


The dotted circle has a radius of $\sqrt{3.6}$ hence distance between any point on the circle and its centre is $\sqrt{3.6}$. The two blue lines are drawn so as to intersect on a point on the circle so that the red line from the centre can connect the intersection. The length of the red line is indeed $\sqrt{3.6}$. There should be possibly many solutions to this problem.
Note: one of the blue lines are overlapped by the red line


With the centre of the circle as the origin, the two lines can be written as
$y= \frac{1}{2}x + 2$ and $y = x$
They intersect at $(\frac{4}{3}, \frac{4}{3})$
Distance from the the origin to the intersection point is
$\sqrt{\frac{4}{3}^2 + \frac{4}{3}^2}$
$= \sqrt{\frac{32}{9}}$
$\approx \sqrt{3.6}$

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  • $\begingroup$ Well, how to get the circle in the first hand? $\endgroup$ – athin Feb 20 at 7:25
  • $\begingroup$ Sorry, @athin, I took advantage of a computer as there is no no-computer tag for this puzzle :) $\endgroup$ – Ébe Isaac Feb 20 at 7:28
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    $\begingroup$ @athin Ignore the circle. The intersection of those two blue lines defines a point that is the required distance from a gridpoint. $\endgroup$ – Jaap Scherphuis Feb 20 at 7:33
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    $\begingroup$ $\frac{32}{9}=3.5555$ so it is only very approximately $3.6$. $\endgroup$ – Jaap Scherphuis Feb 20 at 10:30
  • $\begingroup$ I agree, @JaapScherphuis, nonetheless it's a close approximation when considering the value of its square root. $\endgroup$ – Ébe Isaac Feb 20 at 11:24

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