# Mark Two Points Which Have a Distance of $\sqrt{3.6}$

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 $$\sqrt{3.6}$$? • By points, do you mean the intersection of these lines? Or anywhere on the lines? – Ébe Isaac Feb 20 at 6:58
• @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. – 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:

Here is another more compact solution. 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.

• 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? :) – athin Feb 20 at 7:18
• @athin Ébe Isaac's solution would fit. – Jaap Scherphuis Feb 20 at 7:28
• Unfortunately, their solution uses $5$ horizontal lines, so the challenge is still running :) – athin Feb 20 at 7:42
• @athin I have now added a solution to your bonus challenge. – Jaap Scherphuis Feb 20 at 10:12
• But Ebe's solution doesn't actually produce a distance of sqrt(3.6). – Gareth McCaughan Feb 20 at 11:10

A viable solution: Explanation:

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

Proof:

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
$$\sqrt{\frac{4}{3}^2 + \frac{4}{3}^2}$$
$$= \sqrt{\frac{32}{9}}$$
$$\approx \sqrt{3.6}$$

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