Explaining solutions to Pythagorea puzzle 2.16: Drawing a line through a given point parallel to a given line

Question

In this geometry puzzle (Problem 2.16 of the game "Pythagorea"), a 6x6 grid is given. Inside the grid there's a black line and a point A not on the line. The goal is to draw a line through A parallel to the given black line.

The solution must be achieved by drawing lines between points, including the starting grid points, the intersections between the grid and the given black line, and new points/intersections that appear after drawing more lines. Circles cannot be drawn. Everything must be done within the grid, the grid cannot be extended.

There's no limit to the number of steps/lines used before getting to the solution. And there are several ways to arrive at the solution, but the last step is always drawing the intended parallel line, e.g. the second-to-last step creates a point B so that joining A and B creates the intended parallel line.

If from this description and the first image you can figure out a way to the solution, feel free to explain it. Otherwise, see the second and third images. The parallel line shows in yellow and the other construction lines in blue.

This is a 3-step solution:

This is a 2-step solution:

This last image shows both solutions for comparison.

The question is:

How do these solutions work? What theorems are applied in each one?

Thank you

Answer 1

This is not the solution you have asked for. We just want to show you that the angle of the triangle appeared in the first solution is not equal to $90^o$. At the same time, we include another solution for your perusal. It was obtained by drawing a line joining $H$ and $G$ to intersect a vertical grid line at $O$.

Answer 2

You can show that the two solutions are correct by using similar triangles. In the diagram of the first solution below, $\ AB,CD,FG\ $ and $\ HI\ $ are all parallel. Therefore, by similarity of triangles $\ EAB\ $ and $\ ECD\ $ we have $\ EA:EC=$$\,AB:CD=$$\,1:3\ $ and by similarity of triangles $\ EFG\ $ and $\ EHI\ $ we have $\ EG:EI$$\,=EF:EH$$\,=1:3\ .$ Therefore $\ EA:EC=$$\,EG:EI$$\,=1:3\ ,$ triangles $\ EAG\ $ and $\ ECI\ $ must be similar, and $\ AG\ $ must be parallel to $\ CI\ $.

In the diagram of the second solution below $\ GI:HI=$$\,3:5\ $, so the slope of the line $\ HG\ $ is $\ \frac{3}{5}\ .$ Do you see how the similar triangles $\ DEF\ $ and $\ DBC\ $ can be used to show that the slope of the line $\ AF\ $ is the same?

Here's a third solution which seems to me to be simpler than either of the above. In the diagram below, the lengths of the lines $\ BC\ $ and $\ AC\ $ are clearly $\ \frac{3}{2}\ $ units and $\ \frac{5}{2}\ $ units, respectively, making the slope of the line $\ AB\ $ $\ \frac{3}{5}\ ,$ the same as that of the given line.

Additional solutions

As Calvin Lin points out in his answer there are many more solutions besides those listed above. Besides the grid lines through $\ A\ ,$ the target line intersects $5$ other vertical grid lines and $3$ other horizontal grid lines. Each of those $8$ intersections can be obtained in one step, just as in the second of the solutions given in OP's question. All $8$ of the resulting solutions are shown in the diagram below.

Answer 3

• Note that unlike pure compass-and-straightedge construction questions, you actually have a lot more guidance in playing around.
• You have grids, which means that the initial nodes are lattice points. And since you're connecting nodes with lines, those slopes are rational. In fact, the nodes that you can define would all have rational coordinates (I believe, but haven't fully double checked.)
• Because you have grid lines, cutting lines up into $ \frac{1}{3}, \frac{2}{3}, \frac{1}{5}, \frac{2}{5}, \ldots $ can be done easily under certain conditions.

Solution 1 explained - Similar triangles

• Starting off with point $A$ and line $BC$.
• Notice that $AB$ is easily a "multiple of 4", and hence a "multiple of 2". We can easily bisect $AB$ (at point X, unlabelled) and might be able to work with that point.
• If we extend $AB$ out by 1/2 to $D$, and observe that $DC$ is both a "multiple of 2 and 3" which gives us lots of options, we can trisect $DC$ with point E, Y (unlabelled).
• Then, $AE \parallel XY \parallel BC$.

Solution 2 explained - Coordinate geometry

• Observe that $BC$ has a slope of $\frac{3}{5}$.
• $B = (0, 0), C = (5, 3), A = (2, 4)$.
• To find the line parallel through $A$ (which we'd call $\ell$), we just need another point on it.
• If we chose the vertical grid line that $C$ is on ($y=5$), then that will intersect $\ell$ at $H = (5, 4 + 3\times \frac{3}{5}) = (5, 5 \frac{4}{5} )$.
• So we want to snip off $ \frac{1}{5} $ from the top, which can be done by constructing at line of slope $ \frac{ 1}{5}$.
• Hence construct $FG$ intersecting $ y = 5$ at $H$.
• Note: We could take lots of other lines. Solution 1 could be interpreted as choosing the line $x=5$, and then figuring out where and how to cut it up. As such, on your end, you should choose another line (say $x = 3$), and figure out how to construct the point which gives the intersection with $\ell$.

Answer 4

Just for fun, here are a few more two-step solutions to add to the grid from this earlier answer. The added solutions are shown as thin black lines.

One of the solutions comes from yet another answer, one is the same (extended) line as one of the solutions already shown, but there are three ways to choose the two points to draw that line (or at least two, anyway; the rules were not entirely clear). One of the solutions is derived from the "three-step solution" by omitting the construction line through A. (The other construction line connects two grid points and can be constructed immediately without needing the line through A.)