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I recently happened upon a game called pythagorea.

The idea is that you're given a 6x6 grid. You may click at any intersection to create a point and you may join any two points to create a line (that extends in both directions to the end of the grid). And then you're given various tasks to achieve.

Two puzzles stood out for me as being a little more tricky than the others. My solutions ended up being a little... inelegant. I wondered whether anyone would be able to do better.

The challenge then is to solve the puzzles with as few lines as possible (that's my arbitrary measure of elegance!).

Puzzle 1:

pythagorea-22.17

Puzzle 2:

pythagorea-25.16

Please indicate also the logic that you've used to know that your solution is correct.

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    $\begingroup$ Puzzle 2 is apparently possible with just 10 lines. $\endgroup$
    – Jens
    May 11 '20 at 21:51
  • $\begingroup$ @Jens, oh wow! That's waaaaay better than my solution! Neat theorem. With this particular puzzle, 9 will be enough then... $\endgroup$
    – Dr Xorile
    May 11 '20 at 22:09
  • $\begingroup$ the strategy I've tended to appeal to was you use math the compute the coordinates of the answer, and exploit the fact that the coordinates are rational to construct the points I need. $\endgroup$ May 12 '20 at 1:38
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    $\begingroup$ Blast you, @DrXorile! Now I have another time suck on my device! $\endgroup$
    – shoover
    May 12 '20 at 22:02
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    $\begingroup$ But games aren't addictive or anything, right? $\endgroup$
    – Dr Xorile
    May 12 '20 at 22:05
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A very ad-hoc solution to Puzzle 1 in nine lines (technically ten, because you have to extend one past the point you anchor it to): ((0,0) is the origin)

Draw a line from (-2,2) to (-1,-1). Mark (4/3,0). Draw a line from A to that point and mark its intersection with the opposite side of the triangle. Call it B.
Draw a line from (0,0) to (1,-2). Mark (3/2,-3). Draw a line through that point and (1,3), and mark (13/12,2). Call it C.
Draw a line from (0,-2) to (2,-1). Mark (3,-1/2). Draw a line through that point and (1,3), and mark (2,5/4). Draw a line through that point and (-1,1), and mark (0,13/12). Call it D.
BC and BD are the required other sides.
I actually got the app so have a good picture

As for Puzzle 2, let's adapt Jens's idea, but abuse some grid lines to cheapen the construction.

Our B line will be the one through (-1,0), which we don't even need to draw: its intersection points are (-1,0) and (0,1).
Our A line is the one through (0,-1). Draw it, and draw the line connecting the upper intersection to (-1,0). It intersects the y-axis at D.
Our C line is the one through (1,0). Draw it, and draw the line connecting the lower intersection to (0,1). It intersects the x-axis at E.
DE is our Pascal line, and two more lines (for a total of 7) give us the tangents!
Another sloppy demo The Pascal line is in green, and there are no tangents because the lower one would be too close to the other lines (and the thicker lines cover for inaccurate drawing on mobile!)

EDIT: "You can save one more line."

Let's just skip straight to the Pascal line, which is the line from (-1,1) to (2,-1), and save four!
This didn't feel nearly as satisfying as you'd think

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  • $\begingroup$ Puzzle 1 answer is not correct. All sides must have equal length $\endgroup$
    – Dr Xorile
    May 12 '20 at 22:13
  • $\begingroup$ I just remembered this... Oops. Think I'm close to an actual solution, however. $\endgroup$ May 12 '20 at 22:17
  • $\begingroup$ Puzzle 2 is great. You can actually save one of the lines still $\endgroup$
    – Dr Xorile
    May 12 '20 at 22:19
  • $\begingroup$ A line can be saved on Puzzle 1, but I'll leave that for later. $\endgroup$ May 13 '20 at 0:31
  • $\begingroup$ The solution to Puzzle 2 implies a potentially cheaper general method: rot13(Gur Cnfpny yvar pbeerfcbaqvat gb n cbvag vf gur yvar crecraqvphyne gb gur yvar sebz gur pragre gb gur cbvag guebhtu gur vairefr bs gur cbvag.) $\endgroup$ May 14 '20 at 3:56
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Puzzle 2 is a standard Euclidean construction of drawing tangents to a given circle from a point outside the circle. For this construction you must use only a ruler as a straight edge. The steps are as under:

  1. From the point O outside the circle, draw two lines OAB and ODC. A, B, C, D are the points of intersection of the lines with the circle. Thus you get a cyclic quadrilateral ABCD.
  2. Extend the sides BC and AD to meet at P . Draw the diagonals BD, AC to intersect at Q.
  3. Join P and Q (and extend it if needed) to intersect the circle in points M and N.
  4. OM and ON are two tangents to the circle.

If the lines OAB and ODC are drawn carelessly in step 1 you will find that the point P in step 2 moves outside the grid. To keep it within reasonable limits, one of the lines must be drawn as close to the centre of the circle as possible and the other be drawn as far away as possible.

Most shortcuts to such tangent drawing problems involve remembering the coordinates of two points through which the line PQ passes. By thus drawing the line PQ you get the points M and N.

solution per this method

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  • $\begingroup$ Very nice. Thanks for sharing. I've edited in a diagram following your method $\endgroup$
    – Dr Xorile
    Oct 29 '20 at 23:41

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