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In Euclid: The Game - Level 23, we are given two non-concentric circles:

Starting Situation

The goal is to create one of the tangents shown in the following image:

Ending Situation

Note: it suffices to create a line segment on one of these $2$ tangents. This segment need not actually touch both circles, it just needs to be in the right direction.

Part of the challenge in Euclid: The Game is to do this with a minimal number of "moves", each move standing for the usage of one of the tools given. Creating a point or naming an intersection do not count as a move; everything else does.

The "minimum number of moves" to solve this is said to be $6$. Multiple solutions exist (see below). However, the page mentions that "there are rumours that it is possible to do this level in 5 non-primitive moves".

So the question, is: can someone come up with a solution in $5$ (non-primitive) moves?

SPOILERS BELOW!

I know of two different solutions in $6$ moves:

Solution 1:

Solution 1
Define a random point $M$ on the circle $c_B$.
Translate the line $BM$ to $A$, creating a segment $AN$. (move 1)
Define the intersection $O = c_A \cap AN$.
Compass the length $ON$ centered at $B$, this creates a circle $c_B'$. (move 2)
Create the midpoint $P$ of $AB$. (move 3)
Create the circle $c_P$ centered at $P$ with radius $PB$. (move 4)
Define $Q = c_P \cap c_B'$.
Create the ray $BQ$. (move 5)
Translate $AQ$ to $R$, creating a segment $RS$. (move 6)

Solution 2:

Solution 2
Define a random point $M$ on the circle $c_B$.
Translate the line $BM$ to $A$, creating a segment $AN$. (move 1)
Define the intersection $O = c_A \cap AN$.
Create the ray $BA$. (move 2)
Create the ray $MO$. (move 3)
Define $P = BA \cap MO$.
Create the midpoint $Q$ of $AP$. (move 4)
Create the circle $c_Q$ centered at $Q$ with radius $PQ$. (move 5)
Define $R = c_Q \cap c_A$.
Create the ray $PR$. (move 6)

Remarks:

  • Both $6$-moves-solutions start similarly by picking a random point $M$. Perhaps picking a not-so-random point (for example on the intersection of the circles, see next item) to start with can reduce the number of points needed?

  • You can move the centers of the circles, for example creating intersecting circles! In some of the levels, this is necessary to get the minimum number of moves, so perhaps it can be of use here too?

  • 0
    If I had a straightedge and a compass, I would draw a circle centered at $B$, with radius equal to $R-r$, the difference between the radii $R,r$of the two given circles. Then draw a circle with diameter $AB$. The two new circles intersect at $C$ (and $D$, but that doesn't matter). Draw the line $AC$, then parallel transpose the line a distance of $r$, and you have your tangent. I don't know whether that's 5 steps, though. I'm not familiar with the software. – 2017-01-07
  • 0
    @Arthur I think that's pretty much what solution 1 does. Unfortunately, drawing a circle with radius $R-r$ centered at $B$ takes 2 moves. – 2017-01-07

1 Answers 1

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Here is a simple one in 5 moves -

  1. Draw two parallel lines through centers of the circles - 2 moves

  2. Draw a line through two intersections (on same side) - 1 move

  3. Draw the perpendiculars to the chords through the centers, in the two circles - 2 moves

The intersections of these two perpendicular lines with the perspective circles are defining the segment as required in the question - those are actually the tangent points on the two circles.

I do not think a drawing is needed.

  • 0
    You have to draw the segment itself, unfortunately, which counts as a 6th move. Moreover, the line through the intersections you mention, need not be tangent to either of the circles. See http://i.imgur.com/J7bN61M.png for an image of your construction where this is clear. – 2017-01-11
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    Well you could draw only one perpendicular and than at the intersection have a perpendicular rather than have two of them. May be a drawing will help - will post another answer with a drawing. – 2017-01-12