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The number of turns is 4.
First look at two coins, one going around the other as described. There are two things going on, orbitingand rolling. If the coin only orbited the other coinwith the same point facing it, the coin would turn oncefor each orbit. When rolling is added in, the cointurns an extra turn for each orbit for a total of 2turns for each circumference traveled.
Now when 3 coins are placed so they are all touchingand three lines are drawn from each centre to the othertwo coins centres, these lines pass through the pointswhere the coins touch. Now each line is the diameterof the coins long, so the triangle is an equilateral triangle. Now since it is an equilateral triangleyou can place 6 of them with a common vertex at thecentre of one coin, without overlap and each sharing a common side with the next. Putting a coin at each of the outer vertices gives you 6 coins surrounding the one, alltouching and tangent at that point. Now looking at this, each coin from the problem has six positions that another coin might be placed in (either side, above right or left, below right or left). Each of thesepositions is 1/6th a circumference from the next.
Now looking at the triangle of coins in the problem, each of the three corner coins already have 2 coinsin position, so there is 4 positions free. To travel fromthe position on one side to the other would take 3 moves, each of 1/6th of a circumference for a total of 3 cornerstimes 3 moves or 9/6ths of a circumference.Now the other 3 coins have 4 coins around them for a totalof 2 positions free and it takes 1 move to move from oneto the other for a total of another 3 moves.Therefore the total number of moves is 9+3 or 12 moves, fora total of 12/6ths or 2 circumferences. Now each circumference traveled will produce 2 turns, so the total number of turns is 2.
As an aside, each time you add a row to the bottom3 more inner coins are added for an additional turn, so the formula for the number of turns given n rowswould be "# turns = n+1" and in our case n = 3.
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