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How do we ensure that this is the minimum number of moves required?
(A solution's worst nightmare)
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Proof by frequency? I am pretty sure there are no faster solutions, though I could be mistaken. But so far, the best is 9 moves.
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What is proof by frequency?
Never expected to hear a new math term for today...
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@Manuel Kahayon – It's a term I made up. Basically, proof by the number of people that have gotten the same result as me. :P :P
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@Sharky Kesa – HAHA nice term :p
My proof is proof by Euclidea TT.TT
Can't progress past delta. OCD does not allow me to skip a level without getting 3 stars on it.
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@Manuel Kahayon – I know what you mean. I completed the game yesterday. I got a quicker solution for Eta level 3 (construct a 7 5 ∘ angle). The problem has 4L as optimal. I got 3L. Feel really proud. They're gonna change it the next update.
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@Sharky Kesa – WHOAH that's amazing...
Did the game glitch out or what?
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@Manuel Kahayon – No, it's a legit solution.
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@Sharky Kesa – Yeah, draw 2 circles for the 30 degree, then bisect it or something
But, what I meant was, what happened to the game?
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@Manuel Kahayon – Nothing... It works fine for me.
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@Sharky Kesa – Oh... That was disappointing... I thought there was going to be something like:
"Congratulations, you are smarter than the game developers and we have decided to give you a cookie"
Or something :p
Euclidea now lists an eight move solution.
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Let the other vertices of the triangle be B and C .
Draw a circle centred at A of arbitrary radius such that it intersects A B and A C at P and Q . (1 move)
Draw circles centred at P and Q passing through A . Let these circles intersect each other again at R . (3 moves)
Draw A R let it intersect B C at E . (4 moves)
Draw a circle centred at A passing through E . Draw a circle centred at E passing through A . Let these two circles intersect each other at M and N . (6 moves)
Draw M N . Let this line intersect A B and A C at D and F .
Draw D E and E F .
Here, A D E F is a rhombus, sharing an angle with triangle A B C , and it took 9 moves.