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This is a question which is related to the hardest easy questions. Note that the general solution belongs on math.se and is not solved in simple way. This question is a puzzle and you need to prove the answer.

triangle

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closed as off-topic by The Dragonista, Michael Rize, Rand al'Thor, Gamow, Alexis Feb 21 '15 at 17:12

  • This question does not appear to be about creation and solving of puzzles, within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ See math.stackexchange.com/questions/750410/… $\endgroup$ – Macavity Feb 21 '15 at 6:24
  • $\begingroup$ It belongs on Math SE $\endgroup$ – The Dragonista Feb 21 '15 at 6:27
  • $\begingroup$ This diagram appears to be drawn to scale, in which case the answer is also 15°. $\endgroup$ – Joe Z. Feb 21 '15 at 6:37
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    $\begingroup$ The general solution belongs to math and is not solved in simple way. This is a puzzle and you need to find a way to prove that it is 15! $\endgroup$ – Moti Feb 21 '15 at 6:38
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    $\begingroup$ I'm voting to close this question as off-topic because it belongs on math.se $\endgroup$ – Michael Rize Feb 21 '15 at 9:07
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Although this puzzle is a math problem, it involves enough complex thought and nonconventional process in its solution to count as a puzzle, in my opinion.

In particular, it is a challenge to solve this problem without using trigonometry, as simple angle-chasing will get you nowhere. If you try and map out all possible linear equations of variables, you will find that $\angle AKL$ and $\angle BLK$ cannot be solved for. At first glance it appears that you will need to use scale measurements to solve this problem.

However, it is possible, after some thought. A solution that involves no trigonometry follows.

  1. Rotate $\triangle ALK$ 30 degrees around point $K$, so that $A$ is mapped to $B$. Then $L$ is mapped to a point which we will call $M$, and a new triangle $\triangle BMK$ is created.

  2. Note that since $LA = AB$, $MB = AB$ as well. Further note that since $\angle KAL = 15^\circ$, $\angle KBM = 15^\circ$ as well, so that $\angle ABM = \angle ABK - \angle KBM = 75^\circ - 15^\circ = 60^\circ$.

  3. Since $AB$ and $BM$ are of equal length and $\angle ABM = 60^\circ$, $\triangle ABM$ must be an equilateral triangle, and $AM$ must be of the same length as $AB$ and $BM$.

  4. Now, since $\triangle ABM$ is equilateral, it is also isosceles with $M$ as an apex. Since $\triangle ABK$ is also isosceles in this way, both $K$ and $M$ are perpendicular to $AB$ at its midpoint. So $KM$ is a perpendicular bisector of $AB$, and also an angle bisector of $\angle AKB$.

  5. Therefore $\angle BKM = 15^\circ$, and we rotate the triangle back to find that $\angle AKL = 15^\circ$.

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  • $\begingroup$ The solution is OK, but this is a "math" solution. I would challenge you to find a more elegant way - pure basic first years of geometry. $\endgroup$ – Moti Feb 22 '15 at 6:19
  • $\begingroup$ Unfortunately, even if I did have an answer I wouldn't be able to post it - the question has been closed. $\endgroup$ – Joe Z. Feb 22 '15 at 6:43
  • $\begingroup$ And of course, right after I say that I figure out how to do it - it involves rotating $\triangle AKL$ around point $K$. $\endgroup$ – Joe Z. Feb 22 '15 at 6:53
  • $\begingroup$ It is a pity. This one has a really nice solution. If interested my email motib1@selectemail.net $\endgroup$ – Moti Feb 22 '15 at 6:58
  • $\begingroup$ @Moti: I decided to edit my answer into the intended solution, rather than waiting until I could post a new one. $\endgroup$ – Joe Z. Feb 22 '15 at 7:42

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