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.



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


  • $\begingroup$ 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

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 $\angle AKL = 15^\circ$.

  • $\begingroup$ 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
