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[Daily puzzle] Proving a curious equality
1040 sats \ 14 comments \ @south_korea_ln 27 Oct 2024 science
Can you prove this equality?
\frac{1}{\sin \frac{\pi}{7}} = \frac{1}{\sin \frac{2\pi}{7}} + \frac{1}{\sin \frac{3\pi}{7}}.
I would consider this a hard problem and comes from Math Horizons.
I'll provide hints further down the road.
Previous iteration: #741135 (answer in #741780).
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13 sats \ 11 replies \ @Scroogey 27 Oct 2024
I tried with trigonometric identities, including csc(2\theta) and csc(3\theta), but didn't get anywhere yet.
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1 sat \ 1 reply \ @Undisciplined 27 Oct 2024
That's the route I was going to start with.
Have you tried drawing some triangles?
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32 sats \ 0 replies \ @Scroogey 27 Oct 2024
I used the cosecant simulator with angles 25.7°, 51.4°, and 77.1° and tried to arrange the triangles so the addition of the hypotenuses would be visible. Mirroring triangles along the axes. But found no helpful arrangement.
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1 sat \ 8 replies \ @south_korea_ln OP 27 Oct 2024
Please don't spend too much time on this ;)
The first hint will likely save you a lot of time. Without it, I would never have guessed the direction of the solution suggested by the authors of this problem.
I'll try posting it tomorrow after i wake up.
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22 sats \ 6 replies \ @south_korea_ln OP 28 Oct 2024
Here a figure that should help you figure out the method. Or at least, point you in the direction of a possible solution. Points P_i are equally spaced on the circle, and the black shape is a convex quadrilateral.
A first hint would be to look up Ptolemy’s inequality.
I'll wait a bit before giving out the next hints ;)
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500 sats \ 3 replies \ @Scroogey 28 Oct 2024
Angle units are \frac{1}{7}\pi.
Angle sums in triangles must be 7 (=180°), in quadrilateral 14 (=360°).
Note two pairs of similar triangles. One pair is isosceles.
Ptolemy says
c_1*c_2 + c_1*c_3 = c_2*c_3
Hence
sin(\frac{1}{7}\pi)*sin(\frac{2}{7}\pi) + sin(\frac{1}{7}\pi)*sin(\frac{3}{7}\pi) = sin(\frac{2}{7}\pi)*sin(\frac{3}{7}\pi)
Equals
sin(\frac{1}{7}\pi) = \frac{sin(\frac{2}{7}\pi)*sin(\frac{3}{7}\pi)
}{sin(\frac{2}{7}\pi)+sin(\frac{3}{7}\pi)}
Equals
\frac{1}{sin(\frac{1}{7}\pi)} = \frac{1}{sin(\frac{2}{7}\pi)} + \frac{1}{sin(\frac{3}{7}\pi)}
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11 sats \ 2 replies \ @south_korea_ln OP 28 Oct 2024
I guess the Ptolemy hint did the trick. Or maybe with just the figure, you'd have figured it out too...
There are some beautiful other problems in Math Horizons, I'll have to take some time to curate the ones that are easy to post here.
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0 sats \ 1 reply \ @Scroogey 28 Oct 2024
What makes this so tricky is that there is no right triangle involved, yet you look for one because sin() is involved. At least I did for too long. Thank you!
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22 sats \ 1 reply \ @south_korea_ln OP 28 Oct 2024
Next hint: law of sines...
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1 sat \ 0 replies \ @south_korea_ln OP 28 Oct 2024
As an alternative to the solution in #743121, With the law of sines, one can get to the same conclusions. The triangles to be built are each using an edge of the quadrilateral, an edge going through the center of the circle, and the remaining edge connecting the two endpoints of the first two edges. This construction always gives a right angle with a sine equal to one, and the R^2 will cancel out. The remaining sine is the one that will remain in the final expression.
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1 sat \ 0 replies \ @Jade001sand 27 Oct 2024 outlawed
stackers have outlawed this. turn on wild west mode in your /settings to see outlawed content.
0 sats \ 1 reply \ @Aardvark 27 Oct 2024
42
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0 sats \ 0 replies \ @south_korea_ln OP 28 Oct 2024
:)
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