[Daily puzzle] Proving a curious equality \ stacker news ~science

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[Daily puzzle] Proving a curious equality

1040 sats \ 14 comments \ @south_korea_ln 27 Oct science

Can you prove this equality?

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

I tried with trigonometric identities, including

and

, but didn't get anywhere yet.

1 sat \ 1 reply \ @Undisciplined 27 Oct

That's the route I was going to start with.

Have you tried drawing some triangles?

32 sats \ 0 replies \ @Scroogey 27 Oct

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.

1 sat \ 8 replies \ @south_korea_ln OP 27 Oct

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.

22 sats \ 6 replies \ @south_korea_ln OP 28 Oct

Here a figure that should help you figure out the method. Or at least, point you in the direction of a possible solution. Points

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 ;)

500 sats \ 3 replies \ @Scroogey 28 Oct

Angle units are

Angle sums in triangles must be 7 (=180°), in quadrilateral 14 (=360°).

Note two pairs of similar triangles. One pair is isosceles.

Ptolemy says

Hence

Equals

11 sats \ 2 replies \ @south_korea_ln OP 28 Oct

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.

0 sats \ 1 reply \ @Scroogey 28 Oct

What makes this so tricky is that there is no right triangle involved, yet you look for one because

is involved. At least I did for too long. Thank you!

1 sat \ 0 replies \ @south_korea_ln OP 28 Oct

Yeah, with the law of sines, one can get such right angles, but this is basically by constructing a new set of triangles. See #743146

22 sats \ 1 reply \ @south_korea_ln OP 28 Oct

Next hint: law of sines...

1 sat \ 0 replies \ @south_korea_ln OP 28 Oct

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

will cancel out. The remaining sine is the one that will remain in the final expression.

1 sat \ 0 replies \ @Jade001sand 27 Oct outlawed

stackers have outlawed this. turn on wild west mode in your /settings to see outlawed content.

0 sats \ 1 reply \ @Aardvark 27 Oct

0 sats \ 0 replies \ @south_korea_ln OP 28 Oct