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>> No.11689351 [View]
File: 181 KB, 750x598, __remilia_scarlet_and_flandre_scarlet_touhou_drawn_by_60mai__12b22ea48909e69512481d854d2eaa18.png [View same] [iqdb] [saucenao] [google]
11689351

>>11687285
There's integrals, obviously.
There's trigonometry.
There's the Shoelace formula.
There's using Pythagoras to compute the sides and then finishing off with Heron's formula.
There's also putting it into a rectangle, calculating the rectangle's are and subtracting the three smaller triangles.

I can't really think of anything else that isn't an absolute meme like Monte Carlo.

>> No.11540467 [View]
File: 181 KB, 750x598, __remilia_scarlet_and_flandre_scarlet_touhou_drawn_by_60mai__12b22ea48909e69512481d854d2eaa18.png [View same] [iqdb] [saucenao] [google]
11540467

>>11540402
Seconding the other anon, the exact t is irrelevant.
The obvious method is the good old split it up, so that [math]s = (s \cap t) \cup (s \cap ([n]-t))= a \cup b[/math], where [math]|a| = 1 \mod 2[/math]. So it's now two classical combinatorics problems, sum along odd k smaller than l, choose k out of l times choose m-k out of n-l.
I'll tell you if I come up with anything better. Probably won't.

>> No.11530533 [View]
File: 181 KB, 750x598, __remilia_scarlet_and_flandre_scarlet_touhou_drawn_by_60mai__12b22ea48909e69512481d854d2eaa18.png [View same] [iqdb] [saucenao] [google]
11530533

>>11530225
Mochizuki-sama? I kneel!

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