/sci/ - Science & Math » Thread #13702531

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Anonymous Thu 30 Sep 21:38:35 2021 No.13702531 View Reply Original Report

Quoted By: >>13702551 >>13702568 >>13705421 >>13705426

Ok, mathlets. Figure out what the height and width are of a pentagon with side lengths 1. Express the answer symbolically. Only use the Pythagorean theorem and algebra, no trig or calculus. Pic related

Anonymous

Anonymous Thu 30 Sep 2021 21:42:13 No.13702551 Report

Quoted By: >>13702594

>>13702531
nice homework

Anonymous

Anonymous Thu 30 Sep 2021 21:44:18 No.13702568 Report

Quoted By: >>13702594

>>13702531
Width is $\phi$ , height is $\sqrt{3 + 4\phi} \over 2$
No, I will not elaborate.

Anonymous

Anonymous Thu 30 Sep 2021 21:52:21 No.13702594 Report

Quoted By: >>13702602 >>13704718 >>13704840

>>13702551
>>13702568
Just admit you can't do it. I'm 34 and trying to relearn math, and been realizing simple stuff can be deceptively hard

Anonymous

Anonymous Thu 30 Sep 2021 21:54:33 No.13702602 Report

Quoted By:

>>13702594
Nah, I did it a couple months ago and just read it off my notes

Anonymous

Anonymous Fri 01 Oct 2021 14:00:43 No.13704718 Report

Quoted By:

>>13702594

What part of:
>Width is ?
don't you understand? It's in the fucking definition of a regular pentagon.

Anonymous

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Anonymous Fri 01 Oct 2021 14:44:11 No.13704840 Report

Quoted By: >>13705031

>>13702594
>no trigo or calc
it's trivial but you made it harder with these arbitrary rules.
you have two trapazoids and two right triangles and two variables so you can probably solve it with algebra
or maybe some similar triangles.
either way it's not "deceptively hard", gramps.

Anonymous

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Anonymous Fri 01 Oct 2021 16:02:24 No.13705031 Report

Quoted By: >>13705034 >>13705130 >>13705789 >>13706414

>>13704840
>it's trivial but you made it harder with these arbitrary rules
>either way it's not "deceptively hard", gramps.
>uses angle measurements
Congratulations, you fail. Pic related is the algebra you refused to do; you can derive the height yourself.

Anonymous

Anonymous Fri 01 Oct 2021 16:03:54 No.13705034 Report

Quoted By:

>>13705031
Shit, ignore that $1 + \left(1 + {1 \over \phi} \right)$ in the last line, Should be $1 + {1 \over \phi}$ .

Anonymous

Anonymous Fri 01 Oct 2021 16:30:07 No.13705102 Report

Quoted By:

Work with complex numbers. Let w = e^(2i pi /5). Then width = | w + w^2| = sqrt( (w + w^2 )( w^4 + w^3) ) = sqrt(1 + w^1+ w + 1) = sqrt(1 + 2cos( 2pi/5))
height = | 1/2 + w + w^2 | = sqrt( (1/2 + w + w^2 )(1/2 + w^-1 + w^-2)) = sqrt( 1/4 + 1 +1 + 1/2 ( w + w^-1) + 1/2(w^2 + w^-2) + (w^-1 + w) ) = sqrt( 9/4 + 3cos(2pi/5) + cos(4pi/5))

Anonymous

Anonymous Fri 01 Oct 2021 16:42:03 No.13705130 Report

Quoted By: >>13705453

>>13705031
>muh rads
doesn't matter in this context. tell me the sum of angles of a pentagon in rads, smart guy.
i used degrees because my immediate thought was that EC=sin(54)*2 which also gives phi

Anonymous

Anonymous Fri 01 Oct 2021 18:35:43 No.13705421 Report

Quoted By:

>>13702531
post solution with calculus

Anonymous

Anonymous Fri 01 Oct 2021 18:36:52 No.13705426 Report

Quoted By:

>>13702531
Don't tell me how to solve my problems. Do your own homework

Anonymous

Anonymous Fri 01 Oct 2021 18:46:37 No.13705453 Report

Quoted By:

>>13705130
You're retarded and dependent on a calculator. OP specifically said no trig (i.e., sine and cosine) because it's what Euclid would have done.
And there are one and a half turns in a regular pentagon.

Anonymous

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Anonymous Fri 01 Oct 2021 21:11:26 No.13705789 Report

Quoted By: >>13705855

>>13705031
based

Anonymous

Anonymous Fri 01 Oct 2021 21:44:24 No.13705855 Report

Quoted By: >>13705916 >>13706283

>>13705789
I've also done the measurements in a dodecahedron from scratch, but I refuse to elaborate further

Anonymous

Anonymous Fri 01 Oct 2021 22:07:01 No.13705916 Report

Quoted By:

>>13705855
I'm struggling just to calculate a tetrahedral's side angles from scratch

Anonymous

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Anonymous Sat 02 Oct 2021 00:58:00 No.13706283 Report

Quoted By:

>>13705855
It turns out Evclide did not only conteyne 13 bookes whence translated into the english'd tongue. Doers of Billingsley's Euclid, imprinted by Iohn Dee, teacheth'd the inscription and circumscriptio~ of the five regular bodyes in 3 additional bookes

have you perhaps set forth the passions and proprieties of the icosi-dodecahedron through the cuboctahedron, and receaue the other five regular bodyes?

Anonymous

Anonymous Sat 02 Oct 2021 02:00:28 No.13706414 Report

Quoted By:

>>13705031
Nice proof for the width.
Here's both with trig:
Clearly $w=2\cos 36^\circ$ .
Use the double-angle formula twice to get
$-\cos 36^\circ=\cos 144^\circ = 2\cos^2 72^\circ - 1 = 2(2\cos ^236^\circ - 1)^2 - 1$ .
Now substituting in w and collecting terms gives
$0=w^4-4w^2+w+2=(w-1)(w+2)(w^2-w-1).$
Since trivially w>0, and $\cos 36^\circ \neq \frac12 = \cos 60^\circ$ , the only possible solution is $w=\phi$ .

For the height, we find $2h=\tan 72^\circ$ . Again we use the double-angle formulas and the trig unit to get
$\tan 72^\circ = \frac{\sin 72^\circ}{\cos 72^\circ} = \frac{2\sin 36^\circ \cos 36^\circ}{2\cos^2 36^\circ - 1} = \frac{\phi\sqrt{1-\frac14\phi^2}}{\frac12\phi^2 - 1}=\frac{\phi\sqrt{4-\phi^2}}{\phi^2-2}=\frac{\phi\sqrt{2-\phi^{-1}}}{\phi^{-1}}=\sqrt{2\phi^4-\phi^3}=\sqrt{4\phi+3}.$

El Arcón

El Arcón Sat 02 Oct 2021 02:05:07 No.13706417 Report

Quoted By:

>heidth

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