/sci/ - Science & Math » Thread #13984694

32KiB, 701x567, area one sixth.png

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Anonymous Sun 19 Dec 23:26:45 2021 No.13984694 View Reply Original Report

Quoted By:

Solve for length A.

Can this problem be solved with pen and paper? I had to resort to Wolfram for the final calculation, but it seems like it should have been doable.

Anonymous

Anonymous Sun 19 Dec 2021 23:28:53 No.13984706 Report

Quoted By: >>13984725

i remember when i used to enjoy doing these in school. now i dont even remember how to do it.

Anonymous

Anonymous Sun 19 Dec 2021 23:31:59 No.13984725 Report

Quoted By: >>13984747

>>13984706
I made this one up, and I’m just mad I couldn’t solve it without a computer. Thought anon here would know a neat trick

Anonymous

Anonymous Sun 19 Dec 2021 23:36:35 No.13984747 Report

Quoted By: >>13984759

>>13984725
>I made this one up
I thought about it for 2 minutes, but it's too much of a hassle for me to open paint and give names to all the lines you necessarily have to label to talk about this.
Take it as a lesson, your next thread will be better

Anonymous

Anonymous Sun 19 Dec 2021 23:38:39 No.13984759 Report

Quoted By:

>>13984747
You're saying I should have labelled them all? I can do that next time, or you could just draw it and snap a photo

Anonymous

Anonymous Mon 20 Dec 2021 02:14:40 No.13985415 Report

Quoted By:

I tried two different approaches, but both of them required solving a cubic equation.

Anonymous

Anonymous Mon 20 Dec 2021 02:21:38 No.13985439 Report

Quoted By: >>13985570

By similar triangles the height of the point on the right hand side is $2A(1-A)$ .

The area of the triangle can be calculated by subtracting two triangles from the trapezoid:
$\frac12 \left[\frac12 + 2A(1-A)\right] - \frac14 A - A (1 - A)^2$
$= -A^3 + A^2 - (1/4) A + 1/4$

This must equal 1/6. So we have to solve the cubic
$A^3 - A^2 + \frac14 A - \frac{1}{12} = 0$ .

As usual, we remove the $A^2$ term first by substituting $A = u + \frac13$ .
$u^3 - \frac{1}{12} u - \frac{2}{27} = 0$

Then let $u = \sqrt[3]{p} + \sqrt[3]{q}$ . We have
$p + 3 \sqrt[3]{pq} (\sqrt[3]{p} + \sqrt[3]{q}) + q - \frac{1}{12} (\sqrt[3]{p} + \sqrt[3]{q}) - \frac{2}{27} = 0$
which is solved when
$p + q = \frac{2}{27}$
$3 \sqrt[3]{pq} = \frac{1}{12}$ .

We obtain
$p = \frac{1}{27} + \frac{\sqrt{7}}{2}$
$q = \frac{1}{27} - \frac{\sqrt{7}}{2}$
and thus
$A = \sqrt[3]{\frac{1}{27} + \frac{\sqrt{7}}{2}} + \sqrt[3]{\frac{1}{27} - \frac{\sqrt{7}}{2}} + \frac13$ .

Anonymous

Anonymous Mon 20 Dec 2021 03:14:11 No.13985570 Report

Quoted By:

>>13985439
Great work, thanks for confirming my suspicions that you have to solve a cubic for the answer. I was hoping there was a way using the quadratic equation, but doesn't look like it.

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