/sci/ - Science & Math

>pasta

>>5717609
kind off.

I posted the same question two days(or so) ago, looking for the answer. But beer happened and I fogot about it so now I try again

lel still unsolvable Need to know angle C.

Pic related two versions with different angle c's. Segment "2a" clearly has different lengths.

Assuming the two black lines which appear to form a right angle do, you can use trig to identify the length from the center of the circle to the point at which the red line meets the black lines - just draw a line from the center of the circle below to where the circle touches the black line beneath it and the solution become obvious; if you're assuming the black lines don't make a right angle, let it be C and you can form a general solution

>>5717627
God dammit....

Pic related, find radius now

Connect the center of the circle to the point where either of the bottom tangents touch it, and you have a right angled triangle with lengths r, r and 2a-r. The rest should be easy enough.

r =2a/(1 + sqrt(2))

edit: red line is perpendicular to tangent to circle, so it must pass through the centre thus the line must bisect C so the black lines form a right angle anyway - if you want to know, radius = 2a/(1+cosec45)

Guys, give me the rest of eternity and I'll have this figured out.

>>5717651
Making the final assumption that the lines are tangent to the circle, here is the solution.

First, note that forms a diameter of a circle. Thus, it goes though the circle's center. I've added lines from the tangent to segment 2a. These lines are perpendicular to the edge of the circle, and go through the circle's center, so they are radius, with length r. These segments form a square, and the diagonal of the square has length <span class="math">r\sqrt{2}[/spoiler].

so, <span class="math">2a = r + r\sqrt{2} [/spoiler], and then

<div class="math"> r = \frac{2a}{(1 + \sqrt2)}</div>

>>5717733
pic

You could do it algebraically. I'll reflect the circle onto the first quadrant to make it simpler. The equation of the circle is:

<div class="math">(x-r)^2+(y-r)^2=r^2</div>
Expanding this out and simplifying gives:
<div class="math">x^2+y^2+r^2-2r(x+y)=0</div>
The distance from the origin to any point in space is:

<div class="math">d^2=x^2+y^2</div>
So we can substitute this into the circle equation to get:

<div class="math">d^2+r^2-2r(x+\sqrt{d^2-x^2})=0</div>
Next we find the dd/dx:

<div class="math">2dd'-2r(1+\frac{dd'-x}{\sqrt{d^2-x^2}})=0</div>
Setting the derivative to zero:

<div class="math">d=\sqrt{2}x</div>
The distance is therefore minimized when it equals the x-coordinate times sqrt(2). Plugging back into the circle equation:

<div class="math">2x^2+r^2-4rx=0</div>
the (interesting) solution to which is:

<div class="math">x=r(1-\frac{1}{\sqrt{2}})</div>
The shortest distance in terms of r is then:

<div class="math">d=r(\sqrt{2}-1)</div>
We also know that 2a-d=2r, so:
<div class="math">a=\frac{1}{2} r (1+\sqrt{2})</div>

>>5717782
>>5717737
>same result from two different methods