Anybody know how to solve this? I have been trying to solve this

Since it's symmetric, you can draw spokes coming out from the middle which divide the figure into 8 identical right triangles. You can just solve one triangle and then extrapolate the answer from there.

Let's consider the one triangle that is formed by drawing a line segment A from the middle point to the left, parallel to the bottom of the square, and the line segment B drawn from the middle point up and to the left so that it hits the top left corner of the square. There is a 45 degree angle between line A and B. We know line segment A has length 1, and B is length sqrt(2) from Pythagorean theorem.

The shaded region of this triangle can be calculated easily using trig because it bisects lines A and B, and we know the angle between them. Note that besides line A and B, the third boundary of this region must be a line because of the properties of the problem, so the shaded region is also a triangle.

>>8727001
I tried that, but isn't the region not a triangle? It is curved, so doesn't that not apply?

>>8726965
>besides line A and B, the third boundary of this region must be a line
whyy
shit doesn't look like a line to me, it all looks curved and there are no break points in the "sides" of the orange figure

>>8727009
op, try defining a function that is 1 when a point in the square is closer to the center and 0 when it's farther away
use it to "count" the points by integrating somehow

can a mathfag confirm or deny if this can be done

The boundary nearest to a side side would be the set of points equidistant from the center and that side.

This is one of the classic definitions of a parabola:
https://en.wikipedia.org/wiki/Parabola#Definition_of_a_parabola_as_locus_of_points

>>8727021
i knew it, literally a square distance function
so i'm basically right here>>8727019

Ahhhhh, I see. We gotta make the two points equidistant. Thus x^2+y^2=(1-y)^2, where x and y are the coordinates of the region. Basically, gotta make the hypotenuses of the triangle equal to the distance of the region to the side of the square

File: tmp_5589-Screenshot_2017-03-06-19-13-03-156489007.jpg (897KB, 2560x1440px) Image search: [Google]

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Have fun.

>>8727112
Messed up the theta limits. It's pi/4 not pi/2

File: tmp_5589-Screenshot_2017-03-06-19-32-24-773950396.jpg (962KB, 2560x1440px) Image search: [Google]

962KB, 2560x1440px

>>8727112
I thought it would be easier in polar coordinates. It's not. I don't think I messed up this time.

>>8727160
I tried that too. Best use what >>8727001 said. Didn't seem intuitive at first, but makes sense now

>>8727192
>Didn't seem intuitive at first, but makes sense now

That's because your initial drawing is plain wrong. Whoever gave this to you was trolling you hard.

>>8727235
fucking TAs my dude. God damn it

>>8727254
Fuck no.
That guy >>8727001 is wrong. I guess the drawing is right.

File: Sans titre.png (73KB, 800x600px) Image search: [Google]

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>>8727254

>>8726965
the shaded area is a square of radius 1

>>8727291
Ok it's done. It was fucking easy in retrospect. The answer (for a square of side 2) is 8/3

Have you guys done any measure theory ?


[math] \displaystyle \int_0^1 \int_0^1 1_{y \leq \frac{1-x^2}{2}} dx dy [/math]

>>8727320
whoops went full faggot. I integrated over a full quadrant, not the half of one.
The result should be something like

Integrating colored part in the top-rtight part :
[math] \displaystyle \int_0^1 \int_0^1 1_{y \leq \frac{1-x^2}{2}} 1_{y \geq x} dx dy = \displaystyle \int_0^1 (\frac{1-x^2}{2} -x) 1_{\frac{1-x^2}{2} \geq x} dx [/math]

The final answer would be [math]\frac{4}{3}(4\sqrt{2} -5) \simeq 0.9[/math] which seems more reasonable.