Find the overlapping area -- not easy!

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[douglashopkins]

I did my angle with arccosine rather than arctangent. That way I could work with the ratio s/r directly. Just had to subtract 2 of these angles from pi/2. The values of s/r that came out nice were (squrt3)/2, pi / 3 and 1/ (squrt2).

[Qermaq]

I think I would have started with these three cases: (1) r <= s, thus the overlapping area is (pi)r^2, (2) r >= sqrt(2)s/2, thus the overlapping area is 4s^2, and (3) sqrt(2)s/2 < r < s, where this video starts.

[TedHopp]

Michael, I've noticed that you lean very heavily on coordinate geometry when solving most geometry problems. It's very powerful, of course, but often leads to overly complicated solutions. You like that hammer, but not every problem is a nail.

Here, for instance, there's an obvious eight-fold symmetry and nice right triangles. The given values just cry out for using the arccosine (or arcsecant) to determine angles, and Pythagoras taught us how to get the height of the triangle from the hypotenuse and one leg.

[stewartcopeland]

the second answer's term can be replaced by: r^2*(pi - 4*arccos(s/r))
HW: r = (2^0.5)*s --> A = area of the square / r = s --> A = circle area

[jimschneider]

@7:30 - you may want to point out that your formula for theta is only defined for s < r. You can extend this to s = r if you take limits, but if s > r, you will get a complex number as the answer.

[MCPhssthpok]

If r <= s then the circle lies completely within the square so the overlap is just the area of the circle and if r >= s*sqrt(2) then the square lies entirely within the circle so the overlap is just the area of the square. Your formula only applies for s < r < s*sqrt(2).

[abramsaustin]

I think the full answer needs to be a piecewise function of r and s. If s >= r, we have the "overlapping area" as the full circle, but some of this expression involves complex numbers. If s < √r, the "overlapping area" is the full square, but there's a negative term in the arctan. In √r <= s < r, this expression works out.

[byronwatkins]

The symmetry about x=y demands that the triangles have equal area. Also, the geometry requires r/sqrt(2)<=s<=r. At 9:07, you seem to be missing s in the denominator of arctan's argument. (LCD for subtraction is s sqrt(r^2-s^2))

[SlidellRobotics]

In addition to @Qermac's division into r<s, s≤r≤s√2, s√2<r, I would have noticed the symmetry about the lines x=y and x=-y, and only had to add two areas and multiply by 8.
Later on, you can calculate θ or θ/2 using vector products, which seems a lot easier than the video.

[polarisator]

one can get a really nice solution with 2 simple identities.
1) Let a and b be two non-zero vectors of the same dimension and theta be the angle between them.
Then: a dot b = b dot a = abs(a)*abs(b)*cos(theta). If a = (s;sqrt(r^2-s^2)) and b (sqrt(r^2-s^2);s), then cos(theta) = 2*s*sqrt(r^2-s^2)/r^2 = 2*s/r*sqrt(1-(s/r)^2)
A_1 = r^2/2*theta
2) If you combine the two right triangles, you get an isosceles triangle, with side lengths r and angle pi/2-theta.
A_2 =1/2*r^2*sin(pi/2-theta) = r^2/2 * cos(theta)

Combine to: A = 4*(A_1+A_1) = 2*r^2*(theta+cos(theta)) = 2*r^2*( 2*s/r*sqrt(1-(s/r)^2)+arccos( 2*s/r*sqrt(1-(s/r)^2)))

[trueriver]

When s > r overlap = pi.r^2.

When s.sqrt2 < r it's s^2.

Do I get 66% for calculating two out of the three cases, please?

[hacatu]

Using two inverse trig functions, especially arctan, is bizarre and makes the result much more complicated than necessary. Instead, we can find the angle from the +x axis to the hypotenuse of one of the right triangles using arccos, which just involves s/r, subtract this angle from pi/4, and multiply the result by 8 to get the total angle of all the "corner" sectors.

[MythOfParadise]

I feel like there were a few ways to make this quicker. First, he could have simply used the Pythagorean theorem to solve for the unknown x and y values instead of seeing where the circle and square intercept. Second an easier way to find theta would be to recognize that the two angles surrounding it are the arctan(sqrt(r^2-s^2)/s) and then double it and then subtract that from pi/2 to get theta. It seems just as complicated but it doesn’t need that fairly unknown trig identity and it proves itself. Overall still enjoyed the video. Obviously this is just some criticism from a rando online, but I think it helps.

[ddognine]

Assuming r is constant so that the intersecting area is A(s), nice values for r and s are for inscribed and circumscribed squares which result in A(s) of 4s^2 and pi*r^2, respectively, which also define the upper and lower bound of A(s) for all s >= 0 with the caveat that that domain for A(s) is strictly for r/sqrt(2) <= s <= r.

[CM_France]

Hi,

7:44 : I stopped the video and said "will he use the formula :

8:32 : I answered to myself : "ok, great",

For fun:

2:30, 3:22, 5:56 : "great",

7:44 : "ok, great",

8:19 : "ok, good".

[bobzarnke]

It's simpler just to observe that the 2 (pink-blue-yellow) triangles are right-angled triangles with hypotenuse r and side s; so their areas are s√(r²-s²)/2.
Also, since the triangles are congruent, α+β = π/2;
so θ = π/2-2β and cos(β) = s/r and the desired area is 4s√(r²-s²) + πr² - 4r²arccos(s/r). (A simpler
justification (after the fact, of course) for the latter two terms is to note that the circle comprises the 4 sectors
with angle θ + 8 sectors with angle β.)

[BSnicks]

It is much simpler by not using tangent, but using cosine to find beta. Divide the whole thing into 8 sections. Then angle half-theta=pi/4-beta. Beta=arccos(s/r). Using symetry, you can calculate the area of the triangle and the area of the half-theta circular area. Then multiply by 8.

[saroshadenwalla]

Shouldn't there be an s in the denominator of the expression in the arctan outside of the square root

[saxbend]

I used calculus to find the area of the parts of the circle that are outside of the square. I integrated rsin(theta) d theta between 0 and pi(r-s)/2r (scaling the range such that sin(r) would be equivalent to sin pi/2 and defining x=0 as the minimum x value on the circle and y=0 such that the circle crosses the x axis at x=0). The other area outside of the square is the same thanks to symmetry. Then I subtracted both from the area of a quarter of the circle and multiplied the whole thing by four and got pi(r)^2 + 8r(cos((pi(r-s)/2r) -1) .

[Hyakurin_]

One could also use the formula arctan(x)+arctan(1/x)=π/2 (for x>0) to simplify the alpha-beta expression