Wikipedia:Reference desk/Archives/Mathematics/2021 March 18

= March 18 =

Circle measure problem.
C is at the center of a circle (call it one unit in radius). A and B are on the circle edge. The points in the circle that are less than one unit from either A or B (or both) make up exactly half of the Circle, what is the measure of the angle ACB?

(The amount of the Circle that is less than one unit from A is slightly more than 1/3, so if A & B are at the same point, the amount would be slightly more than 1/3 and if A & B are opposite, it would be slightly more than 2/3. As B moves around the circle from being equal to A to being opposite from A the amount increases monotonically)Naraht (talk) 21:14, 18 March 2021 (UTC)


 * Another quilting problem?
 * I must be missing something, but as long as ∠ACB ≤ 2π/3, isn't your area easily divided into a sector and two segments, all with easy to calculate areas, with that of the segments fixed and that of the sector a function of your angle? -- ToE 01:59, 19 March 2021 (UTC)


 * Elementary, but not simple. Given a point $$P$$ in the plane, let $$\Theta_P$$ stand for the unit-radius disk with centre $$P$$, and $$\Omicron_P$$ for the corresponding unit circle, the boundary of the disk. Furthermore, for a measurable planar shape $$S$$, let $$|S|$$ denote its measure (area). We have three points $$A, B$$ and $$C$$ for which $$A,B\in\Omicron_C$$ (which is equivalent with $$C\in\Omicron_A\cap\Omicron_B$$). The task is, given $$|(\Theta_A\cup\Theta_B)\cap\Theta_C|=u$$, to determine the angle $$\alpha=\angle ACB$$. More specifically, we are given that $$u$$ has to equal the quantity $$\tfrac{1}{2}|\Theta_C|=\tfrac{1}{2}\pi$$.
 * In general, $$|(S\cup T)\cap U|=|S\cap U|+|T\cap U|-|S\cap T\cap U|$$, so
 * $$u=|\Theta_A\cap\Theta_C|+|\Theta_B\cap\Theta_C|-|\Theta_A\cap\Theta_B\cap\Theta_C|$$.
 * Moreover, for reasons of symmetry,
 * $$|\Theta_A\cap\Theta_C|=|\Theta_B\cap\Theta_C|$$,
 * for which the formula for the area of a circular segment (really, a disk segment) gives us
 * $$v=|\Theta_A\cap\Theta_C|= \frac{2\pi}{3}-\frac{1}{2}\sqrt{3}.$$
 * Note that this quantity is independent of the placements of $$A$$ and $$B$$. If we have an expression for $$w=|\Theta_A\cap\Theta_B\cap\Theta_C|$$ as a function of $$\alpha$$, so that $$u=2v-w$$, we can solve the equation $$u=\tfrac{1}{2}\pi$$ for $$\alpha$$. For the special case that $$\alpha=0$$, so $$A$$ and $$B$$ coincide, we find $$u=w=v\approx 1.22837 < \tfrac{1}{2}\pi$$, so these points are distinct. For the special case that $$\alpha=\pi$$, so $$A$$ and $$B$$ are diametrically opposite, $$w=0$$, so $$u=2v\approx 2.45674 > \tfrac{1}{2}\pi$$, so we may assume that $$\alpha<\pi$$. Without loss of generally, we put $$C$$ at the origin of the Cartesian plane, and $$A$$ and $$B$$ symmetrically with respect to the x-axis, so $$A=(c,s),B=(c,{-}s)$$ for values $$c$$ and $$s$$ such that $$c^2+s^2=1$$. Assume, still without loss of generality, that $$c>0$$ and $$s>0$$, so $$A$$ is above $$B$$, in the first quadrant. Putting $$\varphi=\tfrac{1}{2}\alpha$$, we have, of course, that $$c=\cos\varphi,s=\sin\varphi$$. $$\Omicron_C$$ intersects the x-axis in $$M=(1,0)$$, and the line $$\overline{CM}$$ is perpendicular to the line $$\overline{AB}$$. $$\Omicron_C$$ intersects $$\Omicron_A$$ in two points; the lower one is $$U=(\tfrac{1}{2}c+\tfrac{1}{2}\sqrt{3}\,s,-\tfrac{1}{2}\sqrt{3}\,c+\tfrac{1}{2}s)$$. The upper point of intersection of $$\Omicron_C$$ and $$\Omicron_B$$ is the mirror point $$V=(\tfrac{1}{2}c+\tfrac{1}{2}\sqrt{3}\,s,\tfrac{1}{2}\sqrt{3}\,c-\tfrac{1}{2}s)$$. $$U$$ and $$V$$ coincide when $$\varphi=\tfrac{1}{3}\pi,c=\tfrac{1}{2},s=\tfrac{1}{2}\sqrt{3}$$, and then $$U=V=M$$. For this case, using again the disk segment area formula, we find $$w=\tfrac{1}{3}-\tfrac{1}{2}\sqrt{3}$$, so $$u\approx 2.27557 >\tfrac{1}{2}\pi$$, from which we may conclude that $$\varphi<\tfrac{1}{3}\pi$$. $$\Omicron_A$$ and $$\Omicron_B$$ intersect in $$C$$; their second point of intersection is at $$(2c,0)$$, which is outside $$\Theta_C$$ if $$0<\varphi<\tfrac{1}{3}\pi$$. The x-axis divides $$\Theta_A\cap\Theta_B\cap\Theta_C$$ into two equal parts, so it suffices to find the area of the upper part, which is bounded by the line segment $$\overline{CM}$$ and the circular arcs from $$M$$ to $$V$$ along $$\Omicron_C$$ and from $$V$$ back to $$C$$ along $$\Omicron_B$$. The triangle area $$|\triangle CMV|=\tfrac{1}{2}(\tfrac{1}{2}\sqrt{3}\,c-\tfrac{1}{2}s)$$. To this we need to add the areas of the disk segments for the arcs from $$M$$ to $$V$$ and from $$V$$ to $$C$$. Angle $$\angle MCV = \angle BCV-\angle BCM = \tfrac{1}{3}\pi{-}\varphi$$, so for the first of these two disk segments we find area $$\tfrac{1}{2}(\tfrac{1}{3}\pi{-}\varphi-\sin(\tfrac{1}{3}\pi{-}\varphi))$$. Triangle $$\triangle VBC$$ is equilateral, so $$\angle VBC=\tfrac{1}{3}\pi$$, and the area of the remaining disk segment is $$\tfrac{1}{2}(\tfrac{1}{3}\pi-\tfrac{1}{2}\sqrt{3})$$. Combining all, we have
 * $$u=2v-\left((\tfrac{1}{2}\sqrt{3}\,c-\tfrac{1}{2}s)+(\tfrac{1}{3}\pi{-}\varphi-\sin(\tfrac{1}{3}\pi{-}\varphi))+(\tfrac{1}{3}\pi-\tfrac{1}{2}\sqrt{3})\right)$$,
 * where $$c=\cos\varphi,s=\sin\varphi,\varphi=\tfrac{1}{2}\alpha$$. The transcendental equation $$u=\tfrac{1}{2}\pi$$ can only be solved numerically, so there is no point in attempting to massage this into a more pleasing form. The numeric solution is $$\alpha\approx 0.68485$$, putting $$A$$ at approximately $$(0.94194,0.33577)$$. --Lambiam 11:40, 19 March 2021 (UTC)
 * Wow! That's a lot of work to determine α = √3 - π/3 ≈ 0.68485 or 39.239°.  It makes me wonder if I've oversimplified things, but since our answers match, here goes.  I'd already typed this with θ = ∠ACB, so I'll keep my notation.
 * If θ ≤ 2π/3, then the your area can be easily divided into a sector (of circle C) and two segments (of Circles A & B).
 * The sector is of a unit circle with central angle 2π/3 + θ, and thus has area of (12/2)(2π/3 + θ) = π/3 + θ/2.
 * The segments are of a unit circle with central angle π/3, and thus each have an area of (12/2)(π/3 - sin(π/3)) = (π/3 - √3/2)/2. (Each segment is ~ 0.029 · π or 2.9% the area of your circle.)
 * Summing the three areas yields A(θ) = 2π/3 + θ/2 - √3/2, for 0 ≤ θ ≤ 2π/3.
 * Sanity checks:
 * A(0) = 2π/3 - √3/2 ≈ 0.39 · π, or "slightly more than 1/3" the area of your circle.
 * Also: A(2π/3) = π - √3/2 ≈ 0.72 · π, or more than 1/2 your circle, which is good as the area becomes more complicate for larger angles where we have to deal with the intersection of segments.
 * Note that the area becomes simple again at θ = π, with A(π) = 2A(0) = 4π/3 - √3 ≈ 0.78 · π.
 * Also: A(π) - A(2π/3) = π/3 - √3/2 should be the area of two segments, which it is.
 * Solving for A = π/2 yields θ = (√3 - π/3) radians or ≈ 39.2°
 * -- ToE 12:45, 19 March 2021 (UTC)
 * Yeah, that is a lot simpler. I guess I did not fully understand your approach. It requires θ not to exceed 2π/3; otherwise the decomposition into a sector and two segments is not possible. --Lambiam 13:45, 19 March 2021 (UTC)
 * FWIW I concur. If θ is between 2π/3 and π, then the formula for the area covered is different, but I didn't think it was worth working out because it's clear that it increasing as a function of θ (since the overlap decreases) and the value is already greater than π/2 when θ is 2π/3. According to my calculations the area covered ranges from .391π at θ=0 and .782π at θ=π. If the question was to find θ so that the area covered is 3/4 π then you'd need (I assume) the other formula. --RDBury (talk) 14:30, 19 March 2021 (UTC)
 * FWIW I concur. If θ is between 2π/3 and π, then the formula for the area covered is different, but I didn't think it was worth working out because it's clear that it increasing as a function of θ (since the overlap decreases) and the value is already greater than π/2 when θ is 2π/3. According to my calculations the area covered ranges from .391π at θ=0 and .782π at θ=π. If the question was to find θ so that the area covered is 3/4 π then you'd need (I assume) the other formula. --RDBury (talk) 14:30, 19 March 2021 (UTC)