# Separating a disk right into xx_math_0 equivalent items with xx_math_1 line sectors

Can you separate a disk right into xx_math_2 items of equivalent location, with xx_math_3 line sectors? (You can undoubtedly separate it right into xx_math_4 items, yet could those have equivalent locations?)

(This inquiry was left unanswered at an additional discussion forum. I can see with some aesthetic debates that the solution need to be no, yet I could not locate a wonderful means to write it down.)

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2019-11-24 13:27:04
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Answers: 4

(One even more effort at a straightforward and also full remedy.)

Each chord has to separate the circle right into the proportion 4 :3, so each chord has to represent an arc of action about 2.91624. Take into consideration 2 such chords, one in a dealt with area, and also the various other relocating via all feasible arrangements in which it converges the first chord. For simpleness, allow the location of the circle be 7, to make sure that the dealt with chord separates the circle right into areas with location 3 and also 4.

(Note that if the chord were to proceed revolving counterclockwise, what is presently the "base" of the chord would certainly converge the dealt with chord, yet any kind of such arrangement is the representation photo over the vertical bisector of the dealt with chord of an arrangement in the computer animation revealed.)

As the chord revolves counterclockwise, the factor of junction of both chords and also the endpoint of the relocating chord on the upper fifty percent of the circle both action only leftward, so the location of heaven area is monotonically raising and also the location of the red area is monotonically lowering as the chord revolves counterclockwise. The locations of the tinted areas range 0 and also 3, and also are continual features of the turning of the chord. There are 2 placements of the revolving chord for which among both tinted areas has location 1:

In each arrangement, we can establish the locations of the areas by proportion (exchange the relocating chord and also the dealt with chord). In the left layout, the locations are (counterclockwise, beginning with heaven area) 1 :2 :1 :3. In the appropriate layout, the locations are 2 :1 :2 :2. Given that the last chord has to separate 3 of the existing 4 areas in order to have 7 areas, the arrangement with locations 1 :2 :1 :3 (over left) will not function, given that separating either area with location 1 will not offer us 7 areas with location 1.

Currently, collaborating with the arrangement with locations 2 :1 :2 :2 (over right), add a 3rd chord, revolving via all feasible placements that separate the lower appropriate area.

As in the past, the location of the purple area is monotone and also continual as the chord revolves and also is in between 0 and also 2, so there is an one-of-a-kind arrangement where the location of the purple area is 1.

Considering the placement of the purple chord about the initial set chord, the tiny wedge area at the right can not have location 1, and also this can be validated computationally. (It is additionally the instance that the location of the triangular in the facility can not have are 1 in this arrangement.) So, it is not feasible to separate the circle right into 7 areas of equivalent location making use of 3 chords.

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2019-05-09 10:52:45
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Each line sector is a chord of the circle, with 3 of the 7 areas in the sector of the circle established by the chord (the sector is the area in between the chord and also the small arc in between its endpoints). The location of a sector with arc action $\theta$ radians of a circle with distance r is $\frac{r^2}{2}(\theta-\sin\theta)$.

If each of the 7 areas has equivalent location, after that the location of the sector established by each chord has to be $\frac{3}{7}$ of the location of the circle, so for each and every chord, $\frac{r^2}{2}(\theta-\sin\theta)=\frac{3\pi r^2}{7}\Rightarrow\theta\approx 2.91624$. Moving forward, I'll make use of $\omega$ to describe this value.

Currently allow's collaborate with the device circle focused at the beginning and also area one chord to make sure that its small arc would certainly be purged by the turning photos of (1,0) concerning the beginning by angles in the interval $[0,\omega]$. Allow the various other 2 chords be positioned such that their small arcs would certainly be purged by the turning photos of (1,0) concerning the beginning by angles in the periods $[\alpha,\alpha+\omega]$ and also $[\beta,\beta+\omega]$ where $0<\alpha<\beta<2\pi$.

In any kind of setup of chords that generates 7 areas, there will certainly be a main triangular area. The lines having the chords have formulas $y=\frac{\sin\omega}{\cos\omega-1}(x-1)$, $y-\sin\alpha=\frac{\sin\alpha-\sin(\alpha+\omega)}{\cos\alpha-\cos(\alpha+\omega)}(x-\cos\alpha)$, and also $y-\sin\beta=\frac{\sin\beta-\sin(\beta+\omega)}{\cos\beta-\cos(\beta+\omega)}(x-\cos\beta)$. The vertices of the main triangular area go to the pairwise factors of junction of those 3 chords : $\ left (\ cos \ frac \ omega 2 \ cos \ frac \ alpha+\ omega 2 \ sec \ frac \ alpha 2 ,. \ cos \ frac \ omega 2 \ wrong \ frac \ alpha+\ omega 2 \ sec \ frac \ alpha 2 \ right)$, $\ left (\ cos \ frac \ omega 2 \ cos \ frac \ beta+\ omega 2 \ sec \ frac \ beta 2 ,. \ cos \ frac \ omega 2 \ wrong \ frac \ beta+\ omega 2 \ sec \ frac \ beta 2 \ right)$, and also $\left(\cos\frac{\omega}{2}\cos\frac{\alpha+\beta+\omega}{2}\sec\frac{\alpha-\beta}{2},\frac{\csc(\alpha-\beta)}{2}(-\cos\alpha+\cos\beta-\cos(\alpha+\omega)+\cos(\alpha-\beta))\right)$. The location of this triangular is $\frac{1}{2}\left|\left(\cos\left(\alpha-\frac{\beta}{2}\right)-\cos\frac{\beta}{2}\right)\cos^2\frac{\omega}{2}\sec\frac{\alpha}{2}\sec\frac{\alpha-\beta}{2}\tan\frac{\beta}{2}\right|$ (limited to just values of $\alpha$ and also $\beta$ for which the 3 factors of junction are within the circle).

A shape story of the location is revealed listed below (white locations do not please the restraints defined over ; darker shades stand for smaller sized values of the location).

The best feasible location for the triangular is about 0.226346 when $\alpha\approx 2.91624$ and also $\beta\approx 3.36694$ (mathematical estimate from Mathematica), which is well except one 7th the location of the circle, $\frac{\pi}{7}\approx 0.448799$. Consequently, it is not feasible to separate the circle right into 7 areas of equivalent location making use of 3 chords.

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2019-05-08 19:40:37
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Let's concentrate on 2 of the chords. These need to divide the circle in the proportion $4:3$ so the small arc subtended by each is equivalent. Call one $AB$ and also respect it as "dealt with" and also the various other $CD$ which we assume of as relocating. As I can not attract layouts, consider Isaac's left layout. Allow both chords start in the placement offered by Isaac's 2 chords conference on the area. Allow $AB$ be the one left wing and also $CD$ be the one on the right with $B=C$.

Currently revolve $CD$ clockwise. Allow $P$ be the factor of junction. After that the angle $APD$ rises, the factor $P$ relocates monotonely from $B$ in the direction of $A$ on the line $AB$, and also $C$ actions monotonely from $B$ in the direction of $A$ on the small arc. Hence the location of the "field" (bounded by 2 lines and also an arc of the circle) $PBC$ continuously raises, given that at a later on time it will certainly have the field at an earlier time. There is an one-of-a-kind time at which the "field" has location $1/7$ of the circle. Allow $\theta_0$ be the angle $APD$ right now.

In the looked for arrangement, all 3 angles of the main triangular have location $\pi/3$ therefore it exists if and also just if $\theta_0=\pi/3$. So determining whether the arrangement exists lowers to a solitary estimation : what is the location of "field" $PBC$ when angle $APD$ amounts to $\pi/3$ ; this is an estimation that I'm not mosting likely to do :-)

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2019-05-08 02:42:42
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Each line sector is a chord of the circle, with 3 of the 7 areas in the sector of the circle established by the chord (the sector is the area in between the chord and also the small arc in between its endpoints). The location of a sector with arc action α radians of a circle with distance r is $\frac{r^2}{2}(\alpha-\sin\alpha)$-- for simpleness, allow $r=\sqrt{2}$ so the sector location is $\alpha-\sin\alpha$ and also the location of the circle is $2\pi$.

If each of the 7 areas has equivalent location, after that the location of the sector established by each chord has to be $\frac{3}{7}$ of the location of the circle, so for each and every chord, $\alpha-\sin\alpha=\frac{6\pi}{7}\Rightarrow\alpha\approx2.91624$. With 3 such chords, there is no other way to get the main triangular area to have location greater than concerning $\frac{1}{10}$ of the circle, so no other way to get 7 areas of equivalent location.

Listed below left is nearly the biggest the main triangular area can be ; below appropriate is an extra symmetrical arrangement, which in fact does not make the main triangular a lot smaller sized.

modify : When I claimed "empirically" in a comment listed below, that was a little bit inaccurate : offered an angle of action 2.91624 radians, I created (compass and also straightedge) a circle with one chord fulfilling the offered standards. I after that created a couple of loads feasible areas for the 2nd chord and also for each and every of those, a couple of hundred areas for the 3rd chord, gauging the location of the main triangular, when it existed. There are 4 distinctive arrangements where it exists, representing the 4 areas in the shape story in my various other solution. The location function is continual on these areas, so a well-distributed huge example of values throughout each of the areas offers an exact representation of the actions of the function. When a triangular has some facets free and also some facets constricted, the maximum location is commonly at a severe value of the free facets or at an arrangement of topmost proportion. The plethora of building and constructions showed neighborhood extrema for the arrangement over left (narrowest yet highest) and also for an arrangement where the side of the triangular on the various other chord is virtually the entire chord (fastest yet best). The arrangement revealed over left had the biggest location (this is validated computationally in my various other solution). Neither of these severe arrangements, nonetheless, are anywhere near 7 equal-area areas-- in each instance, at the very least one area is lowered to the degenerate instance of a factor. Over right is one of the most balanced arrangement, which is with ease one of the most likely to create 7 equal-area areas. This represents the center of the lighter-colored area in the center of the shape story in my various other solution and also goes to or near a neighborhood minimum of the location function.

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2019-05-08 02:04:45
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