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Can you divide a disk into 7 pieces of equal area, with 3 line segments? (You can surely divide it into 7 pieces, but could those have equal areas?)

(This question was left unanswered at another forum. I can see with some visual arguments that the answer should be no, but I couldn't find a nice way to write it down.)

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4 Answers 4

up vote 4 down vote accepted

Let's focus on two of the chords. These have to split the circle in the ratio $4:3$ so the minor arc subtended by each is equal. Call one $AB$ and regard it as "fixed" and the other $CD$ which we think of as moving. As I can't draw diagrams, look at Isaac's left diagram. Let the two chords start in the position given by Isaac's two chords meeting on the circumference. Let $AB$ be the one on the left and $CD$ be the one on the right with $B=C$.

Now rotate $CD$ clockwise. Let $P$ be the point of intersection. Then the angle $APD$ increases, the point $P$ moves monotonely from $B$ towards $A$ on the line $AB$, and $C$ moves monotonely from $B$ towards $A$ on the minor arc. Thus the area of the "sector" (bounded by two lines and an arc of the circle) $PBC$ steadily increases, since at a later time it will contain the sector at an earlier time. There is a unique time at which the "sector" has area $1/7$ of the circle. Let $\theta_0$ be the angle $APD$ at this time.

In the sought configuration, all three angles of the central triangle have area $\pi/3$ and so it exists if and only if $\theta_0=\pi/3$. So deciding whether the configuration exists reduces to a single calculation: what is the area of "sector" $PBC$ when angle $APD$ equals $\pi/3$; this is a calculation that I'm not going to do :-)

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$\theta_0\approx 1.31$. The problem that I have with this argument is that I can't quite justify that the sought configuration must have an equilateral triangle at the center. If I could justify that, I'd have just computed the area for that configuration. What is the argument that justifies that the central triangle is equilateral? –  Isaac Aug 7 '10 at 17:57
    
Reflect with axis the perpendicular bisector of one of the chords. One must get back the same configuration: thus the centre of the circle lies on the perpendicular bisector of the side of the central triangle. –  Robin Chapman Aug 7 '10 at 18:06
    
If this diagram is useful, feel free to add it: imgftw.net/img/929412773.png What I can't quite convince myself of is that the configuration has to be symmetric, which I think is the justification that reflection over the perpendicular bisector of one chord should give the same configuration. –  Isaac Aug 7 '10 at 18:17
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The angle at which two chords must meet is uniquely determined by the monotonicity Robin mentioned. Since each pair of chords meet at that angle, the figure is necessarily symmetric. (For completeness wrt monotonicity: If the chords make angle $\theta$, the distance between the point of intersection and the midpoint of a chord is $d \cot(\theta/2)$, where $d$ measures chord midpoint to circle center. Thus, the point of intersection moves monotonically along the "fixed" chord's endpoint to its midpoint as $\theta$ increases; the overlapped circular segment area increases correspondingly.) –  Blue Aug 7 '10 at 18:42
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An even more simple-minded argument: if one has various chords in the circle having equal length, the centre of the circle is equidistant from them. Therefore if one has three chords of equal length, making a triangle inside the circle, then the centre of the circle is the incentre of that triangle –  Robin Chapman Aug 7 '10 at 18:55

Each line segment is a chord of the circle, with 3 of the 7 regions in the segment of the circle determined by the chord (the segment is the region between the chord and the minor arc between its endpoints). The area of a segment with arc measure α radians of a circle with radius r is $\frac{r^2}{2}(\alpha-\sin\alpha)$—for simplicity, let $r=\sqrt{2}$ so the segment area is $\alpha-\sin\alpha$ and the area of the circle is $2\pi$.

If each of the 7 regions has equal area, then the area of the segment determined by each chord must be $\frac{3}{7}$ of the area of the circle, so for each chord, $\alpha-\sin\alpha=\frac{6\pi}{7}\Rightarrow\alpha\approx2.91624$. With 3 such chords, there is no way to get the central triangular region to have area more than about $\frac{1}{10}$ of the circle, so no way to get 7 regions of equal area.

Below left is just about the largest the central triangular region can be; below right is a more symmetric configuration, which actually doesn't make the central triangle much smaller.

largest central triangle somewhat symmetric configuration

edit: When I said "empirically" in a comment below, that was a bit imprecise: given an angle of measure 2.91624 radians, I constructed (compass and straightedge) a circle with one chord meeting the given criteria. I then constructed a few dozen possible locations for the second chord and for each of those, a few hundred locations for the third chord, measuring the area of the central triangle, when it existed. There are 4 distinct configurations where it exists, corresponding to the 4 regions in the contour plot in my other answer. The area function is continuous on these regions, so a well-distributed large sample of values across each of the regions gives an accurate depiction of the behavior of the function. When a triangle has some aspects free and some aspects constrained, the maximum area is typically at an extreme value of the free aspects or at a configuration of maximal symmetry. The multitude of constructions indicated local extrema for the configuration above left (narrowest but tallest) and for a configuration where the side of the triangle on the other chord is nearly the whole chord (shortest but widest). The configuration shown above left had the largest area (this is verified computationally in my other answer). Neither of these extreme configurations, however, are anywhere near 7 equal-area regions—in each case, at least one region is reduced to the degenerate case of a point. Above right is the most symmetrical configuration, which is intuitively the most likely to generate 7 equal-area regions. This corresponds to the middle of the lighter-colored region in the middle of the contour plot in my other answer and is at or near a local minimum of the area function.

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Can you clarify why "With 3 such chords, there is no way to get the central triangular region to have area more than about 1/10 of the circle"? –  Tomer Vromen Aug 6 '10 at 23:15
    
@Tom: Purely empirically--the two diagrams in my answer are from dynamic geometry software and the setup was a circle with three chords of the type described (or very nearly so) that I could freely rotate. The configuration in the left diagram (approximately) yielded the largest area for the triangular region and it was in the vicinity of 1/10 the area of the circle. For a more technical argument, I'd look at what happens to the 4/7 side of the first chord when the second and third chords must divide the 3/7 side into 3 equal-area regions (there is probably only one such configuration). –  Isaac Aug 6 '10 at 23:35
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I don't see how this answer says more that the original poster "I can see with some visual arguments that the answer should be no". –  Andrea Ferretti Aug 7 '10 at 0:10
    
@Andrea, indeed. -1 –  Heath Hunnicutt Aug 7 '10 at 1:58
    
@Isaac: When your solution is only empirical, it should be clearly stated –  Casebash Aug 7 '10 at 9:35

Each line segment is a chord of the circle, with 3 of the 7 regions in the segment of the circle determined by the chord (the segment is the region between the chord and the minor arc between its endpoints). The area of a segment with arc measure $\theta$ radians of a circle with radius r is $\frac{r^2}{2}(\theta-\sin\theta)$.

If each of the 7 regions has equal area, then the area of the segment determined by each chord must be $\frac{3}{7}$ of the area of the circle, so for each chord, $\frac{r^2}{2}(\theta-\sin\theta)=\frac{3\pi r^2}{7}\Rightarrow\theta\approx 2.91624$. Going forward, I'll use $\omega$ to refer to this value.

Now let's work with the unit circle centered at the origin and place one chord so that its minor arc would be swept out by the rotation images of (1,0) about the origin by angles in the interval $[0,\omega]$. Let the other two chords be placed such that their minor arcs would be swept out by the rotation images of (1,0) about the origin by angles in the intervals $[\alpha,\alpha+\omega]$ and $[\beta,\beta+\omega]$ where $0<\alpha<\beta<2\pi$.

In any arrangement of chords that yields 7 regions, there will be a central triangular region. The lines containing the chords have equations $y=\frac{\sin\omega}{\cos\omega-1}(x-1)$, $y-\sin\alpha=\frac{\sin\alpha-\sin(\alpha+\omega)}{\cos\alpha-\cos(\alpha+\omega)}(x-\cos\alpha)$, and $y-\sin\beta=\frac{\sin\beta-\sin(\beta+\omega)}{\cos\beta-\cos(\beta+\omega)}(x-\cos\beta)$. The vertices of the central triangular region are at the pairwise points of intersection of those three chords: $\left(\cos\frac{\omega}{2}\cos\frac{\alpha+\omega}{2}\sec\frac{\alpha}{2}, \cos\frac{\omega}{2}\sin\frac{\alpha+\omega}{2}\sec\frac{\alpha}{2}\right)$, $\left(\cos\frac{\omega}{2}\cos\frac{\beta+\omega}{2}\sec\frac{\beta}{2}, \cos\frac{\omega}{2}\sin\frac{\beta+\omega}{2}\sec\frac{\beta}{2}\right)$, and $\left(\cos\frac{\omega}{2}\cos\frac{\alpha+\beta+\omega}{2}\sec\frac{\alpha-\beta}{2},\frac{\csc(\alpha-\beta)}{2}(-\cos\alpha+\cos\beta-\cos(\alpha+\omega)+\cos(\alpha-\beta))\right)$. The area of this triangle is $\frac{1}{2}\left|\left(\cos\left(\alpha-\frac{\beta}{2}\right)-\cos\frac{\beta}{2}\right)\cos^2\frac{\omega}{2}\sec\frac{\alpha}{2}\sec\frac{\alpha-\beta}{2}\tan\frac{\beta}{2}\right|$ (restricted to only values of $\alpha$ and $\beta$ for which the three points of intersection are within the circle).

A contour plot of the area is shown below (white areas do not satisfy the constraints described above; darker colors represent smaller values of the area).

contour plot

The greatest possible area for the triangle is approximately 0.226346 when $\alpha\approx 2.91624$ and $\beta\approx 3.36694$ (numerical approximation from Mathematica), which is well short of one seventh the area of the circle, $\frac{\pi}{7}\approx 0.448799$. Therefore, it is not possible to divide the circle into 7 regions of equal area using 3 chords.

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(One more attempt at a simple and complete solution.)

Each chord must divide the circle into the ratio 4:3, so each chord must correspond to an arc of measure approximately 2.91624. Consider two such chords, one in a fixed location, and the other moving through all possible configurations in which it intersects the first chord. For simplicity, let the area of the circle be 7, so that the fixed chord divides the circle into regions with area 3 and 4.

animation of moving chord

(Note that if the chord were to continue rotating counterclockwise, what is currently the "bottom" of the chord would intersect the fixed chord, but any such configuration is the reflection image over the perpendicular bisector of the fixed chord of a configuration in the animation shown.)

As the chord rotates counterclockwise, the point of intersection of the two chords and the endpoint of the moving chord on the upper half of the circle both move solely leftward, so the area of the blue region is monotonically increasing and the area of the red region is monotonically decreasing as the chord rotates counterclockwise. The areas of the colored regions vary between 0 and 3, and are continuous functions of the rotation of the chord. There are 2 positions of the rotating chord for which one of the two colored regions has area 1:

1:2:1:3 configuration 2:1:2:2 configuration

In each configuration, we can determine the areas of the regions by symmetry (exchange the moving chord and the fixed chord). In the left diagram, the areas are (counterclockwise, starting with the blue region) 1:2:1:3. In the right diagram, the areas are 2:1:2:2. Since the final chord must divide 3 of the existing 4 regions in order to have 7 regions, the configuration with areas 1:2:1:3 (above left) won't work, since dividing either region with area 1 won't give us 7 regions with area 1.

Now, working with the configuration with areas 2:1:2:2 (above right), add a third chord, rotating through all possible positions that divide the lower right region.

adding the third chord

As before, the area of the purple region is monotone and continuous as the chord rotates and is between 0 and 2, so there is a unique configuration where the area of the purple region is 1.

purple region has area 1

Considering the position of the purple chord relative to the original fixed chord, the small wedge region at the right cannot have area 1, and this can be verified computationally. (It is also the case that the area of the triangle in the center cannot have are 1 in this configuration.) So, it is not possible to divide the circle into 7 regions of equal area using 3 chords.

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In any final three-chord, seven-region drawing, the chords determine three identical circular sectors of area 3 (assuming circle area 7). The OVERLAP of any two of these sectors must have area 1: that overlap is one of the 7 regions! Your left-hand red-blue image shows an area-2 overlap, so we can dismiss this case right away. From there, the uniqueness of the angle of overlap, plus Robin's symmetry argument, get us to the impossibility conclusion about as cleanly as we might hope for. –  Blue Aug 10 '10 at 2:35
    
Whoops: "circular segments" and "these segments", not "circular sectors" and "these sectors". (Why can't we edit comments? (Hmmmm ... looks like we can if we act quickly enough.)) –  Blue Aug 10 '10 at 23:49
    
your images are all gone. –  I. J. Kennedy Dec 24 '10 at 4:40

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