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The King of Geometro nation has 2 very smart wives. On the Geometro Wives day he gets a nice heart shaped cake made. It has a number of icing flowers on it.

The King wants to split the cake in half so each wife gets half the cake with half the flowers. Even though everyone in Geometro are mathematically inclined, including the King and the Queens, this problem is not easy for the King.

So he calls upon his trusted advisor Sir Mathy Angle to solve it.

Sir Mathy noticed:

The heart shape is formed by a perfect square ABCD with each side 2 units and two semicircles BC and CD with a radius of 1 unit. So the semicircles are on the two sides (BC and CD) of the square. The flowers appeared randomly placed, but embeded in the cake. Moving the flowers was not an option. But cutting through them was OK as long as each Queen got her half.

And one more thing. The King told Sir Mathy-just to make it harder-that he can only design ONE straight line cut through the cake. No multiple or fancy cuts.

Can Sir Mathy come up with a geometric solution and break the heart in two without breaking the hearts of the King and the Queens? What would the cut look like?

enter image description here

Hint

The answer is in but needs a geometric proof

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    $\begingroup$ If this is a geometry problem, can you also define the shape of the flowers? And what does a flower that has been cut through count as for each queen? $\endgroup$
    – noedne
    Mar 8 at 17:23
  • $\begingroup$ Do the flowers appear in the exact configuration shown, or is that just an example of "random placement"? $\endgroup$
    – noedne
    Mar 8 at 18:06
  • $\begingroup$ @noedne I suppose this sentence 'But cutting through them [flowers] was OK as long as each Queen got her half.' means the flowers can be divided, too/ However, from this one: 'The King wants [...] each wife gets half the cake with half the flowers.' I understand if one flower is cut in one-third, another one should be cut in two-thirds. This would require an assumption that flowers are essentially circles—but for fair (equal) cutting their size may be necessary. $\endgroup$
    – CiaPan
    Mar 8 at 19:58
  • $\begingroup$ sorry for the late response. Flower positions are fixed. And @CiaPan's comment is correct too. $\endgroup$
    – DrD
    Mar 9 at 0:53
  • $\begingroup$ I believe flower geometry is important. otherwise how can I be sure that I will cut the flower in exact half actually? I agree with @noedne. it is also needed to find out where I need to cut the cake exactly not to touch flowers. $\endgroup$
    – Oray
    Mar 12 at 7:41

6 Answers 6

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Following up from Amoz suggestion. If we draw a line from

point B on the left side of the cake, and a second point that is 0.009226129921 'units' directly towards point C from the mid point of the line CD.

Then it divides the cake in half. The area of the bottom half of the cake can be calculated by summing the yellow, pink, bright green, and blue areas shown below (where p is the offset of the crossing point of the cut line from the middle of line CD. The areas are 1+p, 2, theta/2, and p * cos(theta)/2 respectively. Theta can be calculated using the 'sin rule' for the blue triangle as theta = phi+asin(p * sin(phi)), where phi is equal to atan((1-p)/2).

enter image description here

If you do the maths then, with the 'magic' value for p, the areas are approximately

1.0092261299, 2.00000, 0.55955668475, and 0.00201351212284 respectively,

which add up to 2+Pi/2 (half the area of the cake). The line cuts between the stars as shown below

enter image description here

clearly dividing the stars into two sets of seven.

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  • $\begingroup$ I found a little bit different result than yours. Are you sure for your results? if so, I will redo it :) $\endgroup$
    – Oray
    Mar 13 at 23:32
  • $\begingroup$ There will definitely be other solutions in the same regions - i.e. lines at slightly different angles through slightly different end-points. Or I may have stuffed upo the algebra or teh spread-sheet. $\endgroup$
    – Penguino
    Mar 15 at 4:08
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Sir Mathy scribbled for hours and hours on his desk. Lady Fizz Higgs, who was passing by his office, noticed he was in distress.

― Hey, Matt, you seem more troubled than the usual. What's up.

― It's this cake for the king and queens ― he said, pointing at the round table in the office ― I must cut it in two equal parts with one straight cut, but first I must do all the calculations. So far, I've written a demonstration that the cake cut will require a knife of finite length, and that the length has an unique solution that has a lower bound of 2.

― So you have to find out how to cut it in half?

― Yes.

― In two equal parts?

― Yes.

― Well, you just have to cut it through

its center of mass

― What? It can't be as simple as that. I need perfect mathematical proof!

― Stop overthinking things, you silly! As you sure know,

any line that passes through the center of mass of an (planar) area divides that area into two halves of equal mass,

and since I guess we're dealing with a cake of uniform density, that will make things easy.

― Okay, but in that case I should start calculating the double integral of the cake surface.

― Oh, Mat, you have to stop thinking of everything as a field on a R-squared infinite plane, and start thinking of stuff as spherical points of negligible electrical charge in a perfect vacuum with no friction. You see, the cake is not a cake, it's

a collection of three points with mass: one for the square and two for the semicircles

― What a preposterous oversimplification! The cake must be continuous, derivable, and non-porous!

― You're overcomplicating things again. We just have to rotate the cake 45 degrees...

― Pi fourths!

― Okay, pi fourths so that corner A is at the [0,0] coordinate, B at [0,2], C at [2,2] and D at [2,0]

Mathy's face turned red for a second.

― That's minus pi fourths! ― he complained, loudly

― Fiiiiiine ― she let out with a sigh ― that's rotating it minus pi fourths. But you're gonna like the next part, because we're calculating

the center of mass as the weighted average of the three centers of mass, averaged by their mass. The square is trivial, and I can look up the centroid and area of a semicircle in my books, so: $$c_{square} = [1,1]$$ $$m_{square} = 4$$ $$c_{bc} = [1, 2 + \frac{4}{3\pi}]$$ $$c_{cd} = [2 + \frac{4}{3\pi}, 1]$$ $$m_{bc} = m_{cd} = \frac{\pi}{2}$$

― I don't like where this is going, Fizz. You're taking shortcuts instead of letting me calculate those nice and beautiful double integrals.

― Because that's just wasting time. Now we can

calculate the weighted average of the vectors for the centers of mass: $$c_{cake} = \frac{c_{square} \times m_{square} + c_{bc}\times m_{bc} + c_{cd} \times m_{cd}}{m_{square}+m_{bc}+m_{cd}}$$ $$c_{cake} = \frac{4\times \left[1,1\right] + \frac{\pi}{2} \times \left[1, 2 + \frac{4}{3\pi}\right]+ \frac{\pi}{2} \times \left[2 + \frac{4}{3\pi}, 1\right]}{4 + \frac{\pi}{2} + \frac{\pi}{2}}$$ $$c_{cake} = \frac{\left[4,4\right] + \left[\frac{\pi}{2}, \frac{\pi}{2}\left(2 + \frac{4}{3\pi}\right)\right] + \left[\frac{\pi}{2}\left(2 + \frac{4}{3\pi}\right), \frac{\pi}{2}\right]}{4 + \pi}$$ $$c_{cake} = \frac{\left[4 + \frac{\pi}{2} + \frac{\pi}{2}\left(2 + \frac{4}{3\pi}\right), 4 + \frac{\pi}{2} + \frac{\pi}{2}\left(2 + \frac{4}{3\pi}\right)\right]}{4 + \pi}$$ $$c_{cake} = \left[\frac{4 + \frac{\pi}{2} + \frac{\pi}{2}\left(2 + \frac{4}{3\pi}\right)}{4 + \pi},\frac{4 + \frac{\pi}{2} + \frac{\pi}{2}\left(2 + \frac{4}{3\pi}\right)}{4 + \pi}\right]$$ $$c_{cake} ≃ [1.3133, 1.3133]$$

― What? Four decimal digits in your solution??!! The king will have my head for such an imprecision!

― You have to relax, Matt. You've been so focused in your theorems that you haven't realized you haven't got a cake here, you have a grainy pixelated picture of a cake. I mean, when you look at it closely, points A and C don't even align vertically, and one unit is like 102.5 pixels long. We don't need precision.

― Nonsense! If we're using a numerical solution, I'll better use a value of

1.3133002821628142080298404581712377945997291477458964137553166227

instead.

― Wait. How many digits of pi did you use for that?

― Not enough.

Grabbing a nearby pencil, Fizz continued:

― Anyway. We have a grainy picture, so that would be over... here...

And leaning on the picture of the cake, Fizz drew:

cake with center of mass

Mathy sighed.

― Fiine, fiiiiiiiine. But that doesn't solve the problem. We haven't even talked about the flowers, and that will require me at least another 7 hours of calculations.

― For the King's sake, it's fourteen flowers on a heart, not a density function of infinite flowers on a fractal shape. We can brute-force the problem!!

And, suddenly, Fizz drew a straight line through the picture of the cake:

Solution 1

Mathy's left eyebrow started twitching uncontrollably.

― W-What? You can't just-

― And look, there's even a second obvious solution!

Solution 2

Seconds after Fizz drew that second line, Mathy collapsed on the floor and had to be rushed to the hospital.

When he woke up the day after, there was a letter on the nightstand next to him: it was from the King and Queens. They were congratulating Sir Mathy on finding a solution to their satisfaction.

And with that, Mathy and Fizz kept having academic squabbles forever after.

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    $\begingroup$ I believe that only for centrally symmetric shapes is it true that all bisectors pass through the center of mass; and the heart shape is not centrally symmetric. $\endgroup$ Mar 9 at 4:11
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    $\begingroup$ I second 2012rcampion. For example a cut through the center of mass of a triangle and parallel to a side cuts the triangle in parts of 4/9 and 5/9 of the surface. The King will kill you. $\endgroup$
    – Florian F
    Mar 9 at 6:50
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    $\begingroup$ Lady Fizz Higgs must have forgot a lot of geometry,,,, She is wrong in 'any line that passes through the center of mass of an (planar) area divides that area into two halves of equal mass,' For example, in an equilateral triangle the center of mass is at $1/3$, that is approx. $0.3333$ of the height from the base, but a line through it and parallel to the base divides the area into parts being $4/9$ (top triangular part) and $5/9$ (base part) of the whole triangle's area. A line that divides area in half is $(1-1/\sqrt 2)h$ from the base, which is approx. $0.2929$ of the height. $\endgroup$
    – CiaPan
    Mar 9 at 7:32
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    $\begingroup$ Gaaahhhh, I feel like such an idiot right now. I based the reasoning on being able to balance a solid on a point vertically aligned to the CoM and then fell into the wrong intuitive assumption that mass at either side is equal - but it's mass times distance that is equal. $\endgroup$ Mar 9 at 11:26
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There are some answers already but I want to point out how many solutions would be possible for such a problem with this solution;

Firstly, it is obvious that the answer should be passing through as many others noticed shown as below;

enter image description here

so there are infinite line/cut that you can draw/cut in the green area without touching the flowers and it divides flowers into two.

First let's find the total volume of the cake;

$W=4+\pi$

So how now? Let's think the question in terms of a rule of thumb and take a straight line from the corner of the square passing through the center of the one of the half circle as shown below;

enter image description here

It is not that hard to find the area $A_1+A_2$ from this point;

$A_1+A_2=3+\frac{\arctan(2)}{2}$

but if you try to calculate in numbers, it is not actually the exact half of $W$ but there is a little difference inbetween them;

$d=\frac{1}{2}W-\left(A_1+A_2 \right)=2+\frac{\pi}{2}-3-\frac{\arctan(2)}{2}=\frac{\pi-\arctan(2)}{2}-1\cong0.0172$

which is very very small value compare the whole area.

As a result, instead of that cut, we can do some other cuts to make the cake into half without touching any flowers around this position such as;

enter image description here

so if $A_3+A_4=d$, we got our answer, but the cut is actually exaggerated.

or let's think differently;

enter image description here

so if $A_5+A_6+A_7=d$, we got our answer. but

Did you notice that playing around that area, we can find infinite other cuts?

or another way;

enter image description here

so if $A_8+A_9-A_{10}=d$, we got our answer.

these 3 can be geometrically calculated and but

it should be noted that there are infinite other solutions as well in the green area shown on the first figure.

For simplicity I will calculate only one of them from now on.

You can find a result for each of them, but I took the one below which was not shown on previous results;

enter image description here

so if we do calculation you can easily notice that;

$A_{10}=\frac{\alpha}{2}$

and

$A_{9}=1-\frac{\tan(\arctan(2)-\alpha)}{2}$

and

$A_{8}$ is quite messy so I did not want to bother you with it. but if it is requested, I can show the details of the calculation.

As a result $\alpha$ becomes (Wolfram Alpha);

around x ≈ 0.00862 in radian

and

0.494 in degree form

so quite small and it makes areas equal in each other.

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Cutting the cake:

is possible.

The proof is:

the two-dimensional case of the ham sandwich theorem. You can watch a video about it here, but I will summarize the main points below. The proof makes heavy use of the intermediate value theorem (IVT).

Proposition 1:

The cake can be cut in half (ignoring the flowers).

Proof:

Imagine holding a knife over the cake, straight out in front of you. Now imagine moving the knife left and right without rotating it, so that the knife always remains parallel to its original orientation. Consider the difference between the amount of cake to the left and to the right of the knife. If you move the knife all the way to the right, this difference will be positive, and if you move the knife all the way to the left, the difference will be negative. Also, because the cake has finite, nonzero measure, this difference will vary continuously as you move the knife from side to side. Due to the intermediate value theorem, there is guaranteed to be some position of the knife where the cake is cut exactly in half.

Proposition 2:

The position of the cut varies continuously as the cut is rotated.

Proof:

Now imagine walking around the cake and cutting it in half from various angles. Because the previous proof did not depend on the exact shape of the cake, we know that the cake can be cut in half no matter how it is rotated. Consider the cuts that would be made at two slightly different angles. The lines formed by these cuts are not parallel, so they must intersect at some point. The two cuts therefore divide the plane into four regions, two of which we are interested in: the region to the left of the first cut but to the right of the second, and vice versa. Since the cake is cut in half in both cases the amount of cake in these two regions must be the same. But if the intersection point was above or below the cake, one of these regions would be empty: therefore the intersection point must be within the (convex hull of) the cake. Since the cake is bounded, this means that the cut cannot jump discontinuously from one position to another as the cut is rotated.

Proposition 3:

There is at least one cut that divides both the cake and flowers in half (a "fair division").

Proof:

Similar to the proof of proposition 1, consider the difference between the amount of flowers on the left side of the cut and right side of the cut, starting with the cut at some arbitrary angle. If this difference is zero, we have found a fair division. Otherwise, rotate the cut by 180 degrees. As the cut is rotated, its motion is continuous and therefore the difference in the amount of flowers between the left and right side must vary continuously (since we are allowed to cut individual flowers in half). However, after rotating by 180 degrees the left and right sides have swapped places: if the difference was positive, it is now negative, and vice versa. By the intermediate value theorem, there must have been some angle at which the amount of flowers on the left and right sides were equal; and because the whole time the cake was cut exactly in half, that cut was a fair division.

Of course, the proof has one deficiency:

It is nonconstructive: although it proves that a fair division exists, it does not show what it is.

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    $\begingroup$ The ham sandwich theorem shows that a solution exists, but it does not show whether Sir Mathy can find it, as the question asks. $\endgroup$
    – noedne
    Mar 8 at 21:48
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    $\begingroup$ @noedne I addressed the nonconstructive nature of the proof at the end of my answer. Interpreting the question as "can this particular person find a solution" makes it very hard to answer, since it depends on the person, their experience with problems like this, how they're feeling that day... I think the more typical interpretation of questions like this is: "is there a solution that, in principle, this person could find," which I do answer. $\endgroup$ Mar 8 at 22:08
  • $\begingroup$ Yes of course, I interpret "can Sir Mathy find a solution" to ask whether a solution can be constructed, which as you point out, the IVT does not do. $\endgroup$
    – noedne
    Mar 8 at 22:52
  • $\begingroup$ @noedne I think there is some misunderstanding. What do you mean by "constructed?" $\endgroup$ Mar 8 at 22:54
  • $\begingroup$ @2012rcampion Possibly something like en.wikipedia.org/wiki/Geometric_Construction $\endgroup$
    – CiaPan
    Mar 8 at 23:42
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I might have a conceptual solution. It is a lot of calculations and I am unlikely to carry them out correctly to the actual expected final answer (which is known), but if these do have merit (I have not done trigonometry in a long time, so no promises), someone can run with it or edit this answer to do the actual calculations, which I summarized. Or maybe this is a sufficient proof of feasibility.

Consider the line from B to F (extending to G), which DrD found:
heart

The line splits the heart into a top (call it "TT") and bottom ("BB").
The top two semicircles together form a circle of area π (since radius squared is 1).
The area of FGD is (as demonstrated by DrD) tan-1(2)* π/360, ~.553. Call this X. Area of BEF is clearly 1 (1/4 of 4), and rect AEFD has area 2, so the total area of the bottom “BB” is 2+1+X = 3+X, ~3.553
Total area of the top “TT” is π + 1 – X, ~3.588.
TT is larger than BB by a small amount, call it ɛ.
We need to rotate line BG around point B, counter clockwise to shrink TT by ½ ɛ. BB then grows by ½ ɛ and the heart is evenly divided. ɛ is small enough that we assume we will not cut into any flowers (to verify once calculated maybe, but it seems clear it is a very small adjustment and there is likely room for it).
Our final cut is BH:

heart cut

Call the tiny angle HBG “a”.
By the Pythagorean theorem, length of BF is √5 (right?)
We have drawn line IF perpendicular to BH forming right triangles BIF, HIF.
Length of IF is then available as relative to “a” (I think sin(a)* √5 or some other trigonometric ratio)
We know length of FH is 1 as F is the center of a circle.
We now know lengths of two sides of each right triangle BIF and HIF so it is possible to calculate all sides and angles relative to “a”.
This means we can calculate both areas of BIF, HIF relative to “a”.
Angle HFG can be calculated relative to “a” as it is 180 – angle IFB – angle IFH. Call this angle “b”.
The area of semicircle HFG is known relative to “a”, as it is (b/360)* π.
The sum of the areas of polygons BIF, HIF, HFG are now all known relative to “a”. Call their sum “J”.
Our goal was to set the area of J to ½ ɛ, which is a known number. Therefore “a” can be calculated and we can now cut the cake with a straightedge and protractor, unless I made an error somewhere.

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  • $\begingroup$ "Can be calculated" does not imply "can be performed with a straightedge and protractor" (see en.wikipedia.org/wiki/…). $\endgroup$
    – noedne
    Mar 12 at 6:52
  • $\begingroup$ I think it does here because the angle a is simply added to the obvious square cut we started with. $\endgroup$
    – Amoz
    Mar 12 at 11:14
  • $\begingroup$ Although yes a is likely irrational so exact precision is not possible. I only meant that we can make a cut to an arbitrary precision of our choice. I don't know if a precise decimal cut is likely to even exist given the circles. $\endgroup$
    – Amoz
    Mar 12 at 14:32
  • $\begingroup$ I see, I misread as "straightedge and compass," when really you're measuring with a protractor. $\endgroup$
    – noedne
    Mar 12 at 16:28
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Answers and comments above confirm a solution for Sir Mathy.

Here is my argument. The Cake needs to be cut in half area or volume wise as well as equal number of flowers

The total area of Heart is

2x2 (square) + 2 Semicircles of radius 1 = 4 + 3.142 x 1^2 = 7.142

Look at the diagram below. If you draw a line from B through the center point of line CD all the way to G then this line clearly disects the heart in 2 parts with 7 flowers for each portion.

To prove that the area ABFGDA is half of total Heart area (7.142)

The area = Rectangle AEFD + Triangle EBF + Sector DFG

= 2x1 + 1/2 x 2 x 1 + Angle DFG in radians/2 * Radius sqared

= 2+1 + 1.117/2 since angle DFG is same as angle CFB. BCF is a right angled triangle with BF as hypotenuse. So Tan of angle CFB is 2/1. So the angle is about 64 degrees or 1.117 radians.

= 3.558

So the area ABFGDA is very close to 1/2 of 7.142enter image description here

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    $\begingroup$ I think there is a mistake: arctan(2) is 1.10715, not 1.117. This makes the areas of the two pieces 3.55357 and 3.58802, so it is not cut in half. $\endgroup$ Mar 11 at 16:48
  • $\begingroup$ Even the conclusion as stated does not add up: 1/2 of 7.142 is 3.571, which is not very close to 3.558. $\endgroup$
    – noedne
    Mar 11 at 17:14

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