You're Only Getting Cake Because My Mum Forced Me To Give It To You

Area of regular hexagon with side length s is: $A = {3\sqrt{3} \over 2} s^2$

Note: Internal angles of a regular hexagon are $120^{\circ}$ .

Using this we can derive the area $A_s$ of each section cut: $A_s = {\sqrt{3} \over 16} s^2$

Total area of the sections cut is: $6A_s = {3\sqrt{3} \over 8} s^2$

Ratio between the sections cut and the rest of the cake is: ${6A_s \over A} = {{3\sqrt{3} \over 8}s^2 \over {3\sqrt{3} \over 2}s^2} = {1\over4}$

This means the proportion that Garisht keeps is $\boxed{\frac{3}{4}}$

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WLOG let sidelength of the hexagon equal $1$ , then note that joining the midpoints will make hexagon that is regular because for each side length $a$ of this new hexagon we get by the law of cosines:

$a^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \cos \left(\frac{2\pi}{3}\right) = \frac{3}{4}$ and the ratio of areas is $\frac{a^2}{1^2} = \boxed{\frac{3}{4}}.$

Using Trig and the fact that the interior angles of a regular hexagon with side $s$ are $120^\circ$ . We find that the area of a single piece of cake is $\frac{s^2 \sqrt{3}}{16}$ hence the sum of the areas of all the pieces is $\frac{s^2 3\sqrt{3}}{8}$ . Again,using trig ,we can derive the area of a regular hexagon to be $\frac{s^{2}3\sqrt{3} }{2}$ . The ratio of the two formulae gives $\frac{\frac{s^2 \sqrt{3}}{16}}{\frac{s^{2}3\sqrt{3} }{2}} = \frac{1}{4}$ . This is the ratio of slices of cake to total cake and we subtract this ratio from 1 to get the remainder of the cake. $1-\frac{1}{4}=\frac{3}{4}$

The area of the cake is proportional to $s^2$ where $s$ is the length of the side hexagon. We can write this as $A = ks^2$ , $k$ being the proportionality constant. The resulting shape, right after Greedy Garisht sliced his cake, is also a regular hexagon with area defined by $A_{new} = k(s_{new})^2$ . To obtain the ratio $s_{new}:s$ , we consider one of the sliced portion he gave to his friends. It is shaped as an isosceles triangle, with vertex angle equal to $120^o$ and leg length equal to $\frac{s}{2}\}$ . One can verify that the base length is equal to $\{\frac{\sqrt{3}}{2}s\}$ by using either cosine law or dividing the triangle into two $30^o-60^o-90^o$ right triangles and use sine law to obtain half of the base length. The base length is also equal to the side length of the new hexagon, thus we have: $A_{new} = k(\frac{\sqrt{3}}{2}s)^2 = \frac{3}{4}ks^2 = \frac{3}{4} A$ . Thereby the proportion of the cake left to himself is: $\frac{A_{new}}{A} = \frac{3}{4}$ .

Non-trig and easy-to-understand solution:

We get another hexagon by cutting midpoints of the adjecent sides.

If side of hexagonal cake is a, then side of this hexagon becoms ((a/2)^2+(a/2)^2-2(a/2)(a/2)cos(120))^(1/2)=(3^(1/2))a/2

Now, area is proportional to (side)^2

So, fraction is equal to 3/4

Let the side of the hexagon be x .

Area of a regular polygon equals $\frac{S^2N}{4tan(\frac{180}{N})}$

$A_{hexagon}=\frac{x^2\times6}{4tan(\frac{180}{6})}$

which simplifies to $\frac{9{x}^2}{2\sqrt{3}}$

The some of interior angles in a regular polygon is given by ${180(n-2)}$

And this equals ${180(6-2)}$

and ${720}$

Thus each interior angle of the hexagon equals 120

With the manner in which he divided the cake, he created 6 isosceles triangles of sides x/2 around the hexagon .

Area of the isosceles triangles equals $6\times{\frac{1}{2}}\times{\frac{x}{2}}\times{\frac{x}{2}}\times{sin120}$

And this simplifies to $\frac{{x}^2\sqrt{3}}{16}$

Expressing this area as a fraction of the entire area of the hexagon we get $\frac{3{x}^2\sqrt{3}}{8}\times\frac{2\sqrt{3}}{9x^2}$

which simplifies to $\frac{1}{4}$

Thus the proportion of cake that he has left to himself equals ${1}-\frac{1}{4}$

Which gives $\boxed{\frac{3}{4}}$

when Garisht is done with giving slices to his firneds, the hexagonal cake becomes a smaller hexagon. the side of new hexagon is \sqrt{3} /2 times the legth of side of the old hexagon. The area of new hexagon will be 3/4 times of that of old hexagon