4 corrected math
source | link

For $n \geq 8:$

Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $x$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice.

$A_{cake} = \pi r^2$
$A_{slice} = \frac{A_{cake}}{n}$
$\theta_{slice} = \frac{360^\circ}{n}$

$A_{halfSlice} = xy$$A_{halfSlice} = \frac{xy}{2}$, where $x$ is the distance from the origin, and $y$ is the length of the cut. After finding $y$, we can then put $A_{halfSlice}$ in terms of $x$:

$y = xtan(\theta_{slice})$
$A_{halfSlice} = x^2tan(\theta_{slice})$$A_{halfSlice} = \frac{x^2tan(\theta_{slice})}{2}$

Finally, we can solve for $x$, given that we know that $A_{halfSlice} = \frac{A_{slice}}{2}$:

$\frac{\pi r^2}{2n} = x^2tan(\theta_{slice})$$\frac{\pi r^2}{n} = x^2tan(\theta_{slice})$

$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$$x = \frac{\sqrt{\pi}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$

So for $n=8$ and $r=15cm$, we get $x \approx 6.6467cm$$x \approx 9.399cm$. Still unsure about $n$ below 8 though...

For $n \geq 8:$

Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $x$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice.

$A_{cake} = \pi r^2$
$A_{slice} = \frac{A_{cake}}{n}$
$\theta_{slice} = \frac{360^\circ}{n}$

$A_{halfSlice} = xy$, where $x$ is the distance from the origin, and $y$ is the length of the cut. After finding $y$, we can then put $A_{halfSlice}$ in terms of $x$:

$y = xtan(\theta_{slice})$
$A_{halfSlice} = x^2tan(\theta_{slice})$

Finally, we can solve for $x$, given that we know that $A_{halfSlice} = \frac{A_{slice}}{2}$:

$\frac{\pi r^2}{2n} = x^2tan(\theta_{slice})$

$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$

So for $n=8$ and $r=15cm$, we get $x \approx 6.6467cm$. Still unsure about $n$ below 8 though...

For $n \geq 8:$

Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $x$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice.

$A_{cake} = \pi r^2$
$A_{slice} = \frac{A_{cake}}{n}$
$\theta_{slice} = \frac{360^\circ}{n}$

$A_{halfSlice} = \frac{xy}{2}$, where $x$ is the distance from the origin, and $y$ is the length of the cut. After finding $y$, we can then put $A_{halfSlice}$ in terms of $x$:

$y = xtan(\theta_{slice})$
$A_{halfSlice} = \frac{x^2tan(\theta_{slice})}{2}$

Finally, we can solve for $x$, given that we know that $A_{halfSlice} = \frac{A_{slice}}{2}$:

$\frac{\pi r^2}{n} = x^2tan(\theta_{slice})$

$x = \frac{\sqrt{\pi}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$

So for $n=8$ and $r=15cm$, we get $x \approx 9.399cm$. Still unsure about $n$ below 8 though...

3 added 67 characters in body
source | link

For $n \geq 8:$

Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $x$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice.

$A_{cake} = \pi r^2$
$A_{slice} = \frac{A_{cake}}{n}$
$\theta_{slice} = \frac{360^\circ}{n}$

$A_{halfSlice} = xy$, where $x$ is the distance from the origin, and $y$ is the length of the cut. After finding $y$, we can then put $A_{halfSlice}$ in terms of $x$:

$y = xtan(\theta_{slice})$
$A_{halfSlice} = x^2tan(\theta_{slice})$

Finally, we can solve for $x$, given that we know that $A_{halfSlice} = \frac{A_{slice}}{2}$:

$\frac{\pi r^2}{2n} = x^2tan(\theta_{slice})$

$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$

So for $n=8$ and $r=15cm$, we get $x \approx 6.6467cm$. Still unsure about $n$ below 8 though...

For $n \geq 8:$

Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $x$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice.

$A_{cake} = \pi r^2$
$A_{slice} = \frac{A_{cake}}{n}$
$\theta_{slice} = \frac{360^\circ}{n}$

$A_{halfSlice} = xy$, where $x$ is the distance from the origin, and $y$ is the length of the cut. After finding $y$, we can then put $A_{halfSlice}$ in terms of $x$:

$y = xtan(\theta_{slice})$
$A_{halfSlice} = x^2tan(\theta_{slice})$

Finally, we can solve for $x$, given that we know that $A_{halfSlice} = \frac{A_{slice}}{2}$:

$\frac{\pi r^2}{2n} = x^2tan(\theta_{slice})$

$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$

Still unsure about $n$ below 8...

For $n \geq 8:$

Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $x$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice.

$A_{cake} = \pi r^2$
$A_{slice} = \frac{A_{cake}}{n}$
$\theta_{slice} = \frac{360^\circ}{n}$

$A_{halfSlice} = xy$, where $x$ is the distance from the origin, and $y$ is the length of the cut. After finding $y$, we can then put $A_{halfSlice}$ in terms of $x$:

$y = xtan(\theta_{slice})$
$A_{halfSlice} = x^2tan(\theta_{slice})$

Finally, we can solve for $x$, given that we know that $A_{halfSlice} = \frac{A_{slice}}{2}$:

$\frac{\pi r^2}{2n} = x^2tan(\theta_{slice})$

$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$

So for $n=8$ and $r=15cm$, we get $x \approx 6.6467cm$. Still unsure about $n$ below 8 though...

2 added 9 characters in body
source | link

For $n \geq 8:$

Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $x$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice.

$A_{cake} = \pi r^2$
$A_{slice} = \frac{A_{cake}}{n}$
$\theta_{slice} = \frac{360^\circ}{n}$

$A_{halfSlice} = xy$, where $x$ is the distance from the origin, and $y$ is the length of the cut. After finding $y$, we can then put $A_{halfSlice}$ in terms of $x$:

$y = xtan(\theta_{slice})$
$A_{halfSlice} = x^2tan(\theta_{slice})$

Finally, we can solve for $x$, given that we know that $A_{halfSlice} = \frac{A_{slice}}{2}$:

$\frac{\pi r^2}{2} = x^2tan(\theta_{slice})$$\frac{\pi r^2}{2n} = x^2tan(\theta_{slice})$

$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{tan(\theta_{slice})}}$$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$

Still unsure about $n$ below 8...

For $n \geq 8:$

Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $x$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice.

$A_{cake} = \pi r^2$
$A_{slice} = \frac{A_{cake}}{n}$
$\theta_{slice} = \frac{360^\circ}{n}$

$A_{halfSlice} = xy$, where $x$ is the distance from the origin, and $y$ is the length of the cut. After finding $y$, we can then put $A_{halfSlice}$ in terms of $x$:

$y = xtan(\theta_{slice})$
$A_{halfSlice} = x^2tan(\theta_{slice})$

Finally, we can solve for $x$, given that we know that $A_{halfSlice} = \frac{A_{slice}}{2}$:

$\frac{\pi r^2}{2} = x^2tan(\theta_{slice})$

$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{tan(\theta_{slice})}}$

Still unsure about $n$ below 8...

For $n \geq 8:$

Viewing the tip of the slice as the origin and following one edge of the slice, we are trying the find the distance $x$ from the origin in which a perpendicular cut would give two pieces with same area. A perpendicular cut on the slice would result in one piece being a right triangle, so the question is now to find the cut which would give a triangle with area that is half of the slice.

$A_{cake} = \pi r^2$
$A_{slice} = \frac{A_{cake}}{n}$
$\theta_{slice} = \frac{360^\circ}{n}$

$A_{halfSlice} = xy$, where $x$ is the distance from the origin, and $y$ is the length of the cut. After finding $y$, we can then put $A_{halfSlice}$ in terms of $x$:

$y = xtan(\theta_{slice})$
$A_{halfSlice} = x^2tan(\theta_{slice})$

Finally, we can solve for $x$, given that we know that $A_{halfSlice} = \frac{A_{slice}}{2}$:

$\frac{\pi r^2}{2n} = x^2tan(\theta_{slice})$

$x = \frac{\sqrt{\frac{\pi}{2}}r}{\sqrt{n}\sqrt{tan(\theta_{slice})}}$

Still unsure about $n$ below 8...

1
source | link