### Building Tetrahedra

Can you make a tetrahedron whose faces all have the same perimeter?

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load.

# Curvy Areas Poster

##### Age 14 to 16Challenge Level

3 pieces: each piece is $\frac13$ of the total area

4 pieces: each piece is $\frac14$ of the total area

5 pieces: each piece is $\frac15$ of the total area

$m$ pieces: each piece is $\frac1m$ of the total area

Why?

Red, orange and yellow

The top half is made of 3 semicircles on top of each other:

Say the smallest semicircle has radius 1, the medium semicircle has radius 2 and the largest has radius 3

Smallest semicircle has area $\frac12\pi\times1^2 = \frac{\pi}2$

Medium semicircle has area $\frac12\pi\times2^2 = 2\pi$

Largest semicircle has area $\frac12\pi\times3^2 = \frac{9\pi}2$

Whole circle has area $9\pi$

Red area: smallest + (largest $-$ medium) $=\frac{\pi}2 + \frac{9\pi}2 - 2\pi = 3\pi$ which is $\frac13$ of $9\pi$

Yellow area is equal to red area so the orange area must also be $\frac13$ of the total.

Red, orange, yellow and green

Include also extra large semicircle (XL), area $\frac12\pi\times4^2 = 8\pi$

Whole circle has area $16\pi$

Red area: smallest + (XL $-$ large) $= \frac{\pi}2 + 8\pi - \frac{9\pi}2 = 4\pi$ which is $\frac14$ of $16\pi$

Orange area: (medium $-$ smallest) + (large $-$ medium) $=$ large $-$ smallest $= \frac{9\pi}2 - \frac{\pi}2 = 4\pi$

So each piece is $\frac14$ of the total area.

Red, orange, yellow, green and blue

XXL semicircle area $\frac12\pi\times5^2 = \frac{25\pi}2$

Whole circle has area $25\pi$

Red area: smallest + (XXL $-$ XL) $=\frac{\pi}2 + \frac{25\pi}2 - 8\pi = 5\pi$ which is $\frac15$ of the total area

Orange area: (medium $-$ smallest) + (XL - large) $= 2\pi - \frac{\pi}2 + 8\pi - \frac{9\pi}2 = 5\pi,$ also $\frac15$ of the total area

Yellow area: (large $-$ medium)$\times2 = \left(\frac{9\pi}2 - 2\pi\right)\times2 = 5\pi$

So each piece is $\frac15$ of the total area.

$m$ pieces

Largest piece has area $\frac{m^2\pi}2$ and whole circle has area $m^2\pi$

'Red' piece:
\begin{align}\frac{\pi}2 + \frac{m^2\pi}2 - \frac{(m-1)^2\pi}2 &= \tfrac{\pi}2\left(1+m^2 -(m^1-2m+1\right) \\ &= m\pi\end{align}

'Orange' piece:
\begin{align} \left(2\pi - \frac{\pi}2\right) + \left(\frac{(m-1)^2\pi}2 - \frac{(m-2)^2\pi}2\right) &= \frac{\pi}2\left(4-1+\left(m^2-2m+1\right)-\left(m^2-4m+4\right)\right)\\ &= \tfrac{\pi}2\left(4m-2m\right)\\&=m\pi\end{align}

General piece:
\left(\frac{(r+1)^2\pi}2 - \frac{r^2\pi}2\right) + \left(\frac{(m-r)^2\pi}2 - \frac{(m-(r+1))^2\pi}2\right) \\ \begin{align} &= \tfrac{\pi}2\left((r+1)^2-r^2+\left(m^2-2mr+r^2\right)-\left(m^2-2m(r+1)+(r+1)^2\right)\right)\\ &= \tfrac{\pi}2\left(2m(r+1)-2mr\right)\\&=m\pi\end{align}

$\therefore$ all of the pieces have the same area, $\frac1m$ of the total.