Origami Shapes

Ian and Luke of Flegg High School worked out could see that the shape has two pairs of adjacent equal sides, so it is a kite,

I worked through the sequence of folds calculating the sides and angles of the shapes we are making. In part 1, the angles are always obvious $(45^{\circ}, 90^{\circ} or 135^{\circ})$, but in part 2 they are not so obvious.

I'll go through the solution to part 1 to give an idea of how to approach this problem; when you've read this, have a go at solving part 2 in a similar way. This is the shape we get when we fold the first corner up:

This is what we get when we fold the second corner up:

How did we get the values for $x_1$ and $x_2$?

$x_1 = 1 - (\sqrt{2} - 1) = 2 - \sqrt{2}$. (Subtract the length $s$ from the shorter side of the original rectangle.) \par $x_2 = \sqrt{2} (\sqrt{2} - 1) = 2 - \sqrt{2}$. (Use Pythagoras' theorem on the small right-angled triangle.)

What are the area and perimeter of this shape? From the above diagram, the perimeter is obviously 4. The easiest way to calculate the area is to divide the shape into two identical right-angled triangles:

Area = 2.	1	$\sqrt{2} (2 - \sqrt{2}) = 2\sqrt{2} - 2$
	2

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Congratulations to Jason of Laramie Senior High School, Wyoming, USA, and to Ling of Tao Nan School, Singapore, for their excellent solutions to the second part of the problem.

Part I We begin with a rectangle $ABCD$, where $AB$ is 1 unit and $BC$ is $\sqrt{2}$ units (this is the ratio of the sides of A4 paper). The first fold is along $BE$ (see Figure 1), so that $A \to A'$ on $BC$. Then $BA'=1$ and $A'C = \sqrt{2}-1$. The second fold is along $EF$ so that $D\to D'$ on $A'E$. Then $$ D'E=D'F= A'C=\sqrt{2}-1, $$ and by Pythagoras' Theorem, $$ EF= 2-\sqrt{2}. $$ Also, $$ CF= A'D'= 1-D'E=1-(\sqrt{2}-1)=2-\sqrt{2}.$$ Hence $EF=CF$, and $BE=BC$, so that $BCFE$ is a kite. The perimeter is $2\sqrt{2}+2(2-\sqrt{2})=4$. The area of the kite is $$ {\rm area}(ABCD) - {\rm area}(AEB) - {\rm area}(EDF) = 2(\sqrt{2} -1).$$

Part II Taking another sheet of paper (see Figure 2), the first fold is along $DE$ so that $A\to A'$ on $DC$. The second fold is along $A'B$, where $C\to C'$. The third fold is perpendicular to $A'B$ (through the midpoint $R$ of the diagonal) and takes $B \to A'$, folding the diagonal $A'B$ in half. We shall show that the shape obtained, that is $DPRA'$, is a kite by showing that $DP=DA'$ and $RP=RA'$. Step 1 When we make the second fold it appears that $\angle EBC' = 45^o$, and $\angle A'BC'= 22\textstyle{1\over 2}^o$. We shall prove this below, but you may assume it if you wish and go on now to Step 2.

Let $\angle A'BC = \angle A'BC' =x^o$; we shall show that $x^o = 22\textstyle{1\over 2}^o$ (see Figure 3) by showing that $\tan 2x^o = 1$ (so that $2x^o=45^o$); see Figure 3. Now

$$ \tan x^o = {A'C\over BC} = \sqrt{2}-1,$$

and

$$ \tan 2x^o = {2\tan x^o \over 1- (\tan x^o)^2} ={2(\sqrt{2}-1)\over 1-(\sqrt{2}-1)^2} = 1.$$

A further challenge
Can you prove that the fold $PR$ must go through the point $A$?

$\bf{Step 2}$

We leave you to show that the following angles are as given (this needs simple geometry but no trigonometry):

$$\eqalign{ \angle A'BC' &=& 22{1\over 2}^o \\ \angle EBQ &=& 45^o \\ \angle REB &=& 67{1\over 2}^o \\ \angle QEC &=&22{1\over 2}^o \\ \angle AED &=&45^o \\ \angle PEQ &=&45^o\\ \angle ERB &=& 45^o \\ \angle QRE &=& 45^o }$$

This shows that the diagonal $RE$ of the quadilateral $RBEP$ bisects the angles at $E$ and $R$, so that $RBEP$ is a kite. As this is a kite, we have $RP=RB$, and as $R$ is the midpoint of $A'B$, we have

$$ RA'= RP.$$

Also, $EP=EB=\sqrt{2}-1$, so that

$$ DP=DE-PE = \sqrt{2}-(\sqrt{2}-1) = 1 = DA'.$$

Finally, this gives us yet another kite $DPRA'$ because, as we have just shown, $DP=DA'$ and $RA'=RP$.

The perimeter of the kite $DPRA'$ is

$$ 2DA'+2A'R = 2+A'B= 2+\sqrt{1+ (\sqrt{2}-1)^2}= 2+ \sqrt{4-2\sqrt{2}}.$$

The area of the kite $DPRA'$ can be found (with quite a lot of computation) to be ${\textstyle {1\over 2}}$ as we have plenty of information about all lengths and angles in Figure 3, but there is another way to do this.

We notice that the area of the kite $DPRA'$ is \begin{eqnarray} {\rm area}(DPQA') + {\rm area}(QRA') &=& {\rm area}(DEA')-{\rm area}(PEQ) + {\rm area}(QRA') \\ &=& {\rm area}(DEA')-{\rm area}(PEQ) + {\textstyle {1\over 2}} {\rm area}(A'QS). \end{eqnarray} Now notice that the triangles $PEQ$ and $A'QS$ are similar, with linear scale factor

$$\frac{\sqrt{2}-1}{2-\sqrt{2}} = \frac{1}{\sqrt{2}}$$

(where $A'QS$ is the larger).

$$ {\rm area}(PEQ)= {\textstyle {1\over 2}} {\rm area}(A'QS)$$

so that

$$ {\rm area}(DPQA')= {\rm area}(DEA') = {\textstyle {1\over 2}} .$$