Towering trapeziums

T. Hakim from Annie Gale School in Canada defined a new point, $M$, as the midpoint of $CD$.

You had to understand the pattern of the triangles and the relationship between triangles $ABO$ and $ODM$.

This relation states that because the hypotenuse of $ODM$ is the same as $ABO$, and because it is an isosceles right triangle (so the other sides are the same), 

$$ABO = ODM$$

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With this information, I was able to figure out the rest of the triangle, and even the $n^\text{th}$ trapezium with these steps!

1) last full smaller-looking triangle = half of the next triangle (in the first case, $OAB=\frac12OCD$)

2) multiply half-triangle by two to create full triangle (therefore $2\times$ size of the previous triangle) (in the first case, $OCD$ has area $2\times$ area of $OAB$ which is $2\times1=2$)

3) subtract larger from smaller to get trapezium (in the first case, $2-1=1$)

After solving 4 trapeziums, I noticed a pattern, which was doubling the last trapezium's size, which made it a lot easier to solve.

next trapezium = $2\times$ last trapeizum

Matt from Swindon Academy in the UK, Amrit from Hymers College in the UK and Pablo from King's School Alicante in Spain considered the sides $OA$ and $OB$ as the base and height of triangle OAB, since the angle between them is $90^\text o$.

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Amrit said:

Let $OA=OB=x$, then

$\frac12x^2=1$

Therefore $x^2=2\Rightarrow x=\sqrt2$

Let $AB=OC=y$, then by the Pythagorean theorem, 

$2x^2=y^2$

Plugging in the value of $x$, we have $2\times2=y^2\Rightarrow y=2$

The area of triangle $OCD$ is $\frac12y^2=2$. Therefore the area of the trapezium is $2-1=1$.

Pablo and Matt continued using the same methods to find the areas of the next trapeziums and generalise. This is Pablo's work:

TRAPEZIUM 2

$OC = OD = 2$

$CD^2 = 2^2 + 2^2$

$CD^2 = 8$

$CD =OE= 2\sqrt2$

Area of $OEF = 2\sqrt2 \times2\sqrt2\times \frac12= 4$

Area of trapezium $CDFE = $Area$ OEF - $Area$ OCD = 4 - 2 = 2 $ units$^2$

TRAPEZIUM 3

$OE = OF = 2\sqrt2$

$EF^2 = OE^2+ OF^2 = (2\sqrt2)^2+(2\sqrt2)^2$

$EF^2 = 8 + 8$

$EF = 4$

$OG = 4$

Area $OGH = 4\times4\times\frac12 = 8$ units$^2$

Area of trapezium $EFHG = $Area$ OGH - $Area$ OEF = 8 - 4 = 4 $ units$^2$

NTH TRAPEZIUM

For the first trapezium, $ABDC$, we had to subtract the area of the triangle $OAB$ from triangle $OCD$. If we call $OA$ the 'first' length, $OC$ is the 'second' length.

$OA = \sqrt{2^1}$ [first]

$OC = 2 = \sqrt{2^2}$ [second]

this continues for $OE$ and $OG$

So the area of the trapezium was $\frac12 OC^2  - \frac12 OA^2$

If instead of $OA$ and $OC$ we had the $n^\text{th}$ and the $(n+1)^\text{th}$ lengths, we can

say that:

the area of the $n^\text{th}$ trapezium

$= \frac12\left(\sqrt{2^{n+1}}^2\right) - \frac12\left(\sqrt{2^n}^2\right)$

$= \frac12 \times 2^{n+1} - \frac12\times 2^n$

$= 2^n - 2^{n-1}$

$= 2^n\left(1 - 2^{-1}\right)$

$= 2^n\left(1 - \frac12\right)$

$= 2^n\left(\frac12\right)$

$= 2^n\left(2^{-1}\right)$

$= 2^{n-1}$

Amrit used similarity to find the area of the $n^\text{th}$ trapezium:

It is clear to see that all the trapeziums are similar.

Say $OA$ was length $y_0$ and $OC$ was length $y_1$, then $y_1=\sqrt2y_0$                                                                  

If $CD=OE$ was $y_2$, then $y_2=\sqrt2y_1$

If we call $y_n$ the length of the $n^\text{th}$ parallel line, then $y_{n+1}=\sqrt2 y_n$

So $\dfrac{y_{n+1}}{y_n}=\sqrt2$

This is the scale factor between the sides of adjacent trapezia, so the area factor between them is $2$. We now need to find the area of the first trapezia to find a formula for the others.

The area of the first trapezium was $1$, so the area of the second trapezium is $2$, and the area of the third trapezium is $4$, and the area of te fourth trapezium is $8$, so

the area of the $n^\text{th}$ trapezium is $2^{n-1}$.

Guruvignesh and Adithya, also from Hymers College, considered the base and height of the triangle as horizontal and vertical.