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Matter of Scale

Prove Pythagoras Theorem using enlargements and scale factors.

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Conical Bottle

A right circular cone is filled with liquid to a depth of half its vertical height. The cone is inverted. How high up the vertical height of the cone will the liquid rise?

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Triangle ABC is equilateral. D, the midpoint of BC, is the centre of the semi-circle whose radius is R which touches AB and AC, as well as a smaller circle with radius r which also touches AB and AC. What is the value of r/R?

Fit for Photocopying

Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Sam from Appleton Thorn school made a start by noting the following:
If we start of by enlarging an A4 sheet to an A2 sheet we can say:
2 A4 sheets go into an A3 sheet.
2 A3 sheets go into an A2 sheet.

From this information we can gather that the number of A4 sheets that fit into a A2 sheet is $2\times2 = 4$.

Many more of you including Niharika (Leicester High School for Girls) and Declan (Thomas Keble) worked out that:

A4 ->  A3 = 2 = $2^1$
A4 -> A2 = 4 = $2^2$
A4 -> A1 = 8 = $2^3$
A4 -> A0 = 16 = $2^4$
and also:

A5 -> A4 = 2 = $2^1$
A5 -> A3 = 4 = $2^2$
A5 -> A2 = 8 = $2^3$
A5 -> A1 = 16 = $2^4$
A5 -> A0 = 32 = $2^5$
Jonathan (Najing International School), Muntej (Wilson's School) and someone who didnt leave their name so we will call Anon then noticed a formula for getting from A(n) -> A(m): $2^{n-m} $
Anon showed that this works for n< m as well as n> m:
Take for example the transition of A3 to A5. This is a scale factor of ${1\over2} \times {1\over2} = {1\over4}$. Using the formula, $3-5=-2$ so $2^{-2} = {1\over{2^2}} = {1\over4}$.
Using these equations, Chensheng from Wells Cathedral School calculated that the percentage needed to scale by in order to photocopy an A3 poster to an A4 sheet is about 70.7%.
To go from an A3 sheet to an A4 sheet you need to halve the area. Because area is quadratic and length is linear the length is decreased by a scale factor $\sqrt{1\over2}$ which is 0.707 so an A3 sheet needs to be reduced to 70.7% to fit an A4 sheet. 

We received lots of solutions for expressing the length of the longer side of a sheet of paper in terms of its shorter side. Well done to William (Barton Community School), Peter (Torquay Boys' Grammar School), Chensheng, Muntej, Niharika and Anon for getting it right. Here is Anon's method.

To express the longer side of paper A(n) in terms of its short side, we must consider that the longer side of paper A(n) is the same length as the shorter side of paper (n-1). Using the above formula we can see that the area of the larger sheet is twice that of the smaller sheet and we must square root the whole ratio to find the length. $1:2$ becomes $\sqrt{1} : \sqrt{2}$. Therefore, as the shorter side of paper A(n-1) is $\sqrt{2}$ times longer than the shorter side of paper A(n) and equal to length of the longer side of paper A(n), the longer side, L, can be expressed as:
L = S x $\sqrt{2}$, where S is the shorter side.
Niharika, Anon and Declan used this formula to work out the dimensions of an A0 sheet of paper with area $1m^2$. Niharika used simultaneous equations.
$l\times s=1$ and $l=s\times\sqrt{2}$
$(s\times\sqrt{2})\times s = 1$
$s^2\times\sqrt{2} = 1$
${s^2} = {1\over{\sqrt{2}}}$
${s}={1\over{2^{1\over4}}}$ so $l={1\over{2^{1\over4}}}\times 2^{1\over2} = 2^{1\over4}$
This helped them find the exact dimensions of an A4 sheet. This is Anon's answer.

The area of an A4 sheet can be calculated as $1\times{2^{0-4}} = 2^{-4} = 0.0625m^2$.
$a = l\times s$
$0.0625 = s\times\sqrt{2}\times s$
$0.0625 = s^2\times\sqrt{2}$
${s^2} = {0.0625\over{\sqrt{2}}}$
$s = \sqrt{0.0625\over{\sqrt{2}}}$
$l = {0.0625\over{\sqrt{0.0625\over{\sqrt{2}}}}}$
Can you simplify these equations?

Now see if you can use them to define A(-1) and A($1\over2$)