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## Monday, 24 March 2014

### Sunday Afternoon Maths V Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### Two Lines

Let A have the equation y = mx + c. B will have the equation y = cx + m.
Therefore, mx + c = cx + m.
Which rearranges to x(m - c) = m - c.
So x = 1.
Substituting back in, we find y=m+c.
The co-ordinates of the point of intersection are (1,m+c).
##### Extension
Let a, b and c be three distinct numbers. What can you say about the points of intersection of the parabolas:
y = ax2 + bx + c,
y = bx2 + cx + a,
and y = cx2 + ax + b?

#### Odd Sums

They are all equal to one third.
The sum of the first n odd numbers is n2 (this can be proved by induction). This means that (sum of the first n odd numbers) ÷ (sum of the next n odd numbers) is n2/(2n)2-n2 = n2/4n2-n2 = n2/3n2 = 1/3
##### Extension
What is (sum of the first n odd numbers) ÷ (sum of the first n even numbers)?

#### xxxxxx... Again

y = xxxxxx... so y = xy = ey ln x.
By the chain rule and the product rule, dx/dy = ey ln x(dx/dy ln x + y/x).
Rearranging, we get dx/dy = yey ln x/x(1-ey ln x ln x).
This simplifies to dx/dy = xxxxxx...xxxxxx.../x(1-xxxxxx... ln x)
##### Extension
What would the graph of y = xxxxxx... look like?

#### Folding Tube Maps

Once the map is folded, it will look like this:
For the final tetrahedron to be regular, the red lengths must be equal. Let each red length be 2 (this will get rid of halves in the upcoming calculations). By drawing a vertical line in we can work out the width and height of the rectangle:
The width of the rectangle is 3 (one and a half red lengths). Using Pythagoras' Theorem in the blue triangle, we find that the height of the rectangle is √3. Therefore, the ratio of the rectangle is √3:3 or 1:√3.
##### Extension
If the ratio of the rectangle is 1:a, what is the ratio of the lengths of the sides of the tetrahedron?