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)?

x^xxxxx... Again

y = x^xxxxx... so y = x^y = e^{y ln x}.
By the chain rule and the product rule, ^dx/_dy = e^{y ln x}(^dx/_dy ln x + ^y/_x).
Rearranging, we get ^dx/_dy = ^{yey ln x}/_{x(1-e^{y ln x} ln x)}.
This simplifies to ^dx/_dy = ^{xxxxxx...xxxxxx...}/_{x(1-x^xxxxx... ln x)}

Extension

What would the graph of y = x^xxxxx... 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?