## Monday, 21 July 2014

### Sunday Afternoon Maths XXI Answers & Extensions

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

#### Sum Equals Product

If $$a+b=a\times b$$, then:
$$ab-b=a$$ $$b(a-1)=a$$ $$b=\frac{a}{a-1}$$
This will work for any $$a \not= 1$$ ($$a=1$$ will not work as this will mean division by zero).
##### Extension
(i) Given a number $$a$$, can you find a number $$b$$ such that $$b-a =\frac{b}{a}$$?
(ii) Given a number $$a$$, can you find a number $$b$$ such that $$b-a =\frac{a}{b}$$?
(iii) Given a number $$a$$, can you find a number $$b$$ such that $$a-b =\frac{b}{a}$$?
(iv) Given a number $$a$$, can you find a number $$b$$ such that $$a-b =\frac{a}{b}$$?

#### Wool Circles

Starting with the person who starts with the wool and going anti-clockwise, number the people $$0,1,2,3,4,...$$. As the wool is passed, it will be held by people with numbers:
$$0,a+1,2(a+1),3(a+1),...,k(a+1),...$$
For example, if $$n=10$$ and $$a=3$$:
The first person will have the wool again when:
$$k(a+1)\equiv 0 \mod n$$
or
$$k(a+1)=ln$$
This will first occur when (hcf is highest common factor):
$$l=\frac{a+1}{\mbox{hcf}(a+1,n)} \mbox{ and } k=\frac{n}{\mbox{hcf}(a+1,n)}$$
$$k$$ is also the number of people who are holding the wool. So the number of different coloured balls needed is:
$$\frac{n}{\left(\frac{n}{\mbox{hcf}(a+1,n)}\right)}$$$$= \mbox{hcf}(a+1,n)$$
The ball is passed around the circle of $$n$$ people again. This time, the number of people missed alternates between $$a$$ and $$b). How many different coloured balls of wool are now needed? y=(1-\sqrt{2x})^2 ## Sunday, 20 July 2014 ### Sunday Afternoon Maths XXI Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit. This Tuesday is Maths Jam, make sure you're there! #### Sum Equals Product \(3$$ and $$1.5$$ are a special pair of numbers, as $$3+1.5=4.5$$ and $$3\times 1.5=4.5$$ so $$3+1.5=3\times 1.5$$.
Given a number $$a$$, can you find a number $$b$$ such that $$a+b=a\times b$$?

#### Wool Circles

$$n$$ people stand in a circle. The first person takes a ball of wool, holds the end and passes the ball to his right, missing $$a$$ people. Each person who receives the wool holds it and passes the ball on to their right, missing $$a$$ people. Once the ball returns to the first person, a different coloured ball of wool is given to someone who isn't holding anything and the process is repeated. This is done until everyone is holding wool.
For example, if $$n=10$$ and $$a=3$$:
In this example, two different coloured balls of wool are needed.

In terms of $$n$$ and $$a$$, how many different coloured balls of wool are needed?

## Monday, 14 July 2014

### Sunday Afternoon Maths XX Answers & Extensions

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

#### Folding A4 Paper

The sides of A4 paper are in the ratio $$1:\sqrt{2}$$. Let the width of the paper be 1 unit. This means that the height of the paper is $$\sqrt{2}$$ units.
Therefore on the diagram, $$AE=1, BE=1, AC=\sqrt{2}$$. By Pythagoras' Theorem, $$AB=\sqrt{2}$$, so $$AB=AC$$.
$$BF=DF=\sqrt{2}-1$$ so by Pythagoras' Theorem again, $$BD=2-\sqrt{2}$$. $$CD=1-(\sqrt{2}-1)=2-\sqrt{2}$$. Hence, $$CD=BD$$ and so the shape is a kite.
##### Extension
Prove that for a starting rectangle with the sides in any ratio, the resulting shape is a cyclic quadrilateral.

#### Bézier Curves

If
$$P_0=\left(0,1\right)$$ $$P_1=\left(0,0\right)$$ $$P_2=\left(1,0\right)$$
then
$$Q_0=t\left(0,0\right)+(1-t)\left(0,1\right)=\left(0,1-t\right)$$ $$Q_1=t\left(1,0\right)+(1-t)\left(0,0\right)=\left(t,0\right)$$
and so
$$R_0=t\left(t,0\right)+(1-t)\left(0,1-t\right)=\left(t^2,(1-t)^2\right)$$
which means that
$$x=t^2$$$$y=(1-t)^2$$
or
$$y=(1-\sqrt{x})^2$$
##### Extension
What should $$P_0$$, $$P_1$$ and $$P_2$$ be to get a curve with Cartesian equation
$$y=(1-\sqrt{2x})^2$$

## Sunday, 13 July 2014

### Sunday Afternoon Maths XX

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

#### Folding A4 Paper

A Piece of A4 paper is folded as shown:

#### Bézier Curve

A Bézier curve is created as follows:

Image source: WikiPedia
1) A set of points $$P_0$$, ..., $$P_n$$ are chosen (in the example $$n=4$$).
2) A set of points $$Q_0$$, ..., $$Q_{n-1}$$ are defined by $$Q_i=t P_{i+1}+(1-t) P_i$$ (shown in green).
3) A set of points $$R_0$$, ..., $$R_{n-2}$$ are defined by $$R_i=t Q_{i+1}+(1-t) Q_i$$ (shown in blue).
.
.
.
$$n$$) After repeating the process $$n$$ times, there will be one point. The Bézier curve is the path traced by this point at $$t$$ varies between 0 and 1.

What is the Cartesian equation of the curve formed when:
$$P_0=\left(0,1\right)$$ $$P_1=\left(0,0\right)$$ $$P_2=\left(1,0\right)$$

## Monday, 30 June 2014

### Sunday Afternoon Maths XIX Answers & Extensions

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

#### Dartboard

$$\pi (1)^2 - \pi (\frac{1}{2})^2 + \pi (\frac{1}{3})^2 - \pi (\frac{1}{4})^2 + \pi (\frac{1}{5})^2 - ...$$ $$=\sum_{i=1}^\infty \frac{\pi (-1)^{i-1}}{i^2}$$ $$=\pi\sum_{i=1}^\infty \frac{(-1)^{i-1}}{i^2}$$ $$=\pi\left(\frac{\pi^2}{12}\right)$$ $$=\frac{\pi^3}{12}$$
##### Extension
Prove that
$$=\sum_{i=1}^\infty \frac{(-1)^{i-1}}{i^2}=\frac{\pi^2}{12}$$

#### Multiples of Three

If $$10A+B=3n$$, then:
$$100A+B = 100A-10A+10A+B$$$$=90A+3n$$$$=3(30A+n)$$
So $$A0B\div3=30A+n$$.
##### Extension
What are $$A00B\div3$$, $$A000B\div3$$, $$A0000B\div3$$, etc?

## Sunday, 29 June 2014

### Sunday Afternoon Maths XIX

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

#### Dartboard

Concentric circles with radii 1, $$\frac{1}{2}$$, $$\frac{1}{3}$$, $$\frac{1}{4}$$, ... are drawn. Alternate donut-shaped regions are shaded.
What is the total shaded area?

#### Multiples of Three

If the digits of a number add up to a multiple of three, then the number is a multiple of three. Therefore if a two digit number, $$AB$$ (first digit $$A$$, second digit $$B$$; not $$A\times B$$), is a multiple of three, then $$A0B$$ is also a multiple of three.
If $$AB\div 3=n$$, then what is $$A0B\div 3$$?

## Monday, 23 June 2014

### Sunday Afternoon Maths XVIII Answers & Extensions

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

#### Parabola

The co-ordinates of the points where the lines intersect the parabola are $$(a,a^2)$$ and $$(-b,b^2)$$. Hence the gradient of the line between them is:
$$\frac{a^2-b^2}{a-(-b)}=\frac{(a+b)(a-b)}{a+b}=a-b$$
Therefore the y-coordinate is:
$$b^2 + b(a-b) = ba$$
Ferdinand Möbius, who discovered this property called the curve a Multiplicationsmaschine or 'multipliction machine' as it could be used to perform multiplication.
##### Extension
How could you use the graph of $$y=x^2$$ to divide 100 by 7?

#### Seven Digits

Let's call Dr. Dingo's number $$n$$. If the number is squared twice then multiplied by $$n$$, we get $$n^5$$.
For all integers $$n$$, the final digit of $$n^5$$ is the same as the final digit of $$n$$. In other words:
$$n^5\equiv n \mod 10$$
Therefore, the final digit of Dr. Dingo's number is 7.
$$7^5=16807$$ $$17^5=1419857$$ $$27^5=14348907$$
So, in order for the answer to have seven digits, Dr. Dingo's number was 17.
##### Extension
For which integers $$m$$ does there exist an integer $$n$$ such that for all integers $$x$$:
$$x^n\equiv x \mod m$$