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:
$$\frac{n}{\left(\frac{n}{\mbox{hcf}(a+1,n)}\right)}$$$$= \mbox{hcf}(a+1,n)$$
Extension
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:
What shape is made?

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

The shaded area is:
$$\pi (1)^2 - \pi (\frac{1}{2})^2 + \pi (\frac{1}{3})^2 - \pi (\frac{1}{4})^2 + \pi (\frac{1}{5})^2 - ...$$ $$=\sum_{i=1}^\infty \frac{\pi (-1)^{i-1}}{i^2}$$ $$=\pi\sum_{i=1}^\infty \frac{(-1)^{i-1}}{i^2}$$ $$=\pi\left(\frac{\pi^2}{12}\right)$$ $$=\frac{\pi^3}{12}$$
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 \(x\):
$$x^n\equiv x \mod m$$