New Machine Unfriendly £1 Coin

Vending machines identify coins by measuring their width. Circular coins have the same width in every direction, so designers of vending machines do not need to worry about incorrectly rotated coins causing a blockage or being misidentified. But what about seven-sided 20p and 50p coins?

Perhaps surprisingly, 20p and 50p coins also have a constant width, as show by this video. In fact, the sides of any regular shape with an odd number of sides can be curved to give the shape a constant width.

3, 5, 7 and 9 sided shapes of constant width.
Source: Wikipedia

Today, a new 12-sided £1 coin was unveiled. One reason for the number of sides was to make the coin easily identified by touch. However, as only polygons with an odd number of sides can be made into shapes of constant width, this new coin will have a different width when measured corner to corner or side to side. This could lead to vending machines not recognising coins unless a new mechanism is added to correctly align the coin before measuring.

Perhaps an 11-sided or 13-sided design would be a better idea, as this would be easily distinguishable from other coins by touch which being a constant width to allow machines to identify it.

Sunday Afternoon Maths IV Answers & Extensions

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

Arccos + Arcsin

Let y = arccos(x)
This means that x = cos(y)
This gives us this triangle, as cosine is adjacent ÷ hypotenuse.

Call the angle at the top of the triangle z
Sine is opposite ÷ hypotenuse, so sin(z) = x
So z = arcsin(x)
Angles in triangle add to π, so z = ^π/₂ - y
Hence, arcsin(x) = ^π/₂ - y
Therefore arccos(x) + arcsin(x) = y + ^π/₂ - y = ^π/₂

Extension

What are the values of arcsec(x) + arccosec(x) and arctan(x) + arccot(x)?

i ⁱ

By Euler's formula, i = e^{i π/₂}.
This means that i ⁱ = (e^{i π/₂})ⁱ.
    = e^{i 2 π/₂}.
    = e^-π/₂.

It is notable that this is a real number.

Extension

There are other values of i to the power of i. What are they?

x^xxxxx...

2 = x^xxxxx... = x².
So x = √2.

Extension

If x^xxxxx... is equal to four, what is the value of x?

4 and 5 digit numbers

Source: UKMT Grey Kangaroo 2013 Answers

Adam must have removed the final digit of his number before adding or the final digit of the sum would have been an even number. If his original number was ABCDE then, using this, we have:

52713 = ABCDE + ABCD = 11 × ABCD + E.

52713 ÷ 11 = 4792 remainder 1 so Adam's original number was 47921 which has a digit sum of 23.

Sunday Afternoon Maths IV

Here's this week's collection. Answers & extensions can be found here. Why not discuss the problems on Twitter using #SundayAfternoonMaths. This Tuesday is Maths Jam, although I won't be there as it is also parents' evening at my school.

Arccos + Arcsin

Source: EdExcel A-level C3 paper January 2007

What is the value of arccos(x) + arcsin(x)?

i ⁱ

What is the value of i to the power of i?

x^xxxxx...

If x^xxxxx... [x to the power of (x to the power of (x to the power of (x to the power of (...)))) with an infinite number of xs] is equal to two, what is the value of x?

4 and 5 digit numbers

Source: UKMT Grey Kangaroo 2013

Adam chooses a 5-digit positive integer and deletes one of its digits to form a 4-digit integer. The sum of this 4-digit integer and the original 5-digit integer is 52713. What is the sum of the digits of the original 5-digit integer?

Sunday Afternoon Maths III Answers & Extensions

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

Colliding Parallel People

By walking in a straight line, the people will follow a great circle. They begin by travelling parallel to each other, so the 1km they are apart at the start is the furthest they will be apart. They will be 1km apart again after walking half way around the world and will collide at one quarter and three quarters of the way around.

Hence, each person will walk one quarter of the circumference of the Earth before colliding, which will be 10,018km.

Another way to see that this is the answer is to assume that the people stand on the equator and walk North. They will meet at the North Pole, one quarter of the way round.

Extension

If two people stand 10km apart and walk in the same direction, how far will the have to walk until they collide due to the curvature of the Earth? (diameter of Earth = 12,742km)

Twenty

There are two ways to make 20 by multiplying three digits: 2×2×5 and 1×4×5. Listing all the possible orderings of these, we have:
145
154
415
451
514
541
225
252
522
Therefore, there are 9 different three digit numbers where the product of the digits is 20.

Extension

How many 4 digit numbers are there where the product of the digits is 20?
5 digit?
n digit?

Sunday Afternoon Maths III

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

Colliding Parallel People

If two people stand 1km apart and walk in the same direction, how far will the have to walk until they collide due to the curvature of the Earth? (diameter of Earth = 12,742km)

Twenty

Source: UKMT Junior Mathematical Olympiad 2013

How many three digit integers are there for which the product of the digits is 20?

Sunday Afternoon Maths II Answers & Extensions

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

Guess the Diagonal

CG is equal to AF as they are both diagonals in a rectangle.
AF is a radius of the triangle, so AF = 10.
Therefore CG = 10.

Extension

Calculate the lengths of CI and CJ.

Calculate the length of HG.

Sunday Afternoon Maths II

Here's this week's collection. Solutions will be posted on Monday.

Guess the Daigonal

My Best Mathematical and Logic Puzzles by Martin Gardner

Calculate the length of the diagonal CG.