This post was part of the Chalkdust 2016 Advent Calendar.

Welcome to the sixth day of the 2016 Chalkdust Advent Calendar. Today, we have another two puzzles for you to enjoy, plus the answer to the puzzle from the 02 December.

If you’ve spent time browsing the internet recently, you will have noticed that we’re not the only site running an advent calendar. Today’s puzzles are taken from two of our rival calendars, run by Matthew Scroggs (who?) and nrich.

First a puzzle from my own advent calendar.

### Digital sums

Source: mscroggs.co.uk Advent calendar, day 6
When you add up the digits of a number, the result is called the digital sum.

How many different digital sums do the numbers from 1 to 1091 have?

Second, a puzzle from the excellent nrich Advent calendar.

### Wipeout

Source: nrich Secondary Advent calendar, day 10
You are given the numbers 1,2,3,4,5,6 and are allowed to erase one. If you erase 5, the mean of the remaining numbers will be 3.2. Is it possible to erase a number so that the mean of the remaining number is an integer?

If you are given the numbers 1,2,3,4,…,N, can you erase one number so that the mean of the remaining numbers is an integer?

Back on 02 December, I posted a longer version of the following puzzle:

### Decorations

You love big equilateral triangles but hate small equilateral triangles. Can you arrange ten red and blue baubles in a triangle so that no three baubles of the same colour form the vertices of an equilateral triangle?

This is not possible. To see this, first pick a colour for the central bauble. I’ve picked red.

Now we try to colour the rest without making a triangle. One of the three baubles on the following triangles must be red (otherwise there is a blue triangle). Pick one of them to make red. If a different one is red, rotate the triangle to make this one red.

The baubles must be coloured as follows. In each step, the colour is chosen to avoid a triangle.

Now, the bauble shown in green below cannot by either colour, as in each case it makes a triangle.

Hence, it is impossible to find a triangle without a smaller triangle.

This post was part of the Chalkdust 2016 Advent Calendar.

Welcome to the second day of the 2016 Chalkdust Advent Calendar. Today, we have a lovely Christmas-themed puzzle for you to enjoy. Like so many good puzzles, this one is inspired by a puzzle that I found in a Martin Gardner book.

### Decorations

It’s the start of Advent, so you decide to decorate your flat with some homemade decorations. You have a large number of red and blue baubles that you bought in last year’s January sales.

You like equilateral triangles, so you decide to stick ten baubles together into a large equilateral as shown below.

You are not, however, happy with this arrangement of colours as you hate smaller equilateral triangles, and three of the red baubles lie on the vertices of an equilateral triangle (as shown below).
Is it possible to arrange ten baubles into a triangle so that no three baubles of the same colour form the vertices of an equilateral triangle?

I’ll be back later in Advent with the answer to this puzzle and more puzzles…

# Visualising mathematics with 3D printing

There are too many maths books out there that repeat the same old boring stories about Pythagoras drowning people for inventing irrational numbers, Gauss adding up numbers quickly, or Dirichlet eating too many grapes[citation needed]. Luckily, Visualising Mathematics with 3D Printing (Amazon UK, USA) is not one of these: its author, Henry Segerman, has achieved the seemingly impossible and written something refreshingly original and different.

In this book, Segerman takes you on a tour of modern geometry, exploring symmetry, hyperbolic surfaces, four-dimensional shapes, knots and much more. These concepts are hard to visualise. While pictures in a book can help, there is only so much that can be shown in a 2D picture. This is where this book comes into its own: it focuses on using 3D objects to visualise these concepts, with instructions for 3D-printing the objects, as well as interactive simulations, available on the book’s website. For many objects, such as the tiled genus 3 surface shown below, the ability to explore these objects from multiple angles is key to understanding them.

One of my favourite concepts in the book (and one of Segerman’s favourites; perhaps this is why it comes across as the most interesting) is the stereographic projection. By placing a light source at the top point of a sphere, this projection maps the 2D surface of a 3D ball onto the 2D plane. Incredibly, this projection preserves angles: a right angle on the sphere is mapped to a right angle on the plane, as seen in the object below.

Stereographic projection of a dodecahedron

A 3D shape can be represented by lines drawn on a sphere; these lines can then be stereographically projected onto the 2D plane. The result of this projection for a dodecahedron is shown on the left.

The same process can be followed to project a 4D solid onto the 3D “plane”. Below, you can see a model of half of the 120-cell, a regular 4D shape made by connecting dodecahedra. Without a 3D model of this or the simulation below, it is impossible to get an idea of what this projection looks like.

There is so much more in this book that I don’t have space here to show you. My best advice is to go out and buy yourself a copy of the book. Be warned though, that you may find yourself also buying a very expensive 3D printer.

# Maths at EMF camp

Header picture: Simon Arlott, CC-BY-ND 2.0

Recently, I went to a festival. More specifically, I went to Electromagnetic Field (EMF), where I spent three days surrounded by hackers, geeks, scientists, engineers and mathematicians, who had all brought along their coolest toys to show off. In this week’s blog post, I’m going to share my (mostly mathematical) highlights with you.

When learning A-level maths, much time is devoted to learning how to differentiate and integrate. For this week’s blog post, I have collected some puzzles based on these skills. They should be fun to solve, present a few surprises and maybe even provide a teacher or two with an extra challenge for capable students.

The answers to these puzzles will appear here from Sunday at 8am.

# Stopping distances in the Highway Code are wrong

To pass your driving theory test in the UK, you need to know how far it will take you to stop if you brakes at a particular speed. But the numbers given in the Highway Code are based on inaccurate calculations that exist only because they formed an easy formula for stopping distances when we thought in feet instead of metres. Simple mechanics shows that the Highway Code systematically underestimates how long it takes to stop. At the end we propose an easy, safer equation for stopping distances.

The stopping distances you need to learn for your driving theory test are given in the Highway Code as:

Speed Stopping distance
20mph 6 + 6 = 12m
30mph 9 + 14 = 23m
40mph 12 + 24 = 36m
Speed Stopping distance
50mph 15 + 38 = 53m
60mph 18 + 55 = 73m
70mph 21 + 75 = 96m

Your stopping distance is given by thinking distance + braking distance. These numbers disguise a fascinating fact that you can only see if you write the stopping distances in feet: