# Puzzles

## Archive

Show me a random puzzle**Most recent collections**

#### Sunday Afternoon Maths LXVII

Coloured weightsNot Roman numerals

#### Advent calendar 2018

#### Sunday Afternoon Maths LXVI

Cryptic crossnumber #2#### Sunday Afternoon Maths LXV

Cryptic crossnumber #1Breaking Chocolate

Square and cube endings

List of all puzzles

## Tags

averages square numbers number scales sport prime numbers sequences clocks doubling angles triangle numbers 2d shapes differentiation routes colouring factorials logic crossnumbers cryptic clues algebra rugby dice pascal's triangle means books sums floors quadratics irreducible numbers multiples hexagons advent indices calculus digits crosswords palindromes cryptic crossnumbers balancing dodecagons probabilty multiplication probability lines division integration sum to infinity square roots spheres trigonometry cards mean regular shapes 3d shapes wordplay star numbers perfect numbers grids factors percentages addition functions squares volume circles coordinates partitions integers planes folding tube maps games chocolate arrows surds complex numbers ave proportion remainders cube numbers perimeter bases shapes money numbers taxicab geometry menace triangles speed time chalkdust crossnumber geometry parabolas fractions ellipses odd numbers coins people maths area polygons rectangles chess christmas unit fractions shape dates graphs symmetry## 12 December

There are 2600 different ways to pick three vertices of a regular 26-sided shape. Sometime the three vertices you pick form a right angled triangle.

Today's number is the number of different ways to pick three vertices of a regular 26-sided shape so that the three vertices make a right angled triangle.

## Is it equilateral?

Source: Chalkdust issue 07

In the diagram below, \(ABDC\) is a square. Angles \(ACE\) and \(BDE\) are both 75°.

Is triangle \(ABE\) equilateral? Why/why not?

## 20 December

Earlier this year, I wrote a blog post about different ways to prove Pythagoras' theorem. Today's puzzle uses Pythagoras' theorem.

Start with a line of length 2. Draw a line of length 17 perpendicular to it. Connect the ends to make a right-angled triangle.
The length of the hypotenuse of this triangle will be a non-integer.

Draw a line of length 17 perpendicular to the hypotenuse and make another right-angled triangle. Again the new hypotenuse will have a non-integer length.
Repeat this until you get a hypotenuse of integer length. What is the length of this hypotenuse?

## Cutting corners

Source: New Scientist Enigma 1773

The diagram below shows a triangle \(ABC\). The line \(CE\) is perpendicular to \(AB\) and the line \(AD\) is perpedicular to \(BC\).

The side \(AC\) is 6.5cm long and the lines \(CE\) and \(AD\) are 5.6cm and 6.0cm respectively.

How long are the other two sides of the triangle?

## Two triangles

Source: Maths Jam

The three sides of this triangle have been split into three equal parts and three lines have been added.

What is the area of the smaller blue triangle as a fraction of the area of the original large triangle?

## Equal side and angle

Source: Jim Noble on Twitter

In the diagram shown, the lengths \(AD = CD\) and the angles \(ABD=CBD\).

Prove that the lengths \(AB=BC\).

## Square deal

Source: Futility Closet

This unit square is divided into four regions by a diagonal and a line that connects a vertex to the midpoint of an opposite side. What are the areas of the four regions?