Skip over navigation
Cambridge University Faculty of Mathematics NRich logo
menu search
  • Teachers expand_more
    • Early years
    • Primary
    • Secondary
    • Post-16
    • Events
    • Professional development
  • Students expand_more
    • Primary
    • Secondary
    • Post-16
  • Parents expand_more
    • Early Years
    • Primary
    • Secondary
    • Post-16
  • Problem-Solving Schools
  • About NRICH expand_more
    • About us
    • Impact stories
    • Support us
    • Our funders
    • Contact us
  • search

Or search by topic

Number and algebra

  • The Number System and Place Value
  • Calculations and Numerical Methods
  • Fractions, Decimals, Percentages, Ratio and Proportion
  • Properties of Numbers
  • Patterns, Sequences and Structure
  • Algebraic expressions, equations and formulae
  • Coordinates, Functions and Graphs

Geometry and measure

  • Angles, Polygons, and Geometrical Proof
  • 3D Geometry, Shape and Space
  • Measuring and calculating with units
  • Transformations and constructions
  • Pythagoras and Trigonometry
  • Vectors and Matrices

Probability and statistics

  • Handling, Processing and Representing Data
  • Probability

Working mathematically

  • Thinking mathematically
  • Developing positive attitudes
  • Cross-curricular contexts

Advanced mathematics

  • Decision Mathematics and Combinatorics
  • Advanced Probability and Statistics
  • Mechanics
  • Calculus

For younger learners

  • Early Years Foundation Stage

Counting Triangles

Age 11 to 14
Challenge Level Yellow star
  • Problem
  • Getting Started
  • Student Solutions

This solution was submitted by Andrei Lazanu from School 205, Bucharest, Romania. Congratulations Andrei.

To calculate the total number of triangles, I use the combinations without repetition, because in a triangle it doesn't matter about the order of the vertices and without repetition I can't put one vertex of the cube in a triangle more than once. Now, I apply the formula for the combinations without repetition: $${C_n^k}= {{n!} \over {k!(n-k)!}}$$I now apply the formula for $k=3$ and $n=8$: $${C_8^3} = {{8!} \over {3!(8-3)!}} = {{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1} \over {3\times 2\times 1\times 5\times 4\times 3\times 2\times 1}} = {56}$$

So, the total number of triangles is 56.

Now, I search for all the different types of triangles, knowing that the sides of the triangle could be sides of the cube (a), diagonals of the sides of the cube (b) and diagonals of the cube (c).

I write all the possibilities:

Combination Observation
aaa
Impossible
aab
OK
aac
Impossible
abb
Impossible
abc
OK
acc
Impossible
bbb
OK
ccc
Impossible
bbc
Impossible
bcc
Impossible

So, there are only three different types of triangle.


You may also like

All in the Mind

Imagine you are suspending a cube from one vertex and allowing it to hang freely. What shape does the surface of the water make around the cube?

Instant Insanity

Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.

Is There a Theorem?

Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?

  • Tech help
  • Accessibility Statement
  • Sign up to our newsletter
  • Twitter X logo

The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice.

NRICH is part of the family of activities in the Millennium Mathematics Project.

University of Cambridge logo NRICH logo