Advent calendar 2016

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?


Show me a random puzzle
 Most recent collections 

Sunday Afternoon Maths LXVII

Coloured weights
Not Roman numerals

Advent calendar 2018

Sunday Afternoon Maths LXVI

Cryptic crossnumber #2

Sunday Afternoon Maths LXV

Cryptic crossnumber #1
Breaking Chocolate
Square and cube endings

List of all puzzles


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


Show me a random puzzle
▼ show ▼
© Matthew Scroggs 2019