Earlier today I set you the following two problems. Here they are again with solutions.
1. Tricky triangle
What is the length of AD, the dashed line?
Solution: x = 1
The reason that most people fail to solve this one is because the triangle is degenerate: it is, actually, a line, not a triangle.
The horizontal side has length 8. The other two sides add up to 8, so these two sides will lie on top of the horizontal line.
People who spot that the triangle is degenerate still get the puzzle wrong, as they immediately assume that the distance between A and D is 0. A moment’s reflection, and you realize it is 1.
2. Rockin’ rectangles
It’s easy enough to divide a square into five rectangles.
But can you:
i) Divide a square into five rectangles such that no two rectangles share an entire side in common – i.e. the sides of abutting rectangles never start and finish at the same points? (The rectangles do not need to be the same size or shape.)
ii) Divide an 11 x 11 square into five rectangles in such a way that the ten side lengths are the whole numbers from 1 to 10? Hint: if you have solved part i) use this arrangement of rectangles.
Solution
i) You can change the size of the rectangles, and flip the ones around the side to go the other way, but this is the only pattern that works.
ii) There are two solutions, both have a kernel of 4×7. You will get there by trial and error, and by observing the constraints. The side length is 11, so there must be four sets of two side lengths that add up to 11. The area is 121, so the sum of the areas of the 5 rectangles adds up to 121. In other words, how can you divvy up the numbers from 1 to 10 into pairs, the sum of whose products is 121?
I hope you enjoyed these puzzles, I will be back in two weeks time.
I’ve been setting a puzzle here on alternate Mondays since 2015. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
