Earlier today, I presented you with these challenges from the Mathigon puzzle advent calendar. Here they are once more, along with their solutions.

The knights are approaching.

Solution

The round where numbers are missing.

Solution

Every number is equal to the sum of the two numbers before it. For example, 1 + 3 = 4, 3 + 4 = 7, and so on. This pattern is similar to the Fibonacci sequence, where the same rule applies but it starts with 0, 1.

3. Square are you?

Solution

4. Hexa-going, hexa-going, hexagon.

Solution 3/16

5. Satsuma segment

Solution 25/16

This problem involves more steps. You will need to apply Pythagoras’s theorem for right triangles as well: the square of the longest side is equal to the sum of the squares of the other two sides. To start, extend the semicircle to create a full circle and draw a triangle underneath it. The circle intersects with the left side of the square at its midpoint. The left angle of the triangle is a right angle because any point on a circle forms a right angle with the diameter.

Thus d² = a² + (1/2)² + 5/4

and d² = (1 – a)² + 1

Thus a² + 6/4 = 1 – 2a + a² + 1

2a = 2 – 6/4 = 1/2

= 3

The value of d equals 3 when a equals 1/4.² = 25/16, so d = 5/4

I trust you had a great time solving today’s puzzles. I’ll return in 14 days…on December 25th!

Mathigon, owned by the global educational technology company Amplify, is a highly interactive and innovative online platform for mathematics. Best of all, it’s completely free! Be sure to give it a try.

Since 2015, I have been posting a puzzle every other Monday. I am constantly searching for new and interesting puzzles. If you have a suggestion, please email me.

Source: theguardian.com