Students will learn three ways of solving the three squares geometry puzzle. That is, in the figure below, what is \(\alpha + \beta + \gamma?\)
Students should be given time to think about how to solve the problem, but most students won't be able to solve it, and after a while, you can give them the hint shown here. After students have received the hint, the puzzle is very doable. After students have found the solution which follows from the hint, show them these additional methods of solution.
Conclude by giving your students these challenges:
In this problem, every line is horizontal or vertical, and every line is unique. If there are 2 horizontal lines, and 2 vertical lines, then there are 4 crossings. If there are 3 horizontal lines, and 2 vertical lines, how many crossings are there? If there are at least 2 horizontal lines, and 7 lines in total, what is the least number of crossings? the most? Can you find all possible crossings with 7 lines? Can you find the least/greatest number of crossings for 10 lines? 15? 50? any number of lines?
Make the equation below true by replacing each letter with a unique digit (0-9).$$W + O = OF$$
Here's the solution:
The largest sum of two digits is \(9 + 9 = 18,\) so \(O = 1.\) Now we have \(W + 1 = 1F.\) The only way to make \(W + 1 \ge 10\) is if \(W = 9.\) Now we have \(9 + 1 = 1F,\) so \(F = 0.\) In conclusion, the equation is \(9 + 1 = 10.\) Note that it doesn't matter whether we accept leading zeros, such as \(05,\) because if \(O = 0,\) then \(W = F,\) which violates the condition that \(W\) and \(F\) must be unique.