Christmas Math Problem Set 2017 - A Triangle in a Square Tree

We can start by setting all of the sides equal to variables. We take the picture of the tree and label it like this:

From the note, we can then set up some equations. We know that the area of the top right-most sector is a square, so we can express the area as $b^2$ . We know that area of the bottom left-most sector is equal to that of the top right-most sector, and the shorter sides of these two sectors are equal. We can set up this equation as: $y(x \ + \ 6) \ = \ by$ , therefore $(x \ + \ 6) \ = \ b$ . We also know that the area of the top left-most sector is $67$ units more than that of the bottom right-most sector, therefore $b^2 \ = \ ax \ + \ 67$ . Since $b \ = \ (x \ + \ 6)$ , we can substitute this into $b^2 \ = \ ax \ + \ 67$ , and get $(x \ + \ 6)^2 \ = \ ax \ + \ 67$ . Expanding and bringing all of the terms to the left side of the equation: $x^2 \ + \ (12 \ - \ a)x \ - \ 31 \ = \ 0$ . We know that $x$ has a positive integer solution. If we wanted to solve this equation, we could realize that $31$ is a prime number, and the only factorable forms of this equation, that would yield at least one integer solution to $x$ are $(x \ - \ 1)(x \ + \ 31) \ = \ 0$ and $(x \ + \ 1)(x \ - \ 31) \ = \ 0$ . This means that in order for $a$ to be an integer, $12 \ - \ a$ has to be $30$ or $-30$ , and since $a$ must be a positive integer, $12 \ - \ a \ = \ -30$ , therefore $a \ = \ 42$ .