Tessellate S.T.E.M.S. (2019) - Mathematics - Category B - Set 1 - Objective Problem 1

Let the length of $AB$ be expressed as $10a + b$ , where $a$ and $b$ are natural numbers less than $10$ . Let $O$ be the center of the circle, and let $P$ be the intersection point of the two chords. Then, $OP = \sqrt{OC^2-CP^2}$ ,and hence, the condition is equivalent to $\sqrt{OC^2-CP^2}$ being a perfect square. Substituting values and taking the square terms out, we see that this is equivalent to $11(a-b)(a + b )$ being a perfect square. Taking into account that $a$ and $b$ are naturals less than $10$ and the fact that $11$ must divide $(a - b)(a + b)$ for the expression to be a perfect square, we see that this can occur only when $a=6$ and $b=5$ , thus giving us the length of diameter $AB$ as $65$ .