Q1

Let 𝑃𝑄𝑅𝑆 be a trapezium with 𝑄𝑃 parallel to 𝑅𝑆 and with 𝑄𝑃 = 20, 𝑅𝑆 = 26 and 𝑃𝑆 = 𝑄𝑅 = 𝑛, for some positive integer 𝑛. The length of the diagonal 𝑄𝑆 is also a positive integer. What is the maximum possible value of 𝑛?

Consider the two right-angled triangles shown, with height ℎ and vertex 𝑄. Applying Pythagoras’ Theorem, first to the triangle with hypotenuse 𝑄𝑅 gives 3 2 + ℎ 2 = 𝑛 2 (1) and then to the triangle with hypotenuse 𝑄𝑆 gives 232 + ℎ 2 = 𝑚 2 (2). Subtracting, (2) − (1) gives 232 − 3 2 = 𝑚 2 − 𝑛 2 . Therefore 520 = (𝑚 + 𝑛) (𝑚 − 𝑛). As both 𝑚 + 𝑛 and 𝑚 − 𝑛 are integers, we can consider the factor pairs of 520. Taking first 𝑚 + 𝑛 = 520 and 𝑚−𝑛 = 1 produces non-integer values of 𝑚 and 𝑛. Moving to 𝑚+𝑛 = 260 (3) and 𝑚 − 𝑛 = 2 (4) however leads to 2𝑛 = 258 and therefore 𝑛 = 129. Any other factor pairs will have values closer together than 260 and 2 and so will produce a smaller value of 𝑛.