PRMO 2014 question 15

15. Let \( XOY \) be a triangle with \( \angle XOY = 90^\circ \). Let \( M \) and \( N \) be the midpoints of legs \( OX \) and \( OY \), respectively. Suppose that \( XN = 19 \) and \( YM = 22 \). What is \( XY \)?

Solution:

Let \( a = OX^2 \) and \( b = OY^2 \). Since \( \angle XOY = 90^\circ \), by the Pythagorean Theorem: \[ XY^2 = OX^2 + OY^2 = a + b \] Since \( N \) is the midpoint of \( OY \), \[ XN^2 = OX^2 + \left(\frac{OY}{2}\right)^2 = a + \frac{b}{4} \] \[ XN = 19 \Rightarrow XN^2 = 361 \Rightarrow a + \frac{b}{4} = 361 \quad \text{(1)} \] Since \( M \) is the midpoint of \( OX \), \[ YM^2 = OY^2 + \left(\frac{OX}{2}\right)^2 = b + \frac{a}{4} \] \[ YM = 22 \Rightarrow YM^2 = 484 \Rightarrow b + \frac{a}{4} = 484 \quad \text{(2)} \] Multiply equation (1) by 4: \[ 4a + b = 1444 \quad \text{(3)} \] Multiply equation (2) by 4: \[ a + 4b = 1936 \quad \text{(4)} \] Solve equations (3) and (4) simultaneously: Multiply equation (3) by 4: \[ 16a + 4b = 5776 \] Subtract equation (4): \[ 16a + 4b - (a + 4b) = 5776 - 1936 \Rightarrow 15a = 3840 \Rightarrow a = 256 \] Substitute back into (3): \[ 4(256) + b = 1444 \Rightarrow 1024 + b = 1444 \Rightarrow b = 420 \] Now compute: \[ XY^2 = a + b = 256 + 420 = 676 \Rightarrow XY = \sqrt{676} = 26 \] **Answer: \( \boxed{26} \)**