+1768 In the diagram, ABCD is a square. Find PR.
L, M, N, O, are midpoints of sides.
Side AB = 12
+1632 This diagram is not to scale and misleading. Because ABCD is a square, PR is collinear with NL. NL also bisects AM and OC, so AP : PM = 1 : 1, and thus the center of the circle, say point X, XP : PL = 1 : 1, likewise XR : RN = 1 : 1. Thus, PR = NL/2 = 12/2 = 6.
+1632 A pretty simple problem using coordinate geometry:
Let ABCD be a square with side length 12, and let A be (0,0).
Line AM: y = x/2
Line DL: y = -2x + 12
The intersection of these two lines, P, is calculated by setting the two lines equal to each other:
x/2 = -2x + 12, 5x/2 = 12, x = 24/5
y = 24/10 = 12/5, so P has coordinates (24/5, 12/5)
Line OC: y = x/2 + 6
Line NB: y = -2x + 24
The intersection of these two lines, Q, is likewise calculated by setting OC = NB:
x/2 + 6 = -2x + 24, 5x/2 = 18, x = 36/5
y = 36/10 + 6= 48/5, so Q has coordinates (36/5, 48/5)
Using the distance formula between P and Q, PQ = sqrt[(36/5 - 24/5)^2 + (48/5 - 12/5)^2] = sqrt(144/25 + 1296/25) = sqrt(1440/25).
Simplifying, your final answer is \(12\sqrt{10}\over5\)
