x/2, x + rÖ2/2 and r

By pythagoras,

( 1 2 x)2 + ( rÖ2 2 + x)2 = r2

x2 4 + 2r2 4 + xrÖ2 + x2 = r2

x2 + 2r2 + 4xrÖ2 + 4x2 = 4r2

5x2 + 4xrÖ2 = 2r2

Substitute y = rÖ2 (length of the side of the large square)

5x2 + 4xy = y2

5x2 + 4xy - y2 = 0

x = -4y ± Ö (4y)2 + 20y2 10

x = -4y ±   ____ Ö36y2   10

x = -4y ±6y 10

10x = 2y or -10y

ratio must be positive, therefore 10x = 2y, therefore y = 5x
QED

Solution to part B.

Submitted by Samantha Gooneratne, Colombo International School, Sri Lanka. Well done Samantha! Her teacher also solved the problem using similar triangles, which I have include below Samantha's solution.

Let X be the mid point of PQ, C the center of the circle and r the radius.

RX + XC = r Hence PR sin 60 o + CM cos 60 o = r Ö3 2 + 1 2 r = r Therefore Ö3.PQ = r Now LM = 2(r sin 60) so LM = Ö3.r LM = Ö3.(Ö3.PQ) LM = 3PQ

Solution using similar triangles:

NR is a diamter so NLR = 90 ^o
but NLM = 60 ^o so RLP = 30 ^o
Now LPR = 120 ^o so LRP = 30 ^o
Hence LP = pr but PR = PQ
so LP = PQ = QM
LM = 3 PQ