Maclaurin 2016 Q3

Let the three squares with centres \(A\), \(B\) and \(C\) have sides of length \(2a\),\(2b\) and \(2c\) respectively. Then

Introduce coordinate axes as shown.

Then \(A = (a, a)\), \(B = (b, 2a + b)\) and \(C = (-c, c)\).

Now consider the right-angled triangles \(OBF\) and \(ACP\) shown in the next diagram, where \(PA\) is parallel to the \(x\)-axis.

We have \(BF = b\) and \(OF = 2a + b\). Also \(CP = c - a\) and \(PA = c + a\). Using equation (∗), we get \(CP = b\) and \(PA = 2a + b\). Thus \(CP = BF\) and \(PA = OF\). Using either congruent triangles (SAS) or Pythagoras' Theorem, we therefore obtain \(CA = OB\).

Now the gradient of \(OB\) is \(\dfrac{FO}{BF}\) and the gradient of \(AC\) is \(-\dfrac{CP}{PA}\). It follows that the product of these gradients is \(-1\), and hence \(OB\) and \(AC\) are perpendicular.