Well done all of you who sent in good solutions to this problem! Many different methods were used. Saul Foresta and Julia Collins found and used the exact value of tan 22.5 degrees which they found using trig formulae; Andre Lazanu used similar triangles; Kamen Marinov used the Sine Rule and Pythagoras Theorem; Hyeyoun Chung, Arun Ayer, Ngoc Tran and Yatir Halevi used the angle bisector theorem (BA:AO=BN:NO); Robert Goudie used an ingenious construction; and Dorothy Winn used several applications of the Sine Rule.

Robert Goudie's solution.

If you take triangle AON and double its size you form triangle ACR since AC is double AO.

Original square with line AP extended outside the square to R.

So we now have two similar triangles AON and ACR and it is clearly the case that CR is double ON so CR = 48.

Since AR is the bisector of angle OAB we know that angle CAR = 22.5 ^o . Therefore angle CRA = 67.5 ^o .

Angle RCP must be 45 ^o since angle ACB is 45 ^o and angle ACR is 90 ^o . This implies that angle CPR is 67.5 ^o since 180 - 45 - 67.5 = 67.5.

Therefore triangle PCR is isosceles and so PC = CR and hence PC = 48 units.

Andre Lazanu's solution using similar triangles.

I know that in a square the diagonals are also bisectors (in fact, this is true for the rhombus, and the square is a particular case of rhombus), so angle

BAC

and angle

DAC

have both the same measure: 45^. The angles formed by the bisector of angle

BAC

with the sides of the angle have the following measure:

22^{\circ} 30 `

I noted with $L$ the side of the square, using the Pythagorean Theorem, $AC = L \sqrt{2}$

Now, I observe that $AO$ is half $AC$ : $AO = (L \sqrt{2}) / 2$

Triangles $AON$ and $ABP$ are similar, because each one has a right angle, and there is another pair of equal angles (the ones formed by the bisector). So, I have:

$\frac{AO}{AB} = \frac{AN}{AP} = \frac{ON}{BP}$

So, I obtain the following proportion:

$(L \sqrt{2} / 2) / L = 24 / BP$

$BP = 48 / \sqrt{2} = 24 \sqrt{2}$ (units)

Now, I see that angle $BPA$ has $67^{\circ} 30 `$ and angle $BNP$ has the same measure, so triangle $BNP$ is isosceles with $BP = BN$ .

Here I observe that triangles $ABN$ and $ACP$ are similar, because their angles are respectively equal. So

$\frac{AB}{AC} = \frac{BN}{PC} = \frac{AN}{AP}$

As $AB / AC = 1 / \sqrt{2}$ it follows that $PC = \sqrt{2} BN$ .

$PC = \sqrt{2} BN = \sqrt{2} BP = 48$