NZMO Round One

Solution:

Wlog let $AB = 1$ and $BC = a$. Also let $BD = x$. We will try to find all the lengths in the diagram in terms of $a$.

By Pythagoras in $\triangle ABC$ we get $AC = \sqrt{a^2 - 1}$. By the angle-bisector theorem we get $\frac{AD}{DC} = \frac{AB}{BC} = \frac{1}{a}$, and so $DC = a \times AD$. This can be substituted into $AD + DC = AC = \sqrt{a^2 - 1}$ to get $AD(1 + a) = \sqrt{a^2 - 1}$. Therefore
$$AD = \frac{AC}{1 + a} = \frac{\sqrt{a^2 - 1}}{a + 1}.$$
Hence $AD^2 = \frac{a^2 - 1}{(a + 1)^2} = \frac{a - 1}{a + 1}$. Now consider Pythagoras in triangle $\triangle BAD$.
$$x^{2} = 1^{2} + AD^{2} = 1 + \frac{a - 1}{a + 1} = \frac{2a}{a + 1}.$$
Now we do the two directions of the 'if and only if' separately:

First, assuming $\frac{1}{x} - \frac{1}{a} = \frac{1}{2}$ we get $x = \frac{2a}{a + 2}$. Hence
$$\frac{2a}{a + 1} = x^2 = \left(\frac{2a}{a + 2}\right)^2 = \frac{4a^2}{a^2 + 4a + 4}$$
$$a^{2} + 4a + 4 = 2a(a + 1)$$
$$0 = a^{2} - 2a - 4$$
So by the quadratic formula we get $a = 1 \pm \sqrt{5}$ but since $a > 0$ we must have $a = 1 + \sqrt{5}$.
Therefore $x = \frac{2a}{a + 2} = \frac{2(1 + \sqrt{5})}{(1 + \sqrt{5}) + 2} = \sqrt{5} - 1$.
$$\therefore a - x = (1 + \sqrt{5}) - (\sqrt{5} -1) = 2.$$
So $\frac{1}{x} - \frac{1}{a} = \frac{1}{2}$ implies $a - x = 2$.

Second, assuming $a - x = 2$ we get $x = a - 2$. Hence
$$\frac{2a}{a + 1} = x^2 = (a - 2)^2 = a^2 -4a + 4$$
$$2a = (a^2 -4a + 4)(a + 1)$$
$$0 = a^3 -3a^2 -2a + 4$$
$$0 = (a - 1)(a^2 -2a - 4)$$
Since $a - x = 2$ we have $a > 2$ so $(a - 1) \neq 0$ and thus $a^2 - 2a - 4 = 0$. This gives us $a = 1 + \sqrt{5}$ and so again $x = a - 2 = \sqrt{5} - 1$.
$$\therefore \frac{1}{x} - \frac{1}{a} = \frac{1}{\sqrt{5} - 1} - \frac{1}{\sqrt{5} + 1} = \frac{1}{2}.$$
So $a - x = 2$ implies $\frac{1}{x} - \frac{1}{a} = \frac{1}{2}$.

In summary $\frac{1}{x} - \frac{1}{a} = \frac{1}{2}$ is equivalent to $a - x = 2$.

Solution B:

Let $E$ be the foot of the perpendicular from $D$ to $BC$. Note that $E$ is the reflection of $A$ about $BD$ (because $BD$ is an angle bisector) and so we have congruent triangles $\triangle ABD \equiv \triangle EBD$. Also construct point $N$ on side $BC$ to be the reflection of $B$ about line $DE$. So we have three congruent triangles:
$$\triangle ABD\equiv \triangle EBD\equiv \triangle END.$$
Therefore $AB = EB = EN$ and hence $2AB = BE + EN = BN$. We also get $BD = DN$. Let $F$ be the point on line $BC$ such $BD = BF$.

Assume $BC - BD = 2AB$. Since $BD = DN$ and $2AB = BN$ we can get $BC - DN = BN$. Therefore
$$DN = BC - BN = CN$$
and so triangle $DNC$ is isosceles. Now let $\theta = \angle BCA$. Since $\triangle DNC$ is isosceles, we get $\angle CDN = \theta$ and thus $\angle END = 2\theta$. Therefore $\angle EBD = 2\theta$ and $\angle DBA = 2\theta$ and thus $\angle CBA = 4\theta$. Thus
$$90^{\circ} = \angle BCA + \angle CBA = \theta +4\theta .$$
Hence $\theta = 18^{\circ}$. Therefore
$$\frac{AB}{BD} = \cos (\angle ABD) = \cos (2\theta) = \cos (36^{\circ}) = \frac{1 + \sqrt{5}}{4}$$
$$\mathrm{and}\quad \frac{BC}{BD} = \frac{\sin(\angle CDB)}{\sin(\angle BCD)} = \frac{\sin(126^{\circ})}{\sin(\angle 18^{\circ})} = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}$$
Therefore $1 - \frac{BD}{BC} = 1 - \frac{\sqrt{5} - 1}{\sqrt{5} + 1} = \frac{2}{\sqrt{5} + 1} = \frac{BD}{2AB}$.
Hence
$$\frac{1}{BD} -\frac{1}{BC} = \frac{1}{2AB}$$
as required.

- If $BC - BD > 2AB$ then using similar logic we can get $\angle BCA < 18^{\circ}$ which implies
$$\frac{1}{BD} -\frac{1}{BC} < \frac{1}{2AB}$$
- Conversely if $BC - BD < 2AB$ then we get $\angle BCA > 18^{\circ}$ which implies
$$\frac{1}{BD} -\frac{1}{BC} >\frac{1}{2AB}$$
Therefore we conclude that $BC - BD > 2AB$ if and only if $\frac{1}{BD} - \frac{1}{BC} < \frac{1}{2AB}$.