South African Mathematical Olympiad, Third Round

Figure 1

Put $x = OD$ and $y = OE$. Let $BO$ extended meet the circle again in $F$ (so that $DF = 1-x$). Since $\angle BFC = 45^\circ$ and $BF$ is a diameter of the circle, we have $\angle BCF = 90^\circ$ and $BC = CF = \sqrt{2}$. Similarly, if $CE$ extended meets the circle again in $G$, we have $GE = 1-y$, and $CG$ and $BF$ are the diagonals of square $BCFG$ that intersect perpendicularly at $O$. Therefore, from the right triangles $BOE$ and $COD$, we have $CD = \sqrt{1+x^2}$ and $BE = \sqrt{1+y^2}$.
By the similarity of triangles $BDA$ and $CDF$, we have $\frac{AD}{DF} = \frac{BD}{CD}$, i.e., $\frac{AD}{1-x} = \frac{1+x}{\sqrt{1+x^2}}$, so that
$$AD = \frac{1-x^2}{\sqrt{1+x^2}}.$$
Then, by the similarity of triangles $BEG$ and $CEA$, $\frac{AD+DC}{GB} = \frac{CE}{BE}$, i.e.,
$$\frac{\frac{1-x^2}{\sqrt{1+x^2}} + \sqrt{1+x^2}}{\sqrt{2}} = \frac{1+y}{\sqrt{1+y^2}}.$$
Simplifying and solving for $x$ gives $x = \frac{1-y}{1+y}$, so that
$$BD \cdot CE = (1+x)(1+y) = \left(1 + \frac{1-y}{1+y}\right) (1+y) = 2.$$
Note that we have solved here the case where $A$ lies strictly between $G$ and $F$ on the short arc $\widehat{FG}$. The case where $A = F$ or $A = G$ follows trivially, because here $BD \cdot CE$ reduces to either $BF \cdot CO = 2 \cdot 1 = 2$ or $CG \cdot BO = 2$. The case where $A$ lies between $C$ and $F$ on the short arc $\widehat{CF}$ is shown in Figure 2.

Figure 2

In a way similar to the first case, using the similarity between triangles $BDA$ and $CDF$, we see that $\frac{CD}{2+x} = \frac{\sqrt{1+(1+x)^2}}{2+x} = \frac{x}{AD}$, so that $AD = \frac{x(2+x)}{\sqrt{1+(1+x)^2}}$. The similarity between triangles $BEG$ and $CEA$ gives
$$\frac{1-y}{\sqrt{1+y^2}} = \frac{CA}{\sqrt{2}} = \frac{CD-AD}{\sqrt{2}} = \frac{\sqrt{1+(1+x)^2} - \frac{x(2+x)}{\sqrt{1+(1+x)^2}}}{\sqrt{2}}$$
From this equation we solve $x = \frac{2y}{1-y}$ and it follows that $BD \cdot CE = (2+x)(1-y) = (2+\frac{2y}{1-y})(1-y) = 2$.
Finally, the case where $A$ lies between $B$ and $G$ on the short arc $\overset{\sim}{B}G$ follows symmetrically from the above.

We can also follow a trigonometric approach: We first consider the case where $O$ is inside triangle $ABC$. See Figure 3.

Figure 3

We have $\angle BOC = 90^\circ$ (twice $\angle BAC = 45^\circ$). Put $x = OD$, $y = OE$, $\alpha = \angle ACE$, $\beta = \angle ABD$. Then $\tan \alpha = x$ and $\tan \beta = y$. Furthermore, $\alpha + \beta = 45^\circ$ (using $180^\circ = \angle BAD + \angle ABD + \angle BDA = 45^\circ + \beta + (90^\circ + \alpha)$), so that
$$1 = \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{x+y}{1-xy}.$$
Hence, $x+y = 1-xy$, and we have $BD \cdot CE = (1+x)(1+y) = 1+x+y+xy = 1+1 = 2$.

In Figure 4 we have $O$ outside triangle $ABC$.

Figure 4

In this case we have $\beta - \alpha = 45^\circ$ (using $180^\circ = \angle BAC + \angle ABC + \angle BCA = 45^\circ + (\beta + 45^\circ) + (45^\circ - \alpha)$), so that
$$1 = \tan(\beta - \alpha) = \frac{\tan \beta - \tan \alpha}{1 + \tan \alpha \tan \beta} = \frac{y-x}{1+xy}.$$
Hence, $y-x = 1+xy$, and we have $BD \cdot CE = (1-x)(1+y) = 1-x+y-xy = 1+1 = 2$.
Finally, there are the cases where $O$ is either on $AB$ or on $AC$. In the former case we have $D = A$ and $E = O$, so that $BD \cdot CE = BA \cdot CO = 2 \cdot 1 = 2$, and in the latter case, $D = O$ and $E = A$, so that $BD \cdot CE = BO \cdot CA = 1 \cdot 2 = 2$.