Olimpiade Italiana di Matematica

Solution:

With reference to the figure on the left, let us denote by $\theta$ the measure of the angle $\angle B C D$.

The solution of the problem can be broken down into the following six steps.

- The points $B, O, E, D$ lie on a common circle if and only if $D$ is the midpoint of the arc $A B$ on which it lies.
Indeed, it suffices to observe that the four points lie on a common circle if and only if $\angle D O B=\angle D E B$. Moreover, $\angle D E B=90^\circ$ by construction, so the four points lie on a common circle if and only if $\angle D O B=90^\circ$. This last equality holds if and only if the radius $O D$ is perpendicular to the diameter $A B$, which is true if and only if $D$ is the midpoint of the arc $A B$.

- The point $D$ is the midpoint of the arc $A B$ on which it lies if and only if $\theta=45^\circ$.
Indeed, it suffices to observe that the two angles $\angle A C D$ and $\angle D C B$ both have measure $\theta$, since they subtend the arcs $A D$ and $D B$ which have the same length, and the sum of the two angles equals $\angle A C B$, which is a right angle because $A B$ is a diameter.

- The point $E$ belongs to the semicircle with diameter $C B$ lying on the same side of the line $B C$ as $A$.
Indeed, it suffices to observe that the point $E$, by construction, sees the segment $C B$ under an angle of $90^\circ$ (and lies on the same side of the line $B C$ as $A$).

- Maximizing the product $C E \cdot E D$ is equivalent to minimizing the distance from $E$ to $O$.
Indeed, we claim that the product $C E \cdot E D$ equals $R^2-O E^2$, where $R$ is the length $O A$ of the radius of the circle circumscribed about the triangle $A B C$. This can be deduced by applying the intersecting chords theorem to the chord $C D$ and the chord passing through $O$ and $E$, divided by $E$ into segments of length $R+O E$ and $R-O E$. In other words, the product $C E \cdot E D$ is by definition the power of the point $E$ with respect to the initial circle, and the power of any point $E$ interior to a circle of radius $R$ and centre $O$ equals $R^2-O E^2$.

- Let us denote by $O'$ the midpoint of the segment $B C$, which as already seen is also the centre of a semicircle on which $E$ lies. The point of this semicircle closest to $O$ is obtained by intersecting the semicircle with the line $O O'$.
Indeed, with reference to the figure on the right, let us denote by $E$ this intersection, and by $E'$ any other point of the semicircle. Then from the triangle inequality (in every triangle one side is less than the sum of the other two) we deduce that
$$O' O+O E=O' E=O' E'<O' O+O E'$$
from which, simplifying $O' O$, we conclude that $O E<O E'$, as required.

- The point $E$ coincides with the point previously defined, and hence maximizes the product $C E \cdot E D$, if and only if $\theta=45^\circ$.
Indeed, from the previous construction it follows that the triangle $E O' C$ is a right isosceles triangle with the right angle at $O'$.

Alternative solution

As in the first two points of the previous solution, we obtain that the points $B, O, E, D$ lie on a common circle if and only if $\theta=45^\circ$.
Now we show in an alternative way that $E$ maximizes the product $C E \cdot E D$ if and only if $\theta=45^\circ$.
First, we observe that in the right triangle $B E C$ the relation $C E=C B \cdot \cos \theta$ holds. Next, we observe that $\angle C D B=\angle C A B$, since they subtend the same chord $C B$. Consequently, denoting by $\alpha$ the common measure of these angles, in the right triangle $B D E$ the relation $E D=B D \cdot \cos \alpha$ holds. Finally, from the law of sines we know that $D B=2 R \cdot \sin \theta$, where $R$ is the radius of the initial circle.

Combining these three equalities, and applying the double-angle formula for the sine, we deduce that
$$C E \cdot E D=(C B \cdot \cos \theta) \cdot(2 R \cdot \sin \theta \cdot \cos \alpha)=C B \cdot R \cdot \cos \alpha \cdot \sin (2 \theta) .$$
Since $C B, R$ and $\alpha$ are fixed, it is clear that the product is maximal when $2 \theta=90^\circ$, that is, $\theta=45^\circ$.