Question 1
In $\triangle A B C$, let $D, E, F$ be the midpoints of $B C, A C, A B$ respectively and let $G$ be the centroid of the triangle.
For each value of $\angle B A C$, how many non-similar triangles are there in which $A E G F$ is a cyclic quadrilateral?
FIRST SOLUTION
Let $I$ be the intersection of $A G$ and $E F$.
Let $\delta=A I . I G-F I . I E$. Then
$$
A I=A D / 2, \quad I G=A D / 6, \quad F I=B C / 4=I E .
$$

Further, applying the cosine rule to triangles $A B D, A C D$ we get
$$
\begin{aligned}
A B^{2} & =B C^{2} / 4+A D^{2}-A D \cdot B C \cdot \cos \angle B D A, \\
A C^{2} & =B C^{2} / 4+A D^{2}+A D \cdot B C \cdot \cos \angle B D A, \\
\text { so } A D^{2} & =\left(A B^{2}+A C^{2}-B C^{2} / 2\right) / 2 .
\end{aligned}
$$

Hence
$$
\begin{aligned}
\delta & =\left(A B^{2}+A C^{2}-2 B C^{2}\right) / 24 \\
& =\left(4 A B \cdot A C \cdot \cos \angle B A C-A B^{2}-A C^{2}\right) .
\end{aligned}
$$

Now $A E F G$ is a cyclic quadrilateral if and only if $\delta=0$, i.e. if and only if
$$
\begin{aligned}
\cos \angle B A C & =\left(A B^{2}+A B^{2}\right) /(4 . A B . A C) \\
& =(A B / A C+A C / A B) / 4 .
\end{aligned}
$$

Now $A B / A C+A C / A B \geq 2$. Hence $\cos \angle B A C \geq 1 / 2$ and so $\angle B A C \leq 60^{\circ}$.
For $\angle B A C>60^{\circ}$ there is no triangle with the required property.
For $\angle B A C=60^{\circ}$ there exists, within similarity, precisely one triangle (which is equilateral) having the required property.
For $\angle B A C<60^{\circ}$ there exists, within similarity, again precisely one triangle having the required property (even though for fixed median $A E$ there are two, but one arises from the other by interchanging point $B$ with point $C$, thus proving them to be similar).
SECOND SOLUTION (Mr Marcus Brazil, La Trobe University, Bundoora, Melbourne, Australia): $\square$
We require, as above,
$$
A I . I G=E I . I F,
$$
(which by (1) is equivalent to $A D^{2} / 3=C D^{2}$, i.e. $A D=C D \cdot \sqrt{3}$ ).
Let, without loss of generality, $C D=1$. Then $A$ lies on the circle of radius $\sqrt{3}$ with centre $D$. If $C D$ and $D A$ are perpendicular, the angle $B A C$ is $60^{\circ}$, otherwise it must be les.
In this case, for each angle $B A C$ there are two solutions, which are congruent.
THIRD SOLUTION (provided by the Canadian Problems Committee)
${ }^{\prime n}$ the figure as shown below, we first show that it is necessary that $\angle A$ is less than $90^{\circ}$ if the quadriliateral $A E G F$ syclic.
Now, since $E F \| B C$, we get
$$
\begin{aligned}
\angle E G F & =180^{\circ}-\left(B_{1}+C_{1}\right) \\
& \geq 180^{\circ}-(B+C) \\
& =A .
\end{aligned}
$$

Thus, if $A E G F$ is cyclic, we would have $\angle E G F+\angle A=180^{\circ}$. Therefore it is necessary that $0<\angle A \leq 90^{\circ}$.

Continuation "A"
Let $O$ be the circumcentre of $\triangle A F E$. Without loss of generality, let the radius of this circle be 1.
We then let $A=1, F=z=e^{i \theta}$ and $E=z e^{2 i \alpha}=e^{i(\theta+2 \alpha)}$.
Then $\angle A=\alpha, 0<\alpha \leq 90^{\circ}$, and $0<\theta<360^{\circ}-2 \alpha$.
Thus,
$$
B=2 z-1
$$
and
$$
\begin{aligned}
G & =\frac{1}{3}(2 z-1)+\frac{2}{3}\left(z e^{2 i \alpha}\right) \\
& =\frac{1}{3}\left(2 e^{i \theta}+2 e^{i(+2 \alpha)}-1\right) .
\end{aligned}
$$

For quadriateral $A F G E$ to be cyclic, it is now necessary that
7
$$
|G|=1 .
$$

For $|G|=1$, we must have
$$
\begin{aligned}
9= & (2 \cos (\theta)+2 \cos (\theta+2 \alpha)-1)^{2}+(2 \sin (\theta)+2 \sin (\theta+2 \alpha))^{2} \\
= & 4\left(\cos (\theta)+\sin ^{2}(\theta)\right)+4\left(\cos (\theta+2 \alpha)+\sin ^{2}(\theta+2 \alpha)\right)+1 \\
& +8(\cos (\theta) \cos (\theta+2 \alpha)+\sin (\theta) \sin (\theta+2 \alpha))-4 \cos (\theta)-4 \cos (\theta+2 \alpha) \\
= & 9+8 \cos (2 \alpha)-8 \cos (\alpha) \cos (\theta+\alpha) \\
& \cos (\theta+\alpha)=\frac{\cos (2 \alpha)}{\cos (\alpha)} .
\end{aligned}
$$
so that
Now,, $\left.\frac{\cos (2 \alpha)}{\cos (\alpha)} \right\rvert\, \leq 1$ if and only if $\alpha \in\left(0,60^{\circ}\right)$ in the range of $\alpha$ under consideration, that is $\alpha \in\left(0,00^{\circ}\right)$. There is equality if and only if $\alpha=60^{\circ}$.
Note there is only one solution. The apparent other solution is the mirror image of the first. We are solvng tor $\checkmark+\theta$. The other solution is $360^{\circ}-\alpha-\theta$.
Continuation "B"
Let $O$ be the circumcentre of triangle $A E F$. Let $A P$ be a diameter of this circle. Construct the circle with centre $P$ and radius $A P$. Then $B$ and $C$ lie on this circle.

It is clear that the problem is solved if we allow the angle $\angle B A C=\alpha$ to vary and restrict $B$ and $C$ to the constructed circle.

Let $\theta$ be the angle from the drawn axis. Then $\theta$ lies in the range $\left(0,180^{\circ}-\alpha\right)$. We must not forget the necessary restriction of $\alpha$, that is $\alpha \in\left(0,90^{\circ}\right)$.
Now, $D$ lies on an arc of a circle, centre $P$, radius $P D$ exterior to the circle, centre $O$, radius $A O$.
By similarity, $G$, lies on an arc of a circle, centre $Q$, radius $Q G$ where $A Q=\frac{2}{3} A P$ and $Q G=\frac{2}{3} P D$.
For the quadrilateral $A F G E$ to be cyclic, we must have that the radius $Q G$ is greater than or equal to $Q P$.
The easiest way to calulate these radii is to consider the case in which the diameter $A P$ bisects the angle $\angle B A C$.
Thus we re-draw the diagram as below. Let $A H$ be a diameter of the larger circle.
Thus we have $A H=4$ and by similar triangles,
$$
\frac{A D}{A B}=\frac{A B}{A H}=\cos \left(\frac{\alpha}{2}\right),
$$
so that
$$
\begin{aligned}
A D & =4 \cos ^{2}\left(\frac{\alpha}{2}\right) \\
& =2+2 \cos (\alpha) .
\end{aligned}
$$

Thus $P D=2 \cos (\alpha)$ and $Q G=\frac{2}{3} 2 \cos (\alpha)=\frac{4}{3} \cos (\alpha)$.
The necessary condition for a cyclic quadrilateral is then
$$
\frac{4}{3}(1+\cos (\alpha)) \geq 2,
$$
which reduces to
$$
\cos (\alpha) \geq \frac{1}{2} .
$$

Thus it is clear that there is precisely one (up to similarity) solution for $0<\alpha \leq 60^{\circ}$ and no solutions otherwise.