shobber wrote: Given triangle $ ABC$, let $ D$, $ E$, $ F$ be the midpoints of $ BC$, $ AC$, $ AB$ respectively and let $ G$ be the centroid of the triangle. For each value of $ \angle BAC$, how many non-similar triangles are there in which $ AEGF$ is a cyclic quadrilateral? Sorry to awaken an old topic, but no solution has been posted for this problem:

The QuattoMaster 6000 wrote: Clearly, these two figures meet at $ \boxed{2}$ points, so there are two non-similar triangles. [/hide] But both loci a symmetrical with respect of the perpendicular bisector of BC, so I think the two triangles are congruent. Also they intersect only when $ \angle BAC \le 60^\circ$ (it is easy to show with cos th. that $ \cos \angle BAC \ge 1/2$).

From INMO 2018 #1 we have $b^2+c^2=2a^2$ or $2\sin^2 A = \sin ^2B+\sin^2 C$. Thus we have \[2\sin^2 A = \sin ^2B+\sin^2 (A+B).\] Let $\alpha = A+B$, thus \begin{align*} 2\sin^2 A &= \sin^2 \alpha + \sin^2(\alpha - A)\\ \implies 2(1-\cos 2A) &= 1-\cos 2\alpha + 1-\cos (2\alpha - 2A)\\\\ \implies 2\cos 2A &= \cos 2\alpha + \cos (2\alpha - 2A)\\ &= 2\cos(2\alpha - A)\cos A\\ \implies \frac{\cos 2A}{\cos A} &= \cos (2\alpha - A). \end{align*} Thus we have two solutions for $\alpha$ say $\beta, \gamma$ where $2\gamma- A = A-2\beta$ or \[A = \beta + \gamma = A+B_1+\gamma\]which gives $B_1=C_2$ or the number of triangles is $1$.

The QuattoMaster 6000 wrote: First, notice that $ EF\parallel BC$. Now, since $ AEGF$ is cyclic, we have that $ \angle DGB = \angle AGE = \angle AFE = \angle ABD$. From this, coupled with the fact that $ \angle GDB = \angle BDA$, gives us that $ \triangle GBD\sim \triangle BAD$. Hence, \[ \frac {BD}{AD} = \frac {GD}{BD}\implies BD^2 = AD\cdot GD = 3GD\cdot GD = 3GD^2\implies BC^2 = 12DG^2 = \frac {12AD^2}{9} = \frac {4AD^2}{3}\implies AD = BC\sin 60 \]Similarly, we can go backwards, so if \[ AD = BC\sin 60\implies \frac {BD}{AD} = \frac {GD}{BD}\implies \triangle GBD\sim \triangle BAD\implies \angle AGE = \angle DGB = \angle ABD = \angle AFE \]so $ AFGE$ is cyclic if and only if $ AD = BC\sin 60$. Now, we find how many non-similar triangles exist with the property that $ AD = BC\sin 60$ and $ \angle BAC$ is fixed or given. Draw $ BC$, so the locus of $ A$ must be the points that are equidistant from the midpoint of $ BC$ (since $ AD = BC\sin 60$) and a sector of a circle (since $ \angle BAC$ is fixed.) Clearly, these two figures meet at $ \boxed{2}$ points, so there are two non-similar triangles. But those two triangles are symmetric with respect to the perpendicular bisector of $BC$. Therefore, they are similar.

Let $BD=DC=x$ and $AD=y$. Then, \[\frac{x}{2} \cdot \frac{x}{2} = PE \cdot PF = PA \cdot PG = \frac{y}{2} \cdot \frac{y}{6},\]\[\implies y = x\sqrt{3}.\] It is easy to see that $\angle BAC$ is maximized when $\overline{BD} \perp \overline{AC}$. Thus, for $\angle BAC > 60^\circ$, there is no such triangle. If $\angle BAC = 60^\circ$, then $\triangle ABC$ is equilateral, and if $\angle BAC < 60^\circ$, there are multiple configurations of a congruent triangle. Thus, if $\angle BAC \le 60^\circ$, there is one triangle, and otherwise, there is none. $\square$