As $A M=A P$, we have

$$\angle M B A=\frac{1}{2} \operatorname{arcAM}=\frac{1}{2} \operatorname{arc} A P=\angle A B P$$

and likewise

$$\angle Q B A=\frac{1}{2} \operatorname{arc} A Q=\frac{1}{2} \operatorname{arc} c A N=\angle A B N$$

Summing these equalities yields $\angle M B Q=\angle N B P$ as needed.

Q2. Let $E$ and $F$ be two distinct points inside of a parallelogram $A B C D$. Find the maximum number of triangles with the same area and having the vertices in three of the following five points: $A, B, C, D, E, F$.

Solution. We shall use the following two well known results:

Lemma 1. Let $A, B, C, D$ be four points lying in the same plane such that the line $A B$ do not intersect the segment $C D$ (in particular $\triangle B C D$ is a convex quadrilateral). If $[A B C]=[A B D]$, then $A B \| C D$.

Lemma 2. Let $X$ be a point inside of a parallelogram $A B C D$. Then $[A C X]<[A B C]$ and $[B D X]<[4 B D]$. (Here and below the notation $[S]$ stands for the area of the surface of $S$.

With the points $A, B, C, D, E, F$ we can form 20 triangles. We will show that at most ten of them can have the same area. In that sense three cases may occur.

Case 1. $E F$ is parallel with one side of the parallelogram $\triangle B C D$.

We can assume that $E F \| A D, E$ lies inside of the triangle $A B F$ and $F$ lies inside of the triangle $C D E$. With the points $A, B, C, D, E$ we can form ten pairs of triangles as follows:

$(\triangle A D E, \triangle A E F) ;(\triangle A D F, \triangle D E F) ;(\triangle B C E, \triangle B E F) ;(\triangle B C F, \triangle C E F) ;(\triangle C D E, \triangle C D F):$

$(\triangle A B E, \triangle A B F) ;(\triangle A D C, \triangle A C E) ;(\triangle A B C, \triangle A C F) ;(\triangle A B D, \triangle B D E) ;(\triangle C B D, \triangle B D F)$.

Using Lemmas $1-2$ one can easily prove that any two triangles that belong to the same pair have distinct area, so there exists at most ten triangles having the same area.

Case 2. $E F$ is parallel with a diagonal of the parallelogram $\triangle B C D$.

Let us assume $E F \| A C$ and that $E, F$ lie inside of the triangle $A B C$. We consider the following pairs of triangles:

$(\triangle A B D, \triangle B D E) ;(\triangle A B C, \triangle B C F) ;(\triangle A C D, \triangle B C E) ;(\triangle A B F, \triangle A B E) ;(\triangle B E F, \triangle D E F) ;$

$(\triangle A E F, \triangle A C E) ;(\triangle C E F, \triangle A C F) ;(\triangle A D E, \triangle A D F) ;(\triangle D C E, \triangle D C F) ;(\triangle C B D, \triangle B D F)$.

With the same idea as above we deduce that any two triangles that belong to the same pair have distinct area and the conclusion follows.

We also note that if $E$ and $F$ lie on $A C$ then only 16 of 20 triangles are nondegenerate. In this case we consider the following pairs:

$$\begin{aligned} & (\triangle A B E, \triangle A B F) ;(\triangle A B C, \triangle B C F) ;(\triangle B C E, \triangle B E F) ;(\triangle A D E, \triangle A D F) \\ & (\triangle A C D, \triangle D C F) ;(\triangle C D E, \triangle E D F) ;(\triangle B D E, \triangle A B D) ;(\triangle B D F, \triangle B D C) \end{aligned}$$

Case 3. $E F$ is not parallel with any side or diagonal of $\triangle B C D$.

We claim that at most two of the triangles $A E F, B E F, C E F, D E F$ can have the same area. Indeed, supposing the contrary, we may have $[A E F]=[B E F]=[C E F]$. We remark first that $A, B, C$ do not belong to $E F$ (elsewhere, exactly one of the above triangles is degenerate, contradiction!). Hence at least two of the points $A, B, C$ belong to the same side of the line $E F$. Using now Lemma 1 we get that $E F$ is parallel with $A B$ or $B C$ or $A C$. This is clearly a contradiction and our claim follows. With the remaining 16 triangles we form 8 pairs as follows:

$(\triangle A B D, \triangle B D E) ;(\triangle C D B, \triangle B D F) ;(\triangle A D C, \triangle A C E) ;(\triangle A B C, \triangle A C F) ;$

$(\triangle A B E, \triangle A B F) ;(\triangle B C E, \triangle B C F) ;(\triangle A D E, \triangle A D F) ;(\triangle D C E, \triangle D C F)$.

With the same arguments as above, we get at most ten triangles with the same area.

To conclude the proof, it remains only to give an example of points $E, F$ inside of the parallelogram $A B C D$ such that exactly ten of the triangles that can be formed with the vertices $A, B, C, D, E, F$ have the same area.

Denote $A C \cap B D=\{O\}$ and let $M, N$ be the midpoints of $A B$ and $C D$ respectively. Consider $E$ and $F$ the midpoints of $M O$ and $N O$. Then $O, M, N, E, F$ are collinear and $M E=E O=F O=N F$. Since $E$ and $F$ are the centroids of the triangles $A B F$ and $C D E$ we get $[A B E]=[A E F]=[B E F]$ and $[C E F]=[D E F]=[C D F]$. On the other hand, taking into account that $A E C F$ and $B E D F$ are parallelograms we deduce $[A E F]=[C E F]=[A C E]=[A C F]$ and $[B E F]=[D E F]=[B D E]=[B D F]$. From the above equalities we conclude that the triangles

## $\triangle A B E, \triangle C D F, \triangle A C E, \triangle A C F, \triangle B D E, \triangle B D F, \triangle A E F, \triangle B E F, \triangle C E F, \triangle D E F$

have the same area. This finishes our proof.