[b]p1.[/b] Solve for $x$ and $y$ if $\frac{1}{x^2}+\frac{1}{xy}=\frac{1}{9}$ and $\frac{1}{y^2}+\frac{1}{xy}=\frac{1}{16}$


[b]p2.[/b] Find positive integers $p$ and $q$, with $q$ as small as possible, such that $\frac{7}{10} <\frac{p}{q} <\frac{11}{15}$.


[b]p3.[/b] Define $a_1 = 2$ and $a_{n+1} = a^2_n -a_n + 1$ for all positive integers $n$. If $i > j$, prove that  $a_i$ and $a_j$ have no common prime factor.


[b]p4.[/b] A number of points are given in the interior of a triangle. Connect these points, as well as the vertices of the triangle, by segments that do not cross each other until the interior is subdivided into smaller disjoint regions that are all triangles. It is required that each of the givien points is always a vertex of any triangle containing it.
Prove that the number of these smaller triangular regions is always odd.


[b]p5.[/b] In triangle $ABC$, let $\angle ABC=\angle ACB=40^o$ is extended to $D$ such that $AD=BC$. Prove that $\angle BCD=10^o$.
[img]https://cdn.artofproblemsolving.com/attachments/6/c/8abfbf0dc38b76f017b12fa3ec040849e7b2cd.png[/img]


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. Given that $\angle ABC = \angle ACB = 40^\circ$ and $AD = BC$, we need to prove that $\angle BCD = 10^\circ$.

2. First, we use the Law of Sines in $\triangle ABC$:
   \[
   \frac{BC}{\sin 80^\circ} = \frac{AC}{\sin 40^\circ}
   \]
   Since $\sin 80^\circ = \cos 10^\circ$ and $\sin 40^\circ = \cos 50^\circ$, we can rewrite the equation as:
   \[
   BC = 2 \cos 40^\circ \cdot AC
   \]

3. Given that $AD = BC = 2 \cos 40^\circ \cdot AC$, we now apply the Law of Sines in $\triangle ACD$:
   \[
   \frac{AD}{\sin(40^\circ + x)} = \frac{AC}{\sin(40^\circ - x)}
   \]
   Substituting $AD = 2 \cos 40^\circ \cdot AC$ into the equation, we get:
   \[
   \frac{2 \cos 40^\circ \cdot AC}{\sin(40^\circ + x)} = \frac{AC}{\sin(40^\circ - x)}
   \]
   Simplifying, we obtain:
   \[
   2 \cos 40^\circ \cdot \sin(40^\circ - x) = \sin(40^\circ + x)
   \]

4. Using the angle sum and difference identities for sine, we have:
   \[
   2 \cos 40^\circ \left( \sin 40^\circ \cos x - \cos 40^\circ \sin x \right) = \sin 40^\circ \cos x + \cos 40^\circ \sin x
   \]
   Expanding and simplifying, we get:
   \[
   2 \sin 40^\circ \cos 40^\circ \cos x - 2 \cos^2 40^\circ \sin x = \sin 40^\circ \cos x + \cos 40^\circ \sin x
   \]
   \[
   \sin 80^\circ \cos x - 2 \cos^2 40^\circ \sin x = \sin 40^\circ \cos x + \cos 40^\circ \sin x
   \]
   \[
   \cos x (\sin 80^\circ - \sin 40^\circ) = \sin x (\cos 40^\circ + 2 \cos^2 40^\circ)
   \]

5. Using the double angle identities, we know that $\sin 80^\circ = 2 \sin 40^\circ \cos 40^\circ$ and $\cos 80^\circ = \sin 10^\circ$. Thus, we can rewrite the equation as:
   \[
   \cos x (2 \sin 40^\circ \cos 40^\circ - \sin 40^\circ) = \sin x (\cos 40^\circ + \cos 80^\circ + 1)
   \]
   \[
   \cos x (\sin 80^\circ - \sin 40^\circ) = \sin x (\cos 40^\circ + \cos 80^\circ + 1)
   \]
   \[
   \cos x (2 \sin 20^\circ \cos 60^\circ) = \sin x (2 \cos 60^\circ \cos 20^\circ + 1)
   \]
   \[
   \cos x (\sin 20^\circ) = \sin x (\cos 20^\circ + 1)
   \]
   \[
   \cos x (2 \sin 10^\circ \cos 10^\circ) = \sin x (2 \cos^2 10^\circ)
   \]

6. Solving for $x$, we find that $x = 10^\circ$.

Therefore, $\angle BCD = 10^\circ$.

The final answer is $\boxed{10^\circ}$.