[u]Round 5 [/u]


[b]p13.[/b] Circles $\omega_1$, $\omega_2$, and $\omega_3$ have radii $8$, $5$, and $5$, respectively, and each is externally tangent to the other two. Circle $\omega_4$ is internally tangent to $\omega_1$, $\omega_2$, and $\omega_3$, and circle $\omega_5$ is externally tangent to the same three circles. Find the product of the radii of $\omega_4$ and $\omega_5$.


[b]p14.[/b] Pythagoras has a regular pentagon with area $1$. He connects each pair of non-adjacent vertices with a line segment, which divides the pentagon into ten triangular regions and one pentagonal region. He colors in all of the obtuse triangles. He then repeats this process using the smaller pentagon. If he continues this process an infinite number of times, what is the total area that he colors in? Please rationalize the denominator of your answer.


p15. Maisy arranges $61$ ordinary yellow tennis balls and $3$ special purple tennis balls into a $4 \times 4  \times 4$ cube. (All tennis balls are the same size.) If she chooses the tennis balls’ positions in the cube randomly, what is the probability that no two purple tennis balls are touching?


[u]Round 6 [/u]


[b]p16.[/b] Points $A, B, C$, and $D$ lie on a line (in that order), and $\vartriangle BCE$ is isosceles with $\overline{BE} = \overline{CE}$. Furthermore, $F$ lies on $\overline{BE}$ and $G$ lies on $\overline{CE}$ such that $\vartriangle BFD$ and $\vartriangle CGA$ are both congruent to $\vartriangle BCE$. Let $H$ be the intersection of $\overline{DF}$ and $\overline{AG}$, and let $I$ be the intersection of $\overline{BE}$ and $\overline{AG}$. If $m \angle BCE = arcsin \left( \frac{12}{13} \right)$, what is $\frac{\overline{HI}}{\overline{FI}}$ ?


[b]p17.[/b] Three states are said to form a tri-state area if each state borders the other two. What is the maximum possible number of tri-state areas in a country with fifty states? Note that states must be contiguous and that states touching only at “corners” do not count as bordering.


[b]p18.[/b] Let $a, b, c, d$, and $e$ be integers satisfying $$2(\sqrt[3]{2})^2 + \sqrt[3]{2}a + 2b + (\sqrt[3]{2})^2c +\sqrt[3]{2}d + e = 0$$ and $$25\sqrt5 i + 25a - 5\sqrt5 ib - 5c + \sqrt5 id + e = 0$$ where $i =\sqrt{-1}$. Find $|a + b + c + d + e|$.


[u]Round 7[/u]


[b]p19.[/b] What is the greatest number of regions that $100$ ellipses can divide the plane into? Include the unbounded region.


[b]p20.[/b] All of the faces of the convex polyhedron $P$ are congruent isosceles (but NOT equilateral) triangles that meet in such a way that each vertex of the polyhedron is the meeting point of either ten base angles of the faces or three vertex angles of the faces. (An isosceles triangle has two base angles and one vertex angle.) Find the sum of the numbers of faces, edges, and vertices of $P$.


[b]p21.[/b] Find the number of ordered $2018$-tuples of integers $(x_1, x_2, .... x_{2018})$, where each integer is between $-2018^2$ and $2018^2$ (inclusive), satisfying $$6(1x_1 + 2x_2 +...· + 2018x_{2018})^2 \ge  (2018)(2019)(4037)(x^2_1 + x^2_2 + ... + x^2_{2018}).$$


PS. You should use hide for answers. Rounds 1-4 have been posted [url=https://artofproblemsolving.com/community/c4h2784936p24472982]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
Given the equations:
\[
2(\sqrt[3]{2})^2 + \sqrt[3]{2}a + 2b + (\sqrt[3]{2})^2c + \sqrt[3]{2}d + e = 0
\]
\[
25\sqrt{5}i + 25a - 5\sqrt{5}ib - 5c + \sqrt{5}id + e = 0
\]

Let's introduce the substitutions:
\[
m = (\sqrt[3]{2})^2, \quad n = \sqrt[3]{2}, \quad p = \sqrt{5}i
\]

This transforms the system into:
\[
2m + na + 2b + mc + nd + e = 0
\]
\[
25p + 25a - 5pb - 5c + pd + e = 0
\]

We need to solve this system of equations. Let's rewrite them for clarity:
\[
2m + mc + na + nd + 2b + e = 0
\]
\[
25p - 5pb + 25a + pd - 5c + e = 0
\]

To eliminate the terms involving \(m\), \(n\), and \(p\), we need to set their coefficients to zero. This gives us the following system of equations:
\[
2 + c = 0 \quad \Rightarrow \quad c = -2
\]
\[
a + d = 0 \quad \Rightarrow \quad d = -a
\]
\[
25 - 5b + d = 0 \quad \Rightarrow \quad 25 - 5b - a = 0 \quad \Rightarrow \quad a = 25 - 5b
\]

Substituting \(d = -a\) into the second equation:
\[
a + (-a) = 0 \quad \Rightarrow \quad 0 = 0 \quad \text{(which is always true)}
\]

Now, substituting \(a = 25 - 5b\) into the first equation:
\[
2b + e = 0 \quad \Rightarrow \quad e = -2b
\]

We now have:
\[
a = 25 - 5b
\]
\[
c = -2
\]
\[
d = -a = -(25 - 5b) = -25 + 5b
\]
\[
e = -2b
\]

To find the values of \(a\), \(b\), \(c\), \(d\), and \(e\), we need to ensure they satisfy the original equations. Let's check the consistency:
\[
a = 0, \quad b = 5, \quad c = -2, \quad d = 0, \quad e = -10
\]

Substituting these values back into the original equations:
\[
2(\sqrt[3]{2})^2 + \sqrt[3]{2}(0) + 2(5) + (\sqrt[3]{2})^2(-2) + \sqrt[3]{2}(0) + (-10) = 0
\]
\[
2m + 2b - 2m - 10 = 0 \quad \Rightarrow \quad 10 - 10 = 0 \quad \Rightarrow \quad 0 = 0
\]

\[
25\sqrt{5}i + 25(0) - 5\sqrt{5}i(5) - 5(-2) + \sqrt{5}i(0) + (-10) = 0
\]
\[
25p - 25p + 10 - 10 = 0 \quad \Rightarrow \quad 0 = 0
\]

Both equations are satisfied. Therefore, the values are correct.

The sum of \(a + b + c + d + e\) is:
\[
0 + 5 - 2 + 0 - 10 = -7
\]

The absolute value is:
\[
|a + b + c + d + e| = |-7| = 7
\]

The final answer is \(\boxed{7}\)