## Mission Math Tournament: Regular Round Part 2 Solutions

11. Find the sum of all real values of x that satisfy the equation $x(x-1)=1$.

Solution: $x(x-1)=1 \implies x^2-x=1 \implies x^2-x-1=0$. For any quadratic equation $ax^2+bx+c=0$, the two roots can be found via the quadratic formula to be $\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\frac{-b-\sqrt{b^2-4ac}}{2a}$. Adding these two together gives that the sum of the roots of any quadratic equation is $-\frac{b}{a}$. This is one of a set of useful equations regarding sums and products of roots of polynomials known as Vieta’s Formulae. Applying this sum equation here, we find the sum of the values of x that satisfy the given equation to be $-\frac{-1}{1}=\boxed{1}$.

12. Find the area of the smallest square that can contain a line segment of length 2.

Solution: The longest line segment that can be contained in any square is clearly the diagonal. By properties of a 45º-45º-90º right triangle formed by dividing a square along a diagonal, the side length of the square is $\sqrt{2}$. The area is therefore $(\sqrt{2})^2=\boxed{2}$.

13. Find all real solutions to $(x^{10}+30)^{(x^9+27)^{(x^8+24)^{.^{.^{. ^{(x^3+9)^{(x^2+6)^{(x+3)}}}}}}}}=1$.

Solution: When a real number a is raised to the power of a real number b, there are three ways a result of 1 could occur: the base is 1, the base is -1 and the exponent is even, or the base is nonzero and the exponent is zero. As $x^{10}+30\geq30$ for any real x, only the third case mentioned above could occur. Thus, we seek values of x that make the exponent 0. The only way a real number raised to a real number could equal 0 is if the base is zero and the exponent is nonzero. Thus, we set the base of this quantity, $x^9+27$, to be 0. Solving, we find that the only real solution to $x^9+27=0$ is $\boxed{-\sqrt[3]{3}}$. Now, all we need to do is to make sure this term’s exponent is nonzero. Again, we use the fact that a real number raised to a real number is 0 if and only if the base is zero and the exponent is nonzero. For any real x, $x^8+24\geq24$, so the  exponent of $x^9+27$ cannot be zero, and thus we have found the one and only solution.

14. What is the smallest possible number of sides a polygon could have if its centroid is outside of itself?

Solution: This would not be a triangle, as a triangle can’t even be concave. It is possible, however, for a $\boxed{4}$-sided polygon to have a centroid outside of it, and an example is depicted below.

15. If $\frac{x^2}{x^3+1}=\frac{2}{7}$ and $\frac{x^4}{x^6+1}=\frac{4}{65}$, find all possible values of x.

Solution: Squaring the first equation, we find that $\frac{x^4}{x^6+2x^3+1}=\frac{4}{49}$. Cross multiplying, we find that $49x^4=4x^6+8x^3+4$. Cross multiplying the other equation, we find that $65x^4=4x^6+4$. Subtracting the second equation from the squared first equation gives $-16x^4=8x^3$. This has two solutions, 0 and $\boxed{-\frac{1}{2}}$. The first solution is extraneous.

Problem 16 was dropped.

17. An ant starts at $(1,1,1,1,1,1)$ in a six-dimensional world consisting of the 729 points $(u,v,w,x,y,z)$ such that $u,v,w,x,y,z,\in \{-1,0,1\}$. Each move the ant takes a stroll from its current location to another point in its world exactly n units of distance away. Find the product of all n such that given unlimited moves the ant can eventually visit all 729 points in its world.

Solution: For a move to take the ant from a lattice point to a different lattice point, it must be of distance $\sqrt{n}$ for some positive integer n. For any n greater than 6, the ant will never be able to reach the point in the center of the six-dimensional cube, so we now only have six cases for n to check. Note that whenever n is even, The parity of the sum of the coordinates of the location of the ant is conserved through any legal move, so no even n works. If we prove that the ant can move to an adjacent point given an arbitrary location and we can prove that the ant can get to the center, we have proven that the value of n works. Proving this for n=1 is trivial, as the ant is simply moving to any adjacent square. For n=3, we find $(1,1,1,1,1,1)\rightarrow(1,1,1,0,0,0)\rightarrow(0,0,0,0,0,0)$ brings the ant to the center and $(1,1,1,1,1,1)\rightarrow(1,1,1,0,0,0)\rightarrow(1,1,0,-1,1,0)\rightarrow(0,0,0,-1,0,0)\rightarrow(0,1,1,0,0,0)\rightarrow(1,1,1,1,1,0)$ brings the ant to an adjacent square. For n=5, we take the complement of any move in n=1. Thus, all odd values of n under 6 are valid, and multiplying the square roots gives $\sqrt{1}\sqrt{3}\sqrt{5}=\boxed{\sqrt{15}}$.

18. Participants in the regional lottery pick two of the numbers 1, 2, 3, 4, and 5. A ticket wins if neither of the two numbers matches the two numbers randomly drawn by the lottery oﬃcial. What is the smallest number of tickets a participant should have in order to guarantee that one of the tickets wins?

Solution: The minimum number of tickets is $\boxed{4}$, for example if one chooses the tickets 1 and 2, 1 and 3, 2 and 3, and 4 and 5. Since any ticket wins for only $\binom{5-2}{2}=3$ outcomes, and there are $\binom{5}{2}=10>3(3)$ possible outcomes, 3 tickets cannot cover all possibilities.

19. How many complex numbers are roots of $f(z)=\frac{(z-1)(z^2-1)(z^3-1)...(z^{14}-1)(z^{15}-1)(z^{16}-1)}{(z+1)(z^2+1)(z^3+1)...(z^{14}+1)(z^{15}+1)(z^{16}+1)}$?

Solution: The numerator consists of the nth roots of unity, up to n=16, and the denominator consists of the nth roots of minus unity, up to n=16. Thus, a number is a root if it is a root of unity while not being a root of minus unity, when considering all degrees of roots up to 16. Thinking in polar coordinates using De Moivre’s Theorem, one can form a Farey Sequence representing the roots of unity in terms of the fraction of the away around the ring of unity at which they are located; in this case, we’re looking for the 16th Farey Sequence, with the last term omitted due to being a doubling of the first case: $\{0,\frac{1}{16},\frac{1}{15},\frac{1}{14},\frac{1}{13}...\frac{1}{8},\frac{2}{15},\frac{1}{7}...\frac{14}{15},\frac{15}{16}\}$. In the case of the denominator, we double denominators and take only odd numerators of terms in the sequence.

We now perform parity casework on the fraction of a circle at which each potential root is located: specifically, we note the parity of both the numerator and the denominator. When the parity of both is even, the fraction will eventually simplify into one of the other cases, and thus all of these cases can be discounted. Whenever the denominator is odd, the fraction is not equivalent to any with an even denominator and an odd numerator. Whenever the numerator is odd and the denominator is even, the corresponding complex number is not a root. Thus, we count, for each odd number less than 16, the number of simplified fractions that have that number as the denominator. There are are 49 such fractions, and thus the equation has $\boxed{49}$ roots.

20. In a certain game, you pick two relatively prime numbers less than or equal to 1000. Then, you subtract the smaller number from the larger number. You earn one point if the resulting difference is less than the original smaller number, and repeat the process using these two numbers. The game ends when the difference of 1 appears. Find the two numbers you should pick at the start to achieve the highest possible score. Here’s a sample run of this game’s process using starting values of 77 and 111:

111-77=34, 34<77, score 1 point

77-34=43, 43>34, score 0 points

43-34=9, 9<34, score 1 point

34-9=25, 25>9, score 0 points

25-9=16, 16>9, score 0 points

16-9=7, 7<9, score 1 point

9-7=2, 2<7, score 1 point

7-2=5, 5>2, score 0 points

5-2=3, 3>2 score 0 points

3-2=1, 1<2, score 1 point

Difference of 1 appears, game ends, score=5

Solution: Think from the bottom up. Whenever one adds a term to another, one gets the previous term in one of these sequences. Thus, we try to maximize the number of addings while staying under a certain value, and thus we add starting from 1 and 2. By adding only once, we ensure that no moves coming down are “wasted,” or in other words, producing 0 points. By adding starting with 1 and 2, we move up the Fibonacci sequence, stopping at the given value of 1000, to find the two numbers we want to be $\boxed{610 \text{ and } 987}$.