$10^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament
Saturday 24 February 2007
Individual Round: Algebra Test
- [3] Compute
(Note that $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.) Answer: 2006. We have
- [3] Two reals $x$ and $y$ are such that $x-y=4$ and $x^{3}-y^{3}=28$. Compute $x y$.
Answer: $-\mathbf{3}$. We have $28=x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)=(x-y)\left((x-y)^{2}+3 x y\right)=4 \cdot(16+3 x y)$, from which $x y=-3$. 3. [4] Three real numbers $x, y$, and $z$ are such that $(x+4) / 2=(y+9) /(z-3)=(x+5) /(z-5)$. Determine the value of $x / y$. Answer: 1/2. Because the first and third fractions are equal, adding their numerators and denominators produces another fraction equal to the others: $((x+4)+(x+5)) /(2+(z-5))=(2 x+9) /(z-3)$. Then $y+9=2 x+9$, etc. 4. [4] Compute
Answer: $\frac{\mathbf{4 3} \text {. }}{63}$ Use the factorizations $n^{3}-1=(n-1)\left(n^{2}+n+1\right)$ and $n^{3}+1=(n+1)\left(n^{2}-n+1\right)$ to write
- [5] A convex quadrilateral is determined by the points of intersection of the curves $x^{4}+y^{4}=100$ and $x y=4$; determine its area. Answer: $\mathbf{4 \sqrt { 1 7 }}$. By symmetry, the quadrilateral is a rectangle having $x=y$ and $x=-y$ as axes of symmetry. Let $(a, b)$ with $a>b>0$ be one of the vertices. Then the desired area is
- [5] Consider the polynomial $P(x)=x^{3}+x^{2}-x+2$. Determine all real numbers $r$ for which there exists a complex number $z$ not in the reals such that $P(z)=r$. Answer: $\mathbf{r}>\mathbf{3}, \mathbf{r}<\frac{\mathbf{4 9}}{\mathbf{2 7}}$. Because such roots to polynomial equations come in conjugate pairs, we seek the values $r$ such that $P(x)=r$ has just one real root $x$. Considering the shape of a cubic, we are interested in the boundary values $r$ such that $P(x)-r$ has a repeated zero. Thus, we write
Then $q=-2 p-1$ and $1=p(p+2 q)=p(-3 p-2)$ so that $p=1 / 3$ or $p=-1$. It follows that the graph of $P(x)$ is horizontal at $x=1 / 3$ (a maximum) and $x=-1$ (a minimum), so the desired values $r$ are $r>P(-1)=3$ and $r<P(1 / 3)=1 / 27+1 / 9-1 / 3+2=49 / 27$. 7. [5] An infinite sequence of positive real numbers is defined by $a_{0}=1$ and $a_{n+2}=6 a_{n}-a_{n+1}$ for $n=0,1,2, \cdots$ Find the possible value(s) of $a_{2007}$. Answer: $\mathbf{2}^{\mathbf{2 0 0 7}}$. The characteristic equation of the linear homogeneous equation is $m^{2}+m-6=$ $(m+3)(m-2)=0$ with solutions $m=-3$ and $m=2$. Hence the general solution is given by $a_{n}=A(2)^{n}+B(-3)^{n}$ where $A$ and $B$ are constants to be determined. Then we have $a_{n}>0$ for $n \geq 0$, so necessarily $B=0$, and $a_{0}=1 \Rightarrow A=1$. Therefore, the unique solution to the recurrence is $a_{n}=2^{n}$ for all n . 8. [6] Let $A:=\mathbb{Q} \backslash{0,1}$ denote the set of all rationals other than 0 and 1. A function $f: A \rightarrow \mathbb{R}$ has the property that for all $x \in A$,
Compute the value of $f(2007)$. Answer: $\log (\mathbf{2 0 0 7} / \mathbf{2 0 0 6})$. Let $g: A \rightarrow A$ be defined by $g(x):=1-1 / x$; the key property is that
The given equation rewrites as $f(x)+f(g(x))=\log |x|$. Substituting $x=g(y)$ and $x=g(g(z))$ gives the further equations $f(g(y))+f(g(g(y)))=\log |g(x)|$ and $f(g(g(z)))+f(z)=\log |g(g(x))|$. Setting $y$ and $z$ to $x$ and solving the system of three equations for $f(x)$ gives
For $x=2007$, we have $g(x)=\frac{2006}{2007}$ and $g(g(x))=\frac{-1}{2006}$, so that
- [7] The complex numbers $\alpha_{1}, \alpha_{2}, \alpha_{3}$, and $\alpha_{4}$ are the four distinct roots of the equation $x^{4}+2 x^{3}+2=0$. Determine the unordered set
Answer: ${\mathbf{1} \pm \sqrt{\mathbf{5}}, \mathbf{- 2}}$. Employing the elementary symmetric polynomials $\left(s_{1}=\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=\right.$ $-2, s_{2}=\alpha_{1} \alpha_{2}+\alpha_{1} \alpha_{3}+\alpha_{1} \alpha_{4}+\alpha_{2} \alpha_{3}+\alpha_{2} \alpha_{4}+\alpha_{3} \alpha_{4}=0, s_{3}=\alpha_{1} \alpha_{2} \alpha_{3}+\alpha_{2} \alpha_{3} \alpha_{4}+\alpha_{3} \alpha_{4} \alpha_{1}+\alpha_{4} \alpha_{1} \alpha_{2}=0$, and $s_{4}=\alpha_{1} \alpha_{2} \alpha_{3} \alpha_{4}=2$ ) we consider the polynomial
Because $P$ is symmetric with respect to $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$, we can express the coefficients of its expanded form in terms of the elementary symmetric polynomials. We compute
The roots of $P(x)$ are -2 and $1 \pm \sqrt{5}$, so the answer is ${1 \pm \sqrt{5},-2}$. Remarks. It is easy to find the coefficients of $x^{2}$ and $x$ by expansion, and the constant term can be computed without the complete expansion and decomposition of $\left(\alpha_{1} \alpha_{2}+\alpha_{3} \alpha_{4}\right)\left(\alpha_{1} \alpha_{3}+\alpha_{2} \alpha_{4}\right)\left(\alpha_{1} \alpha_{4}+\right.$ $\alpha_{2} \alpha_{3}$ ) by noting that the only nonzero 6 th degree expressions in $s_{1}, s_{2}, s_{3}$, and $s_{4}$ are $s_{1}^{6}$ and $s_{4} s_{1}^{2}$. The general polynomial $P$ constructed here is called the cubic resolvent and arises in Galois theory. 10. [8] The polynomial $f(x)=x^{2007}+17 x^{2006}+1$ has distinct zeroes $r_{1}, \ldots, r_{2007}$. A polynomial $P$ of degree 2007 has the property that $P\left(r_{j}+\frac{1}{r_{j}}\right)=0$ for $j=1, \ldots, 2007$. Determine the value of $P(1) / P(-1)$.
Answer: | $\mathbf{2 8 9}$. | | :---: | | For some constant $k$, we have |
Now writing $\omega^{3}=1$ with $\omega \neq 1$, we have $\omega^{2}+\omega=-1$. Then