problem_type: Algebra

exam: HMMT

problem:
Compute

$$
\left\lfloor\frac{2007!+2004!}{2006!+2005!}\right\rfloor
$$

(Note that $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.)

solution:
2006. We have

$$
\left\lfloor\frac{2007!+2004!}{2006!+2005!}\right\rfloor=\left\lfloor\frac{\left(2007 \cdot 2006+\frac{1}{2005}\right) \cdot 2005!}{(2006+1) \cdot 2005!}\right\rfloor=\left\lfloor\frac{2007 \cdot 2006+\frac{1}{2005}}{2007}\right\rfloor=\left\lfloor 2006+\frac{1}{2005 \cdot 2007}\right\rfloor
$$

solution_match: Answer:


problem: Two reals $x$ and $y$ are such that $x-y=4$ and $x^{3}-y^{3}=28$. Compute $x y$.

solution: $-\mathbf{3}$. 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 $x y=-3$.


problem: 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 $x / y$.

solution: 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.


solution:
$\frac{\mathbf{4 3} \text {. }}{63}$ 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

$$
\frac{1 \cdot 7}{3 \cdot 3} \cdot \frac{2 \cdot 13}{4 \cdot 7} \cdot \frac{3 \cdot 21}{5 \cdot 13} \cdot \frac{4 \cdot 31}{6 \cdot 21} \cdot \frac{5 \cdot 43}{7 \cdot 31}=\frac{1 \cdot 2 \cdot 43}{3 \cdot 6 \cdot 7}=\frac{43}{63}
$$


problem: 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.

solution:
$\mathbf{4 \sqrt { 1 7 }}$. 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

$$
(\sqrt{2}(a-b)) \cdot(\sqrt{2}(a+b))=2\left(a^{2}-b^{2}\right)=2 \sqrt{a^{4}-2 a^{2} b^{2}+b^{4}}=2 \sqrt{100-2 \cdot 4^{2}}=4 \sqrt{17}
$$


problem: 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 $P(z)=r$.

solution:
$\mathbf{r}>\mathbf{3}, \mathbf{r}<\frac{\mathbf{4 9}}{\mathbf{2 7}}$. 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

$$
P(x)-r=x^{3}+x^{2}-x+(2-r)=(x-p)^{2}(x-q)=x^{3}-(2 p+q) x^{2}+p(p+2 q) x-p^{2} q
$$

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 $r<P(1 / 3)=1 / 27+1 / 9-1 / 3+2=49 / 27$.


problem: 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 $a_{2007}$.

solution: $\mathbf{2}^{\mathbf{2 0 0 7}}$. 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 .


problem:
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 $x \in A$,

$$
f(x)+f\left(1-\frac{1}{x}\right)=\log |x|
$$

Compute the value of $f(2007)$.

solution:
$\log (\mathbf{2 0 0 7} / \mathbf{2 0 0 6})$. Let $g: A \rightarrow A$ be defined by $g(x):=1-1 / x$; the key property is that

$$
g(g(g(x)))=1-\frac{1}{1-\frac{1}{1-\frac{1}{x}}}=x
$$

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

$$
f(x)=\frac{1}{2} \cdot(\log |x|-\log |g(x)|+\log |g(g(x))|)
$$

For $x=2007$, we have $g(x)=\frac{2006}{2007}$ and $g(g(x))=\frac{-1}{2006}$, so that

$$
f(2007)=\frac{\log |2007|-\log \left|\frac{2006}{2007}\right|+\log \left|\frac{-1}{2006}\right|}{2}=\log (2007 / 2006)
$$


problem:
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

$$
\left\{\alpha_{1} \alpha_{2}+\alpha_{3} \alpha_{4}, \alpha_{1} \alpha_{3}+\alpha_{2} \alpha_{4}, \alpha_{1} \alpha_{4}+\alpha_{2} \alpha_{3}\right\} .
$$

solution:
$\{\mathbf{1} \pm \sqrt{\mathbf{5}}, \mathbf{- 2}\}$. 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

$$
P(x)=\left(x-\left(\alpha_{1} \alpha_{2}+\alpha_{3} \alpha_{4}\right)\right)\left(x-\left(\alpha_{1} \alpha_{3}+\alpha_{2} \alpha_{4}\right)\right)\left(x-\left(\alpha_{1} \alpha_{4}+\alpha_{2} \alpha_{3}\right)\right)
$$

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

$$
\begin{aligned}
P(x) & =x^{3}-s_{2} x^{2}+\left(s_{3} s_{1}-4 s_{4}\right) x+\left(-s_{3}^{2}-s_{4} s_{1}^{2}+s_{4} s_{2}\right) \\
& =x^{3}-8 x-8 \\
& =(x+2)\left(x^{2}-2 x-4\right)
\end{aligned}
$$

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.


problem: 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 $P(1) / P(-1)$.

solution:
| $\mathbf{2 8 9}$. |
| :---: |
| For some constant $k$, we have |

$$
P(z)=k \prod_{j=1}^{2007}\left(z-\left(r_{j}+\frac{1}{r_{j}}\right)\right)
$$

Now writing $\omega^{3}=1$ with $\omega \neq 1$, we have $\omega^{2}+\omega=-1$. Then

$$
\begin{gathered}
P(1) / P(-1)=\frac{k \prod_{j=1}^{2007}\left(1-\left(r_{j}+\frac{1}{r_{j}}\right)\right)}{k \prod_{j=1}^{2007}\left(-1-\left(r_{j}+\frac{1}{r_{j}}\right)\right)}=\prod_{j=1}^{2007} \frac{r_{j}^{2}-r_{j}+1}{r_{j}^{2}+r_{j}+1}=\prod_{j=1}^{2007} \frac{\left(-\omega-r_{j}\right)\left(-\omega^{2}-r_{j}\right)}{\left(\omega-r_{j}\right)\left(\omega^{2}-r_{j}\right)} \\
=\frac{f(-\omega) f\left(-\omega^{2}\right)}{f(\omega) f\left(\omega^{2}\right)}=\frac{\left(-\omega^{2007}+17 \omega^{2006}+1\right)\left(-\left(\omega^{2}\right)^{2007}+17\left(\omega^{2}\right)^{2006}+1\right)}{\left(\omega^{2007}+17 \omega^{2006}+1\right)\left(\left(\omega^{2}\right)^{2007}+17\left(\omega^{2}\right)^{2006}+1\right)}=\frac{\left(17 \omega^{2}\right)(17 \omega)}{\left(2+17 \omega^{2}\right)(2+17 \omega)} \\
=\frac{289}{4+34\left(\omega+\omega^{2}\right)+289}=\frac{289}{259} .
\end{gathered}
$$