# $11^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament 

## Saturday 23 February 2008

## Individual Round: Algebra Test

1. [3] Positive real numbers $x, y$ satisfy the equations $x^{2}+y^{2}=1$ and $x^{4}+y^{4}=\frac{17}{18}$. Find $x y$.

Answer: $\sqrt[\frac{1}{6}]{ }$ We have $2 x^{2} y^{2}=\left(x^{2}+y^{2}\right)^{2}-\left(x^{4}+y^{4}\right)=\frac{1}{18}$, so $x y=\frac{1}{6}$.
2. [3] Let $f(n)$ be the number of times you have to hit the $\sqrt{ }$ key on a calculator to get a number less than 2 starting from $n$. For instance, $f(2)=1, f(5)=2$. For how many $1<m<2008$ is $f(m)$ odd?
Answer: 242 This is $\left[2^{1}, 2^{2}\right) \cup\left[2^{4}, 2^{8}\right) \cup\left[2^{16}, 2^{32}\right) \ldots$, and $2^{8}<2008<2^{16}$ so we have exactly the first two intervals.
3. [4] Determine all real numbers $a$ such that the inequality $\left|x^{2}+2 a x+3 a\right| \leq 2$ has exactly one solution in $x$.
Answer: 1,2 Let $f(x)=x^{2}+2 a x+3 a$. Note that $f(-3 / 2)=9 / 4$, so the graph of $f$ is a parabola that goes through $(-3 / 2,9 / 4)$. Then, the condition that $\left|x^{2}+2 a x+3 a\right| \leq 2$ has exactly one solution means that the parabola has exactly one point in the strip $-1 \leq y \leq 1$, which is possible if and only if the parabola is tangent to $y=1$. That is, $x^{2}+2 a x+3 a=2$ has exactly one solution. Then, the discriminant $\Delta=4 a^{2}-4(3 a-2)=4 a^{2}-12 a+8$ must be zero. Solving the equation yields $a=1,2$.
4. [4] The function $f$ satisfies

$$
f(x)+f(2 x+y)+5 x y=f(3 x-y)+2 x^{2}+1
$$

for all real numbers $x, y$. Determine the value of $f(10)$.
Answer: $\quad-49$ Setting $x=10$ and $y=5$ gives $f(10)+f(25)+250=f(25)+200+1$, from which we get $f(10)=-49$.

Remark: By setting $y=\frac{x}{2}$, we see that the function is $f(x)=-\frac{1}{2} x^{2}+1$, and it can be checked that this function indeed satisfies the given equation.
5. [5] Let $f(x)=x^{3}+x+1$. Suppose $g$ is a cubic polynomial such that $g(0)=-1$, and the roots of $g$ are the squares of the roots of $f$. Find $g(9)$.
Answer: 899 Let $a, b, c$ be the zeros of $f$. Then $f(x)=(x-a)(x-b)(x-c)$. Then, the roots of $g$ are $a^{2}, b^{2}, c^{2}$, so $g(x)=k\left(x-a^{2}\right)\left(x-b^{2}\right)\left(x-c^{2}\right)$ for some constant $k$. Since $a b c=-f(0)=-1$, we have $k=k a^{2} b^{2} c^{2}=-g(0)=1$. Thus,

$$
g\left(x^{2}\right)=\left(x^{2}-a^{2}\right)\left(x^{2}-b^{2}\right)\left(x^{2}-c^{2}\right)=(x-a)(x-b)(x-c)(x+a)(x+b)(x+c)=-f(x) f(-x)
$$

Setting $x=3$ gives $g(9)=-f(3) f(-3)=-(31)(-29)=899$.
6. [5] A root of unity is a complex number that is a solution to $z^{n}=1$ for some positive integer $n$. Determine the number of roots of unity that are also roots of $z^{2}+a z+b=0$ for some integers $a$ and $b$.
Answer: 8 The only real roots of unity are 1 and -1 . If $\zeta$ is a complex root of unity that is also a root of the equation $z^{2}+a z+b$, then its conjugate $\bar{\zeta}$ must also be a root. In this case, $|a|=|\zeta+\bar{\zeta}| \leq|\zeta|+|\bar{\zeta}|=$ 2 and $b=\zeta \bar{\zeta}=1$. So we only need to check the quadratics $z^{2}+2 z+1, z^{2}+z+1, z^{2}+1, z^{2}-z+1, z^{2}-2 z+1$. We find 8 roots of unity: $\pm 1, \pm i, \frac{1}{2}( \pm 1 \pm \sqrt{3} i)$.
7. [5] Compute $\sum_{n=1}^{\infty} \sum_{k=1}^{n-1} \frac{k}{2^{n+k}}$.

Answer: $\frac{4}{9}$ We change the order of summation:

$$
\sum_{n=1}^{\infty} \sum_{k=1}^{n-1} \frac{k}{2^{n+k}}=\sum_{k=1}^{\infty} \frac{k}{2^{k}} \sum_{n=k+1}^{\infty} \frac{1}{2^{n}}=\sum_{k=1}^{\infty} \frac{k}{4^{k}}=\frac{4}{9} .
$$

(The last two steps involve the summation of an infinite geometric series, and what is sometimes called an infinite arithmetico-geometric series. These summations are quite standard, and thus we omit the details here.)
8. [6] Compute $\arctan \left(\tan 65^{\circ}-2 \tan 40^{\circ}\right)$. (Express your answer in degrees as an angle between $0^{\circ}$ and $180^{\circ}$.)
Answer: $25^{\circ}$ First Solution: We have

$$
\tan 65^{\circ}-2 \tan 40^{\circ}=\cot 25^{\circ}-2 \cot 50^{\circ}=\cot 25^{\circ}-\frac{\cot ^{2} 25^{\circ}-1}{\cot 25^{\circ}}=\frac{1}{\cot 25^{\circ}}=\tan 25^{\circ} .
$$

Therefore, the answer is $25^{\circ}$.
Second Solution: We have
$\tan 65^{\circ}-2 \tan 40^{\circ}=\frac{1+\tan 20^{\circ}}{1-\tan 20^{\circ}}-\frac{4 \tan 20^{\circ}}{1-\tan ^{2} 20^{\circ}}=\frac{\left(1-\tan 20^{\circ}\right)^{2}}{\left(1-\tan 20^{\circ}\right)\left(1+\tan 20^{\circ}\right)}=\tan \left(45^{\circ}-20^{\circ}\right)=\tan 25^{\circ}$.
Again, the answer is $25^{\circ}$.
9. [7] Let $S$ be the set of points $(a, b)$ with $0 \leq a, b \leq 1$ such that the equation

$$
x^{4}+a x^{3}-b x^{2}+a x+1=0
$$

has at least one real root. Determine the area of the graph of $S$.
Answer: $\frac{1}{4}$ After dividing the equation by $x^{2}$, we can rearrange it as

$$
\left(x+\frac{1}{x}\right)^{2}+a\left(x+\frac{1}{x}\right)-b-2=0
$$

Let $y=x+\frac{1}{x}$. We can check that the range of $x+\frac{1}{x}$ as $x$ varies over the nonzero reals is $(-\infty,-2] \cup[2, \infty)$. Thus, the following equation needs to have a real root:

$$
y^{2}+a y-b-2=0 .
$$

Its discriminant, $a^{2}+4(b+2)$, is always positive since $a, b \geq 0$. Then, the maximum absolute value of the two roots is

$$
\frac{a+\sqrt{a^{2}+4(b+2)}}{2} .
$$

We need this value to be at least 2 . This is equivalent to

$$
\sqrt{a^{2}+4(b+2)} \geq 4-a .
$$

We can square both sides and simplify to obtain

$$
2 a \geq 2-b
$$

This equation defines the region inside $[0,1] \times[0,1]$ that is occupied by $S$, from which we deduce that the desired area is $1 / 4$.
10. [8] Evaluate the infinite sum

$$
\sum_{n=0}^{\infty}\binom{2 n}{n} \frac{1}{5^{n}}
$$

Answer: $\sqrt{5}$ First Solution: Note that

$$
\begin{aligned}
\binom{2 n}{n} & =\frac{(2 n)!}{n!\cdot n!}=\frac{(2 n)(2 n-2)(2 n-4) \cdots(2)}{n!} \cdot \frac{(2 n-1)(2 n-3)(2 n-5) \cdots(1)}{n!} \\
& =2^{n} \cdot \frac{(-2)^{n}}{n!}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right) \cdots\left(-\frac{1}{2}-n+1\right) \\
& =(-4)^{n}\binom{-\frac{1}{2}}{n}
\end{aligned}
$$

Then, by the binomial theorem, for any real $x$ with $|x|<\frac{1}{4}$, we have

$$
(1-4 x)^{-1 / 2}=\sum_{n=0}^{\infty}\binom{-\frac{1}{2}}{n}(-4 x)^{n}=\sum_{n=0}^{\infty}\binom{2 n}{n} x^{n} .
$$

Therefore,

$$
\sum_{n=0}^{\infty}\binom{2 n}{n}\left(\frac{1}{5}\right)^{n}=\frac{1}{\sqrt{1-\frac{4}{5}}}=\sqrt{5}
$$

Second Solution: Consider the generating function

$$
f(x)=\sum_{n=0}^{\infty}\binom{2 n}{n} x^{n}
$$

It has formal integral given by

$$
g(x)=I(f(x))=\sum_{n=0}^{\infty} \frac{1}{n+1}\binom{2 n}{n} x^{n+1}=\sum_{n=0}^{\infty} C_{n} x^{n+1}=x \sum_{n=0}^{\infty} C_{n} x^{n}
$$

where $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$ is the $n$th Catalan number. Let $h(x)=\sum_{n=0}^{\infty} C_{n} x^{n}$; it suffices to compute this generating function. Note that

$$
1+x h(x)^{2}=1+x \sum_{i, j \geq 0} C_{i} C_{j} x^{i+j}=1+x \sum_{k \geq 0}\left(\sum_{i=0}^{k} C_{i} C_{k-i}\right) x^{k}=1+\sum_{k \geq 0} C_{k+1} x^{k+1}=h(x)
$$

where we've used the recurrence relation for the Catalan numbers. We now solve for $h(x)$ with the quadratic equation to obtain

$$
h(x)=\frac{1 / x \pm \sqrt{1 / x^{2}-4 / x}}{2}=\frac{1 \pm \sqrt{1-4 x}}{2 x}
$$

Note that we must choose the - sign in the $\pm$, since the + would lead to a leading term of $\frac{1}{x}$ for $h$ (by expanding $\sqrt{1-4 x}$ into a power series). Therefore, we see that

$$
f(x)=D(g(x))=D(x h(x))=D\left(\frac{1-\sqrt{1-4 x}}{2}\right)=\frac{1}{\sqrt{1-4 x}}
$$

and our answer is hence $f(1 / 5)=\sqrt{5}$.