2007 USA TST Problems

Problem 1

Circles $\omega_{1}$ and $\omega_{2}$ intersect at $P$ and $Q$ . $AC$ and $BD$ are chords of $\omega_{1}$ and $\omega_{2}$ , respectively, such that $P$ is on segment $AB$ and on ray $CD$ . Lines $AC$ and $BD$ intersect at $X$ . Let the line through $P$ parallel to $AC$ intersect $\omega_{2}$ again at $Y$ , and let the line through $P$ parallel to $BD$ intersect $\omega_{1}$ again at $Z$ . Prove $Q, X, Y, Z$ are collinear.

Problem 2

Let $a_{1}\le a_{2}\le ...\le a_{n}$ , $b_{1}\le b_{2}\le ... \le b_{n}$ be two nonincreasing sequences of reals such that $a_{1}\le b_{1}$ , $a_{1}+a_{2}\le b_{1}+b_{2}$ , $\displaystyle \ldots$ , $\displaystyle a_{1}+a_{2}+\ldots+a_{n-1}\le b_{1}+b_{2}+\ldots+b_{n-1}$ and $\displaystyle a_{1}+a_{2}+\ldots+a_{n}=b_{1}+b_{2}+\ldots+b_{n}$ For any real number $m$ , the number of pairs $(i, j)$ such that $a_{i}-a_{j}=m$ is equal to the number of pairs $(k, l)$ such that $b_{k}-b_{l}=m$ . Prove that $a_{i}=b_{i}$ for $i=1, 2, ..., n$ .

Problem 3

For some $\theta \in (0, \frac{\pi}{2})$ , $\displaystyle \cos{\theta}$ is irrational. If, for some positive integer $k$ , $\displaystyle \cos{(k\theta)}$ and $\displaystyle \cos{([k+1]\theta)}$ are both rational, then show $\theta=\frac{\pi}{6}$ .

Problem 4

Are there two positive integers $(a, b)$ such that, for each positive integer $n$ , $b^{n}-n$ is not divisible by $a$ ?

Problem 5

Let the tangents at $B$ and $C$ to the circumcircle of $\triangle{ABC}$ meet at $T$ . Let the perpendicular to $AT$ at $A$ meet ray $BC$ at $S$ . Let $B_{1}, C_{1}$ lie on $ST$ such that $B_{1}T=C_{1}T=BT$ and so that $T$ lies between $S$ and $B_{1}$ . Prove that $\triangle{AB_{1}C_{1}}\sim \triangle{ABC}$ .

Problem 6

For any polynomial $P$ , let $r(2i-1)$ be the remainder mod $1024$ from 0 to 1023, inclusive, of $P(2i-1)$ for $i=1, 2, ..., 512$ . Call the set $\{r(1), r(3), r(5), ..., r(1023)\}$ the remainder sequence of $P$ . Call a remaidner sequence complete if it is a permutation of $\{1, 3, 5, ..., 1023\}$ . Show that the number of complete remainder sequences is at most $2^{35}$ .

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