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Difference between revisions of "2007 USA TST Problems"

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22007 USA TST Problems
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== Problem 1 ==
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Circles <math>\omega_{1}</math> and <math>\omega_{2}</math> intersect at <math>P</math> and <math>Q</math>.  <math>AC</math> and <math>BD</math> are chords of <math>\omega_{1}</math> and <math>\omega_{2}</math>, respectively, such that <math>P</math> is on segment <math>AB</math> and on ray <math>CD</math>.  Lines <math>AC</math> and <math>BD</math> intersect at <math>X</math>.  Let the line through <math>P</math> parallel to <math>AC</math> intersect <math>\omega_{2}</math> again at <math>Y</math>, and let the line through <math>P</math> parallel to <math>BD</math> intersect <math>\omega_{1}</math> again at <math>Z</math>.  Prove <math>Q, X, Y, Z</math> are collinear.
  
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{{wikify}}
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[[2007 USA TST Problems/Problem 1|Solution]]
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== Problem 2 ==
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1. Circles <math>\omega_{1}</math> and <math>\omega_{2}</math> intersect at <math>P</math> and <math>Q</math>.  <math>AC</math> and <math>BD</math> are chords of <math>\omega_{1}</math> and <math>\omega_{2}</math>, respectively, such that <math>P</math> is on segment <math>AB</math> and on ray <math>CD</math>.  Lines <math>AC</math> and <math>BD</math> intersect at <math>X</math>.  Let the line through <math>P</math> parallel to <math>AC</math> intersect <math>\omega_{2}</math> again at <math>Y</math>, and let the line through <math>P</math> parallel to <math>BD</math> intersect <math>\omega_{1}</math> again at <math>Z</math>.  Prove <math>Q, X, Y, Z</math> are collinear.
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Let <math>a_{1}\le a_{2}\le ...\le a_{n}</math>, <math>b_{1}\le b_{2}\le ... \le b_{n}</math> be two nonincreasing sequences of reals such that
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2. Let <math>a_{1}\le a_{2}\le ...\le a_{n}</math>, <math>b_{1}\le b_{2}\le ... \le b_{n}</math> be two nonincreasing sequences of reals such that
 
 
<math>
 
<math>
 
a_{1}\le b_{1}
 
a_{1}\le b_{1}
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</math>
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</math>, <math>a_{1}+a_{2}\le b_{1}+b_{2}</math>, <math>\displaystyle \ldots</math>, <math>
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\displaystyle a_{1}+a_{2}+\ldots+a_{n-1}\le b_{1}+b_{2}+\ldots+b_{n-1}
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<math>
 
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a_{1}+a_{2}\le b_{1}+b_{2}
 
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</math>
 
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...
 
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<math>
 
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a_{1}+a_{2}+...+a_{n-1}\le b_{1}+b_{2}+...+b_{n-1}
 
 
</math>
 
</math>
 
and
 
and
 
<math>
 
<math>
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a_{1}+a_{2}+...+a_{n}=b_{1}+b_{2}+...+b_{n}
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\displaystyle a_{1}+a_{2}+\ldots+a_{n}=b_{1}+b_{2}+\ldots+b_{n}
 
</math>
 
</math>
 
For any real number <math>m</math>, the number of pairs <math>(i, j)</math> such that <math>a_{i}-a_{j}=m</math> is equal to the number of pairs <math>(k, l)</math> such that <math>b_{k}-b_{l}=m</math>.  Prove that <math>a_{i}=b_{i}</math> for <math>i=1, 2, ..., n</math>.
 
For any real number <math>m</math>, the number of pairs <math>(i, j)</math> such that <math>a_{i}-a_{j}=m</math> is equal to the number of pairs <math>(k, l)</math> such that <math>b_{k}-b_{l}=m</math>.  Prove that <math>a_{i}=b_{i}</math> for <math>i=1, 2, ..., n</math>.
  
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[[2007 USA TST Problems/Problem 2|Solution]]
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== Problem 3 ==
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For some <math>\theta \in (0, \frac{\pi}{2})</math>, <math>\displaystyle  \cos{\theta}</math> is irrational.  If, for some positive integer <math>k</math>, <math>\displaystyle \cos{(k\theta)}</math> and <math>\displaystyle \cos{([k+1]\theta)}</math> are both rational, then show <math>\theta=\frac{\pi}{6}</math>.
  
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3. For some <math>\theta \in (0, \frac{\pi}{2})</math>, <math>\cos{\theta}</math> is irrational.  If, for some positive integer <math>k</math>, <math>\cos{(k\theta)}</math> and <math>\cos{([k+1]\theta)}</math> are both rational, then show <math>\theta=\frac{\pi}{6}</math>.
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[[2007 USA TST Problems/Problem 3|Solution]]
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== Problem 4 ==
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Are there two positive integers <math>(a, b)</math> such that, for each positive integer <math>n</math>, <math>b^{n}-n</math> is not divisible by <math>a</math>?
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4. Are there two positive integers <math>(a, b)</math> such that, for each positive integer <math>n</math>, <math>b^{n}-n</math> is not divisible by <math>a</math>?
 
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5. Let the tangents at <math>B</math> and <math>C</math> to the circumcircle of <math>\triangle{ABC}</math> meet at <math>T</math>.  Let the perpendicular to <math>AT</math> at <math>A</math> meet ray <math>BC</math> at <math>S</math>.  Let <math>B_{1}, C_{1}</math> lie on <math>ST</math> such that <math>B_{1}T=C_{1}T=BT</math> and so that <math>T</math> lies between <math>S</math> and <math>B_{1}</math>.  Prove that <math>\triangle{AB_{1}C_{1}}\sim \triangle{ABC}</math>.
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[[2007 USA TST Problems/Problem 4|Solution]]
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== Problem 5 ==
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Let the tangents at <math>B</math> and <math>C</math> to the circumcircle of <math>\triangle{ABC}</math> meet at <math>T</math>.  Let the perpendicular to <math>AT</math> at <math>A</math> meet ray <math>BC</math> at <math>S</math>.  Let <math>B_{1}, C_{1}</math> lie on <math>ST</math> such that <math>B_{1}T=C_{1}T=BT</math> and so that <math>T</math> lies between <math>S</math> and <math>B_{1}</math>.  Prove that <math>\triangle{AB_{1}C_{1}}\sim \triangle{ABC}</math>.
  
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[[2007 USA TST Problems/Problem 5|Solution]]
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== Problem 6 ==
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For any polynomial <math>P</math>, let <math>r(2i-1)</math> be the remainder mod <math>1024</math> from 0 to 1023, inclusive, of <math>P(2i-1)</math> for <math>i=1, 2, ..., 512</math>.  Call the set <math>\{r(1),\ r(3),\ r(5),\ ...,\ r(1023)\}</math> the ''remainder sequence'' of <math>P</math>.  Call a remaidner sequence '''complete''' if it is a permutation of <math>\{1, 3, 5, ..., 1023\}</math>.  Show that the number of complete remainder sequences is at most <math>2^{35}</math>.
  
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6. For any polynomial <math>P</math>, let <math>r(2i-1)</math> be the remainder mod <math>1024</math> from 0 to 1023, inclusive, of <math>P(2i-1)</math> for <math>i=1, 2, ..., 512</math>.  Call the set <math>\{r(1), r(3), r(5), ..., r(1023)\}</math> the [i]remainder sequence[/i] of <math>P</math>.  Call a remaidner sequence [b]complete[/b] if it is a permutation of <math>\{1, 3, 5, ..., 1023\}</math>.  Show that the number of complete remainder sequences is at most <math>2^{35}</math>.
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[[2007 USA TST Problems/Problem 6|Solution]]
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== See also ==
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*[[TST]]

Latest revision as of 19:59, 24 May 2007

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.

Solution

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 $i=1, 2, ..., n$.

Solution

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 $\theta=\frac{\pi}{6}$.

Solution

Problem 4

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

Solution

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 $\triangle{AB_{1}C_{1}}\sim \triangle{ABC}$.

Solution

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 $2^{35}$.

Solution

See also

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