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Difference between revisions of "1998 USAMO Problems"

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[[1998 USAMO Problems/Problem 5|Solution]]
 
[[1998 USAMO Problems/Problem 5|Solution]]
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Proof by induction.  For n=2, the proof is trivial, since <math>S = (1,2)</math> satisfies the condition.  Assume now that there is such a set S of n elements, <math>a_1, a_2,...a_n</math> which satisfy the condition.  The key is to note that if <math>m=a_1a_2...a_n</math>, then if we define <math>b_i=a_i + km</math> for all <math>i\le n</math>, where k is a positive integer, then <math>a_i \mid b_i</math> and <math>b_i - b_j = a_i - a_j</math>, and so <math>(b_i - b_j)^2 = (a_i - a_j)^2 \mid a_ia_j \mid b_ib_j</math>.
 
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Let <math>b_{n+1}=m +km</math>.  Consider the set <math>T = (b_1,b_2,...,b_n,b_{n+1})</math>. To finish the proof, we simply need to choose a k such that <math>(b_{n+1}-b_i)^2 \mid b_{n+1}b_i</math> for all <math>i\le n</math>.  Since <math>(b_{n+1}-b_i)^2 = (m-a_i)^2</math>, simply choose k so that <math>k+1 = (m-a_1)^2(m-a_2)^2...(m-a_n)^2</math>.
 
  
 
==Problem 6==
 
==Problem 6==

Revision as of 12:17, 3 June 2011

Problems of the 1998 USAMO.

Problem 1

Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1\leq i\leq 999$) so that for all $i$, $|a_i-b_i|$ equals $1$ or $6$. Prove that the sum \[|a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|\] ends in the digit $9$.

Solution

Problem 2

Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$.

Solution

Problem 3

Let $a_0,a_1,\cdots ,a_n$ be numbers from the interval $(0,\pi/2)$ such that \[\tan (a_0-\frac{\pi}{4})+ \tan (a_1-\frac{\pi}{4})+\cdots +\tan (a_n-\frac{\pi}{4})\geq n-1.\] Prove that \[\tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}.\] Solution

Problem 4

A computer screen shows a $98 \times 98$ chessboard, colored in the usual way. One can select with a mouse any rectangle with sides on the lines of the chessboard and click the mouse button: as a result, the colors in the selected rectangle switch (black becomes white, white becomes black). Find, with proof, the minimum number of mouse clicks needed to make the chessboard all one color.

Solution

Problem 5

Prove that for each $n\geq 2$, there is a set $S$ of $n$ integers such that $(a-b)^2$ divides $ab$ for every distinct $a,b\in S$.

Solution

Problem 6

Let $n \geq 5$ be an integer. Find the largest integer $k$ (as a function of $n$) such that there exists a convex $n$-gon $A_{1}A_{2}\dots A_{n}$ for which exactly $k$ of the quadrilaterals $A_{i}A_{i+1}A_{i+2}A_{i+3}$ have an inscribed circle. (Here $A_{n+j} = A_{j}$.)

Solution Proof by induction. For n=2, the proof is trivial, since $S = (1,2)$ satisfies the condition. Assume now that there is such a set S of n elements, $a_1, a_2,...a_n$ which satisfy the condition. The key is to note that if $m=a_1a_2...a_n$, then if we define $b_i=a_i + km$ for all $i\le n$, where k is a positive integer, then $a_i \mid b_i$ and $b_i - b_j = a_i - a_j$, and so $(b_i - b_j)^2 = (a_i - a_j)^2 \mid a_ia_j \mid b_ib_j$.

Let $b_{n+1}=m +km$. Consider the set $T = (b_1,b_2,...,b_n,b_{n+1})$. To finish the proof, we simply need to choose a k such that $(b_{n+1}-b_i)^2 \mid b_{n+1}b_i$ for all $i\le n$. Since $(b_{n+1}-b_i)^2 = (m-a_i)^2$, simply choose k so that $k+1 = (m-a_1)^2(m-a_2)^2...(m-a_n)^2$.

Resources

1998 USAMO (Problems • Resources)
Preceded by
1997 USAMO 		Followed by
1999 USAMO
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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