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Difference between revisions of "1998 USAMO Problems"

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Problems of the [[1998 USAMO | 1998]] [[USAMO]].
 
Problems of the [[1998 USAMO | 1998]] [[USAMO]].
  
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==Problem 1==
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==Day 1==
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===Problem 1===
 
Suppose that the set <math>\{1,2,\cdots, 1998\}</math> has been partitioned into disjoint pairs <math>\{a_i,b_i\}</math> (<math>1\leq i\leq 999</math>) so that for all <math>i</math>, <math>|a_i-b_i|</math> equals <math>1</math> or <math>6</math>. Prove that the sum
 
Suppose that the set <math>\{1,2,\cdots, 1998\}</math> has been partitioned into disjoint pairs <math>\{a_i,b_i\}</math> (<math>1\leq i\leq 999</math>) so that for all <math>i</math>, <math>|a_i-b_i|</math> equals <math>1</math> or <math>6</math>. Prove that the sum
 
<cmath> |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|  </cmath>
 
<cmath> |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|  </cmath>
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[[1998 USAMO Problems/Problem 1|Solution]]
 
[[1998 USAMO Problems/Problem 1|Solution]]
  
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==Problem 2==
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===Problem 2===
 
Let <math>{\cal C}_1</math> and <math>{\cal C}_2</math> be  concentric circles, with <math>{\cal C}_2</math> in the interior of  <math>{\cal C}_1</math>. From a point <math>A</math> on <math>{\cal C}_1</math> one draws the tangent <math>AB</math> to <math>{\cal C}_2</math> (<math>B\in {\cal C}_2</math>). Let <math>C</math> be the second point of intersection of <math>AB</math> and <math>{\cal C}_1</math>, and let <math>D</math> be the midpoint of <math>AB</math>. A line passing through <math>A</math> intersects <math>{\cal C}_2</math> at <math>E</math> and <math>F</math> in such a way that the perpendicular  bisectors of <math>DE</math> and <math>CF</math> intersect at a point <math>M</math> on <math>AB</math>. Find, with proof,  the ratio <math>AM/MC</math>.
 
Let <math>{\cal C}_1</math> and <math>{\cal C}_2</math> be  concentric circles, with <math>{\cal C}_2</math> in the interior of  <math>{\cal C}_1</math>. From a point <math>A</math> on <math>{\cal C}_1</math> one draws the tangent <math>AB</math> to <math>{\cal C}_2</math> (<math>B\in {\cal C}_2</math>). Let <math>C</math> be the second point of intersection of <math>AB</math> and <math>{\cal C}_1</math>, and let <math>D</math> be the midpoint of <math>AB</math>. A line passing through <math>A</math> intersects <math>{\cal C}_2</math> at <math>E</math> and <math>F</math> in such a way that the perpendicular  bisectors of <math>DE</math> and <math>CF</math> intersect at a point <math>M</math> on <math>AB</math>. Find, with proof,  the ratio <math>AM/MC</math>.
  
 
[[1998 USAMO Problems/Problem 2|Solution]]
 
[[1998 USAMO Problems/Problem 2|Solution]]
  
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==Problem 3==
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===Problem 3===
 
Let <math>a_0,a_1,\cdots ,a_n</math> be numbers from the interval <math>(0,\pi/2)</math> such that
 
Let <math>a_0,a_1,\cdots ,a_n</math> be numbers from the interval <math>(0,\pi/2)</math> such that
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<cmath> \tan (a_0-\frac{\pi}{4})+ \tan (a_1-\frac{\pi}{4})+\cdots +\tan (a_n-\frac{\pi}{4})\geq n-1.  </cmath>
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<cmath> \tan \left(a_0-\frac{\pi}{4}\right)+ \tan \left(a_1-\frac{\pi}{4}\right)+\cdots +\tan \left(a_n-\frac{\pi}{4}\right)\geq n-1.  </cmath>
 
Prove that
 
Prove that
 
<cmath> \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}.  </cmath>
 
<cmath> \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}.  </cmath>
 
[[1998 USAMO Problems/Problem 3|Solution]]
 
[[1998 USAMO Problems/Problem 3|Solution]]
  
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==Problem 4==
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==Day 2==
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===Problem 4===
 
A computer screen shows a <math>98 \times 98</math> 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.
 
A computer screen shows a <math>98 \times 98</math> 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.
  
 
[[1998 USAMO Problems/Problem 4|Solution]]
 
[[1998 USAMO Problems/Problem 4|Solution]]
  
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==Problem 5==
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===Problem 5===
 
Prove that for each <math>n\geq 2</math>, there is a set <math>S</math> of <math>n</math> integers such that <math>(a-b)^2</math> divides <math>ab</math> for every distinct <math>a,b\in S</math>.
 
Prove that for each <math>n\geq 2</math>, there is a set <math>S</math> of <math>n</math> integers such that <math>(a-b)^2</math> divides <math>ab</math> for every distinct <math>a,b\in S</math>.
  
 
[[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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===Problem 6===
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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>.
 
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==Problem 6==
 
 
Let <math>n \geq 5</math> be an integer. Find the largest integer <math>k</math> (as a function of <math>n</math>) such that there exists a convex <math>n</math>-gon <math>A_{1}A_{2}\dots A_{n}</math> for which exactly <math>k</math> of the quadrilaterals <math>A_{i}A_{i+1}A_{i+2}A_{i+3}</math> have an inscribed circle. (Here <math>A_{n+j} = A_{j}</math>.)
 
Let <math>n \geq 5</math> be an integer. Find the largest integer <math>k</math> (as a function of <math>n</math>) such that there exists a convex <math>n</math>-gon <math>A_{1}A_{2}\dots A_{n}</math> for which exactly <math>k</math> of the quadrilaterals <math>A_{i}A_{i+1}A_{i+2}A_{i+3}</math> have an inscribed circle. (Here <math>A_{n+j} = A_{j}</math>.)
  
 
[[1998 USAMO Problems/Problem 6|Solution]]
 
[[1998 USAMO Problems/Problem 6|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>.
 
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== Resources ==
 
  
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== See Also ==
 
{{USAMO newbox|year=1998|before=[[1997 USAMO]]|after=[[1999 USAMO]]}}
 
{{USAMO newbox|year=1998|before=[[1997 USAMO]]|after=[[1999 USAMO]]}}
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{{MAA Notice}}
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* [http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27&year=1998 1998 USAMO Problems on the resources page]
 

Latest revision as of 05:11, 24 November 2020

Problems of the 1998 USAMO.

Day 1

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 \[\tan \left(a_0-\frac{\pi}{4}\right)+ \tan \left(a_1-\frac{\pi}{4}\right)+\cdots +\tan \left(a_n-\frac{\pi}{4}\right)\geq n-1.\] Prove that \[\tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}.\] Solution

Day 2

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

See Also

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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