Difference between revisions of "2003 USAMO Problems"

Revision as of 16:08, 7 September 2008

Problems of the 2003 USAMO.

Day 1

Problem 1

Prove that for every positive integer $\displaystyle n$ there exists an $\displaystyle n$ -digit number divisible by $\displaystyle 5^n$ all of whose digits are odd.

Solution

Problem 2

A convex polygon $\mathcal{P}$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $\mathcal{P}$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.

Solution

Problem 3

Let $n \neq 0$ . For every sequence of integers

$A = a_0,a_1,a_2,\dots, a_n$

satisfying $0 \le a_i \le i$ , for $i=0,\dots,n$ , define another sequence

$t(A)= t(a_0), t(a_1), t(a_2), \dots, t(a_n)$

by setting $\displaystyle t(a_i)$ to be the number of terms in the sequence $\displaystyle A$ that precede the term $\displaystyle a_i$ and are different from $\displaystyle a_i$ . Show that, starting from any sequence $\displaystyle A$ as above, fewer than $\displaystyle n$ applications of the transformation $\displaystyle t$ lead to a sequence $\displaystyle B$ such that $\displaystyle t(B) = B$ .

Solution

Day 2

Problem 4

Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$ , respectively. Lines $AB$ and $DE$ intersect at $F$ , while lines $BD$ and $CF$ intersect at $M$ . Prove that $MF = MC$ if and only if $MB\cdot MD = MC^2$ .

Solution

Problem 5

Let $a$ , $b$ , $c$ be positive real numbers. Prove that

$\dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \dfrac{(2c + a + b)^2}{2c^2 + (a + b)^2} \le 8.$

Solution

Problem 6

At the vertices of a regular hexagon are written six nonnegative integers whose sum is 2003. Bert is allowed to make moves of the following form: he may pick a vertex and replace the number written there by the absolute value of the difference between the numbers written at the two neighboring vertices. Prove that Bert can make a sequence of moves, after which the number 0 appears at all six vertices.

Solution

@@ Line 1: / Line 1: @@
-Problems of the [[2000 USAMO | 2000]] [[USAMO]].
+Problems of the [[2003 USAMO | 2003]] [[USAMO]].
 == Day 1 ==
@@ Line 5: / Line 5: @@
 === Problem 1 ===
-Call a real-valued function <math>f</math> very convex if
+Prove that for every positive integer <math> \displaystyle n </math> there exists an <math> \displaystyle n </math>-digit number divisible by <math> \displaystyle 5^n </math> all of whose digits are odd.
-<cmath>\frac {f(x) + f(y)}{2} \ge f\left(\frac {x + y}{2}\right) + |x - y|</cmath>
+* [[2003 USAMO Problems/Problem 1 | Solution]]
-holds for all real numbers <math>x</math> and <math>y</math>. Prove that no very convex function exists.
-* [[2000 USAMO Problems/Problem 1 | Solution]]
 === Problem 2 ===
-Let <math>S</math> be the set of all triangles <math>ABC</math> for which
+A convex polygon <math> \mathcal{P} </math> in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon <math> \mathcal{P} </math> are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.
-<cmath>5 \left( \dfrac{1}{AP} + \dfrac{1}{BQ} + \dfrac{1}{CR} \right) - \dfrac{3}{\min\{ AP, BQ, CR \}} = \dfrac{6}{r},</cmath>
-where <math>r</math> is the inradius and <math>P, Q, R</math> are the points of tangency of the incircle with sides <math>AB, BC, CA,</math> respectively. Prove that all triangles in <math>S</math> are isosceles  and similar to one another.
+* [[2003 USAMO Problems/Problem 2 | Solution]]
-* [[2000 USAMO Problems/Problem 2 | Solution]]
 === Problem 3 ===
-A game of solitaire is played with <math>R</math> red cards, <math>W</math> white cards, and <math>B</math> blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of <math>R, W,</math> and <math>B,</math> the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.
+Let <math> n \neq 0 </math>. For every sequence of integers
+<center>
+<math>
+A = a_0,a_1,a_2,\dots, a_n
+</math>
+</center>
+satisfying <math>0 \le a_i \le i</math>, for <math>i=0,\dots,n</math>, define another sequence
+<center>
+<math>
+t(A)= t(a_0), t(a_1), t(a_2), \dots, t(a_n)
+</math>
+</center>
+by setting <math> \displaystyle t(a_i) </math> to be the number of terms in the sequence <math> \displaystyle A </math> that precede the term <math> \displaystyle a_i </math> and are different from <math> \displaystyle a_i </math>. Show that, starting from any sequence <math> \displaystyle A </math> as above, fewer than <math> \displaystyle n </math> applications of the transformation <math> \displaystyle t </math> lead to a sequence <math> \displaystyle B </math> such that <math> \displaystyle t(B) = B </math>.
-* [[2000 USAMO Problems/Problem 3 | Solution]]
+* [[2003 USAMO Problems/Problem 3 | Solution]]
 == Day 2 ==
 === Problem 4 ===
+Let <math>ABC</math> be a triangle. A circle passing through <math>A</math> and <math>B</math> intersects segments <math>AC</math> and <math>BC</math> at <math>D</math> and <math>E</math>, respectively. Lines <math>AB</math> and <math>DE</math> intersect at <math>F</math>, while lines <math>BD</math> and <math>CF</math> intersect at <math>M</math>. Prove that <math>MF = MC</math> if and only if <math>MB\cdot MD = MC^2</math>.
+* [[2003 USAMO Problems/Problem 4 | Solution]]
-* [[2000 USAMO Problems/Problem 4 | Solution]]
 === Problem 5 ===
+Let <math>a</math>, <math>b</math>, <math>c</math> be positive real numbers. Prove that
+<center><math>\dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \dfrac{(2c + a + b)^2}{2c^2 + (a + b)^2} \le 8.</math></center>
+* [[2003 USAMO Problems/Problem 5 | Solution]]
-* [[2000 USAMO Problems/Problem 5 | Solution]]
 === Problem 6 ===
+At the vertices of a regular hexagon are written six nonnegative integers whose sum is 2003. Bert is allowed to make moves of the following form: he may pick a vertex and replace the number written there by the absolute value of the difference between the numbers written at the two neighboring vertices. Prove that Bert can make a sequence of moves, after which the number 0 appears at all six vertices.
+* [[2003 USAMO Problems/Problem 6 | Solution]]
-* [[2000 USAMO Problems/Problem 6 | Solution]]
 == Resources ==
 * [[USAMO Problems and Solutions]]
-* [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2000-ua/03usamo-test.shtml 2000 USAMO Problems and Solutions]
+* [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2003-ua/03usamo-test.shtml 2003 USAMO Problems and Solutions]
-* [http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27&year=2000 2000 USAMO Problems on the Resources page]
+* [http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27&year=2003 2003 USAMO Problems on the Resources page]

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