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Difference between revisions of "2005 USAMO Problems"

 
m (cosmetics)
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=== Problem 1 ===
 
=== Problem 1 ===
  
−
Determine all composite positive integers <math> \displaystyle n</math> for which it is possible to arrange all divisors of <math>n</math> that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.
+
Determine all composite positive integers <math>n</math> for which it is possible to arrange all divisors of <math>n</math> that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.
  
 
[[2005 USAMO Problems/Problem 1 | Solution]]
 
[[2005 USAMO Problems/Problem 1 | Solution]]
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Prove that the
 
Prove that the
 
system
 
system
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<center>
+
<cmath>
−
<math>
+
\begin{align*}
−
\begin{matrix} \qquad x^6+x^3+x^3y+y & = 147^{157} \\[.1in]
+
x^6+x^3+x^3y+y & = 147^{157} \\
 
x^3+x^3y+y^2+y+z^9 & = 157^{147}
 
x^3+x^3y+y^2+y+z^9 & = 157^{147}
−
\end{matrix}
+
\end{align*}
−
</math>
+
</cmath>
−
</center>
+
has no solutions in integers <math>x</math>, <math>y</math>, and <math>z</math>.
−
has no solutions in integers <math> \displaystyle x </math>, <math> \displaystyle y </math>, and <math> \displaystyle z </math>.
+
 
 +
* [[2005 USAMO Problems/Problem 2 | Solution]]
  
 
=== Problem 3 ===
 
=== Problem 3 ===
  
−
Let <math> \displaystyle ABC </math> be an acute-angled triangle, and let <math> \displaystyle P </math> and <math> \displaystyle Q </math> be two points on side <math> \displaystyle BC </math>. Construct point <math> \displaystyle C_1 </math> in such a way that convex quadrilateral <math> \displaystyle APBC_1 </math> is cyclic, <math> \displaystyle QC_1 \mid\mid CA </math>, and <math> \displaystyle C_1 </math> and <math> \displaystyle Q </math> lie on opposite sides of line <math> \displaystyle AB </math>. Construct point <math> \displaystyle B_1 </math> in such a way that convex quadrilateral <math> \displaystyle APCB_1 </math> is cyclic, <math> \displaystyle QB_1 \mid\mid BA </math>, and <math> \displaystyle B_1 </math> and <math> \displaystyle Q </math>  lie on opposite sides of line <math> \displaystyle AC </math>.  Prove that points <math> \displaystyle B_1, C_1,P </math>, and <math> \displaystyle Q </math> lie on a circle.
+
Let <math>ABC</math> be an acute-angled triangle, and let <math>P</math> and <math>Q</math> be two points on side <math>BC</math>. Construct point <math>C_1 </math> in such a way that convex quadrilateral <math>APBC_1</math> is cyclic, <math>QC_1 \parallel CA</math>, and <math>C_1</math> and <math>Q</math> lie on opposite sides of line <math>AB</math>. Construct point <math>B_1</math> in such a way that convex quadrilateral <math>APCB_1</math> is cyclic, <math>QB_1 \parallel BA </math>, and <math>B_1 </math> and <math>Q </math>  lie on opposite sides of line <math>AC</math>.  Prove that points <math>B_1, C_1,P</math>, and <math>Q</math> lie on a circle.
 +
 
 +
* [[2005 USAMO Problems/Problem 3 | Solution]]
  
 
== Day 2 ==
 
== Day 2 ==

Revision as of 21:53, 11 April 2008

Problems from the 2005 USAMO.

Day 1

Problem 1

Determine all composite positive integers $n$ for which it is possible to arrange all divisors of $n$ that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.

Solution

Problem 2

Prove that the system \begin{align*} x^6+x^3+x^3y+y & = 147^{157} \\ x^3+x^3y+y^2+y+z^9 & = 157^{147} \end{align*} has no solutions in integers $x$, $y$, and $z$.

Problem 3

Let $ABC$ be an acute-angled triangle, and let $P$ and $Q$ be two points on side $BC$. Construct point $C_1$ in such a way that convex quadrilateral $APBC_1$ is cyclic, $QC_1 \parallel CA$, and $C_1$ and $Q$ lie on opposite sides of line $AB$. Construct point $B_1$ in such a way that convex quadrilateral $APCB_1$ is cyclic, $QB_1 \parallel BA$, and $B_1$ and $Q$ lie on opposite sides of line $AC$. Prove that points $B_1, C_1,P$, and $Q$ lie on a circle.

Day 2

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

Resources

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