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Difference between revisions of "1969 Canadian MO Problems"

 
Line 1: Line 1:
 
== Problem 1 ==
 
== Problem 1 ==
  
−
Show that if <math>\displaystyle a_1/b_1=a_2/b_2=a_3/b_3</math> and <math>\displaystyle p_1,p_2,p_3</math> are not all zero, then <math>\displaystyle\left(\frac{a_1}{b_1} \right)^n=\frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}</math> for every positive integer <math>\displaystyle n.</math>
+
Show that if <math>a_1/b_1=a_2/b_2=a_3/b_3</math> and <math>p_1,p_2,p_3</math> are not all zero, then <math>\left(\frac{a_1}{b_1} \right)^n=\frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}</math> for every positive integer <math>n.</math>
  
 +
[[1969 Canadian MO Problems/Problem 1 | Solution]]
 
== Problem 2 ==
 
== Problem 2 ==
  
−
Determine which of the two numbers <math>\displaystyle \sqrt{c+1}-\sqrt{c}</math>, <math>\displaystyle\sqrt{c}-\sqrt{c-1}</math> is greater for any <math>\displaystyle c\ge 1</math>.
+
Determine which of the two numbers <math>\sqrt{c+1}-\sqrt{c}</math>, <math>\sqrt{c}-\sqrt{c-1}</math> is greater for any <math>c\ge 1</math>.
  
 +
[[1969 Canadian MO Problems/Problem 2 | Solution]]
 
== Problem 3 ==
 
== Problem 3 ==
  
−
Let <math>\displaystyle c</math> be the length of the [[hypotenuse]] of a [[right triangle]] whose two other sides have lengths <math>\displaystyle a</math> and <math>\displaystyle b</math>. Prove that <math>\displaystyle a+b\le c\sqrt{2}</math>. When does the equality hold?
+
Let <math>c</math> be the length of the [[hypotenuse]] of a [[right triangle]] whose two other sides have lengths <math>a</math> and <math>b</math>. Prove that <math>a+b\le c\sqrt{2}</math>. When does the equality hold?
  
 +
[[1969 Canadian MO Problems/Problem 3 | Solution]]
 
== Problem 4 ==
 
== Problem 4 ==
  
−
Let <math>\displaystyle ABC</math> be an equilateral triangle, and <math>\displaystyle P</math> be an arbitrary point within the triangle. Perpendiculars <math>\displaystyle PD,PE,PF</math> are drawn to the three sides of the triangle. Show that, no matter where <math>\displaystyle P</math> is chosen, <math>\displaystyle \frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}</math>.
+
Let <math>ABC</math> be an equilateral triangle, and <math>P</math> be an arbitrary point within the triangle. Perpendiculars <math>PD,PE,PF</math> are drawn to the three sides of the triangle. Show that, no matter where <math>P</math> is chosen, <math>\frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}</math>.
  
 +
[[1969 Canadian MO Problems/Problem 4 | Solution]]
 
== Problem 5 ==
 
== Problem 5 ==
  
−
Let <math>\displaystyle ABC</math> be a triangle with sides of length <math>\displaystyle a</math>, <math>\displaystyle b</math> and <math>\displaystyle c</math>. Let the bisector of the <math>\displaystyle \angle C</math> cut <math>\displaystyle AB</math> at <math>\displaystyle D</math>. Prove that the length of <math>\displaystyle CD</math> is <math>\displaystyle \frac{2ab\cos \frac{C}{2}}{a+b}.</math>
+
Let <math>ABC</math> be a triangle with sides of length <math>a</math>, <math>b</math> and <math>c</math>. Let the bisector of the <math>\angle C</math> cut <math>AB</math> at <math>D</math>. Prove that the length of <math>CD</math> is <math>\frac{2ab\cos \frac{C}{2}}{a+b}.</math>
  
 +
[[1969 Canadian MO Problems/Problem 5 | Solution]]
 
== Problem 6 ==
 
== Problem 6 ==
  
−
Find the sum of <math>\displaystyle 1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!</math>, where <math>\displaystyle  n!=n(n-1)(n-2)\cdots2\cdot1</math>.
+
Find the sum of <math>1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!</math>, where <math> n!=n(n-1)(n-2)\cdots2\cdot1</math>.
  
 +
[[1969 Canadian MO Problems/Problem 6 | Solution]]
 
== Problem 7 ==
 
== Problem 7 ==
  
−
Show that there are no integers <math>\displaystyle a,b,c</math> for which <math>\displaystyle a^2+b^2-8c=6</math>.
+
Show that there are no integers <math>a,b,c</math> for which <math>a^2+b^2-8c=6</math>.
−
 
 
  
 +
[[1969 Canadian MO Problems/Problem 7 | Solution]]
 
== Problem 8 ==
 
== Problem 8 ==
  
−
Let <math>\displaystyle f</math> be a function with the following properties:
+
Let <math>f</math> be a function with the following properties:
  
−
1) <math>\displaystyle f(n)</math> is defined for every positive integer <math>\displaystyle n</math>;
+
1) <math>f(n)</math> is defined for every positive integer <math>n</math>;
  
−
2) <math>\displaystyle f(n)</math> is an integer;
+
2) <math>f(n)</math> is an integer;
  
−
3) <math>\displaystyle f(2)=2</math>;
+
3) <math>f(2)=2</math>;
  
−
4) <math>\displaystyle f(mn)=f(m)f(n)</math> for all <math>\displaystyle m</math> and <math>\displaystyle n</math>;
+
4) <math>f(mn)=f(m)f(n)</math> for all <math>m</math> and <math>n</math>;
  
−
5) <math>\displaystyle f(m)>f(n)</math> whenever <math>m>n</math>.
+
5) <math>f(m)>f(n)</math> whenever <math>m>n</math>.
  
−
Prove that <math>\displaystyle f(n)=n</math>.
+
Prove that <math>f(n)=n</math>.
  
 +
[[[1969 Canadian MO Problems/Problem 8 | Solution]]
 
== Problem 9 ==
 
== Problem 9 ==
  
−
Show that for any quadrilateral inscribed in a circle of radius <math>\displaystyle 1,</math> the length of the shortest side is less than or equal to <math>\displaystyle \sqrt{2}</math>.
+
Show that for any quadrilateral inscribed in a circle of radius <math>1,</math> the length of the shortest side is less than or equal to <math>\sqrt{2}</math>.
  
 +
[[1969 Canadian MO Problems/Problem 9 | Solution]]
 
== Problem 10 ==
 
== Problem 10 ==
  
−
Let <math>\displaystyle ABC</math> be the right-angled isosceles triangle whose equal sides have length 1. <math>\displaystyle P</math> is a point on the [[hypotenuse]], and the feet of the [[perpendicular]]s from <math>\displaystyle P</math> to the other sides are <math>\displaystyle Q</math> and <math>\displaystyle R</math>. Consider the areas of the triangles <math>\displaystyle APQ</math> and <math>\displaystyle PBR</math>, and the area of the [[rectangle]] <math>\displaystyle QCRP</math>. Prove that regardless of how <math>\displaystyle P</math> is chosen, the largest of these three areas is at least <math>\displaystyle 2/9</math>.
+
Let <math>ABC</math> be the right-angled isosceles triangle whose equal sides have length 1. <math>P</math> is a point on the [[hypotenuse]], and the feet of the [[perpendicular]]s from <math>P</math> to the other sides are <math>Q</math> and <math>R</math>. Consider the areas of the triangles <math>APQ</math> and <math>PBR</math>, and the area of the [[rectangle]] <math>QCRP</math>. Prove that regardless of how <math>P</math> is chosen, the largest of these three areas is at least <math>2/9</math>.
  
 +
[[1969 Canadian MO Problems/Problem 10 | Solution]]
 
== Resources ==
 
== Resources ==
  

Revision as of 12:46, 8 October 2007

Problem 1

Show that if $a_1/b_1=a_2/b_2=a_3/b_3$ and $p_1,p_2,p_3$ are not all zero, then $\left(\frac{a_1}{b_1} \right)^n=\frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}$ for every positive integer $n.$

Solution

Problem 2

Determine which of the two numbers $\sqrt{c+1}-\sqrt{c}$, $\sqrt{c}-\sqrt{c-1}$ is greater for any $c\ge 1$.

Solution

Problem 3

Let $c$ be the length of the hypotenuse of a right triangle whose two other sides have lengths $a$ and $b$. Prove that $a+b\le c\sqrt{2}$. When does the equality hold?

Solution

Problem 4

Let $ABC$ be an equilateral triangle, and $P$ be an arbitrary point within the triangle. Perpendiculars $PD,PE,PF$ are drawn to the three sides of the triangle. Show that, no matter where $P$ is chosen, $\frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}$.

Solution

Problem 5

Let $ABC$ be a triangle with sides of length $a$, $b$ and $c$. Let the bisector of the $\angle C$ cut $AB$ at $D$. Prove that the length of $CD$ is $\frac{2ab\cos \frac{C}{2}}{a+b}.$

Solution

Problem 6

Find the sum of $1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!$, where $n!=n(n-1)(n-2)\cdots2\cdot1$.

Solution

Problem 7

Show that there are no integers $a,b,c$ for which $a^2+b^2-8c=6$.

Solution

Problem 8

Let $f$ be a function with the following properties:

1) $f(n)$ is defined for every positive integer $n$;

2) $f(n)$ is an integer;

3) $f(2)=2$;

4) $f(mn)=f(m)f(n)$ for all $m$ and $n$;

5) $f(m)>f(n)$ whenever $m>n$.

Prove that $f(n)=n$.

[[[1969 Canadian MO Problems/Problem 8 | Solution]]

Problem 9

Show that for any quadrilateral inscribed in a circle of radius $1,$ the length of the shortest side is less than or equal to $\sqrt{2}$.

Solution

Problem 10

Let $ABC$ be the right-angled isosceles triangle whose equal sides have length 1. $P$ is a point on the hypotenuse, and the feet of the perpendiculars from $P$ to the other sides are $Q$ and $R$. Consider the areas of the triangles $APQ$ and $PBR$, and the area of the rectangle $QCRP$. Prove that regardless of how $P$ is chosen, the largest of these three areas is at least $2/9$.

Solution

Resources

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