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Difference between revisions of "1973 Canadian MO Problems"
m (→‎Problem 4)
 
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==Problem 2==
 
==Problem 2==
  
−
Find all real numbers that satisfy the equation <math>|x+3|-|x-1|=x+1</math>. (Note: <math>|a| = a</math> if <math>a\ge 0; |a|=-a if a<0</math>.)  
+
Find all real numbers that satisfy the equation <math>|x+3|-|x-1|=x+1</math>. (Note: <math>|a| = a</math> if <math>a\ge 0; |a|=-a</math> if <math>a<0</math>.)  
  
 
[[1973 Canadian MO Problems/Problem 2 | Solution]]
 
[[1973 Canadian MO Problems/Problem 2 | Solution]]
Line 29: Line 29:
 
==Problem 4==
 
==Problem 4==
  
−
The figure shows a (convex) polygon with nine vertices. The six diagonals which have been drawn dissect the polygon into the seven triangles: <math>P_{0}P_{1}P_{3}, P_{0}P_{3}P_{6}, P_{0}P_{6}P_{7}, P_{0}P_{7}P_{8}, P_{1}P_{2}P_{3}, P_{3}P_{4}P_{6}, P_{4}P_{5}P_{6}</math>. In how many ways can these triangles be labeled with the names <math>\triangle_{1}, \triangle_{2}, \triangle_{3}, \triangle_{4}, \triangle_{5}, \triangle_{6}, \triangle_{7}</math> so that <math>P_{i}</math> is a vertex of triangle <math>\triangle_{i}</math> for <math>i = 1, 2, 3, 4, 5, 6, 7</math>? Justify your answer.  
+
<asy>
 +
size(200);
 +
pair A=dir(120), B=dir(80);
 +
for(int k=0;k<9;++k)
 +
{
 +
pair C=dir(120-(40)*(k+2));
 +
D(A--B);
 +
A=B;B=C;
 +
}
 +
for(int k=0;k<3;++k)
 +
{
 +
pair A1=dir(120-(40)*(3*k));
 +
pair B1=dir(120-(40)*(3*k+2));
 +
pair C1=dir(120-(40)*(3*k+3));
 +
D(A1--B1);
 +
D(A1--C1);
 +
}
 +
for(int k=0;k<9;++k)
 +
{
 +
pair A=dir(120+(40)*(k));
 +
MP("P_{"+string(k)+"}",A,11,A);
 +
}
 +
</asy>
 +
 
 +
The figure shows a (convex) polygon with nine vertices. The six diagonals which have been drawn dissect the polygon into the seven triangles: <math>P_{0}P_{1}P_{3},~ P_{0}P_{3}P_{6},~  P_{0}P_{6}P_{7},~ P_{0}P_{7}P_{8},~  P_{1}P_{2}P_{3}, ~ P_{3}P_{4}P_{6},~ P_{4}P_{5}P_{6}</math>. In how many ways can these triangles be labeled with the names <math>\triangle_{1}, ~ \triangle_{2}, ~ \triangle_{3}, ~ \triangle_{4}, ~ \triangle_{5},~  \triangle_{6},~  \triangle_{7}</math> so that <math>P_{i}</math> is a vertex of triangle <math>\triangle_{i}</math> for <math>i = 1, 2, 3, 4, 5, 6, 7</math>? Justify your answer.
  
 
[[1973 Canadian MO Problems/Problem 4 | Solution]]
 
[[1973 Canadian MO Problems/Problem 4 | Solution]]
Line 35: Line 59:
 
==Problem 5==
 
==Problem 5==
  
−
For every positive integer <math>n</math>, let <math>h(n) = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}</math>. For example, <math>h(1) = 1, h(2) = 1+\frac{1}{2}, h(3) = 1+\frac{1}{2}+\frac{1}{3}</math>. Prove that for <math>n=2,3,4,\ldots n+h(1)+h(2)+h(3)+\cdots+h(n-1) = nh(n)</math>.
+
For every positive integer <math>n</math>, let <math>h(n) = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}</math>.  
 +
 
 +
For example, <math>h(1) = 1, h(2) = 1+\frac{1}{2}, h(3) = 1+\frac{1}{2}+\frac{1}{3}</math>.  
 +
 
 +
Prove that <math>n+h(1)+h(2)+h(3)+\cdots+h(n-1) = nh(n)\qquad</math>  for  <math>n=2,3,4,\ldots</math>
  
 
[[1973 Canadian MO Problems/Problem 5 | Solution]]
 
[[1973 Canadian MO Problems/Problem 5 | Solution]]

Latest revision as of 17:23, 8 October 2014

Problem 1

$\text{(i)}$ Solve the simultaneous inequalities, $x<\frac{1}{4x}$ and $x<0$; i.e. find a single inequality equivalent to the two simultaneous inequalities.

$\text{(ii)}$ What is the greatest integer that satisfies both inequalities $4x+13 < 0$ and $x^{2}+3x > 16$.

$\text{(iii)}$ Give a rational number between $11/24$ and $6/13$.

$\text{(iv)}$ Express $100000$ as a product of two integers neither of which is an integral multiple of $10$.

$\text{(v)}$ Without the use of logarithm tables evaluate $\frac{1}{\log_{2}36}+\frac{1}{\log_{3}36}$.


Solution

Problem 2

Find all real numbers that satisfy the equation $|x+3|-|x-1|=x+1$. (Note: $|a| = a$ if $a\ge 0; |a|=-a$ if $a<0$.)

Solution

Problem 3

Prove that if $p$ and $p+2$ are prime integers greater than $3$, then $6$ is a factor of $p+1$.

Solution

Problem 4

[asy] size(200); pair A=dir(120), B=dir(80); for(int k=0;k<9;++k) { pair C=dir(120-(40)*(k+2)); D(A--B); A=B;B=C; } for(int k=0;k<3;++k) { pair A1=dir(120-(40)*(3*k)); pair B1=dir(120-(40)*(3*k+2)); pair C1=dir(120-(40)*(3*k+3)); D(A1--B1); D(A1--C1); } for(int k=0;k<9;++k) { pair A=dir(120+(40)*(k)); MP("P_{"+string(k)+"}",A,11,A); } [/asy]

The figure shows a (convex) polygon with nine vertices. The six diagonals which have been drawn dissect the polygon into the seven triangles: $P_{0}P_{1}P_{3},~ P_{0}P_{3}P_{6},~ P_{0}P_{6}P_{7},~ P_{0}P_{7}P_{8},~ P_{1}P_{2}P_{3}, ~ P_{3}P_{4}P_{6},~ P_{4}P_{5}P_{6}$. In how many ways can these triangles be labeled with the names $\triangle_{1}, ~ \triangle_{2}, ~ \triangle_{3}, ~ \triangle_{4}, ~ \triangle_{5},~ \triangle_{6},~ \triangle_{7}$ so that $P_{i}$ is a vertex of triangle $\triangle_{i}$ for $i = 1, 2, 3, 4, 5, 6, 7$? Justify your answer.

Solution

Problem 5

For every positive integer $n$, let $h(n) = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$.

For example, $h(1) = 1, h(2) = 1+\frac{1}{2}, h(3) = 1+\frac{1}{2}+\frac{1}{3}$.

Prove that $n+h(1)+h(2)+h(3)+\cdots+h(n-1) = nh(n)\qquad$ for $n=2,3,4,\ldots$

Solution

Problem 6

If $A$ and $B$ are fixed points on a given circle not collinear with center $O$ of the circle, and if $XY$ is a variable diameter, find the locus of $P$ (the intersection of the line through $A$ and $X$ and the line through $B$ and $Y$).

Solution

Problem 7

Observe that $\frac{1}{1}= \frac{1}{2}+\frac{1}{2};\quad \frac{1}{2}=\frac{1}{3}+\frac{1}{6};\quad \frac{1}{3}=\frac{1}{4}+\frac{1}{12};\qu...$ (Error compiling LaTeX. Unknown error_msg) State a general law suggested by these examples, and prove it.

Prove that for any integer $n$ greater than $1$ there exist positive integers $i$ and $j$ such that $\frac{1}{n}= \frac{1}{i(i+1)}+\frac{1}{(i+1)(i+2)}+\frac{1}{(i+2)(i+3)}+\cdots+\frac{1}{j(j+1)}.$

Solution

Resources

1973 Canadian MO

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