Difference between revisions of "1973 Canadian MO Problems"

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

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Difference between revisions of "1973 Canadian MO Problems"

Latest revision as of 17:23, 8 October 2014

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Resources

@@ Line 36: / Line 36: @@
 pair C=dir(120-(40)*(k+2));
 D(A--B);
-MP("P_{"+string(k)+"}",A,11,A);
 A=B;B=C;
 }
 for(int k=0;k<3;++k)
 {
@@ Line 47: / Line 45: @@
 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>