Difference between revisions of "1981 IMO Problems"

Revision as of 21:57, 28 October 2006

Problems of the 22nd IMO 1981 U.S.A.

Day I

Problem 1

$\displaystyle P$ is a point inside a given triangle $\displaystyle ABC$ . $\displaystyle D, E, F$ are the feet of the perpendiculars from $\displaystyle P$ to the lines $\displaystyle BC, CA, AB$ , respectively. Find all $\displaystyle P$ for which

$\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$

is least.

Solution

Problem 2

Let $\displaystyle 1 \le r \le n$ and consider all subsets of $\displaystyle r$ elements of the set $\{ 1, 2, \ldots , n \}$ . Each of these subsets has a smallest member. Let $\displaystyle F(n,r)$ denote the arithmetic mean of these smallest numbers; prove that

$F(n,r) = \frac{n+1}{r+1}.$

Solution

Problem 3

Determine the maximum value of $\displaystyle m^2 + n^2$ , where $\displaystyle m$ and $\displaystyle n$ are integers satisfying $m, n \in \{ 1,2, \ldots , 1981 \}$ and $\displaystyle ( n^2 - mn - m^2 )^2 = 1$ .

Solution

Day II

@@ Line 31: / Line 31: @@
 === Problem 3 ===
-Determine the maximum value of <math> \displaystyle m^3 + n^3 </math>, where <math> \displaystyle m </math> and <math> \displaystyle n </math> are integers satisfying <math> m, n \in \{ 1,2, \ldots , 1981 \} </math> and <math> \displaystyle ( n^2 - mn - m^2 )^2 = 1 </math>.
+Determine the maximum value of <math> \displaystyle m^2 + n^2 </math>, where <math> \displaystyle m </math> and <math> \displaystyle n </math> are integers satisfying <math> m, n \in \{ 1,2, \ldots , 1981 \} </math> and <math> \displaystyle ( n^2 - mn - m^2 )^2 = 1 </math>.
 [[1981 IMO Problems/Problem 3 | Solution]]

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Difference between revisions of "1981 IMO Problems"

Revision as of 21:57, 28 October 2006

Contents

Day I

Problem 1

Problem 2

Problem 3

Day II

Problem 4

Problem 5

Problem 6

Resources