1960 IMO Problems: Difference between revisions

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Problems of the 2nd IMO 1960 Romania.

Day I

Problem 1

Solution

Problem 2

Solution

Problem 3

In a given right triangle $ABC$ , the hypotenuse $BC$ , of length $a$ , is divided into $n$ equal parts ( $n$ and odd integer). Let $\alpha$ be the acute angle subtending, from $A$ , that segment which contains the midpoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse of the triangle. Prove that:

$\displaystyle\tan{\alpha}=\frac{4nh}{(n^2-1)a}.$

Solution

Day II

@@ Line 15: / Line 15: @@
 === Problem 3 ===
+In a given right triangle <math>ABC</math>, the hypotenuse <math>BC</math>, of length <math>a</math>, is divided into <math>n</math> equal parts (<math>n</math> and odd integer). Let <math>\alpha</math> be the acute angle subtending, from <math>A</math>, that segment which contains the midpoint of the hypotenuse. Let <math>h</math> be the length of the altitude to the hypotenuse of the triangle. Prove that:
+<center><math>
+\displaystyle\tan{\alpha}=\frac{4nh}{(n^2-1)a}.
+</math>
+</center>
-In a given right triangle <math>ABC</math>, the hypotenuse <math>BC</math>, of length <math>a</math>, is divided into <math>n</math> equal parts (<math>n</math> and odd integer). Let <math>\alpha</math> be the acute angel subtending, from <math>A</math>, that segment which contains the mdipoint of the hypotenuse. Let <math>h</math> be the length of the altitude to the hypotenuse fo the triangle. Prove that:
-\[
-\tan{\alpha}=\dfrac{4nh}{(n^2-1)a}.
-\]

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1960 IMO Problems: Difference between revisions

Revision as of 00:37, 12 March 2007

Contents

Day I

Problem 1

Problem 2

Problem 3

Day II

Problem 4

Problem 5

Problem 6

Resources