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Difference between revisions of "1959 IMO Problems/Problem 4"

 
(IMO box)
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== Problem ==
 
== Problem ==
  
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( ''Proposed by Hungary'' ) Construct a right triangle with a given hypotenuse <math>\displaystyle c</math> such that the median drawn to the hypotenuse is the [[geometric mean]] of the two legs of the triangle.
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Construct a right triangle with a given hypotenuse <math>c</math> such that the median drawn to the hypotenuse is the [[geometric mean]] of the two legs of the triangle.
  
 
== Solutions ==
 
== Solutions ==
  
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We denote the [[cathetus | catheti]] of the triangle as <math>\displaystyle a</math> and <math>\displaystyle b</math>.  We also observe the well-known fact that in a right triangle, the median to the hypotenuse is of half the length of the hypotenuse.  (This is true because if we inscribe the triangle in a circle, the hypotenuse is the diameter, so a segment from any point on the circle to the midpoint of the hypotenuse is a radius.)
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We denote the [[cathetus | catheti]] of the triangle as <math>a</math> and <math>b</math>.  We also observe the well-known fact that in a right triangle, the median to the hypotenuse is of half the length of the hypotenuse.  (This is true because if we inscribe the triangle in a circle, the hypotenuse is the diameter, so a segment from any point on the circle to the midpoint of the hypotenuse is a radius.)
  
 
=== Solution 1 ===
 
=== Solution 1 ===
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</center>
 
</center>
  
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However, we notice that twice the area of the triangle <math>\displaystyle abc </math> is <math>\displaystyle ab</math>, since <math>\displaystyle a</math> and <math>\displaystyle b</math> form a right angle.  However, twice the area of the triangle is also the product of <math>\displaystyle c</math> and the altitude to <math>\displaystyle c</math>.  Hence the altitude to <math>\displaystyle c</math> must have length <math>\frac{c}{4}</math>.  Therefore if we construct a circle with diameter <math>\displaystyle c</math> and a line parallel to <math>\displaystyle c</math> and of distance <math>\frac{c}{4}</math> from <math>\displaystyle c</math>, either point of intersection between the line and the circle will provide a suitable third vertex for the triangle.  Q.E.D.
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However, we notice that twice the area of the triangle <math>abc </math> is <math>ab</math>, since <math>a</math> and <math>b</math> form a right angle.  However, twice the area of the triangle is also the product of <math>c</math> and the altitude to <math>c</math>.  Hence the altitude to <math>c</math> must have length <math>\frac{c}{4}</math>.  Therefore if we construct a circle with diameter <math>c</math> and a line parallel to <math>c</math> and of distance <math>\frac{c}{4}</math> from <math>c</math>, either point of intersection between the line and the circle will provide a suitable third vertex for the triangle.  Q.E.D.
  
 
=== Solution 2 ===
 
=== Solution 2 ===
  
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We denote the angle between <math>\displaystyle b</math> and <math>\displaystyle c</math> as <math>\displaystyle \alpha</math>.  The problem requires that
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We denote the angle between <math>b</math> and <math>c</math> as <math>\alpha</math>.  The problem requires that
  
 
<center>
 
<center>
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''Note''.  It is not difficult to reconcile these two constructions.  Indeed, we notice that the altitude of the triangle is of length <math>\displaystyle c \sin{\alpha}\cos{\alpha}</math>, which both of the solutions set equal to <math> \frac{c}{4} </math> .
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''Note''.  It is not difficult to reconcile these two constructions.  Indeed, we notice that the altitude of the triangle is of length <math>c \sin{\alpha}\cos{\alpha}</math>, which both of the solutions set equal to <math> \frac{c}{4} </math> .
  
  
 
{{alternate solutions}}
 
{{alternate solutions}}
  
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== Resources ==
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{{IMO box|year=1959|num-b=3|num-a=5}}
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* [[1959 IMO Problems]]
 
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=341522#p341522 Dicussion on AoPS/MathLinks]
 
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[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Revision as of 20:23, 25 October 2007

Problem

Construct a right triangle with a given hypotenuse $c$ such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.

Solutions

We denote the catheti of the triangle as $a$ and $b$. We also observe the well-known fact that in a right triangle, the median to the hypotenuse is of half the length of the hypotenuse. (This is true because if we inscribe the triangle in a circle, the hypotenuse is the diameter, so a segment from any point on the circle to the midpoint of the hypotenuse is a radius.)

Solution 1

The conditions of the problem require that

$ab = \frac{c^2}{4}.$

However, we notice that twice the area of the triangle $abc$ is $ab$, since $a$ and $b$ form a right angle. However, twice the area of the triangle is also the product of $c$ and the altitude to $c$. Hence the altitude to $c$ must have length $\frac{c}{4}$. Therefore if we construct a circle with diameter $c$ and a line parallel to $c$ and of distance $\frac{c}{4}$ from $c$, either point of intersection between the line and the circle will provide a suitable third vertex for the triangle. Q.E.D.

Solution 2

We denote the angle between $b$ and $c$ as $\alpha$. The problem requires that

$ab = \frac{c^2}{4},$

or, equivalently, that

$2 \frac{ab}{c^2} = \frac{1}{2}.$

However, since $\frac{a}{c} = \sin{\alpha};\; \frac{b}{c} = \cos{\alpha}$, we can rewrite the condition as

$2\sin{\alpha}\cos{\alpha} = \frac{1}{2},$

or, equivalently, as

$\sin{2\alpha} = \frac{1}{2}.$

From this it becomes apparent that $2\alpha = \frac{\pi}{6}$ or $\frac{5\pi}{6}$; hence the other two angles in the triangle must be $\frac{ \pi }{12}$ and $\frac{ 5 \pi }{12}$, which are not difficult to construct. Q.E.D.

Note. It is not difficult to reconcile these two constructions. Indeed, we notice that the altitude of the triangle is of length $c \sin{\alpha}\cos{\alpha}$, which both of the solutions set equal to $\frac{c}{4}$ .


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1959 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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