Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(IMO box)
m
Line 51: Line 51:
  
 
{{alternate solutions}}
 
{{alternate solutions}}
 +
 +
== See Also ==
 +
  
 
{{IMO box|year=1959|num-b=3|num-a=5}}
 
{{IMO box|year=1959|num-b=3|num-a=5}}
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Revision as of 10:00, 30 May 2012

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.

See Also

1959 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1959_IMO_Problems/Problem_4&oldid=47283"