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

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Solution 4
 
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Let X be the length of said median. Then by the median length formula, 2x^2=a^2+b^2-2(c/2)^2. This simplifies to c/2. Then we know c/2= √ab. Then ab=(c^2)/4. Also a^2+b^2=c^2. Solving gets us b=c/(√2) and a=c(√2)/4. C is fixed.
 
  
 
== See Also ==
 
== See Also ==

Revision as of 17:01, 18 July 2017

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 $\frac{c}{4}$ .

Solution 3

If we let the legs be $a$ and $b$ with $a < b$, then $c^2 = a^2 + b^2$. Because $4c^2 = a^2 b^2$ as well, we immediately deduce via some short computations that $a = (2 - \sqrt{3})b$. Thus, $\frac{a}{b} = \tan 15^\circ$, and so one of the angles of the triangle must be $15^\circ$. But a $15^\circ$ angle is easily constructed by bisecting a $30^\circ$ angle (which is formed by constructing the altitude of an equilateral triangle), and from there it is not difficult to construct the desired right triangle.

{{alternate solutions}

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
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