Problem 4 from the 1959 IMO: Construction of a Right Triangle

Here's a problem from an old Russian problem collection:

Construct a right-angled triangle whose hypotenuse \(c\) is given if it is known that the median from the right angle equals the geometric mean of the remaining two sides of the triangles.

problem #4 from the 1959 IMO - the very first

Solution

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Copyright © 1996-2017 Alexander Bogomolny

The applet below illustrates the problem and a solution:

15 January 2015, Created with GeoGebra

The solution is based on the observation that the triangles \(ABC\) and \(CBD\) are similar, implying the proportion \(\displaystyle\frac{b}{c}=\frac{h}{a}\). The solution further makes use of the fact that in a right triangle the midpoint of the hypotenuse serves as the circumcenter so that the median from the right angle equals half of the hypotenuse. The construction is then obvious: draw a line parallel to the hypotenuse at distance \(c/4\) from it. The two points where that line intersects the semicircle with \(AB\) as diameter are sought locations of vertex \(C\).

The original solution [The IMO Compendium, p357-358] went in a somewhat different way:

By the Pythagorean theorem \(c^{2}=a^{2}+b^{2}\) which, together with \(c/2=\sqrt{ab}\), implies \(a+b=c\sqrt{3/2}\). Given a desired \(\Delta ABC\), let \(D\) be a point on \(AC\) such that \(CD=CB\). In that case, \(AD=a+b=c\sqrt{3/2}\), and also \(\angle ADB=45^{\circ}\), because \(BC=CD\).

problem #4 from the 1959 IMO - original solution

With this analysis in mind, measure \(AD=c\sqrt{3/2}\) and draw a ray \(DX\) such that \(\angle ADX=45^{\circ}\). Next, construct circle \(C(A,c)\) with center \(A\) and radius \(c\). Take \(B\) to be the point of intersection of the circle and \(DX\) and drop a perpendicular \(BC\) to \(AD\).

References

  1. D. Djukic et al, The IMO Compendium, Springer, 2011 (Second edition)

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Copyright © 1996-2017 Alexander Bogomolny

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