1959 IMO Problems/Problem 4
Construct a right triangle with a given hypotenuse such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.
We denote the catheti of the triangle as and . 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.)
The conditions of the problem require that
However, we notice that twice the area of the triangle is , since and form a right angle. However, twice the area of the triangle is also the product of and the altitude to . Hence the altitude to must have length . Therefore if we construct a circle with diameter and a line parallel to and of distance from , either point of intersection between the line and the circle will provide a suitable third vertex for the triangle. Q.E.D.
We denote the angle between and as . The problem requires that
or, equivalently, that
However, since , we can rewrite the condition as
or, equivalently, as
From this it becomes apparent that or ; hence the other two angles in the triangle must be and , 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 , which both of the solutions set equal to .
If we let the legs be and with , then . Because as well, we immediately deduce via some short computations through quadratic formula that and . Thus, and , and so one of the angles of the triangle must be and the other must be . A angle is easily constructed by bisecting a angle (which is formed by constructing the altitude of an equilateral triangle), and angle is constructed by constructing a 15 degree angle on top of the 60 degree point.
WLOG, let the length of the hypotenuse be 2. The radius of the circle with the hypotenuse as the diameter will be 1. Let the circle has an center of (1,0) on a Cartesian plane. Let be the horizontal length from the origin to the desired point on the circumference which satisfies the question. Since the equation of the circle is , . The desired point which have a horizontal length of from the origin would have a vertical length of when substituted into the previous equation. By using Pythagoras Theorem, the length from the origin to this point is and the length from this point to the point (2,0) is . So, we need to find the which . By solving this quadratic equation, we get , the length of the two sides would be and . With this, we get a conclusion of this triangle should have a length ratio of .
We draw perpendicular on and denote the angle between the median and as . The area of the triangle is or
We can write as So we have,
Which simplifies to be
So, we draw and from it's midpoint, we draw a segment forming an angle Join the vertices to get the required triangle
This is a geometric solution. We start with some observations in the first paragraph and describe the construction in the next paragraph. Consider any right triangle with hypotenuse c. Construct square with side length c on the opposite side of c from the triangle. Construct 3 more triangles congruent to the original triangle on the outside of each of the three other sides of the square in such a way that a larger square with side length a+b is formed. The area of this larger square is c^2+2ab = 1.5c^2 so a+b = (3/2)^(1/2) *c
Now go back to the given c. Construct a square with side length c first. Let two adjacent corners of it be M and N. Construct the square's diagonals and let center of the square be O. Construct a circle around O with radius (3/2)^(1/2) * OM. To do so, take OM, first draw a 2:1 right triangle with OM being the short side to obtain a hypothenuse with length 3^(1/2) OM, then draw a 1:1 right triangle from that side to obtain a hypothenuse with length 6^(1/2)*OM. then bisect it for desired length. Next draw the circular locus of points X such that MXO is 45 degrees. To accomplish this, simply find the point on the perpendicular bisector of OM that is 1/(2OM) from OM. There are two such points, pick the one closer to N. Draw circle around this new point going through O and M. The intersection of the two circles is the desired third vertex of the triangle with the given hypotenuse c.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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