1960 IMO Problems/Problem 7

Problem

An isosceles trapezoid with bases and and altitude is given.

a) On the axis of symmetry of this trapezoid, find all points such that both legs of the trapezoid subtend right angles at ;

b) Calculate the distance of from either base;

c) Determine under what conditions such points actually exist. Discuss various cases that might arise.

Solution

(a) The intersection of the circle with diameter one of the legs with the axis of symmetry.

(b)Let be the distance from to one of the bases; then must be the distance from to the other base. Similar triangles give x^2 - hx + \frac{ac}{4} = 0x = \frac{h \pm \sqrt{h^2 - ac}}{2}.h^2 \ge ac$.

See Also

1960 IMO (Problems)
Preceded by Problem 6	1 • 2 • 3 • 4 • 5 • 6 • 7	Followed by Last Question

[1]

In our above picture, ABCD is our trapezoid with AB=a and CD=c and its height is 'h'. AF and BE are perpendicular to CD such that AF= BE= h. XY is our axis of symmetry and it intersects with CD at a point O. Point O is our origin of reference whose coordinates are (0,0).

Let our point P be on the axis of symmetry at z distance from the origin O.

The coordinates of the points A,B,C,D,E,F and P are given in the figure.

Now,

Slope of the line PC= (z-0)/(0-c/2) = -2z/c Slope of the line PB= (z-h)/(0-a/2) = -2(z-h)/a

Since the leg BC subtends a right angle at P, the angle BPC should be a right angle. This means that the product of the slope of PC and PB is -1.

i.e

4z(z-h)=-ac

or z^2 - zh + ac/4= O

Now, solving for z, we get, z= [(h + ( h^2 - ac ) ^1/2 ]/2 and [(h - ( h^2 - ac ) ^1/2 ]/2

So, z is the distance of the points from the base CD..

Also the points are possible only when , h^2 - ac >= 0.. and doesn't exist for h^2 -ac <0