IMO 1960 · Problem 7

An isosceles trapezoid with bases \(a\) and \(c\) and altitude \(h\) is given.

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

(b) Calculate the distance of \(P\) from either base.

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

Official text of the problem. Official problem paper (PDF)

Answer

Let \(M,N\) be the midpoints of the bases of lengths \(a,c\), respectively. The required points are precisely the intersections of the symmetry axis with the circle having either leg as diameter.

Their number is \[ \begin{cases} 2,&h^2>ac,\\ 1,&h^2=ac,\\ 0,&h^2<ac. \end{cases} \]

When they exist, they lie strictly between \(M\) and \(N\). Writing \[ \Delta=\sqrt{h^2-ac}, \] their distances from the bases of lengths \(a\) and \(c\), respectively, are \[ \boxed{\left(\frac{h+\Delta}{2},\frac{h-\Delta}{2}\right) \quad\text{and}\quad \left(\frac{h-\Delta}{2},\frac{h+\Delta}{2}\right).} \] When \(h^2=ac\), these describe the same point, the midpoint of \(MN\).

Solution

The given trapezoid is nondegenerate, so \(a,c,h>0\). Label its vertices consecutively \(A,B,C,D\), with \(AB=a\) and \(CD=c\).

First, a geometric construction follows from Thales’ theorem: draw the circle with diameter \(AD\) and intersect it with the symmetry axis. Every intersection \(P\) satisfies \(\angle APD=90^\circ\). Reflection in the symmetry axis fixes \(P\) and sends \(A,D\) to \(B,C\), respectively, so it also gives \(\angle BPC=90^\circ\). Conversely, every required point belongs to that circle. Thus this construction finds exactly all the required points.

To calculate their positions and determine when they exist, choose coordinates \[ A=\left(-\frac a2,0\right),\qquad B=\left(\frac a2,0\right),\qquad C=\left(\frac c2,h\right),\qquad D=\left(-\frac c2,h\right). \] The symmetry axis is the \(y\)-axis. Write an arbitrary point on the entire axis as \(P=(0,t)\), where \(t\in\mathbb R\).

Because \(a,c>0\), \(P\) cannot coincide with any vertex, so all angles under consideration are defined. Moreover, \[ \overrightarrow{PA}\cdot\overrightarrow{PD} = \overrightarrow{PB}\cdot\overrightarrow{PC} = \frac{ac}{4}-t(h-t). \] Consequently, both angles are right angles if and only if \[ t(h-t)=\frac{ac}{4}. \] Completing the square gives \[ \left(t-\frac h2\right)^2=\frac{h^2-ac}{4}. \]

This proves the complete classification:

  • If \(h^2<ac\), the right-hand side is negative, so no required point exists.
  • If \(h^2=ac\), the unique solution is \(t=h/2\), the midpoint of \(MN\). The circle in the construction is tangent to the symmetry axis there.
  • If \(h^2>ac\), there are exactly two solutions, \[ t_\pm=\frac{h\pm\sqrt{h^2-ac}}2, \] situated symmetrically about the midpoint of \(MN\).

Whenever solutions exist, positivity of \(ac\) implies \[ 0\le\sqrt{h^2-ac}<h. \] Hence \(0<t_\pm<h\): all required points lie strictly inside the trapezoid. Their perpendicular feet on the bases are \(M\) and \(N\), so their distances from those bases are \(t_\pm\) and \(h-t_\pm\), yielding the stated formulas.

No ordering of \(a,c\) was assumed. If the rectangular case \(a=c\) is admitted, the same result gives two points when \(h>a\), one when \(h=a\) (the center of the square), and none when \(h<a\).

The solution was checked against John Scholes' IMO solution archive (Kalva).