IMO 1959 · Problem 6

Two planes, \(P\) and \(Q\), intersect along the line \(p\). The point \(A\) is given in the plane \(P\), and the point \(C\) in the plane \(Q\); neither of these points lies on the straight line \(p\). Construct an isosceles trapezoid \(ABCD\) (with \(AB\) parallel to \(CD\)) in which a circle can be inscribed, and with vertices \(B\) and \(D\) lying in the planes \(P\) and \(Q\) respectively.

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

Solution

Let \(\ell_A\) and \(\ell_C\) be the lines through \(A\) and \(C\), respectively, parallel to \(p\).

In any solution, \(AB\) and \(CD\) must be parallel to \(p\). Indeed, the line through a point of \(p\) parallel to these two parallel lines belongs to both \(P\) and \(Q\), and therefore is \(p\). Thus necessarily \[ B\in\ell_A,\qquad D\in\ell_C. \] The lines \(\ell_A,\ell_C\) are distinct: otherwise their common line would belong to both given planes and hence equal \(p\), contrary to the hypotheses. They therefore determine a plane \(R\), in which we perform the construction.

Let \(H\) be the perpendicular projection of \(C\) onto \(\ell_A\), and put \[ d=AH,\qquad h=CH>0. \]

We will obtain two proper isosceles trapezoids when \(d>h\), a square when \(d=h\), and no solution when \(d<h\).

Necessary conditions. Suppose the required trapezoid has base lengths \(a,b\) and equal leg length \(s\). Its bases have a common perpendicular bisector. For completeness, in a proper isosceles trapezoid we may write the vertices as \[ (0,0),\quad(a,0),\quad(x+b,h),\quad(x,h). \] Equality of the legs gives \[ x^2=(x+b-a)^2. \] Since \(a\ne b\), this implies \(x=(a-b)/2\), proving the asserted symmetry. The rectangular case has the same symmetry.

Consequently, the projection of diagonal \(AC\) onto the direction of the bases has length \((a+b)/2\), and Pythagoras gives \[ d=\frac{a+b}{2}, \qquad s^2=h^2+\left(\frac{a-b}{2}\right)^2. \] If a circle is inscribed, adding the equal tangent lengths from the four vertices yields \[ a+b=2s. \] Thus \(s=d\), so \[ \left(\frac{a-b}{2}\right)^2=d^2-h^2. \] In particular, \(d\ge h\) is necessary.

Construction. Suppose \(d\ge h\).

  1. In \(R\), draw the circle with centre \(C\) and radius \(d\).
  2. For each intersection \(B\) of this circle with \(\ell_A\), construct the perpendicular bisector \(m\) of \(AB\).
  3. Reflect \(C\) across \(m\) to obtain \(D\).

There are two choices for \(B\) if \(d>h\), and one if \(d=h\).

Verification. Put \[ t=\sqrt{d^2-h^2}. \] Since \(h>0\), we have \(0\le t<d\). Choose coordinates in \(R\) with \[ A=(0,0),\qquad H=(d,0),\qquad C=(d,h). \] For either available choice, with \(\varepsilon\in\{1,-1\}\), \[ B=(d+\varepsilon t,0). \] Reflection in the perpendicular bisector of \(AB\) gives \[ D=(\varepsilon t,h). \] Hence \(D\in\ell_C\), and \[ AB=d+\varepsilon t>0,\qquad CD=d-\varepsilon t>0. \] These coordinates show that \(A,B,C,D\) form a convex quadrilateral in the stated order. Reflection in \(m\) exchanges \(A,B\) and \(D,C\), so the trapezoid is isosceles. Moreover, \[ AD=BC=d,\qquad AB+CD=2d. \] Also, \(B\in P\) and \(D\in Q\), as required.

It remains to prove that a circle can be inscribed; the equality of opposite-side sums alone will not be used as a sufficient condition.

Let \(X,Y\) be the midpoints of \(AB,CD\), and let \(O\) be the midpoint of \(XY\). The segment \(XY\) lies on the symmetry axis and is perpendicular to both bases. Thus \(O\) is inside the trapezoid and has distance \(h/2\) from each base. By symmetry, its distances from the two legs are equal; denote their common value by \(r\).

The area of the trapezoid is \[ K=\frac{(AB+CD)h}{2}=dh. \] Splitting it into the four triangles with vertex \(O\) gives \[ K =\frac{(AB+CD)h}{4} +\frac{(AD+BC)r}{2} =\frac{dh}{2}+dr. \] Therefore \(r=h/2\). The circle with centre \(O\) and radius \(h/2\) is consequently tangent to all four supporting lines. Its disk lies in their four inward half-planes, so it is contained in the trapezoid and touches all four side segments. This is the required incircle.

Finally, every solution has \(BC=d\), so its vertex \(B\) must be one of the circle-line intersections used above; symmetry then determines \(D\). Thus the construction is exhaustive.

When \(d=h\), we have \(t=0\), and the construction gives the square \[ A=(0,0),\quad B=(d,0),\quad C=(d,d),\quad D=(0,d). \] If “trapezoid” is required to have exactly one pair of parallel sides, this square is excluded, and existence requires \(d>h\). When \(d<h\), including \(d=0\), no solution exists.

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