IMO 1960 · Problem 4

Construct triangle \(ABC\), given \(h_a, h_b\) (the altitudes from \(A\) and \(B\)) and \(m_a\), the median from vertex \(A\).

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

Solution

Put \[ p=h_a,\qquad q=\frac{h_b}{2},\qquad m=m_a. \] These lengths must be positive for a nondegenerate triangle.

The key observation is that if \(M\) is the midpoint of \(BC\), then \[ d(M,AC)=\frac12d(B,AC)=q, \] where \(d(P,r)\) denotes the perpendicular distance from a point to a line. Indeed, the homothety with center \(C\) and ratio \(2\) sends \(M\) to \(B\) and preserves the line \(AC\).

Thus we seek two lines: \(BC\), passing through \(M\) at distance \(p\) from \(A\), and \(AC\), passing through \(A\) at distance \(q\) from \(M\).

Construction.

  1. Draw a segment \(AM\) of length \(m\).
  2. Draw the circle with center \(A\) and radius \(p\). Construct all its tangent lines \(\ell\) through \(M\).
  3. Draw the circle with center \(M\) and radius \(q\). Construct all its tangent lines \(t\) through \(A\).
  4. For each nonparallel pair \((\ell,t)\), let \(C=\ell\cap t\), and construct \(B\) as the reflection of \(C\) in \(M\).

These are straightedge-and-compass operations. For either tangent construction, when the given point is outside the circle, the circle with diameter \(AM\) determines the two tangent points by intersection with the relevant circle, using Thales’ theorem. When the given point lies on the circle, the unique tangent is the perpendicular to \(AM\) at that point.

Verification. Since \(p,q>0\), the point \(A\) does not lie on \(\ell\), and \(M\) does not lie on \(t\). Consequently \(C\ne M\), so \(B\ne C\), and \(ABC\) is nondegenerate.

By construction, \(M\) is the midpoint of \(BC\), and therefore the median from \(A\) has length \(AM=m\). Tangency gives \[ d(A,BC)=p. \] The homothety described above gives \[ d(B,AC)=2d(M,t)=2q=h_b. \] Thus every triangle constructed has precisely the prescribed data.

Existence and limiting cases. In any solution, \[ p=d(A,BC)\le AM=m,\qquad q=d(M,AC)\le AM=m. \] Hence \(p\le m\) and \(q\le m\) are necessary.

If \(p=q=m\), both tangent constructions have only one line: \(\ell\) is perpendicular to \(AM\) at \(M\), while \(t\) is perpendicular to \(AM\) at \(A\). They are distinct parallel lines, so no triangle exists.

Conversely, suppose \(p\le m\) and \(q\le m\), with at least one inequality strict. Both tangent families are nonempty, and at least one family contains two distinct lines. Fix any line in the other family. The two distinct lines, which pass through the same point, cannot both be parallel to that fixed line. Hence a nonparallel pair exists and produces a triangle.

Thus the exact existence conditions are \[ \boxed{m_a\ge h_a,\qquad 2m_a\ge h_b, \qquad\text{with simultaneous equality excluded}.} \] This includes the cases where either inequality alone is an equality. Parallel choices are simply discarded.

Finally, every possible solution is obtained: after identifying its vertex \(A\) and midpoint \(M\) with the constructed segment, its lines \(BC\) and \(AC\) must belong to the respective tangent families, and \(B\) must be the reflection of \(C\) in \(M\).

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