IMO 1961 · Problem 5

Construct triangle \(ABC\) if \(AC=b\), \(AB=c\) and \(\angle AMB=\omega\), where \(M\) is the midpoint of segment \(BC\) and \(\omega<90^\circ\). Prove that a solution exists if and only if \[ b\tan\frac{\omega}{2}\le c<b. \] In what case does the equality hold?

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

Answer

A solution exists exactly when \[ b\tan\frac{\omega}{2}\le c<b. \] Equality holds exactly when \(\angle BAC=90^\circ\), equivalently when \(AM=BM=CM\). A straightedge-and-compass construction is given below.

Solution

The given lengths are positive, and the prescribed angle satisfies \(0<\omega<90^\circ\).

Suppose first that a required triangle exists. Put \(x=AM\), \(y=BM=CM\), and \(t=\cos\omega>0\). Since \(\angle AMC=180^\circ-\omega\), the cosine rule gives \[ c^2=x^2+y^2-2xyt,\qquad b^2=x^2+y^2+2xyt. \] Consequently, \[ b^2-c^2=4xyt>0, \] so \(c<b\). Moreover, \[ (1+t)c^2-(1-t)b^2=2t(x-y)^2\ge0. \] Dividing by \(1+t\) and using the half-angle identity yields \[ c^2\ge b^2\frac{1-\cos\omega}{1+\cos\omega} =b^2\tan^2\frac{\omega}{2}. \] Thus the stated inequalities are necessary.

Equality in the lower bound holds exactly when \(x=y\), that is, \[ AM=BM=CM. \] This means that \(A\) lies on the circle with diameter \(BC\), which, by Thales’ theorem and its converse, is equivalent to \(\angle BAC=90^\circ\).

Now assume \[ b\tan\frac{\omega}{2}\le c<b. \] Perform the following construction:

  1. Draw a segment \(CD=2b\), and let \(A\) be its midpoint.
  2. Construct \(O\) on the perpendicular to \(CD\) through \(A\), on either chosen side, so that \[ \angle ACO=90^\circ-\omega. \]
  3. Draw the circle \(\Gamma\) with centre \(O\) through \(C\), and the circle \(\gamma\) with centre \(A\) and radius \(c\). Choose either intersection as \(B\).

We prove that these intersections exist and give the required triangle.

Write \(d=AO\) and \(R=OC\). The right triangle \(AOC\) gives \[ \angle AOC=\omega,\qquad d=b\cot\omega,\qquad R=\frac{b}{\sin\omega}. \] In particular, \(R>d>0\), and \[ R-d =b\frac{1-\cos\omega}{\sin\omega} =b\tan\frac{\omega}{2}\le c, \] whereas \[ c<b<R+d. \] These are precisely the circle-intersection inequalities \[ R-d\le c\le R+d, \] so \(\Gamma\) and \(\gamma\) intersect, including tangency when \(c=R-d\).

Because \(A\) is the midpoint of \(CD\) and \(OA\perp CD\), the circle \(\Gamma\) also passes through \(D\). Its minor arc \(CD\) has central angle \(2\omega<180^\circ\).

For any chosen intersection \(B\), we have \(AB=c<b=AC\). Comparing triangles \(AOB\) and \(AOC\), which share \(AO\) and satisfy \(OB=OC\), the cosine rule gives \[ \angle AOB<\angle AOC=\omega. \] Hence \(B\) lies on the open minor arc \(CD\). By the inscribed-angle theorem, \[ \angle CBD=\frac{360^\circ-2\omega}{2} =180^\circ-\omega. \]

Let \(M\) be the midpoint of \(BC\). In triangle \(BCD\), the segment \(MA\) joins the midpoints of \(BC\) and \(CD\). More precisely, \[ \overrightarrow{MA}=\frac12\overrightarrow{BD}, \qquad \overrightarrow{MB}=-\frac12\overrightarrow{BC}. \] Therefore \[ \angle AMB=180^\circ-\angle CBD=\omega. \] Also, \(AC=b\) and \(AB=c\) by construction. Finally, \(B\ne C,D\) because \(c<b\); since \(C,D\) are the only intersections of \(\Gamma\) with line \(CD\), the triangle \(ABC\) is nondegenerate. This remains true in the tangency case.

Thus the construction proves sufficiency, and the equality case is exactly the right triangle identified above.

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