IMO 1967 · Problem 1

Let \(ABCD\) be a parallelogram with side lengths \(AB = a, AD = 1\), and with \(\angle BAD = \alpha\). If \(\triangle ABD\) is acute, prove that the four circles of radius \(1\) with centers \(A, B, C, D\) cover the parallelogram if and only if \[ a \le \cos \alpha + \sqrt{3}\sin \alpha. \]

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

Solution

We interpret coverage by a circle as coverage by its closed disk. Let \(O\) and \(R\) be the circumcenter and circumradius of \(\triangle ABD\). We will prove that the parallelogram is covered exactly when \(R\le 1\).

First, any triangle is covered by the three closed disks of radius equal to its circumradius centered at its vertices. Indeed, let its vertices be \(X_1,X_2,X_3\), its circumcenter be \(Q\), and its circumradius be \(r\). Every point \(P\) in the triangle can be written \[ P=\lambda_1X_1+\lambda_2X_2+\lambda_3X_3, \qquad \lambda_i\ge0,\qquad \lambda_1+\lambda_2+\lambda_3=1. \] Expanding squared distances gives \[ \sum_{i=1}^3\lambda_i PX_i^2 =\sum_{i=1}^3\lambda_i QX_i^2-QP^2 =r^2-QP^2\le r^2. \] Consequently, at least one of the distances \(PX_i\) is at most \(r\).

Thus, if \(R\le1\), the unit disks centered at \(A,B,D\) cover \(\triangle ABD\). The congruent triangle \(BCD\) has the same circumradius, so it is covered by the unit disks centered at \(B,C,D\). Together these triangles constitute the parallelogram, proving sufficiency.

Conversely, suppose that \(R>1\). Since \(\triangle ABD\) is acute, its circumcenter \(O\) lies inside it, hence inside the parallelogram. We have \[ OA=OB=OD=R>1. \] We also show that \(OC>R\). Take \(A\) as the origin and write \[ B=\mathbf b,\qquad D=\mathbf d,\qquad C=\mathbf b+\mathbf d, \qquad O=\mathbf o. \] The equalities \(OA=OB=OD\) imply \[ 2\mathbf o\cdot\mathbf b=|\mathbf b|^2, \qquad 2\mathbf o\cdot\mathbf d=|\mathbf d|^2. \] Therefore \[ \begin{aligned} OC^2 &=|\mathbf o-\mathbf b-\mathbf d|^2\\ &=|\mathbf o|^2+2\mathbf b\cdot\mathbf d\\ &=R^2+2a\cos\alpha >R^2, \end{aligned} \] because \(\alpha\) is acute. Hence none of the four unit disks contains \(O\), so they do not cover the parallelogram. This proves the claimed equivalence with \(R\le1\).

It remains to express that condition in terms of \(a\) and \(\alpha\). By the cosine rule and the extended sine rule, \[ BD^2=a^2+1-2a\cos\alpha, \qquad R=\frac{BD}{2\sin\alpha}. \] Since \(\sin\alpha>0\), \[ \begin{aligned} R\le1 &\iff a^2+1-2a\cos\alpha\le4\sin^2\alpha\\ &\iff (a-\cos\alpha)^2\le3\sin^2\alpha. \end{aligned} \] Moreover, the acute angle at \(B\) gives \[ a-\cos\alpha=BD\cos\angle ABD>0. \] We may therefore take square roots to obtain \[ R\le1 \iff a-\cos\alpha\le\sqrt3\sin\alpha \iff \boxed{a\le\cos\alpha+\sqrt3\sin\alpha}. \] The disks are closed, so the equality case is included.

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