Consider in a plane $ P$ the points $ O,A_1,A_2,A_3,A_4$ such that \[ \sigma(OA_iA_j) \geq 1 \quad \forall i, j = 1, 2, 3, 4, i \neq j.\] where $ \sigma(OA_iA_j)$ is the area of triangle $ OA_iA_j.$ Prove that there exists at least one pair $ i_0, j_0 \in \{1, 2, 3, 4\}$ such that \[ \sigma(OA_iA_j) \geq \sqrt{2}.\]
Problem
Source: IMO Shortlist 1989, Problem 28, ILL 87
Tags: geometry, area of a triangle, geometric inequality, IMO Shortlist, point set
Pathological
30.06.2019 04:02
Funny result! Let $a, b, c, d$ denote the lengths of $OA_1, OA_2, OA_3, OA_4$. WLOG, let $A_1, A_2, A_3, A_4$ be labelled in counter-clockwise order as viewed from $O$. Let $\alpha = \angle A_1OA_2, \beta = \angle A_2OA_3, \gamma = \angle A_3OA_4$. In this way, we have that $\angle A_4OA_1 = 360 - \alpha - \beta - \gamma.$ Rather than using that ugly $\sigma$ to denote area, we will use the simple $[\triangle ABC]$ to denote the unsigned area of $\triangle ABC.$ The following relations are derived from a well-known triangle area formula: $$[\triangle OA_1A_2] = \left| \frac12 a b \sin \alpha \right| ,$$ $$[\triangle OA_2A_3] = \left | \frac12 bc \sin \beta \right|,$$ $$[\triangle OA_3A_4] = \left|\frac12 cd \sin \gamma\right|,$$ $$[\triangle OA_4A_1] = \left|\frac12 da \sin (\alpha + \beta + \gamma)\right|,$$ $$[\triangle OA_1A_3] = \left|\frac12 ac \sin(\alpha + \beta)\right|,$$and $$[\triangle OA_2A_4] = \left|\frac12 bd \sin(\beta + \gamma)\right|.$$ Hence, we obtain that: $$[\triangle OA_1A_2] \cdot [\triangle OA_3A_4] = \frac14 abcd \left|\sin\alpha \cdot \sin \gamma\right|,$$ $$[\triangle OA_2A_3] \cdot [\triangle OA_4A_1] = \frac14 abcd \left|\sin \beta \cdot \sin (\alpha + \beta + \gamma)\right|,$$and $$[\triangle OA_1A_3] \cdot [\triangle OA_2A_4] = \frac14 abcd \left| \sin (\alpha + \beta) \cdot \sin (\beta + \gamma) \right|.$$ Lemma. $\sin \alpha \cdot \sin \gamma + \sin \beta \cdot \sin (\alpha + \beta + \gamma) = \sin (\alpha + \beta) \cdot \sin (\beta + \gamma).$ Proof. This is a straightforward application of the sum-to-product formulas, and the proof is ommitted. $\blacksquare$ As a result of the lemma, we have that: $$\pm [\triangle OA_1A_2] \cdot [\triangle OA_3A_4] \pm [\triangle OA_2A_3] \cdot [\triangle OA_4A_1] \pm [\triangle OA_1A_3] \cdot [\triangle OA_2A_4] = 0,$$ for some choice of the signs. WLOG, there is one $+$ sign and two $-$ signs. Since all areas are at least one, each term in the above is of absolute value at least $1$. This implies that the term with the $+$ sign before it must have absolute value at least $2$, and therefore one of the multiplicands has absolute value $\ge \sqrt2$, as desired. $\square$