## An IMO problem

July 20, 2010

I spent this weekend working on and off a very interesting IMO problem – problem 6 from IMO 2006. You can find that here . It was a real hard nut and only led to a crushed ego :(. Let me first describe the problem and a result I managed to prove in this direction, which was however much weaker than the problem statement.

IMO 2006 Problem 6 : Assign to each side $b$ of a convex polygon $P$ the maximum area of a triangle that has $b$ as a side and is contained in $P$. Show that the sum of the areas assigned to the sides of $P$ is at least twice the area of $P$.

I managed to prove the following result.

Theorem: Assign to each side $b$ of a convex polygon $P$ the maximum area of a triangle that has $b$ as a side and is contained in $P$. Then the sum of the areas assigned to the sides of $P$ is at least $\frac{3}{2}$ the area of $P$.

The proof is as follows. We prove first a small lemma.

Lemma 1 : Let $P$ be a convex polygon with $n \geq 4$ sides. Then there is a quadrilateral or a triangle with vertices among the vertices of $P$ that has area at least $\frac{1}{2}$ the area of $P$.

Proof: Let the polygon have vertices $A_1, A_2, \dots, A_n$. Without loss of generality assume that $A_1 A_k$ is the longest diagonal. The angles $\angle A_k A_1 A_2, \angle A_k A_1 A_n, \angle A_1 A_k A_{k-1}, \angle A_1 A_k A_{k+1}$ are all less than $\pi/2$ since $A_1 A_k$ is the longest diagonal. See the figure below for an example. For any vertex $A_i$ denote by $h(A_i)$ the distance from the line $A_1 A_k$. We will call the number $h(A_i)$ the height of $A_i$.

If $k = 2$ or $k = n$ then let $A_q$ be the vertex with maximum height. By our earlier observation regarding the angles at vertices $A_1, A_k$ with the line $A_1 A_k$, $P$ is enclosed inside the rectangle with base $A_1 A_k$ and height $h(A_q)$. As such the area of the triangle $A_1 A_k A_q$ is at least $\frac{1}{2}$ the area of $P$.

If $2 < k < n$ consider the vertex $A_p$ that has the maximum height among all vertices $A_i, 1 < i < k$ and the vertex $A_q$ that has maximum height among all vertices $A_i, k < i \leq n$. Clearly as before $P$ is enclosed inside the rectangle $R$ with a side parallel to $A_1 A_k$ through $A_p, A_q$ and the other sides through $A_1$ and $A_k$. Now

$\text{area}(A_1 A_p A_k A_q) = \text{area}(A_1 A_p A_k) + \text{area}(A_1 A_k A_q) \geq \frac{1}{2} \text{area}(R)$

We can now prove the theorem as follows.

Proof of Theorem: Let $A_1 A_p A_k A_q$ be a maximum area quadrilateral whose vertices are also vertices of $P$. We will call this quadrilateral $Q$. Clearly $\text{area}(Q) \geq \frac{1}{2} \text{area}(P)$ as we saw in the
previous lemma. The lemma actually showed that in few cases there
is actually a triangle $T$ with vertices among those of $P$ with area at least $\frac{1}{2}$ of $P$. We will not deal with the case of a triangle here since it is similar but we remark that if there is such a triangle, the theorem actually gets stronger and we can in fact prove what was required in the IMO problem.

Consider the triangles in the following four sets

$\left \{ A_1 A_2 A_q, A_2 A_3 A_q, \dots, A_{p-1} A_p A_q \right \}$

$\left \{ A_p A_{p+1} A_q, \dots, A_{k-1} A_k A_q \right \}$

$\left \{ A_k A_{k+1} A_p, \dots, A_{q-1} A_q A_p \right \}$

$\left \{ A_q A_{q+1} A_p, \dots, A_n A_1 A_p \right \}$

The sum of the areas of the triangles is at most the sum of the numbers on the edges $b$ of $P$. However, the sum of the areas is easily seen to be at least

$\text{area}(P) + \text{area}(Q) \geq \frac{3}{2} \text{area}(P)$

and we are done.

I found the solution to the IMO problem here and it is a real gem. It was a relief to see the solution!