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 \(P\).

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

Solution

Write \([K]\) for the area of a polygon \(K\). If \([P]=0\), the assertion is immediate, so assume \([P]>0\).

For each side \(b\), let \(w_b\) be the distance between its supporting line and the parallel supporting line on the opposite side of \(P\). By convexity, a triangle with base \(b\) and any third vertex in \(P\) is contained in \(P\). Thus the area assigned to \(b\) is \[ M_b=\frac12 |b|w_b. \] Consecutive collinear sides may be merged: their widths are equal, so this does not change the sum of their assigned areas.

Consider the auxiliary polygon \[ D=P-P=\{x-y:x,y\in P\}. \] This is a convex polygon: if \(v_1,\ldots,v_n\) are the vertices of \(P\), then \(D\) is the convex hull of the points \(v_i-v_j\). Indeed, writing \(x=\sum_i\alpha_i v_i\) and \(y=\sum_j\beta_jv_j\) as convex combinations gives \[ x-y=\sum_{i,j}\alpha_i\beta_j(v_i-v_j), \] and the set of differences is convex. Also, \(D=-D\), and the origin is in its interior.

We will prove \[ [D]=2\sum_b M_b \qquad\text{and}\qquad [D]\ge 4[P]. \]

The area of \(D\) in terms of the assigned areas.
Fix a direction parallel to a side of \(P\), and let \(u\) be a perpendicular unit vector. Define \[ h_+=\max_{x\in P}u\cdot x,\qquad h_-=\min_{x\in P}u\cdot x,\qquad w=h_+-h_-. \] Let \(F_+\) and \(F_-\) be the portions of \(P\) on these two supporting lines. Each is either a side or a single vertex. Denote their lengths by \(\ell_+\) and \(\ell_-\), counting a vertex as having length zero.

For \(z=x-y\in D\), we have \[ u\cdot z\le h_+-h_-=w, \] with equality exactly when \(x\in F_+\) and \(y\in F_-\). Consequently, the supporting face of \(D\) in direction \(u\) is \[ F_+-F_-. \] The difference of two parallel segments of lengths \(\ell_+\) and \(\ell_-\) is a segment of length \(\ell_++\ell_-\). Hence \(D\) has an edge of that length at distance \(w\) from the origin, and an equal opposite edge.

This describes every edge of \(D\): if both corresponding supporting faces of \(P\) were vertices, their difference would be a point, not an edge.

Joining the origin to the vertices of \(D\) partitions it into triangles. The two edges associated with our chosen direction contribute total area \[ 2\cdot\frac12(\ell_++\ell_-)w=(\ell_++\ell_-)w. \] Summing over the distinct directions parallel to sides of \(P\), each side of \(P\) is counted exactly once. Therefore \[ [D]=\sum_b |b|w_b=2\sum_b M_b. \tag{1} \]

Comparing the areas of \(D\) and \(P\).
Translate \(P\) so that its lowest and highest horizontal supporting lines have heights \(0\) and \(h\), where \(h>0\). This translation does not change \(D\).

For \(0\le t\le h\), let \[ I_t=\{x\in\mathbb R:(x,t)\in P\}, \] and let \(f(t)\) be its length. Each \(I_t\) is a nonempty closed interval, possibly a point.

For \(-h\le s\le h\), put \[ t=\frac{h+s}{2},\qquad r=\frac{h-s}{2}. \] Since \(t-r=s\), the horizontal section of \(D\) at height \(s\) contains the interval \(I_t-I_r\). The difference of two intervals has length equal to the sum of their lengths, so this section has length at least \[ f\left(\frac{h+s}{2}\right)+f\left(\frac{h-s}{2}\right). \]

The area of a polygon is the integral of its horizontal section lengths; here this is simply the trapezoid area formula applied between successive vertex heights. Therefore \[ \begin{aligned} [D] &\ge \int_{-h}^{h} \left[ f\left(\frac{h+s}{2}\right) + f\left(\frac{h-s}{2}\right) \right]\,ds\\ &=2\int_0^h f(t)\,dt+2\int_0^h f(t)\,dt\\ &=4[P]. \end{aligned} \] Combining this with (1) gives \[ \boxed{\sum_b M_b\ge 2[P]}, \] as required.

The solution was checked against the official IMO shortlist solutions.