IMO 2003 · Problem 3

A convex hexagon has the property that for any pair of opposite sides the distance between their midpoints is \(\sqrt{3}/2\) times the sum of their lengths Show that all the hexagon’s angles are equal.

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

Solution

Let the vertices be \(A_1,\ldots,A_6\) in cyclic order, with indices taken modulo \(6\). Write \[ \mathbf e_i=\overrightarrow{A_iA_{i+1}},\qquad \ell_i=|\mathbf e_i|>0, \] and let \(M_i\) be the midpoint of \(A_iA_{i+1}\). Define \[ \mathbf u=\mathbf e_1-\mathbf e_4,\qquad \mathbf v=\mathbf e_5-\mathbf e_2,\qquad \mathbf w=\mathbf e_3-\mathbf e_6, \] and \[ a=\ell_1+\ell_4,\qquad b=\ell_2+\ell_5,\qquad c=\ell_3+\ell_6. \] The triangle inequality gives \[ |\mathbf u|\le a,\qquad |\mathbf v|\le b,\qquad |\mathbf w|\le c. \tag{1} \]

Since \(\mathbf e_1+\cdots+\mathbf e_6=0\), the midpoint formula gives \[ \begin{aligned} 2\overrightarrow{M_1M_4} &=\mathbf e_1+2\mathbf e_2+2\mathbf e_3+\mathbf e_4 =\mathbf w-\mathbf v,\\ 2\overrightarrow{M_2M_5}&=\mathbf w-\mathbf u,\\ 2\overrightarrow{M_3M_6}&=\mathbf v-\mathbf u. \end{aligned} \] Thus the hypotheses are equivalent to \[ |\mathbf v-\mathbf w|^2=3a^2,\qquad |\mathbf w-\mathbf u|^2=3b^2,\qquad |\mathbf u-\mathbf v|^2=3c^2. \tag{2} \]

Expanding scalar products yields the identity \[ |\mathbf v-\mathbf w|^2+|\mathbf w-\mathbf u|^2+|\mathbf u-\mathbf v|^2 = 3\bigl(|\mathbf u|^2+|\mathbf v|^2+|\mathbf w|^2\bigr) -|\mathbf u+\mathbf v+\mathbf w|^2. \] Together with (2), this becomes \[ 0= 3(a^2-|\mathbf u|^2)+3(b^2-|\mathbf v|^2)+3(c^2-|\mathbf w|^2) +|\mathbf u+\mathbf v+\mathbf w|^2. \] Every term on the right is nonnegative by (1), so every term vanishes. Hence \[ \mathbf u+\mathbf v+\mathbf w=0,\qquad |\mathbf u|=a,\quad |\mathbf v|=b,\quad |\mathbf w|=c. \tag{3} \]

We next show that these three norms are equal. By (2), (3), and the parallelogram identity, \[ 3|\mathbf u|^2 =|\mathbf v-\mathbf w|^2 =2|\mathbf v|^2+2|\mathbf w|^2-|\mathbf v+\mathbf w|^2 =2|\mathbf v|^2+2|\mathbf w|^2-|\mathbf u|^2. \] Consequently, \[ 2|\mathbf u|^2=|\mathbf v|^2+|\mathbf w|^2. \] The corresponding equations hold cyclically. Subtracting them shows that \[ |\mathbf u|=|\mathbf v|=|\mathbf w|=r>0. \] Since their sum is zero, we also obtain \[ \mathbf u\cdot\mathbf v =\mathbf v\cdot\mathbf w =\mathbf w\cdot\mathbf u =-\frac{r^2}{2}. \tag{4} \] For example, \(2\mathbf u\cdot\mathbf v=|\mathbf u+\mathbf v|^2-|\mathbf u|^2-|\mathbf v|^2=-r^2\).

Finally, equality in each triangle inequality in (1) means that the two nonzero vectors being added have the same direction. Therefore the six unit side vectors, in order, are \[ \frac{\mathbf u}{r},\quad -\frac{\mathbf v}{r},\quad \frac{\mathbf w}{r},\quad -\frac{\mathbf u}{r},\quad \frac{\mathbf v}{r},\quad -\frac{\mathbf w}{r}. \] By (4), any two consecutive vectors in this list, including the last and first, have scalar product \(1/2\).

At each vertex, the rays forming the interior angle point opposite to the incoming side vector and along the outgoing side vector. Their unit vectors therefore have scalar product \(-1/2\). Convexity now implies that every interior angle is \[ \arccos(-1/2)=120^\circ. \] This also excludes straight angles if consecutive collinear sides were initially allowed. Thus all six angles are equal. \(\square\)

The solution was checked against Evan Chen's published IMO solution notes.