IMO 1982 Problem 5

Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 13m39s

Problem

The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are divided by inner points $M$ and $N$ respectively, so that $$ {AM\over AC}={CN\over CE}=r. $$ Determine $r$ if $B,M$ and $N$ are collinear.

Exploration

A regular hexagon admits a convenient coordinate realization in which opposite vertices lie on coordinate axes and successive vertices differ by rotations of $60^\circ$. In such a model, diagonal segments become simple linear combinations of roots of unity or planar coordinates, and proportional division points acquire affine expressions.

The condition $AM/AC = CN/CE = r$ places $M$ and $N$ as affine images of $C$ along two distinct diagonals. The structure of a regular hexagon suggests that symmetry should reduce the configuration so that $M$ and $N$ are linearly related through a rotation by $120^\circ$ or $180^\circ$ applied to one of the defining segments.

A direct coordinate placement reduces the problem to a single collinearity condition in the plane. The key expectation is that the determinant formed by vectors from $B$ to $M$ and from $B$ to $N$ becomes a quadratic expression in $r$, forcing a unique positive solution.

Problem Understanding

This is a Type C problem, requiring the determination of a parameter $r$ defined through proportional division of two diagonals of a regular hexagon under a collinearity condition.

A regular hexagon $ABCDEF$ is fixed. Points $M$ and $N$ lie on diagonals $AC$ and $CE$ respectively, with identical ratio conditions

$$\frac{AM}{AC} = \frac{CN}{CE} = r.$$

The point $B$ together with $M$ and $N$ are required to lie on a single line, and the task is to determine the unique value of $r$ satisfying this geometric constraint.

The constraint couples two distinct affine parameters along different diagonals, so the problem reduces to a compatibility condition between two affine constructions inside a rigidly symmetric figure.

The extremal value arises because collinearity imposes a single scalar constraint on a single parameter $r$, leading to a quadratic equation with a unique admissible root in $(0,1)$.

The final result will be

$$\boxed{\frac{1}{\sqrt{3}}}.$$

Proof Architecture

Lemma 1 introduces a coordinate model for the regular hexagon in the Euclidean plane, assigning explicit coordinates to all vertices.

Lemma 2 expresses the coordinates of $M$ and $N$ in terms of $r$ using affine combinations along diagonals $AC$ and $CE$.

Lemma 3 translates the collinearity condition of $B, M, N$ into a vanishing determinant condition involving vectors $\overrightarrow{BM}$ and $\overrightarrow{BN}$.

Lemma 4 solves the resulting algebraic equation and identifies all admissible values of $r$, selecting the one compatible with the geometric constraints $0 < r < 1$.

The most delicate step is Lemma 3, where collinearity must be translated correctly into an algebraic constraint without loss of sign or orientation information.

Solution

Lemma 1

A regular hexagon may be placed in the plane with vertices

$$A = (1,0), \quad B = \left(\frac12, \frac{\sqrt3}{2}\right), \quad C = \left(-\frac12, \frac{\sqrt3}{2}\right),$$

$$D = (-1,0), \quad E = \left(-\frac12, -\frac{\sqrt3}{2}\right), \quad F = \left(\frac12, -\frac{\sqrt3}{2}\right).$$

The distances between consecutive vertices are equal and adjacent edges form angles of $60^\circ$ because each point is obtained by rotating the previous one by $60^\circ$ around the origin.

This establishes a coordinate model consistent with a regular hexagon.

Lemma 2

The point $M$ lies on segment $AC$ with

$$\frac{AM}{AC} = r,$$

hence

$$M = A + r(C-A).$$

Since

$$C-A = \left(-\frac32, \frac{\sqrt3}{2}\right),$$

it follows that

$$M = \left(1 - \frac{3r}{2}, \frac{r\sqrt3}{2}\right).$$

The point $N$ lies on segment $CE$ with

$$\frac{CN}{CE} = r,$$

$$N = C + r(E-C).$$

Since

$$E-C = (0, -\sqrt3),$$

one obtains

$$N = \left(-\frac12, \frac{\sqrt3}{2} - r\sqrt3\right).$$

This expresses both $M$ and $N$ as affine functions of $r$.

The certification is that the affine parameterization correctly encodes proportional division along directed segments.

Lemma 3

The points $B, M, N$ are collinear if and only if the vectors $\overrightarrow{BM}$ and $\overrightarrow{BN}$ are linearly dependent, which is equivalent to vanishing of the determinant

$$\det(\overrightarrow{BM}, \overrightarrow{BN}) = 0.$$

Compute the vectors

$$\overrightarrow{BM} = \left(\frac{1-3r}{2}, \frac{\sqrt3(r-1)}{2}\right),$$

$$\overrightarrow{BN} = (-1, -r\sqrt3).$$

The determinant is

$$\frac{1-3r}{2}(-r\sqrt3) - \frac{\sqrt3(r-1)}{2}(-1).$$

This simplifies to

$$-\frac{r\sqrt3(1-3r)}{2} + \frac{\sqrt3(r-1)}{2}.$$

Collinearity holds exactly when this expression equals zero.

The certification is that determinant vanishing encodes collinearity in the plane without introducing extraneous algebraic conditions.

Lemma 4

Setting the determinant equal to zero and multiplying by $\frac{2}{\sqrt3}$ yields

$$-r(1-3r) + (r-1) = 0.$$

Expanding gives

$$-r + 3r^2 + r - 1 = 0,$$

hence

$$3r^2 - 1 = 0.$$

Solving produces

$$r^2 = \frac{1}{3}.$$

Since $r$ represents a ratio of segment lengths, it satisfies $0 < r < 1$, and therefore

$$r = \frac{1}{\sqrt3}.$$

The certification is that all algebraic transformations preserve equivalence and the geometric constraint selects the positive root.

Assembly

The coordinate model satisfies the structure of a regular hexagon. The points $M$ and $N$ are correctly constructed as affine divisions of diagonals $AC$ and $CE$. The collinearity condition reduces exactly to a quadratic equation in $r$, whose unique admissible solution is $r = 1/\sqrt3$.

Verification of Key Steps

The determinant computation is rederived directly from the coordinates of $B$, $M$, and $N$. Subtracting coordinates yields $\overrightarrow{BM}$ and $\overrightarrow{BN}$ with no ambiguity in sign, and recomputing the determinant produces the same quadratic expression $3r^2-1$.

The affine construction of $M$ and $N$ is rechecked by confirming that both points lie on their respective segments when $0 < r < 1$, since each is expressed as a convex combination of endpoints.

The algebraic reduction from the determinant equation is rechecked by expanding each term separately; cancellation of linear terms in $r$ occurs identically, leaving a purely quadratic constraint.

Alternative Approaches

A complex number formulation places the vertices at sixth roots of unity and expresses $M$ and $N$ as affine combinations in the complex plane. The collinearity condition becomes a real-valued constraint on a ratio of complex differences, ultimately producing the same quadratic equation in $r$.

A purely synthetic approach uses rotations by $60^\circ$ and $120^\circ$ in the regular hexagon to relate triangles $ABM$ and $CBN$, leading again to a proportionality constraint equivalent to the same quadratic condition. The coordinate method remains more direct because it reduces all geometric relations to a single determinant computation without introducing rotational identities.