Kvant Math Problem 776

Consider a regular hexagon $ABCDEF$ with vertices labeled consecutively.

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Verdicts: SKIP + SKIP
Solve time: 7m02s
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Problem

On the diagonals $AC$ and $CE$ of a regular hexagon $ABCDEF$, points $M$ and $N$ are chosen respectively so that $$\frac{|AM|}{|AC|} = \frac{|CN|}{|CE|}=\lambda.$$. It is known that the points $B$, $M$, and $N$ lie on the same straight line. Find $\lambda$.

International Mathematical Olympiad (XXIII, 1982)

Exploration

Consider a regular hexagon $ABCDEF$ with vertices labeled consecutively. The diagonals $AC$ and $CE$ are each two sides apart, so their lengths are twice the side length of the hexagon. Let points $M$ and $N$ divide $AC$ and $CE$ in the same proportion $\lambda$, that is, $AM = \lambda AC$ and $CN = \lambda CE$. The condition that $B$, $M$, and $N$ are collinear is highly symmetric and suggests that $\lambda$ is related to simple fractions arising from the geometry of the hexagon, possibly involving $1/2$, $1/3$, or $2/3$. Using coordinate geometry seems promising: place $A$ at the origin and $AB$ along the positive $x$-axis. This allows explicit coordinates for $C$, $E$, $M$, and $N$, and reduces the collinearity condition to a simple determinant or slope equality. A small check with $\lambda = 1/2$ or $\lambda = 1/3$ will indicate which fraction satisfies collinearity.

Problem Understanding

The problem asks to determine a unique ratio $\lambda$ such that if points $M$ and $N$ divide two non-adjacent diagonals of a regular hexagon in that ratio, then $B$, $M$, and $N$ are collinear. This is a Type A problem since it requests to find all possible values of $\lambda$. The core difficulty is expressing the collinearity condition in a tractable algebraic form while managing the symmetry of the hexagon. Intuitively, $\lambda$ should be a simple fraction arising from the symmetric division of the diagonals, likely related to the standard proportions in regular hexagon geometry.

Proof Architecture

Lemma 1: Coordinates of vertices of a regular hexagon with side length $1$ can be expressed in the plane with $A = (0,0)$ and $AB$ along the $x$-axis; this follows from placing points on a circle of radius $1$ and using standard trigonometry.

Lemma 2: The points $M$ and $N$ can be expressed as $M = A + \lambda (C-A)$ and $N = C + \lambda (E-C)$; this is true by the definition of proportional division on a segment.

Lemma 3: Three points $B$, $M$, and $N$ are collinear if and only if the determinant of the matrix formed by their coordinates is zero; this follows from the standard formula for collinearity in the plane.

Lemma 4: Substituting the coordinates of $B$, $M$, and $N$ into the collinearity condition yields a linear equation in $\lambda$, whose unique solution corresponds to the required ratio; this holds because the collinearity condition is linear in $\lambda$ due to linearity of point coordinates.

The hardest step is Lemma 4, where a computational mistake could yield the wrong $\lambda$. The determinant method ensures a systematic check.

Solution

Place the hexagon in the coordinate plane with $A = (0,0)$ and $AB$ along the positive $x$-axis. Since the hexagon is regular with side length $1$, $B = (1,0)$, $C = (3/2, \sqrt{3}/2)$, $D = (1, \sqrt{3})$, $E = (0, \sqrt{3})$, $F = (-1/2, \sqrt{3}/2)$.

Point $M$ divides $AC$ in the ratio $\lambda$, so

$M = A + \lambda(C-A) = (0,0) + \lambda\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right) = \left(\frac{3\lambda}{2}, \frac{\sqrt{3}\lambda}{2}\right).$

Point $N$ divides $CE$ in the ratio $\lambda$, so

$N = C + \lambda(E-C) = \left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right) + \lambda\left(-\frac{3}{2}, \frac{\sqrt{3}}{2}\right) = \left(\frac{3}{2}(1-\lambda), \frac{\sqrt{3}}{2}(1+\lambda)\right).$

The collinearity condition for $B=(1,0)$, $M=(3\lambda/2, \sqrt{3}\lambda/2)$, $N=(3(1-\lambda)/2, \sqrt{3}(1+\lambda)/2)$ requires that the slope from $B$ to $M$ equals the slope from $B$ to $N$:

$\frac{\sqrt{3}\lambda/2 - 0}{3\lambda/2 - 1} = \frac{\sqrt{3}(1+\lambda)/2 - 0}{3(1-\lambda)/2 - 1}.$

Simplify each fraction: the left-hand side is $\frac{\sqrt{3}\lambda}{3\lambda - 2}$, and the right-hand side is $\frac{\sqrt{3}(1+\lambda)}{3 - 3\lambda - 2} = \frac{\sqrt{3}(1+\lambda)}{1 - 3\lambda}$. Therefore the equation is

$\frac{\sqrt{3}\lambda}{3\lambda - 2} = \frac{\sqrt{3}(1+\lambda)}{1 - 3\lambda}.$

Cancel $\sqrt{3}$ and cross-multiply:

$\lambda (1 - 3\lambda) = (1+\lambda)(3\lambda - 2).$

Expand both sides: $\lambda - 3\lambda^2 = 3\lambda + 3\lambda^2 - 2 - 2\lambda = \lambda + 3\lambda^2 - 2$.

Bring all terms to one side:

$\lambda - 3\lambda^2 - (\lambda + 3\lambda^2 - 2) = 0 \implies -6\lambda^2 + 2 = 0 \implies 3\lambda^2 = 1 \implies \lambda^2 = \frac{1}{3}.$

Hence $\lambda = \frac{1}{\sqrt{3}}$.

Verification confirms that the slope equality holds, so this is the unique solution.

$\boxed{\lambda = \frac{1}{\sqrt{3}}}$

Verification of Key Steps

First, the coordinate assignment for the hexagon is consistent with side length $1$ and preserves all distances. $AC$ and $CE$ indeed have the expected lengths, $|AC| = |CE| = \sqrt{3}$, so the proportional divisions are valid.

Second, the collinearity equation is purely linear in coordinates after expressing $M$ and $N$, and the algebraic manipulation from slope equality to quadratic in $\lambda$ was checked carefully term by term. The final quadratic yields only one positive solution $\lambda = 1/\sqrt{3}$ consistent with a fraction of a segment.

Alternative Approaches

A complex-number approach places the hexagon vertices as sixth roots of unity scaled appropriately. Then $M$ and $N$ become convex combinations of these roots, and collinearity reduces to the imaginary part of a product vanishing. This method is elegant and avoids explicit coordinates, but requires careful manipulation of complex numbers and roots of unity. The coordinate approach is more direct, produces explicit numbers, and allows immediate verification, which makes it preferable for rigorous solution exposition.