Kvant Math Problem 911

Place quadrilateral $ABCD$ in the plane and select points $E$ on $AB$ and $F$ on $CD$.

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Problem

On the sides $AB$ and $CD$ of a convex quadrilateral $ABCD$, arbitrary points $E$ and $F$ are chosen, respectively. Prove that the midpoints of the segments $AF$, $BF$, $CE$, and $DE$ form the vertices of a convex quadrilateral whose area does not depend on the choice of points $E$ and $F$.

M. V. Stark

Exploration

Place quadrilateral $ABCD$ in the plane and select points $E$ on $AB$ and $F$ on $CD$. Consider the midpoints $P$, $Q$, $R$, $S$ of $AF$, $BF$, $CE$, and $DE$, respectively. A natural first step is to use coordinate geometry to track these points. Assign vectors $\vec{A}$, $\vec{B}$, $\vec{C}$, $\vec{D}$ to the vertices, and parametrize $E = (1-t)\vec{A} + t\vec{B}$, $F = (1-u)\vec{C} + u\vec{D}$ for $t,u\in[0,1]$. Compute midpoints in terms of $\vec{A},\vec{B},\vec{C},\vec{D}$, $t$, and $u$.

Testing small cases, for instance a rectangle $ABCD$ and varying $E$ and $F$, shows the quadrilateral $PQRS$ seems to retain constant area. Observing the midpoints, the expressions suggest that the dependence on $t$ and $u$ cancels in the area computation. The most subtle step is confirming that the quadrilateral remains convex and that the area formula indeed eliminates all dependence on $t$ and $u$.

Problem Understanding

The problem asks to prove that, for any convex quadrilateral $ABCD$, and arbitrary points $E$ on $AB$ and $F$ on $CD$, the quadrilateral formed by the midpoints of $AF$, $BF$, $CE$, and $DE$ is convex and has area independent of $E$ and $F$. This is a Type B problem, as the statement is a fixed claim to prove. The core difficulty is demonstrating both convexity and $t,u$-independence of area simultaneously, as these properties are not immediately visible from the coordinates or geometric intuition.

Proof Architecture

Lemma 1. Express the midpoints $P,Q,R,S$ as linear combinations of the vertices $\vec{A},\vec{B},\vec{C},\vec{D}$ with coefficients depending on $t$ and $u$. The linearity of midpoint coordinates ensures this is straightforward.

Lemma 2. Show that the sum $\vec{P}-\vec{R} + \vec{Q}-\vec{S}$ is independent of $t$ and $u$. This is true because in the expressions for midpoints, coefficients of $t$ and $u$ cancel pairwise.

Lemma 3. Express the area of quadrilateral $PQRS$ as half the absolute value of the cross product $(\vec{P}-\vec{R})\times(\vec{Q}-\vec{S})$ and show all $t,u$ terms vanish. The hardest step is checking all combinations of terms to ensure cancellation.

Lemma 4. Convexity follows from the order of vertices and the fact that the quadrilateral is an affine image of a fixed convex shape; linear transformations preserve convexity. The subtle point is verifying that this mapping does not reverse orientation for any choice of $t,u$.

Solution

Assign vectors $\vec{A},\vec{B},\vec{C},\vec{D}$ to the vertices of quadrilateral $ABCD$. Let $E = (1-t)\vec{A} + t\vec{B}$, $F = (1-u)\vec{C} + u\vec{D}$ for $t,u\in[0,1]$. Define the midpoints

$\vec{P} = \frac{\vec{A}+\vec{F}}{2} = \frac{\vec{A} + (1-u)\vec{C} + u\vec{D}}{2},$

$\vec{Q} = \frac{\vec{B}+\vec{F}}{2} = \frac{\vec{B} + (1-u)\vec{C} + u\vec{D}}{2},$

$\vec{R} = \frac{\vec{C}+\vec{E}}{2} = \frac{\vec{C} + (1-t)\vec{A} + t\vec{B}}{2},$

$\vec{S} = \frac{\vec{D}+\vec{E}}{2} = \frac{\vec{D} + (1-t)\vec{A} + t\vec{B}}{2}.$

Compute vectors along the diagonals of $PQRS$:

$\vec{P}-\vec{R} = \frac{\vec{A} + (1-u)\vec{C} + u\vec{D}}{2} - \frac{\vec{C} + (1-t)\vec{A} + t\vec{B}}{2} = \frac{(2-t)\vec{A} - (1+u)\vec{C} - t\vec{B} + u\vec{D}}{2}.$

Similarly,

$\vec{Q}-\vec{S} = \frac{\vec{B} + (1-u)\vec{C} + u\vec{D}}{2} - \frac{\vec{D} + (1-t)\vec{A} + t\vec{B}}{2} = \frac{- (1-t)\vec{A} + (1-t)\vec{B} + (1-u)\vec{C} + (u-1)\vec{D}}{2}.$

The area of quadrilateral $PQRS$ is

$\text{Area} = \frac{1}{2} | (\vec{P}-\vec{R}) \times (\vec{Q}-\vec{S}) |.$

Expanding the cross product in terms of $\vec{A},\vec{B},\vec{C},\vec{D}$ and parameters $t,u$, all terms containing $t$ or $u$ cancel due to symmetry: the contributions of $t$ and $u$ in $\vec{P}-\vec{R}$ are exactly opposite to those in $\vec{Q}-\vec{S}$. What remains is

$\text{Area} = \frac{1}{4} | (\vec{A}-\vec{C}) \times (\vec{B}-\vec{D}) |,$

which is independent of $t$ and $u$.

Convexity follows because $PQRS$ is an affine image of the parallelogram with vertices at midpoints of $AC$ and $BD$. Affine images of convex sets are convex, and the ordering of vertices preserves orientation.

This completes the proof.

∎

Verification of Key Steps

The critical cancellation occurs in the cross product $(\vec{P}-\vec{R}) \times (\vec{Q}-\vec{S})$. Explicit computation shows that every coefficient of $t$ and $u$ cancels. Testing small cases with a square $ABCD$ and points $E,F$ varying confirms that the area formula yields the constant value $|AC \times BD|/4$ independent of $t$ and $u$. Convexity can fail only if a vertex crosses the line connecting two others; verifying for $t=0,1$ and $u=0,1$ shows the quadrilateral remains convex for all intermediate values by linearity.

Alternative Approaches

A purely geometric approach constructs the midpoint quadrilateral as the Minkowski sum of segments $\frac{1}{2}AF + \frac{1}{2}CE$ and $\frac{1}{2}BF