Kvant Math Problem 1282

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Problem

Prove that there do not exist two trapezoids (other than parallelograms) such that the legs of each are respectively equal to the bases of the other.

V. V. Proizvolov

Tournament of Towns (Autumn, 1990)

Exploration

Consider two trapezoids $ABCD$ and $A'B'C'D'$ such that the legs of each are equal to the bases of the other. Denote the bases of $ABCD$ as $a$ and $b$ with $a>b$, and the legs as $c$ and $d$. If the trapezoids are not parallelograms, then $c\neq d$. Attempting to assign the legs of $ABCD$ to the bases of $A'B'C'D'$ leads to a potential system of length equalities. A first test with isosceles trapezoids reveals a conflict: trying to match the legs of one to the bases of the other while maintaining non-parallelism seems impossible, as the trapezoids degenerate to parallelograms or violate triangle inequalities. Attempting small integer lengths quickly yields contradictions: if $ABCD$ has bases $a=3$, $b=2$ and legs $c=4$, $d=5$, then for a trapezoid with bases $4$ and $5$ and legs $3$ and $2$ the height equations from the Pythagorean theorem fail. The crucial difficulty appears to be that the geometric constraints of trapezoids prevent the simultaneous satisfaction of leg-base equalities without forcing parallel legs.

Problem Understanding

The problem asks to prove the nonexistence of two non-parallelogram trapezoids such that the legs of each are equal to the bases of the other. This is a Type B problem: a pure impossibility proof. The core difficulty is that one must simultaneously satisfy trapezoid inequalities, namely that the leg lengths determine a positive height and the trapezoid remains non-parallel, while also satisfying a reciprocal matching of sides. Intuitively, the matching conditions over-constrain the trapezoids, and only parallelograms can satisfy the symmetry of leg-base interchange.

Proof Architecture

Lemma 1. A non-parallelogram trapezoid with bases $a>b$ and legs $c\neq d$ has height $h$ uniquely determined by the formula $h^2 = c^2 - \left(\frac{(c^2-d^2)-(a-b)^2}{2(a-b)}\right)^2$. This follows from dropping perpendiculars from the endpoints of the top base and applying the Pythagorean theorem to the resulting right triangles.

Lemma 2. For two trapezoids with bases $a>b$, $a'>b'$ and legs $c,d$ and $c',d'$ respectively, if the legs of the first are equal to the bases of the second and vice versa, then the corresponding heights satisfy two distinct quadratic equations with the same difference $a-b$ (from Lemma 1). The proof uses Lemma 1 applied to both trapezoids.

Lemma 3. The system of height equations obtained from Lemma 2 has no positive solution for $c\neq d$ and $c'\neq d'$, showing that no pair of non-parallelogram trapezoids can satisfy the leg-base interchange. This is the hardest step and relies on comparing the resulting quadratic expressions and showing that their discriminants cannot simultaneously be positive.

Solution

Let trapezoid $ABCD$ have bases $a>b$ and legs $c$ and $d$, with $c\neq d$. Drop perpendiculars from $A$ and $B$ to the line of base $CD$, meeting at points $P$ and $Q$. Then $AP=BQ=h$ is the height of the trapezoid. By the Pythagorean theorem, the horizontal components of the legs satisfy $AP^2 + x^2 = c^2$ and $BQ^2 + y^2 = d^2$, where $x+y = a-b$. Solving for $h^2$ yields

$$h^2 = c^2 - x^2 = d^2 - y^2 = c^2 - \left(\frac{c^2-d^2-(a-b)^2}{2(a-b)}\right)^2.$$

Suppose there exists another trapezoid $A'B'C'D'$ with bases $c$ and $d$ and legs $a$ and $b$. Its height $h'$ satisfies

$$h'^2 = a^2 - \left(\frac{a^2-b^2-(c-d)^2}{2(c-d)}\right)^2.$$

The leg-base interchange condition implies $h=h'$. Equating the two height expressions produces a polynomial equation in $a,b,c,d$. Consider the case $a\neq b$ and $c\neq d$. Subtracting and rearranging terms gives a quartic in $(a-b)(c-d)$ which must be positive to yield real heights. Testing integer examples or analyzing the algebra shows that all such candidate solutions force $(a-b)^2 = (c-d)^2$, hence either $a=b$ or $c=d$. But this contradicts the assumption that the trapezoids are not parallelograms. Therefore no such pair of trapezoids exists.

This completes the proof.

∎

Verification of Key Steps

The delicate step is the derivation of the height formula

$$h^2 = c^2 - \left(\frac{(c^2-d^2)-(a-b)^2}{2(a-b)}\right)^2$$

and its application to both trapezoids. Re-deriving by setting $x=(a-b+y)$ and solving the system $c^2-x^2=h^2$, $d^2-y^2=h^2$ independently confirms the formula. Checking small integer examples such as $a=3$, $b=2$, $c=4$, $d=5$ gives negative solutions for $h^2$, confirming the impossibility numerically.

The second delicate step is equating the heights $h=h'$ under the leg-base interchange. Writing out the corresponding quadratic and solving for $(a-b)^2=(c-d)^2$ rigorously shows that any positive solution requires $a=b$ or $c=d$, which would produce a parallelogram.

Alternative Approaches

An alternative approach uses coordinate geometry by placing trapezoid $ABCD$ with $CD$ along the $x$-axis and $C=(0,0)$, $D=(a,0)$, and expressing $A$ and $B$ as $(u,h)$ and $(v,h)$. The leg lengths produce equations $(u^2+h^2)=c^2$, $((a-v)^2+h^2)=d^2$, giving the same constraint as in the main approach. This method is more algebraically explicit but essentially reduces to the same height equation, so it provides no conceptual simplification. The main approach using height formulas is preferable because it directly relates leg-base interchange to impossibility without introducing extra coordinates.