Kvant Math Problem 651

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 1m56s
Source on kvant.digital

Problem

A lady checked in a sofa, a suitcase, a valise, a basket, a picture, a cardboard box, and a small dog as luggage. The sofa weighed the same as the suitcase and the valise together, and also the same as the picture, the basket, and the cardboard box together. The picture, the basket, and the cardboard box all weighed the same, and each of them was heavier than the small dog. When the luggage was being unloaded, the lady claimed that the dog was not of the right breed. Upon inspection it turned out that the dog outweighs the sofa when either a valise or a suitcase is added to it on the scale. Prove that the lady’s claim was justified.

A. L. Toom

Correspondence Mathematical Olympiads

Exploration

Let the sofa, suitcase, valise, picture, basket, cardboard box, and dog have weights $S, U, V, P, B, C, D$ respectively.

The given relations are

$$S = U + V, \qquad S = P + B + C, \qquad P = B = C, \qquad P > D, \qquad D + U > S, \qquad D + V > S.$$

From $P = B = C$, the second sofa relation becomes $S = 3P$, hence $P = S/3$. This immediately connects all objects to the sofa.

The inequalities involving the dog and the suitcase/valise suggest comparing $D$ with $U$ and $V$ by substituting $S = U + V$ into $D + U > S$ and $D + V > S$. This produces direct inequalities between $D$ and the luggage components.

A key tension appears between $P > D$ and $P = S/3$, which tends to force $S$ to be large compared to $D$, while the inequalities involving $D + U$ and $D + V$ tend to force $S$ to be small compared to $D$. Reconciling these will likely lead to a contradiction.

Problem Understanding

This is a Type B problem.

We are given a system of weight equalities among several pieces of luggage and a dog. The structure is highly constrained: three items are equal, the sofa is the sum of two different pairs of collections, and additional inequalities involve the dog.

We must prove that the lady’s claim, namely that the dog is not of the right breed, is justified. Interpreting this mathematically, the claim corresponds to the compatibility of the stated measurements; showing that the given conditions force an impossibility establishes that her suspicion is correct.

The core difficulty is extracting a contradiction from the interplay between additive equalities and inequalities involving the dog.

Proof Architecture

First, we prove that $S = 3P$ using $P = B = C$ and $S = P + B + C$.

Second, we rewrite $P > D$ as $S > 3D$.

Third, we derive from $D + U > S$ and $S = U + V$ that $D > V$, and symmetrically that $D > U$.

Fourth, we combine $D > U$ and $D > V$ with $S = U + V$ to obtain $S < 2D$.

Fifth, we show that $S > 3D$ and $S < 2D$ are incompatible, producing a contradiction.

The most delicate step is extracting strict inequalities $D > U$ and $D > V$ from the given additive comparisons.

Solution

Since $P = B = C$, the equality $S = P + B + C$ becomes

$$S = P + P + P = 3P.$$

Hence

$$P = \frac{S}{3}.$$

The condition $P > D$ then implies

$$\frac{S}{3} > D,$$

which is equivalent to

$$S > 3D.$$

Using $S = U + V$, the inequality $D + U > S$ becomes

$$D + U > U + V,$$

from which subtraction of $U$ yields

$$D > V.$$

Similarly, the inequality $D + V > S$ becomes

$$D + V > U + V,$$

and subtraction of $V$ yields

$$D > U.$$

From $D > U$ and $D > V$, adding the inequalities gives

$$2D > U + V.$$

Using $S = U + V$, this becomes

$$2D > S,$$

$$S < 2D.$$

We now compare the two inequalities obtained for $S$:

$$S > 3D \qquad \text{and} \qquad S < 2D.$$

No real number $S$ can satisfy both inequalities simultaneously, since $3D > 2D$ for all positive $D$. This contradiction shows that the assumed configuration of weights is impossible under the stated relations.

Therefore the measurements force an inconsistency, which confirms that the dog cannot be compatible with the claimed setup, and the lady’s assertion that something is wrong with the dog is justified.

This completes the proof. ∎

Verification of Key Steps

The deduction $D > V$ from $D + U > S$ relies only on substituting $S = U + V$ and subtracting $U$ from both sides, which preserves the strict inequality. The same transformation applies symmetrically for $D > U$.

The transition from $D > U$ and $D > V$ to $S < 2D$ uses addition of strict inequalities, which is valid because both inequalities compare real numbers independently.

Finally, the incompatibility of $S > 3D$ and $S < 2D$ follows from the ordering $2D < 3D$ for all real $D$, ensuring no overlap of admissible values of $S$.

Alternative Approaches

One alternative approach is to eliminate all variables except $U, V, D$ by expressing every weight in terms of the dog and two inequalities, then deriving a direct contradiction without introducing $S$ explicitly in the final stage.

Another approach is to interpret the system geometrically in $\mathbb{R}^3$, viewing each object as a vector in a one-dimensional weight space; the constraints then define incompatible linear half-spaces whose intersection is empty. This viewpoint makes the contradiction structurally immediate but is less elementary than the direct algebraic elimination used above.