, 2019

Combinatorics Difficulty 6.8 National olympiad Prove it Baltic Way

The expressions $x+y$ , $x-y$ , $x^2+xy+y^2$ and $x^2-xy+y^2$ are written on the two sides of two cards in such a way that each side of each card contains exactly one of these expressions. The cards are laid on the table on top of each other in such a way that only the top side of the uppermost card is visible. Alice and Bob who know the expressions but not how they distribute on the invisible sides of the cards play the following game. Without inspecting the invisible sides of the cards, Alice picks one card according to her preference, the other card is left to Bob. Now both players may examine both sides of their card. Alice chooses a real value to either $x$ or $y$ according to her preference and tells her choice to Bob; then Bob chooses a real value to the other variable according to his preference. The player with larger product of the values of the expressions on two sides of their card wins. Does either of the players have a winning strategy and if yes then who does?

Solution

Answer: Yes, Alice.

Let Alice choose the card where at least one of the two expressions is a trinomial. She can do it as follows: if the visible side of the topmost card contains a trinomial then she can pick that card, otherwise the bottommost card definitely contains a trinomial and she can pick that one. By case study, we can show that Alice always can choose a value to $y$ in such a way that the product of the expressions on the sides of her card is larger than that the product of the expressions on the sides of Bob's card in the case of any value of $x$ .

* If one card contains expressions $x^2+xy+y^2$ and $x-y$ and the other card contains $x^2-xy+y^2$ and $x+y$ then the product of the expressions on the first card is $x^3-y^3$ and the product of the expressions on the second card is $x^3+y^3$ . If Alice has the first card then she can choose a negative value to $y$ , in which case $x^3-y^3 > x^3+y^3$ for any value of $x$ . If Alice has the other card then she can choose a positive value to $y$ , in which case $x^3+y^3 > x^3-y^3$ for any value of $x$ .

* If one card contains expressions $x^2+xy+y^2$ and $x+y$ and the other card contains $x^2-xy+y^2$ and $x-y$ then the product of the expressions on the first card is $(x^3+2xy^2)+(2x^2y+y^3)$ and the product of the expressions on the second card is $(x^3+2xy^2)-(2x^2y+y^3)$ . Similarly to the previous case, the ordering between the products depends on the value of $y$ solely, because $x$ occurs with even exponent in all terms that have different signs in the expressions. Alice wins by choosing a positive value to $y$ if she has expressions with only plus signs and a negative value to $y$ if she has expressions with minus signs.

* If one card contains expressions $x^2+xy+y^2$ and $x^2-xy+y^2$ and the other card contains $x+y$ and $x-y$ then the product of the expressions on the first card is $x^4+x^2y^2+y^4$ and the product of the expressions on the second card is $x^2-y^2$ . According to the Alice's choice, she has the first card. She can win by assigning a real number whose absolute value is greater than 1 to $y$ . Indeed, if Bob assigns a real number whose absolute value is not greater than 1 to $x$ then his product is negative and her product is positive, and if Bob assigns a real number whose absolute value is greater than 1 to $x$ then his product is less than $x^2$ whereas her product is larger than $x^4$ which is larger than $x^2$ .

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.