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Algebra Difficulty 6.9 National Olympiad Prove it United States

Problem:

While waiting for their food at a restaurant in Harvard Square, Ana and Banana draw 3 squares $\square_{1}, \square_{2}, \square_{3}$ on one of their napkins. Starting with Ana, they take turns filling in the squares with integers from the set $\{1,2,3,4,5\}$ such that no integer is used more than once. Ana's goal is to minimize the minimum value $M$ that the polynomial $a_{1} x^{2}+a_{2} x+a_{3}$ attains over all real $x$ , where $a_{1}, a_{2}, a_{3}$ are the integers written in $\square_{1}, \square_{2}, \square_{3}$ respectively. Banana aims to maximize $M$ . Assuming both play optimally, compute the final value of $100 a_{1}+10 a_{2}+a_{3}$ .

Solution

Solution:

Relabel $a_{1}, a_{2}, a_{3}$ as $a, b, c$ . This is minimized at $x=\frac{-b}{2a}$ , so $M = c - \frac{b^{2}}{4a}$ .

If in the end $a=5$ or $b \in \{1,2\}$ , then $\frac{b^{2}}{4a} \leq 1$ and $M \geq 0$ . The only way for Ana to block this is to set $b=5$ , which will be optimal if we show that it allows Ana to force $M<0$ , which we will now do. At this point, Banana has two choices:

- If Banana fixes a value of $a$ , Ana's best move is to pick $c=1$ , or $c=2$ if it has not already been used. The latter case yields $M < -1$ , while the optimal move in the latter case $(a=4)$ yields $M = 1 - \frac{25}{16} > -1$ .
- If Banana fixes a value of $c$ , then if that value is not 1 Ana can put $a=1$ , yielding $M \leq 4 - \frac{25}{4} < -1$ . On the other hand, if Banana fixes $c=1$ then Ana's best move is to put $a=2$ , yielding $M = 1 - \frac{25}{8} < -1$ .

Thus Banana's best move is to set $a=4$ , eliciting a response of $c=1$ . Since $1 - \frac{25}{16} < 0$ , this validates our earlier claim that $b=5$ was the best first move.

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