Evdokimov M.A.

Tanya took a list of one hundred numbers $1,2,3, \ldots, 100$ and crossed out several of them. It turned out that no matter which two numbers from the remaining ones Tanya took as $\$ a \$$ and $\$ b \$$, the equation $\$ x \wedge 2+a x+b=0 \$$ has at least one real root. What is the maximum number of numbers that could remain uncrossed?
The equation $\$ x^{\wedge} 2+\mathrm{ax}+\mathrm{b}=0 \$$ has real roots if and only if its discriminant $\$ \mathrm{D}=\mathrm{a} \wedge 2-4 \mathrm{~b} \$$ is non-negative. This means that any two of the remaining numbers must satisfy the inequality \$ $\mathrm{2} \wedge 2$ \geqslant $4 b \$$.

And here is the most important idea: the smaller we take $\$ \mathrm{a} \$$ and the larger we take $\$ \mathrm{~b} \$$, the "harder" it is for this inequality to hold! In other words, if it holds when a is the smallest of the numbers and b is the largest, then it will hold for any pair of numbers.

Let $\$ \mathrm{~m} \$$ be the smallest of the selected numbers, and $\$ \mathrm{M} \$$ be the largest. We have proven the inequality \$m $\wedge 2$ \geqslant $4 \mathrm{M} \$$. We can finish the solution in different ways. Let's show one of the paths. If $\$ \mathrm{M}=100 \$$, then \$m 2 \geqslant $400 \$$ and $\$ m \geqslant 20 \$$ - thus, we can leave 81 numbers: from 20 to 100. Suppose there are more than 81 numbers. Then $\$ \mathrm{M}>\mathrm{m}+80 \$$, $\$ \mathrm{~m}<20 \$$ and for the equation $\$ \mathrm{x}^{\wedge} 2+\mathrm{mx}+\mathrm{M}=0 \$ \$ \$ \mathrm{D}=\mathrm{m} \wedge 2-4 \mathrm{M}<\mathrm{m} \wedge 2-4(\mathrm{~m}+80)=$ $\mathrm{m} \wedge 2-4 \mathrm{~m}-320=(\mathrm{m}-2) \wedge 2$ - 324 \leqslant $17 \wedge 2-324<0 \$ \$$, which means the equation does not have real roots. Therefore, there cannot be more than 81 numbers.

## O:

Submit a comment