Library / /380 of 1394

, 2024

Algebra Difficulty 5.0 AIME, harder Prove it United States

Problem:
Let $f(x)$ be a quotient of two quadratic polynomials. Given that $f(n) = n^{3}$ for all $n \in \{1,2,3,4,5\}$ , compute $f(0)$ .

Solution

Solution:
Let $f(x) = \frac{p(x)}{q(x)}$ . Then, $x^{3} q(x) - p(x)$ has $1,2,3,4,5$ as roots. Therefore, WLOG, let

$x^{3} q(x) - p(x) = (x-1)(x-2)(x-3)(x-4)(x-5) = x^{5} - 15 x^{4} + 85 x^{3} - \ldots$

Thus, $q(x) = x^{2} - 15 x + 85$ , so $q(0) = 85$ . Plugging $x = 0$ in the above equation also gives $-p(0) = -120$ . Hence, the answer is $\frac{120}{85} = \frac{24}{17}$ .

Remark. From the solution above, it is not hard to see that the unique $f$ that satisfies the problem is
$f(x) = \frac{225 x^{2} - 274 x + 120}{x^{2} - 15 x + 85}$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.