2020 AIME II Problems/Problem 11

Revision as of 18:24, 7 June 2020 by Topnotchmath (talk | contribs) (Solution)

Problem

Let $P(X) = x^2 - 3x - 7$, and let $Q(x)$ and $R(x)$ be two quadratic polynomials also with the coefficient of $x^2$ equal to $1$. David computes each of the three sums $P + Q$, $P + R$, and $Q + R$ and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If $Q(0) = 2$, then $R(0) = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

Let $Q(x) = x^2 + ax + 2$ and $R(x) = x^2 + bx + c$. We can write the following: \[P + Q = 2x^2 + (a - 3)x - 5\] \[P + R = 2x^2 + (b - 3)x + (c - 7)\] \[Q + R = 2x^2 + (a + b)x + (c + 2)\] Let the common root of $P+Q,P+R$ be $r$; $P+R,Q+R$ be $s$; and $P+Q,Q+R$ be $t$. We then have that the roots of $P+Q$ are $r,t$, the roots of $P + R$ are $r, s$, and the roots of $Q + R$ are $s,t$.

Video Solution

https://youtu.be/BQlab3vjjxw ~ CNCM

See Also