Difference between revisions of "2020 AIME II Problems/Problem 11"

Revision as of 18:31, 7 June 2020

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 $s,t$ .

By Vieta's, we have: $(1) r + t = \dfrac{3 - a}{2}$ $(2) r + s = \dfrac{3 - b}{2}$ $(3) s + t = \dfrac{-a - b}{2}$ $(4) rt = \dfrac{-5}{2}$ $(5) rs = \dfrac{c - 7}{2}$ $(6) st = \dfrac{c + 2}{2}$

Video Solution

https://youtu.be/BQlab3vjjxw ~ CNCM

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 https://youtu.be/BQlab3vjjxw ~ CNCM
 ==See Also==
-{{AIME box|year=2020|n=II|num-b=10|num-a=2}}
+{{AIME box|year=2020|n=II|num-b=10|num-a=12}}
 [[Category:Intermediate Algebra Problems]]
 {{MAA Notice}}

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Difference between revisions of "2020 AIME II Problems/Problem 11"

Revision as of 18:31, 7 June 2020

Contents

Problem

Solution

Video Solution

See Also