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Difference between revisions of "2020 AIME II Problems/Problem 11"
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By Vieta's, we have:
 
By Vieta's, we have:
−
<cmath>(1) r + t = \dfrac{3 - a}{2}</cmath>
+
<cmath> r + t = \dfrac{3 - a}{2}\tag{1}</cmath>
−
<cmath>(2) r + s = \dfrac{3 - b}{2}</cmath>
+
<cmath>r + s = \dfrac{3 - b}{2}\tag{2}</cmath>
−
<cmath>(3) s + t = \dfrac{-a - b}{2}</cmath>
+
<cmath>s + t = \dfrac{-a - b}{2}\tag{3}</cmath>
−
<cmath>(4) rt = \dfrac{-5}{2}</cmath>
+
<cmath>rt = \dfrac{-5}{2}\tag{4}</cmath>
−
<cmath>(5) rs = \dfrac{c - 7}{2}</cmath>
+
<cmath>rs = \dfrac{c - 7}{2}\tag{5}</cmath>
−
<cmath>(6) st = \dfrac{c + 2}{2}</cmath>
+
<cmath>st = \dfrac{c + 2}{2}\tag{6}</cmath>
  
 
Subtracting <math>(3)</math> from <math>(1)</math>, we get <math>r - s = \dfrac{3 + b}{2}</math>.  Adding this to <math>(2)</math>, we get <math>2r = 3 \implies r = \dfrac{3}{2}</math>.  This gives us that <math>t = \dfrac{-5}{3}</math> from <math>(4)</math>.  Substituting these values into <math>(5)</math> and <math>(6)</math>, we get <math>s = \dfrac{c-7}{3}</math> and <math>s = \dfrac{-3c - 6}{10}</math>.  Equating these values, we get <math>\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)</math>.  Thus, our answer is <math>52 + 19 = \boxed{071}</math>. ~ TopNotchMath
 
Subtracting <math>(3)</math> from <math>(1)</math>, we get <math>r - s = \dfrac{3 + b}{2}</math>.  Adding this to <math>(2)</math>, we get <math>2r = 3 \implies r = \dfrac{3}{2}</math>.  This gives us that <math>t = \dfrac{-5}{3}</math> from <math>(4)</math>.  Substituting these values into <math>(5)</math> and <math>(6)</math>, we get <math>s = \dfrac{c-7}{3}</math> and <math>s = \dfrac{-3c - 6}{10}</math>.  Equating these values, we get <math>\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)</math>.  Thus, our answer is <math>52 + 19 = \boxed{071}</math>. ~ TopNotchMath

Revision as of 18:41, 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: \[r + t = \dfrac{3 - a}{2}\tag{1}\] \[r + s = \dfrac{3 - b}{2}\tag{2}\] \[s + t = \dfrac{-a - b}{2}\tag{3}\] \[rt = \dfrac{-5}{2}\tag{4}\] \[rs = \dfrac{c - 7}{2}\tag{5}\] \[st = \dfrac{c + 2}{2}\tag{6}\]

Subtracting $(3)$ from $(1)$, we get $r - s = \dfrac{3 + b}{2}$. Adding this to $(2)$, we get $2r = 3 \implies r = \dfrac{3}{2}$. This gives us that $t = \dfrac{-5}{3}$ from $(4)$. Substituting these values into $(5)$ and $(6)$, we get $s = \dfrac{c-7}{3}$ and $s = \dfrac{-3c - 6}{10}$. Equating these values, we get $\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)$. Thus, our answer is $52 + 19 = \boxed{071}$. ~ TopNotchMath

Video Solution

https://youtu.be/BQlab3vjjxw ~ CNCM

See Also

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