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

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<cmath>Q + R = 2x^2 + (a + b)x + (c + 2)</cmath>
 
<cmath>Q + R = 2x^2 + (a + b)x + (c + 2)</cmath>
 
Let the common root of <math>P+Q,P+R</math> be <math>r</math>; <math>P+R,Q+R</math> be <math>s</math>; and <math>P+Q,Q+R</math> be <math>t</math>.  We then have that the roots of <math>P+Q</math> are <math>r,t</math>, the roots of <math>P + R</math> are <math>r, s</math>, and the roots of <math>Q + R</math> are <math>s,t</math>.
 
Let the common root of <math>P+Q,P+R</math> be <math>r</math>; <math>P+R,Q+R</math> be <math>s</math>; and <math>P+Q,Q+R</math> be <math>t</math>.  We then have that the roots of <math>P+Q</math> are <math>r,t</math>, the roots of <math>P + R</math> are <math>r, s</math>, and the roots of <math>Q + R</math> are <math>s,t</math>.
 +
 +
By Vieta's, we have:
 +
<cmath>(1) r + t = \dfrac{3 - a}{2}</cmath>
 +
<cmath>(2) r + s = \dfrac{3 - b}{2}</cmath>
 +
<cmath>(3) s + t = \dfrac{-a - b}{2}</cmath>
 +
<cmath>(4) rt = \dfrac{-5}{2}</cmath>
 +
<cmath>(5) rs = \dfrac{c - 7}{2}</cmath>
 +
<cmath>(6) st = \dfrac{c + 2}{2}</cmath>
  
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/BQlab3vjjxw ~ CNCM
 
https://youtu.be/BQlab3vjjxw ~ CNCM
 
==See Also==
 
==See Also==

Revision as of 18:28, 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

See Also

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