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− | ==Problem==
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− | Let <math>P(x) = x^2 - 3x - 7</math>, and let <math>Q(x)</math> and <math>R(x)</math> be two quadratic polynomials also with the coefficient of <math>x^2</math> equal to <math>1</math>. David computes each of the three sums <math>P + Q</math>, <math>P + R</math>, and <math>Q + R</math> and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If <math>Q(0) = 2</math>, then <math>R(0) = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
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− | ==Solution 1==
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− | Let <math>Q(x) = x^2 + ax + 2</math> and <math>R(x) = x^2 + bx + c</math>. We can write the following:
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− | <cmath>P + Q = 2x^2 + (a - 3)x - 5</cmath>
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− | <cmath>P + R = 2x^2 + (b - 3)x + (c - 7)</cmath>
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− | <cmath>Q + R = 2x^2 + (a + b)x + (c + 2)</cmath>
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− | 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>.
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− | By Vieta's, we have:
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− | <cmath> r + t = \dfrac{3 - a}{2}\tag{1}</cmath>
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− | <cmath>r + s = \dfrac{3 - b}{2}\tag{2}</cmath>
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− | <cmath>s + t = \dfrac{-a - b}{2}\tag{3}</cmath>
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− | <cmath>rt = \dfrac{-5}{2}\tag{4}</cmath>
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− | <cmath>rs = \dfrac{c - 7}{2}\tag{5}</cmath>
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− | <cmath>st = \dfrac{c + 2}{2}\tag{6}</cmath>
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− | 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
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− | ==Solution 2==
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− | Let <math>P+Q, Q+R</math> have shared root <math>q</math>, <math>Q+R, R+P</math> have shared root <math>r</math>, and the last pair having shared root <math>p</math>. We will now set <math>Q(x) = x^2+ax+2</math>, and <math>R(x) = x^2+bx+c</math>. We wish to find <math>c</math>, and now we compute <math>P+Q,Q+R,R+P</math>.
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− | <cmath>P+Q = 2x^2+(a-3)x-5 = 2(x-p)(x-q)</cmath>
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− | <cmath>Q+R = 2x^2+(a+b)x+(2+c) = 2(x-q)(x-r)</cmath>
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− | <cmath>R+P = 2x^2+(b-3)x+(c-7) = 2(x-r)(x-p)</cmath>
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− | From here, we equate coefficients. This means <math>p+q = \frac{3-a}{2}, p+r = \frac{3-b}{2}, q+r = \frac{-a-b}{2} \implies p = \frac{3}{2}</math>. Now, <math>pq = \frac{-5}{2} \implies q = -\frac{5}{3}</math>. Finally, we know that <math>pr = \frac{c-7}{2}, qr = \frac{c+2}{2} \implies c = \frac{52}{19} = R(0) \implies \boxed{071}.</math>
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| ==Solution 3== | | ==Solution 3== |
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| ~quacker88 | | ~quacker88 |
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− | ==Video Solution==
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− | https://youtu.be/BQlab3vjjxw ~ CNCM
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− | ==See Also==
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− | {{AIME box|year=2020|n=II|num-b=10|num-a=12}}
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− | {{MAA Notice}}
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