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

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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>.
 
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==
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==Solution 1==
 
Let <math>Q(x) = x^2 + ax + 2</math> and <math>R(x) = x^2 + bx + c</math>.  We can write the following:
 
Let <math>Q(x) = x^2 + ax + 2</math> and <math>R(x) = x^2 + bx + c</math>.  We can write the following:
 
<cmath>P + Q = 2x^2 + (a - 3)x - 5</cmath>
 
<cmath>P + Q = 2x^2 + (a - 3)x - 5</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
 
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>
  
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/BQlab3vjjxw ~ CNCM
 
https://youtu.be/BQlab3vjjxw ~ CNCM
 
==See Also==
 
==See Also==

Revision as of 19:46, 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 1

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

Solution 2

Let $P+Q, Q+R$ have shared root $q$, $Q+R, R+P$ have shared root $r$, and the last pair having shared root $p$. We will now set $Q(x) = x^2+ax+2$, and $R(x) = x^2+bx+c$. We wish to find $c$, and now we compute $P+Q,Q+R,R+P$. \[P+Q = 2x^2+(a-3)x-5 = 2(x-p)(x-q)\] \[Q+R = 2x^2+(a+b)x+(2+c) = 2(x-q)(x-r)\] \[R+P = 2x^2+(b-3)x+(c-7) = 2(x-r)(x-p)\] From here, we equate coefficients. This means $p+q = \frac{3-a}{2}, p+r = \frac{3-b}{2}, q+r = \frac{-a-b}{2} \implies p = \frac{3}{2}$. Now, $pq = \frac{-5}{2} \implies q = -\frac{5}{3}$. Finally, we know that $pr = \frac{c-7}{2}, qr = \frac{c+2}{2} \implies c = \frac{52}{19} = R(0) \implies \boxed{071}.$

Video Solution

https://youtu.be/BQlab3vjjxw ~ CNCM

See Also

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