Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2020 AIME II Problems/Problem 11

Revision as of 19:30, 9 June 2020 by Asbodke (talk | contribs) (→‎Solution 3)

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 $pr = \frac{c-7}{2}, qr = \frac{c+2}{2} \implies c = \frac{52}{19} = R(0) \implies \boxed{071}.$

Solution 3

We know that $P(x)=x^2-3x-7$.

Since $Q(0)=2$, the constant term in $Q(x)$ is $2$. Let $Q(x)=x^2+ax+2$.

Finally, let $R(x)=x^2+bx+c$.

$P(x)+Q(x)=2x^2+(a-3)x-5$. Let its roots be $p$ and $q$.

$P(x)+R(x)=2x^2+(b-3)x+(c-7)$ Let its roots be $p$ and $r$.

$Q(x)+R(x)=2x^2+(a+b)x+(c+2)$. Let its roots be $q$ and $r$.

By vietas, $p+q=\frac{3-a}{2}, p+r=\frac{3-b}{2}, q+r=\frac{-(a+b)}{2}$

We could work out the system of equations, but it's pretty easy to see that $p=\frac32, q=-\frac{a}{2}, r=-\frac{b}{2}$.

$\text{Again, by vietas, }pq=-\frac52\text{, } pr=\frac{c-7}{2}\text{, } qr=\frac{c+2}{2}\text{, } \text{multiplying everything together a}\text{nd taking the sqrt of both sides,}$ \[(pqr)^2=\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)\] \[pqr=\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)}\] $\text{Dividing this }\text{equation by }qr=\frac{c+2}{2}$ \[\frac{pqr}{qr}=\frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)}}{\frac{c+2}{2}}\] \[p = \frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}}{\sqrt{\frac{c+2}{2}}}\] $\text{Recall th}\text{at }p=\frac32 \text{ and square both sides}$ \[\frac94=\frac{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}{\frac{c+2}{2}}\] $\text{Solving gives } c=\frac{52}{19}, \text{ so our answer is }\boxed{071}$

~quacker88

Solution 4 (Official MAA)

Let the common root of $P+Q$ and $P+R$ be $p$, the common root of $P+Q$ and $Q+R$ be $q$, and the common root of $Q+R$ and $P+R$ be $r$. Because $p$ and $q$ are both roots of $P+Q$ and $P+Q$ has leading coefficient $2$, it follows that $P(x) + Q(x) = 2(x-p)(x-q).$ Similarly, $P(x) + R(x) = 2(x-p)(x-r)$ and $Q(x) + R(x) = 2(x-q)(x-r)$. Adding these three equations together and dividing by $2$ yields\[P(x) + Q(x) + R(x) = (x-p)(x-q) + (x-p)(x-r) + (x-q)(x-r),\]so \begin{align*} P(x) &= (P(x) + Q(x) + R(x)) - (Q(x) + R(x)) \\ &= (x-p)(x-q) + (x-p)(x-r) - (x-q)(x-r) \\ &= x^2 - 2px + (pq + pr - qr). \end{align*}Similarly, \begin{align*} Q(x) &= x^2 - 2qx + (pq + qr - pr) \text{~ and}\\ R(x) &= x^2 - 2rx + (pr + qr - pq). \end{align*}Comparing the $x$ coefficients yields $p = \tfrac32$, and comparing the constant coefficients yields $-7 = pq + pr - qr = \tfrac32(q+r) - qr$. The fact that $Q(0) = 2$ implies that $\tfrac32(q-r) + qr = 2$. Adding these two equations yields $q = -\tfrac53$, and so substituting back in to solve for $r$ gives $r=-\tfrac{27}{19}$. Finally,\[R(0) = pr + qr - pq = \left(-\frac{27}{19}\right)\left(\frac32-\frac53\right) + \frac52 = \frac{9}{38} + \frac52 = \frac{52}{19}.\]The requested sum is $52 + 19 = 71$. Note that $Q(x) = x^2 + \frac{10}3x + 2$ and $R(x) = x^2 + \frac{54}{19}x + \frac{52}{19}$.

Video Solution

https://youtu.be/BQlab3vjjxw ~ CNCM

See Also

2020 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_11&oldid=124697"