AIME 2020 II · 第 11 题

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

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 3 (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

$$P(x) = (P(x) + Q(x) + R(x)) - (Q(x) + R(x))$$ $$= (x-p)(x-q) + (x-p)(x-r) - (x-q)(x-r)$$ $$= x^2 - 2px + (pq + pr - qr).$$ Similarly,

$$Q(x) = x^2 - 2qx + (pq + qr - pr) \text{~ and}$$ $$R(x) = x^2 - 2rx + (pr + qr - pq).$$ 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

Another one:

https://www.youtube.com/watch?v=AXN9x51KzNI