2020 AIME II Problem 11

Question

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.

Answer

Answer: 71

Concepts:polynomial Vieta’s Formulas system of equations

Difficulty rating: 2920

Small hint:

Each pairwise sum has leading coefficient

2,

so it equals

2

times the product of the two linear factors from its known roots

Big hint:

Adding the root sums of

P+Q

and

P+R

and subtracting that of

Q+R

isolates twice the root shared by the first two sums

Solution:

Let

a

be the common root of

P+Q

and

P+R,

let

b

be that of

P+Q

and

Q+R,

and let

c

be that of

P+R

and

Q+R.

Each sum is quadratic with leading coefficient

2,

so

P+Q = 2(x-a)(x-b),

P+R = 2(x-a)(x-c),

and

Q+R = 2(x-b)(x-c).

Write

Q = x^2 + qx + 2

and

R = x^2 + rx + s.

By Vieta’s formulas, the root sums are

a + b = \frac{3-q}{2},

a + c = \frac{3-r}{2},

and

b + c = -\frac{q+r}{2},

so

\begin{aligned} 2a &= (a+b) + (a+c) - (b+c) \\ &= \frac{3-q}{2} + \frac{3-r}{2} \\ &\quad {}+ \frac{q+r}{2} = 3, \end{aligned}

giving

a = \frac{3}{2}.

The constant term of

P + Q

is

-7 + 2 = -5,

so

2ab = -5

and

b = -\frac{5}{3}.

Also

2ac = P(0) + R(0) = s - 7

and

2bc = Q(0) + R(0) = s + 2.

Subtracting,

2c(b - a) = 9,

so

c = \frac{9}{2\left(-\frac{5}{3} - \frac{3}{2}\right)} = -\frac{27}{19}.

Then

s = 2bc - 2

= 2 \cdot \left(-\frac{5}{3}\right)\left(-\frac{27}{19}\right) - 2

= \frac{90}{19} - \frac{38}{19} = \frac{52}{19},

so

R(0) = \frac{52}{19}

and

m + n = 52 + 19 = 71.