2025 AMC 10A Problems/Problem 18

(Problem goes here)

The $\textit{harmonic\ mean}$ of a collection of numbers is the reciprocal of the arithmetic mean of the reciprocals of the numbers in the collection. For example, the harmonic mean of 4, 4, and 5 is

$\frac{1}{\frac{1}{3}(\frac{1}{4}+\frac{1}{4}+\frac{1}{5})}=\frac{30}{7}$

What is the harmonic mean of all the real roots of the 4050th degree polynomial

$\prod_{k=1}^{2025} (kx^2-4x-3) = (x^2-4x-3)(2x^2-4x-3)(3x^2-4x-3)\dots (2025x^2-4x-3) ?$ $\textbf{(A) } -\frac{5}{3} \qquad\textbf{(B) } -\frac{3}{2} \qquad\textbf{(C) } -\frac{6}{5} \qquad\textbf{(D) } -\frac{5}{6} \qquad\textbf{(E) } -\frac{2}{3}$

Solution 1

Let the polynomial be $f(x),$ and denote the $4050$ roots to be $x_1,x_2,...,x_{4050}.$ Hence, $HM = \dfrac{4050}{\frac{1}{x_1}+\frac{1}{x_2}+...\frac{1}{x_{4050}}}.$ We can multiply the numerator and denominator of this fraction by $x_1x_2...x_{4050}$ to create symmetric sums, which yields $HM = \dfrac{4050(x_1x_2...x_{4050})}{x_2x_3...x_{4050}+x_1x_3...x_{4050}+...+x_1x_2...x_{4049}}.$

By Vieta's Formulas, since $f(x)$ is of even degree, the product of its roots, $x_1x_2...x_{4050},$ is just the constant term of $f(x),$ call it $c_0.$ Likewise, the denominator of our harmonic mean, $x_2x_3...x_{4050}+x_1x_3...x_{4050}+...+x_1x_2...x_{4049},$ is the negated coefficient of $x$ in the standard form of $f(x).$ Let the coefficient of $x$ in the standard form of $f(x)$ be $c_1.$ Note that we do not have to worry about dividing by the coefficient of $x^{4050}$ when using Vieta's Formulas, as they will eventually cancel out in our harmonic mean calculations.

So, $HM = \dfrac{4050c_0}{-c_1}.$

The constant term in $f(x)$ is just $c_0=(-3)^{2025}.$ For the coefficient of the $x$ term in $f(x),$ there are $\dbinom{2025}{2024}=2025$ ways to choose $2024$ of the trinomials to include a $-3,$ and the one trinomial not chosen will include a $-4x.$ Hence, $c_1=2025\cdot (-3)^{2024}\cdot (-4).$

Finally, $HM=\dfrac{4050\cdot(-3)^{2025}}{-2025\cdot (-3)^{2024}\cdot (-4)}=\dfrac{2\cdot(-3)}{-(-4)}=\boxed{\text{(B) }-\dfrac{3}{2}}.$

~Tacos_are_yummy

Note: this solution is massively massive. Solution 2 is faster

Alternate Solution

Let the 4050 roots be in $x_i$ . Each of the 2025 quadratics' roots are, using the quadratic formula, $\frac{2\pm\sqrt{4+3k}}k$ . The harmonic mean is therefore $\frac1{\frac1{4050} \cdot \sum{ \frac{k}{2+\sqrt{4+3k}} + \frac{k}{2-\sqrt{4-3k}} } }$ . Further simplifying, $\frac{k}{2+\sqrt{4+3k}} + \frac{k}{2-\sqrt{4-3k}} = \frac{ k(2+\sqrt{4+3k}) +k(2-\sqrt{4+3k}) }{(2+\sqrt{4+3k})(2-\sqrt{4+3k})} = \frac{4k}{2^2 - (4+3k)} = \frac{4k}{-3k} = -\frac43$ . Substituting back in, we get $\frac{1}{\frac1{4050} \cdot 2025 (-\frac43)}$ , so therefore the resultant answer is $-\frac32$

Note 2: there appears to be an edit conflict... this is my first edit conflict merge, so I hope I didn't mess anything up

~math660

Solution 2 (Fast)

Let the roots of $k x^2 - 4x - 3$ be $r_1$ and $r_2$ . $\dfrac{1}{r_1}+\dfrac{1}{r_2}=\dfrac{r_2+r_1}{r_1 r_2}$ . By Vieta's, $r_2+r_1=\dfrac{4}{k}$ , $r_1 r_2=-\dfrac{3}{k}$ , $\dfrac{r_2+r_1}{r_1 r_2}=-\dfrac{4}{3}$ . Thus, the arithmetic mean of the reciprocal of all real roots is $\dfrac{(-4/3)\cdot 2025}{4050}=\dfrac{-2700}{4050}=-\dfrac{2}{3}$ and therefore the harmonic mean is $\boxed{\text{(B) }-\dfrac{3}{2}}.$

~pigwash

Page

Toolbox

Search

2025 AMC 10A Problems/Problem 18

Solution 1

Alternate Solution

Solution 2 (Fast)