2025 AMC 12A Problems/Problem 12

The following problem is from both the 2025 AMC 10A #18 and 2025 AMC 12A #12, so both problems redirect to this page.

Problem

The harmonic mean of a collection of real numbers is the reciprocal of the arithmetic mean of the reciprocals of the numbers in the collection. What is the harmonic mean of all real roots of the $4050^{\text{th}}$ degree polynomial $\prod_{k = 1}^{2025} (kx^{2} - 4x - 3) = (x^{2} - 4x - 3)(2x^{2} - 4x - 3)(3x^{2} - 4x - 3) \cdots (2025x^{2} - 4x - 3)?$ $\textbf{(A)}~-\frac{5}{3} \qquad \textbf{(B)}~-\frac{3}{2} \qquad \textbf{(C)}~-\frac{6}{5} \qquad \textbf{(D)}~-\frac{5}{6} \qquad \textbf{(E)}~-\frac{2}{3}$

Solution 1

We will need to determine the sum of the reciprocals of the roots. To find the sum of the reciprocals of the roots $p,q$ of the quadratic $ax^2+bx+c$ , we use Vieta's formulas. Recall that $p+q = -b/a$ and $pq = c/a$ . Therefore, $\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{\frac{-b}{a}}{\frac{c}{a}} = \frac{-b}{a} \cdot \frac{a}{c} = \frac{-b}{c},$ which doesn't depend on $a$ .

The sum of the reciprocals of the roots of the quadratic $x^2-4x-3$ is $\frac{-(-4)}{-3} = -4/3.$ The same is true for every quadratic in the form $ax^2-4x-3$ . The sum of all the reciprocals of the roots of $ax^2+bx+c$ is $2025 \cdot \left(-\frac{4}{3}\right).$

Because we have $2025$ quadratics, there are $2 \cdot 2025 = 4050$ total roots. Our answer is $\frac{1}{\frac{1}{4050}\cdot \frac{-4\cdot 2025}{3}} =$ $\boxed{\text{(B) \frac{3}{2}}}.$ (Error compiling LaTeX. Unknown error_msg)

~lprado

~some edits by i_am_not_suk_at_math (saharshdevaraju 13:26, 8 November 2025 (EST)saharshdevaraju)

Solution 2 (similar to solution 1 but with quadratic formula)

We first find the general roots for quadratics in the form $kx^2-4x-3$ . Using the quadratic formula we have \begin{align*} x&=\frac{4 \pm \sqrt{16+12k}}{2k} \\ &= \frac{4\pm 2\sqrt{4+3k}}{2k} \\ &= \frac{2+\sqrt{4+3k}}{k}, \frac{2-\sqrt{4+3k}}{k} \\ \end{align*} Since we are asked to add the reciprocals of all $4050$ roots in the harmonic mean, we will first add the general roots in terms of $k$ . We have, \begin{align*} \frac{k}{2+\sqrt{4+3k}} + \frac{k}{2-\sqrt{4+3k}} &= \frac{k(2-\sqrt{4+3k}) + k(2+\sqrt{4+3k})}{2^2-(\sqrt{4+3k})^2} \\ &= -\frac{4k}{3k}=-\frac{4}{3}. \\ \end{align*} Thus, each pair of roots add up to $-\frac{4}{3}$ , and since there are $2025$ pairs of roots, the harmonic mean of the desired expression is \begin{align*} \frac{1}{\frac{1}{4050} \left (2025 \left (-\frac43 \right ) \right)} &= \frac{1}{\frac12 \left ( -\frac43 \right)} \\ & = \boxed{-\frac32}, \boxed{B}. \\ \end{align*}

~evanhliu2009

Solution 3 (Vieta's only)

We are asked to find $\frac{4050}{\sum_{i=1}^{4050}\frac{1}{i_1}}$ . By Vieta's, note that $r_1 \cdot r_2 \dots r_{4050} = -\frac{3^{2025}}{2025!} = k$ ( $k$ is a constant). Then, note that we are asked to find $\frac{4050}{\sum_{i=1}^n \frac{\frac{k}{r_i}}{k}}$ , and by Vieta's we get that $\sum_{i=1}^n \frac{k}{r_i} = \frac{2025 \cdot 4 \cdot 3^{2024}}{2025!}$ , so substituting this in, we get $-\frac{4050}{\frac{\frac{2025 \cdot 4 \cdot 3^{2024}}{2025!}}{\frac{3^{2025}}{2025!}}} = -\frac{4050 \cdot 3^{2025}}{4050 \cdot 3^{2024} \cdot 2} = \boxed{-\frac{3}{2}}$ .

~ScoutViolet

Solution 4 (Vieta's Theorem)

We have: $\prod_{k = 1}^{2025} (kx^{2} - 4x - 3) = (x^{2} - 4x - 3)(2x^{2} - 4x - 3)(3x^{2} - 4x - 3) \cdots (2025x^{2} - 4x - 3)$

So we can analyze each part of this "snake":

According to Vieta's theorem:

For the first one: $x^{2} - 4x - 3$ : We have: $\begin{cases} x_1+x_2=\tfrac{4}{1}\\ x_1x_2=-\tfrac{3}{1} \end{cases}$ So we have: $\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{x_1x_2}=-\dfrac{4}{3}$

For the Second one: $2x^{2} - 4x - 3$ : We have: $\begin{cases} x_3+x_4=\tfrac{4}{2}\\ x_3x_4=-\tfrac{3}{2} \end{cases}$ So we have: $\dfrac{1}{x_3}+\dfrac{1}{x_4}=\dfrac{x_3+x_4}{x_3x_4}=-\dfrac{4}{3}$

For the Third one: $3x^{2} - 4x - 3$ : We have: $\begin{cases} x_5+x_6=\tfrac{4}{3}\\ x_5x_6=-\tfrac{3}{3} \end{cases}$ So we have: $\dfrac{1}{x_5}+\dfrac{1}{x_6}=\dfrac{x_5+x_6}{x_5x_6}=-\dfrac{4}{3}$

......

Then we should check the last one. For the 2025th. one: $2025x^{2} - 4x - 3$ : We have: $\begin{cases} x_{4049}+x_{4050}=\tfrac{4}{2025}\\ x_{4049}x_{4050}=-\tfrac{3}{2025} \end{cases}$ So we have: $\dfrac{1}{x_{4049}}+\dfrac{1}{x_{4050}}=\dfrac{x_{4049}+x_{4050}}{x_{4049}x_{4050}}=-\dfrac{4}{3}$ $\begin{align*} \therefore H.M.&=\dfrac{1}{\dfrac{1}{4050}\sum_{i=1}^{4050}\dfrac{1}{x_i}}\\ &=\dfrac{4050}{2025\times\left(-\dfrac{4}{3}\right)}\\ &=-\dfrac{3}{2} \end{align*}$ Therefore, the final answer is $\color{red}\boxed{\color{black}-\dfrac{3}{2}}\color{black}$ .

~funkCCP

Solution 5 (Vietas Formulas)

For a general quadratic polynomial $ax^2 + bx + c = 0$ with roots $p$ and $q$ , the sum of the reciprocals is: \begin{align*} \frac{1}{p} + \frac{1}{q} &= \frac{p+q}{pq} \\ \end{align*} Using Vieta's formulas ( $p+q = -\frac{b}{a}$ and $pq = \frac{c}{a}$ ): \begin{align*} \frac{1}{p} + \frac{1}{q} &= \frac{-\frac{b}{a}}{\frac{c}{a}} = -\frac{b}{c} \end{align*} This result demonstrates that the sum of the reciprocals depends only on the coefficients $b$ and $c$ , and is independent of the leading coefficient $a$ . Since the overall polynomial is a product of quadratics, and all these quadratics share the same $b$ and $c$ coefficients, the sum of the reciprocals for the roots of every individual quadratic is the same. We use the coefficients from one of the quadratics, which is $ax^2 - 4x - 3 = 0$ . Thus, $b=-4$ and $c=-3$ .

$\sum (\text{reciprocals}) = -\frac{b}{c} = -\frac{(-4)}{(-3)} = -\frac{4}{3}$

We substitute the calculated sum of the reciprocals into the formula for the harmonic mean of 2 numbers:

$\text{HM} = \frac{2}{-\frac{4}{3}} = 2 \times \left(-\frac{3}{4}\right) = -\frac{6}{4} = -\frac{3}{2}$

The harmonic mean of the roots of any single quadratic is $-\frac{3}{2}$ . Since the sum of the reciprocals is the same for all quadratics, the harmonic mean of all the roots of the overall polynomial is also $\boxed{\textbf{(B) }\dfrac{-3}{2}}$ .

~Voidling

~small type fix by i_am_not_suk_at_math (saharshdevaraju 13:39, 8 November 2025 (EST)saharshdevaraju)

\item{\frac{1}{5} isn’t even a choice!}

Note

It is important to note that the question asks for the sum of all $\textbf{real}$ roots. We must therefore be careful in making sure that all roots are real and distinct. We can show that they are real because $16+12k>0$ for all $k>0$ and we can show they are distinct because, if we assume that $a$ is a root to both $px^2-4x-3$ and $qx^2-4x-3$ we would have $px^2-4x-3=qx^2-4x-3=0$ which implies $px^2=qx^2$ for all $x$ , which is only possible if $p=q$ .

~ Shadowleafy

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