2025 AMC 12A Problems/Problem 12

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, $\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{-b/a}{c/a} = \frac{-b}{c},$ 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 $\frac{1}{\frac{1}{4050}\cdot \frac{-4\cdot 2025}{3}} = \boxed{-\frac{3}{2}}.$

~lprado

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

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

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search