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

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 $p=q$ .

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2025 AMC 12A Problems/Problem 12

Problem

Solution 1

Note