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2025 AMC 10A Problems/Problem 18

(Problem goes here)

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, \[HM=\dfrac{4050\cdot(-3)^{2025}}{-2025\cdot (-3)^{2024}\cdot (-4)}=\dfrac{2\cdot(-3)}{-(-4)}=\boxed{\text{(A) }-\dfrac{3}{2}}.\]

~Tacos_are_yummy_1

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 $\frac1{\frac1{4050} \cdot \sum{ \frac{k}{2+\sqrt{4+3k}} + \frac{k}{2+\sqrt{4-3k}} } }$. When adding the two opposing roots together, we get $-\frac43$ (my elaboration on this is missing, uh oh), so therefore the resultant answer is $-\frac32$

Note: I apologize for the styling so far... I will add more LaTeX and beautify this once I figure out more syntax

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 $\boxed{\text{(A) }-\dfrac{3}{2}}.$

~pigwash

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