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2025 AMC 10A Problems/Problem 18: Difference between revisions

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-jb2015007
-jb2015007
==Solution 7 ==  
==Solution 7 (Simple Concept)==  
To obtain a polynomial whose roots are the reciprocals of the roots of a given polynomial, we reverse the order of the coefficients.   
To obtain a polynomial whose roots are the reciprocals of the roots of a given polynomial, we reverse the order of the coefficients.   
That is, if
That is, if
Line 96: Line 96:
<cmath>-\frac{4}{3}.</cmath>
<cmath>-\frac{4}{3}.</cmath>


The problem asks for the \emph{harmonic mean} of the original roots, which is the reciprocal of this average. Therefore the desired value is
The problem asks for the <imath>\textit{harmonic\ mean}</imath> of the original roots, which is the reciprocal of this average. Therefore the desired value is
<cmath>-\frac{3}{2}.</cmath>
<cmath>-\frac{3}{2}.</cmath>
\[
\boxed{-\frac{3}{2}}
\]
-VedAR
-VedAR


Revision as of 00:37, 8 November 2025

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

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

Video Solution

https://youtu.be/CCYoHk2Af34

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

Solution 2

Let the \(4050\) roots be represented as $x_i$, with $1 \le i \le 2025$. Each of the \(2025\) quadratics' roots are, using the quadratic formula: \[\frac{2\pm\sqrt{4+3k}}k\] The harmonic mean is therefore: \[\frac1{\frac1{4050} \cdot \sum_{k=1}^{2025} { x_{2k-1} + x_{2k} } } = \frac1{\frac1{4050} \cdot \sum_{k=1}^{2025} { \frac{k}{2+\sqrt{4+3k}} + \frac{k}{2-\sqrt{4-3k}} } }\] Further simplifying the expression inside, we get: \[\frac{k}{2+\sqrt{4+3k}} + \frac{k}{2-\sqrt{4-3k}} = \frac{ k(2+\sqrt{4+3k}) +k(2-\sqrt{4+3k}) }{(2+\sqrt{4+3k})(2-\sqrt{4+3k})} = \frac{4k}{2^2 - (4+3k)} = \frac{4k}{-3k} = -\frac43\] Substituting back in, we get $\frac{1}{\frac1{4050} \cdot 2025 (-\frac43)}$, so therefore the resultant answer is $\boxed{\textbf{(B) }-\frac32}$

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

~math660

~minor $\LaTeX$ edits by i_am_not_suk_at_math (saharshdevaraju 07:53, 7 November 2025 (EST)saharshdevaraju)

Solution 3

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 $\boxed{\textbf{(B) }-\dfrac{3}{2}}.$

~pigwash

~minor $\LaTeX$ edits by i_am_not_suk_at_math (saharshdevaraju 07:54, 7 November 2025 (EST)saharshdevaraju)

Solution 4

We will use a fact. To find the sum of the reciprocals of the roots of a polynomial, we reverse it and find the sum of roots. So we have $-3x^2-4x+k$ where k is any value from 1 to 2025. Finding the sum of roots, we get $-4/3$ is the sum of roots.

HM Formula: There are 4050 roots so this is our numerator. The denominator is the sum of possible values, which is $2025 * \frac{-4}{3}$ So, using HM formula we get $\frac{4050}{2025 \cdot (-\frac{4}{3})} = 2 \cdot -(3/4) = \boxed{\text{(B) }-\dfrac{3}{2}}$

~Aarav22

~Minor edits for latex by JerryZYang

Solution 5

Note that when we expand this, the sum of the reciprocals of the roots is $-\text{x coefficient}/\text{constant}$. The coefficient of $x^2$ does not affect either of these, so we can assume that all of them are just $x^2-4x+3$, then just plug into formula to get $-\frac{3}{2}$

Solution 6 (5 Second Cheese)

The harmonic mean of $2$ numbers, $a$ and $b$ is $\dfrac{2ab}{a+b}$. So, in terms of the roots of the polynomial $k^2-4x-3$, we have the roots, $a$ and $b$ to multiply to $-\dfrac{3}{k}$ so $-\dfrac{6}{k}$, and they sum to $\dfrac{4}{k}$. So, our answer is simply \[-\dfrac{\dfrac{6}{k}}{\dfrac{4}{k}} = -\dfrac{3}{2} \Longrightarrow \boxed{\text{(B)}}.\]

-jb2015007

Solution 7 (Simple Concept)

To obtain a polynomial whose roots are the reciprocals of the roots of a given polynomial, we reverse the order of the coefficients. That is, if \[a_nx^2 + a_{n-1}x + a_0 = 0\] has roots \(r_1, r_2\), then \[a_0x^2 + a_{n-1}x + a_n = 0\] has roots \(\frac{1}{r_1}, \frac{1}{r_2}\).

In the problem, each quadratic becomes \[-3x^2 - 4x + k = 0.\]

By Vieta’s Formula, the sum of the roots is \[-\frac{-4}{-3} = -\frac{4}{3}.\]

There are \(2025\) such quadratics, and each one has the same sum of roots. Thus, the average of these sums is also \[-\frac{4}{3}.\]

The problem asks for the $\textit{harmonic\ mean}$ of the original roots, which is the reciprocal of this average. Therefore the desired value is \[-\frac{3}{2}.\] -VedAR

Video Solution

https://youtu.be/hobODkqQG_s?si=WNv3sfXjIONcJh-M ~ Pi Academy

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution

https://youtu.be/gWSZeCKrOfU?si=MOaTfgfdEf74aRdq&t=2834 ~MK

See Also

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 10 Problems and Solutions

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