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| The <imath>\textit{harmonic\ mean}</imath> 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
| | #redirect [[2025 AMC 12A Problems/Problem 12]] |
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| <cmath>\frac{1}{\frac{1}{3}(\frac{1}{4}+\frac{1}{4}+\frac{1}{5})}=\frac{30}{7}</cmath>
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| What is the harmonic mean of all the real roots of the 4050th degree polynomial
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| <cmath> \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) ?</cmath>
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| <imath>\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}</imath>
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| ==Video Solution==
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| https://youtu.be/CCYoHk2Af34
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| ==Solution 1==
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| Let the polynomial be <imath>f(x),</imath> and denote the <imath>4050</imath> roots to be <imath>x_1,x_2,...,x_{4050}.</imath> Hence, <cmath>HM = \dfrac{4050}{\frac{1}{x_1}+\frac{1}{x_2}+...\frac{1}{x_{4050}}}.</cmath>
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| We can multiply the numerator and denominator of this fraction by <imath>x_1x_2...x_{4050}</imath> to create symmetric sums, which yields <cmath>HM = \dfrac{4050(x_1x_2...x_{4050})}{x_2x_3...x_{4050}+x_1x_3...x_{4050}+...+x_1x_2...x_{4049}}.</cmath>
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| By Vieta's Formulas, since <imath>f(x)</imath> is of even degree, the product of its roots, <imath>x_1x_2...x_{4050},</imath> is just the constant term of <imath>f(x),</imath> call it <imath>c_0.</imath>
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| Likewise, the denominator of our harmonic mean, <imath>x_2x_3...x_{4050}+x_1x_3...x_{4050}+...+x_1x_2...x_{4049},</imath> is the negated coefficient of <imath>x</imath> in the standard form of <imath>f(x).</imath> Let the coefficient of <imath>x</imath> in the standard form of <imath>f(x)</imath> be <imath>c_1.</imath>
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| Note that we do not have to worry about dividing by the coefficient of <imath>x^{4050}</imath> when using Vieta's Formulas, as they will eventually cancel out in our harmonic mean calculations.
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| So, <cmath>HM = \dfrac{4050c_0}{-c_1}.</cmath>
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| The constant term in <imath>f(x)</imath> is just <imath>c_0=(-3)^{2025}.</imath> For the coefficient of the <imath>x</imath> term in <imath>f(x),</imath> there are <imath>\dbinom{2025}{2024}=2025</imath> ways to choose <imath>2024</imath> of the trinomials to include a <imath>-3,</imath> and the one trinomial not chosen will include a <imath>-4x.</imath> Hence, <imath>c_1=2025\cdot (-3)^{2024}\cdot (-4).</imath>
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| Finally, <cmath>HM=\dfrac{4050\cdot(-3)^{2025}}{-2025\cdot (-3)^{2024}\cdot (-4)}=\dfrac{2\cdot(-3)}{-(-4)}=\boxed{\text{(B) }-\dfrac{3}{2}}.</cmath>
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| ~Tacos_are_yummy
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| ==Solution 2==
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| Let the \(4050\) roots be represented as <imath> x_i </imath>, with <imath>1 \le i \le 2025</imath>. Each of the \(2025\) quadratics' roots are, using the quadratic formula:
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| <cmath>\frac{2\pm\sqrt{4+3k}}k</cmath>
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| The harmonic mean is therefore:
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| <cmath>\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}} } }</cmath>
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| Further simplifying the expression inside, we get:
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| <cmath>\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</cmath>
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| Substituting back in, we get <imath>\frac{1}{\frac1{4050} \cdot 2025 (-\frac43)}</imath>, so therefore the resultant answer is <imath>\boxed{\textbf{(B) }-\frac32}</imath>
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| Note: there appears to be an edit conflict... this is my first edit conflict merge, so I hope I didn't mess anything up.
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| ~math660
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| ~minor <imath>\LaTeX</imath> edits by i_am_not_suk_at_math (saharshdevaraju 07:53, 7 November 2025 (EST)saharshdevaraju)
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| ==Solution 3 ==
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| Let the roots of <imath>k x^2 - 4x - 3</imath> be <imath>r_1</imath> and <imath>r_2</imath>.
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| <imath>\dfrac{1}{r_1}+\dfrac{1}{r_2}=\dfrac{r_2+r_1}{r_1 r_2}</imath>.
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| By Vieta's, <imath>r_2+r_1=\dfrac{4}{k}</imath>, <imath>r_1 r_2=-\dfrac{3}{k}</imath>, <imath>\dfrac{r_2+r_1}{r_1 r_2}=-\dfrac{4}{3}</imath>.
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| Thus, the arithmetic mean of the reciprocal of all real roots is <imath>\dfrac{(-4/3)\cdot 2025}{4050}=\dfrac{-2700}{4050}=-\dfrac{2}{3}</imath> and therefore the harmonic mean is <imath>\boxed{\textbf{(B) }-\dfrac{3}{2}}.</imath>
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| ~pigwash
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| ~minor <imath>\LaTeX</imath> edits by i_am_not_suk_at_math (saharshdevaraju 07:54, 7 November 2025 (EST)saharshdevaraju)
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| ==Solution 4 ==
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| We will use a fact.
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| To find the sum of the reciprocals of the roots of a polynomial, we reverse it and find the sum of roots.
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| So we have <imath>-3x^2-4x+k</imath> where k is any value from 1 to 2025.
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| Finding the sum of roots, we get <imath>-4/3</imath> is the sum of roots.
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| HM Formula:
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| There are 4050 roots so this is our numerator.
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| The denominator is the sum of possible values, which is <imath>2025 * \frac{-4}{3}</imath>
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| So, using HM formula we get <imath>\frac{4050}{2025 \cdot (-\frac{4}{3})} = 2 \cdot -(3/4) = \boxed{\text{(B) }-\dfrac{3}{2}}</imath>
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| ~Aarav22
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| ~Minor edits for latex by JerryZYang
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| ==Solution 5 ==
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| Note that when we expand this, the sum of the reciprocals of the roots is <imath>-\text{x coefficient}/\text{constant}</imath>. The coefficient of <imath>x^2</imath> does not affect either of these, so we can assume that all of them are just <imath>x^2-4x+3</imath>, then just plug into formula to get <imath>-\frac{3}{2}</imath>
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| ==Solution 6 (5 Second Cheese)==
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| The harmonic mean of <imath>2</imath> numbers, <imath>a</imath> and <imath>b</imath> is <imath>\dfrac{2ab}{a+b}</imath>. So, in terms of the roots of the polynomial <imath>k^2-4x-3</imath>, we have the roots, <imath>a</imath> and <imath>b</imath> to multiply to <imath>-\dfrac{3}{k}</imath> so <imath>-\dfrac{6}{k}</imath>, and they sum to <imath>\dfrac{4}{k}</imath>. So, our answer is simply <cmath>
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| -\dfrac{\dfrac{6}{k}}{\dfrac{4}{k}} = -\dfrac{3}{2} \Longrightarrow \boxed{\text{(B)}}.
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| </cmath>
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| -jb2015007
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| == Solution 7 ==
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| To find the reciprocals of the roots of a polynomial, we can reverse the order of its coefficients.
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| More formally, if
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| \[
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| a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0
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| \]
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| has roots \( r_1, r_2, \ldots, r_n \), then
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| \[
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| a_0x^n + a_1x^{n-1} + \cdots + a_{n-1}x + a_n = 0
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| \]
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| has roots \( \frac{1}{r_1}, \frac{1}{r_2}, \ldots, \frac{1}{r_n} \).
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| Applying this to the given quadratic, each polynomial becomes
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| \[
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| -3x^2 - 4x + k = 0.
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| \]
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| By Vieta's formulas, the sum of the roots of this quadratic is
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| \[
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| \frac{-(-4)}{-3} = -\frac{4}{3}.
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| \]
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| Since each of the \(2025\) quadratics has the same sum of roots, the average of these sums is also
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| \[
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| -\frac{4}{3}.
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| \]
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| The problem asks for the harmonic mean of the original roots, which requires taking the reciprocal of this average. Thus, the final answer is
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| \[
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| -\frac{3}{2}.
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| -VedAR
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| \]
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| == Video Solution ==
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| https://youtu.be/hobODkqQG_s?si=WNv3sfXjIONcJh-M ~ Pi Academy
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| ==Video Solution by SpreadTheMathLove==
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| https://www.youtube.com/watch?v=dAeyV60Hu5c
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| ==Video Solution==
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| https://youtu.be/gWSZeCKrOfU?si=MOaTfgfdEf74aRdq&t=2834
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| ~MK
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| ==See Also==
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| {{AMC10 box|year=2025|ab=A|num-b=17|num-a=19}}
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| * [[AMC 10]]
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| * [[AMC 10 Problems and Solutions]]
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| * [[Mathematics competitions]]
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| * [[Mathematics competition resources]]
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| {{MAA Notice}}
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