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| (Problem goes here)
| | #redirect [[2025 AMC 12A Problems/Problem 12]] |
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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{(A) }-\dfrac{3}{2}}.</cmath>
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| ~Tacos_are_yummy_1
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| ==Alternate Solution==
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| Let the 4050 roots be in <imath>x_i</imath>. Each of the 2025 quadratics' roots are, using the quadratic formula, <imath>\frac{2\pm\sqrt{4+3k}}k</imath>. The harmonic mean is therefore <imath>\frac1{\frac1{4050} \cdot \sum{ \frac{k}{2+\sqrt{4+3k}} + \frac{k}{2+\sqrt{4-3k}} } }</imath>. When adding the two opposing roots together, we get <imath>-\frac43</imath> (my elaboration on this is missing, uh oh), so therefore the resultant answer is <imath>-\frac32</imath>
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| Note: I apologize for the styling so far... I will add more LaTeX and beautify this once I figure out more syntax
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| 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
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| ~math660
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| ==Solution 2==
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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 real roots is <imath>\dfrac{(-4/3)\cdot 2025}{4050}=\dfrac{-2700}{4050}=-\dfrac{2}{3}</imath> and the harmonic mean is <imath>\boxed{\text{(A) }-\dfrac{3}{2}}.</imath>
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| ~pigwash
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