2025 AMC 12A Problems/Problem 12: Difference between revisions
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== Solution 3 (Vieta's only) == | == Solution 3 (Vieta's only) == | ||
We are asked to find <imath>\frac{4050}{\sum_{i=1}^{4050}\frac{1}{i_1}}</imath>. By Vieta's, note that <imath>r_1 \cdot r_2 \dots r_{4050} = -\frac{3^{2025}}{2025!} = k</imath> (<imath>k</imath> is a constant). Then, note that we are asked to find <imath>\frac{4050}{\sum_{i=1}^n \frac{\frac{k}{r_i}}{k}}</imath>, and by Vieta's we get that <imath>sum_{i=1}^n \frac{k}{r_i} = \frac{2025 \cdot 4 \cdot 3^{2024}}{2025!}</imath>, so substituting this in, we get <imath>-\frac{4050}{\frac{\frac{2025 \cdot 4 \cdot 3^{2024}}{2025!}}{\frac{3^{2025}}{2025!}}} = -\frac{4050 \cdot 3^{2025}}{4050 \cdot 3^{2024} \cdot 2} = -\frac{3}{2}</imath>. | |||
~ScoutViolet | |||
== Note == | == Note == | ||
Revision as of 19:31, 6 November 2025
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 
degree polynomial![\[\prod_{k = 1}^{2025} (kx^{2} - 4x - 3) = (x^{2} - 4x - 3)(2x^{2} - 4x - 3)(3x^{2} - 4x - 3) \cdots (2025x^{2} - 4x - 3)?\]](http://latex-new.aopstest.com/d/3/8/d389b7c2925219f8cd8d7f9b0791c9bf4c1aebf0.png)
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 
of the quadratic 
, we use Vieta's formulas. Recall that 
and 
. Therefore, ![\[\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{-b/a}{c/a} = \frac{-b}{c},\]](http://latex-new.aopstest.com/e/f/6/ef6e050c17f57e240479ac3a1290e33129dafa38.png)
which doesn't depend on 
.
The sum of the reciprocals of the roots of the quadratic 
is 
The same is true for every quadratic in the form 
. The sum of all the reciprocals of the roots of is 
Because we have 
quadratics, there are 
total roots. Our answer is ![\[\frac{1}{\frac{1}{4050}\cdot \frac{-4\cdot 2025}{3}} = \boxed{-\frac{3}{2}}.\]](http://latex-new.aopstest.com/b/5/d/b5d88b2eeaea06eb0abee83bf056178d6076d0a3.png)
~lprado
Solution 2 (similar to solution 1 but with quadratic formula)
We first find the general roots for quadratics in the form 
. Using the quadratic formula we have
\begin{align*}
x&=\frac{4 \pm \sqrt{16+12k}}{2k} \\
&= \frac{4\pm 2\sqrt{4+3k}}{2k} \\
&= \frac{2+\sqrt{4+3k}}{k}, \frac{2-\sqrt{4+3k}}{k} \\
\end{align*}
Since we are asked to add the reciprocals of all 
roots in the harmonic mean, we will first add the general roots in terms of 
. We have,
\begin{align*}
\frac{k}{2+\sqrt{4+3k}} + \frac{k}{2-\sqrt{4+3k}} &= \frac{k(2-\sqrt{4+3k}) + k(2+\sqrt{4+3k})}{2^2-(\sqrt{4+3k})^2} \\
&= -\frac{4k}{3k}=-\frac{4}{3}. \\
\end{align*}
Thus, each pair of roots add up to 
, and since there are pairs of roots, the harmonic mean of the desired expression is
\begin{align*}
\frac{1}{\frac{1}{4050} \left (2025 \left (-\frac43 \right ) \right)} &= \frac{1}{\frac12 \left ( -\frac43 \right)} \\
& = \boxed{-\frac32}, \boxed{B}. \\
\end{align*}
~evanhliu2009
Solution 3 (Vieta's only)
We are asked to find 
. By Vieta's, note that 
( is a constant). Then, note that we are asked to find 
, and by Vieta's we get that 
, so substituting this in, we get 
.
~ScoutViolet
Note
It is important to note that the question asks for the sum of all 
roots. We must therefore be careful in making sure that all roots are real and distinct. We can show that they are real because 
for all 
and we can show they are distinct because, if we assume that is a root to both 
and 
we would have 
which implies 
for all 
, which is only possible if 
.
~ Shadowleafy
