2025 AMC 10A Problems/Problem 18: Difference between revisions
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== Solution 7 == | |||
To find the reciprocals of the roots of a polynomial, we can reverse the order of its coefficients. | |||
More formally, if | |||
\[ | |||
a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0 | |||
\] | |||
has roots \( r_1, r_2, \ldots, r_n \), then | |||
\[ | |||
a_0x^n + a_1x^{n-1} + \cdots + a_{n-1}x + a_n = 0 | |||
\] | |||
has roots \( \frac{1}{r_1}, \frac{1}{r_2}, \ldots, \frac{1}{r_n} \). | |||
Applying this to the given quadratic, each polynomial becomes | |||
\[ | |||
-3x^2 - 4x + k = 0. | |||
\] | |||
By Vieta's formulas, the sum of the roots of this quadratic is | |||
\[ | |||
\frac{-(-4)}{-3} = -\frac{4}{3}. | |||
\] | |||
- | Since each of the \(2025\) quadratics has the same sum of roots, the average of these sums is also | ||
\[ | |||
-\frac{4}{3}. | |||
\] | |||
The problem asks for the harmonic mean of the original roots, which requires taking the reciprocal of this average. Thus, the final answer is | |||
\[ | |||
-\frac{3}{2}. | |||
-VedAR | |||
\] | |||
== Video Solution == | == Video Solution == | ||
https://youtu.be/hobODkqQG_s?si=WNv3sfXjIONcJh-M ~ Pi Academy | https://youtu.be/hobODkqQG_s?si=WNv3sfXjIONcJh-M ~ Pi Academy | ||
Revision as of 00:28, 8 November 2025
The 
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
![\[\frac{1}{\frac{1}{3}(\frac{1}{4}+\frac{1}{4}+\frac{1}{5})}=\frac{30}{7}\]](http://latex-new.aopstest.com/5/f/d/5fd6cd73ccdeb73722a54567e68c74cc24250e6b.png)
What is the harmonic mean of all the real roots of the 4050th degree polynomial
![\[\prod_{k=1}^{2025} (kx^2-4x-3) = (x^2-4x-3)(2x^2-4x-3)(3x^2-4x-3)\dots (2025x^2-4x-3) ?\]](http://latex-new.aopstest.com/1/f/5/1f596b6d7721a11d1a1baf9895bb679f92572423.png)
Video Solution
Solution 1
Let the polynomial be 
and denote the 
roots to be 
Hence, ![\[HM = \dfrac{4050}{\frac{1}{x_1}+\frac{1}{x_2}+...\frac{1}{x_{4050}}}.\]](http://latex-new.aopstest.com/d/a/a/daa3d1403405d1515bdc59c10988a3643a2e797d.png)
We can multiply the numerator and denominator of this fraction by 
to create symmetric sums, which yields ![\[HM = \dfrac{4050(x_1x_2...x_{4050})}{x_2x_3...x_{4050}+x_1x_3...x_{4050}+...+x_1x_2...x_{4049}}.\]](http://latex-new.aopstest.com/4/3/a/43a5e85a32979fe56d89a686f267e7d77ec04c4f.png)
By Vieta's Formulas, since 
is of even degree, the product of its roots, 
is just the constant term of call it 
Likewise, the denominator of our harmonic mean, 
is the negated coefficient of 
in the standard form of 
Let the coefficient of in the standard form of be 
Note that we do not have to worry about dividing by the coefficient of 
when using Vieta's Formulas, as they will eventually cancel out in our harmonic mean calculations.
So, ![\[HM = \dfrac{4050c_0}{-c_1}.\]](http://latex-new.aopstest.com/8/6/e/86ed69da139cb77370f68b265193eb501e4e6431.png)
The constant term in is just 
For the coefficient of the term in there are 
ways to choose 
of the trinomials to include a 
and the one trinomial not chosen will include a 
Hence, 
Finally, ![\[HM=\dfrac{4050\cdot(-3)^{2025}}{-2025\cdot (-3)^{2024}\cdot (-4)}=\dfrac{2\cdot(-3)}{-(-4)}=\boxed{\text{(B) }-\dfrac{3}{2}}.\]](http://latex-new.aopstest.com/d/3/8/d383798941c4cdaf85e3e0fdc93143d50da7c5a3.png)
~Tacos_are_yummy
Solution 2
Let the \(4050\) roots be represented as 
, with 
. Each of the \(2025\) quadratics' roots are, using the quadratic formula:
![\[\frac{2\pm\sqrt{4+3k}}k\]](http://latex-new.aopstest.com/2/c/e/2ce316355840760fcb28a616a6890e5edda6a03e.png)
The harmonic mean is therefore:
![\[\frac1{\frac1{4050} \cdot \sum_{k=1}^{2025} { x_{2k-1} + x_{2k} } } = \frac1{\frac1{4050} \cdot \sum_{k=1}^{2025} { \frac{k}{2+\sqrt{4+3k}} + \frac{k}{2-\sqrt{4-3k}} } }\]](http://latex-new.aopstest.com/1/c/5/1c5c922914ff92e255f1478bfcd9a6f05fccd415.png)
Further simplifying the expression inside, we get:
![\[\frac{k}{2+\sqrt{4+3k}} + \frac{k}{2-\sqrt{4-3k}} = \frac{ k(2+\sqrt{4+3k}) +k(2-\sqrt{4+3k}) }{(2+\sqrt{4+3k})(2-\sqrt{4+3k})} = \frac{4k}{2^2 - (4+3k)} = \frac{4k}{-3k} = -\frac43\]](http://latex-new.aopstest.com/c/1/e/c1e1ade3c2ad14b40411a710cae09ab745c4f19d.png)
Substituting back in, we get 
, so therefore the resultant answer is 
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 
edits by i_am_not_suk_at_math (saharshdevaraju 07:53, 7 November 2025 (EST)saharshdevaraju)
Solution 3
Let the roots of 
be 
and 
.
.
By Vieta's, 
, 
, 
.
Thus, the arithmetic mean of the reciprocal of all real roots is 
and therefore the harmonic mean is 
~pigwash
~minor 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 
where k is any value from 1 to 2025.
Finding the sum of roots, we get 
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 
So, using HM formula we get 
~Aarav22
~Minor edits for latex by JerryZYang
Solution 5
Note that when we expand this, the sum of the reciprocals of the roots is 
. The coefficient of 
does not affect either of these, so we can assume that all of them are just 
, then just plug into formula to get 
Solution 6 (5 Second Cheese)
The harmonic mean of 
numbers, 
and 
is 
. So, in terms of the roots of the polynomial 
, we have the roots, and to multiply to 
so 
, and they sum to 
. So, our answer is simply ![\[-\dfrac{\dfrac{6}{k}}{\dfrac{4}{k}} = -\dfrac{3}{2} \Longrightarrow \boxed{\text{(B)}}.\]](http://latex-new.aopstest.com/e/e/3/ee3650047001bcde27a3c1ded6ce1275ce6efd86.png)
-jb2015007
Solution 7
To find the reciprocals of the roots of a polynomial, we can reverse the order of its coefficients. More formally, if \[ a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0 \] has roots \( r_1, r_2, \ldots, r_n \), then \[ a_0x^n + a_1x^{n-1} + \cdots + a_{n-1}x + a_n = 0 \] has roots \( \frac{1}{r_1}, \frac{1}{r_2}, \ldots, \frac{1}{r_n} \).
Applying this to the given quadratic, each polynomial becomes \[ -3x^2 - 4x + k = 0. \] By Vieta's formulas, the sum of the roots of this quadratic is \[ \frac{-(-4)}{-3} = -\frac{4}{3}. \]
Since each of the \(2025\) quadratics has the same sum of roots, the average of these sums is also \[ -\frac{4}{3}. \]
The problem asks for the harmonic mean of the original roots, which requires taking the reciprocal of this average. Thus, the final answer 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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