2025 AMC 10A Problems/Problem 18: Difference between revisions
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==Alternate Solution== | ==Alternate Solution== | ||
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>. Further simplifying, <imath>\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</imath>. Substituting back in, we get <imath>\frac{1}{\frac1{4050} \cdot 2025 (-\frac43)}</imath>, so therefore the resultant answer is <imath>-\frac32</imath> | |||
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>. Further simplifying, <imath>\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</imath>. Substituting back in, we get <imath>\frac{1}{\frac1{4050} \cdot 2025 (-\frac43)}</imath>, so therefore the resultant answer is <imath>-\frac32</imath> | |||
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 | 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 | ||
Revision as of 14:09, 6 November 2025
(Problem goes here)
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_1
Alternate Solution
Let the 4050 roots be in 
. Each of the 2025 quadratics' roots are, using the quadratic formula, 
. The harmonic mean is therefore 
. Further simplifying, 
. Substituting back in, we get 
, so therefore the resultant answer is 
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
~math660
Solution 2 (Fast)
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
