Difference between revisions of "2025 AIME II Problems/Problem 15"
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However, we are not finished, trying <math>k=512</math> into the equation <math>(1)</math> from earlier does not give us equality, thus it is an extraneous root. The sum of all <math>k</math> then must be <math>8+32+200=\boxed{240}</math>. | However, we are not finished, trying <math>k=512</math> into the equation <math>(1)</math> from earlier does not give us equality, thus it is an extraneous root. The sum of all <math>k</math> then must be <math>8+32+200=\boxed{240}</math>. | ||
| − | + | ~ [https://artofproblemsolving.com/wiki/index.php/User:Lisztepos lisztepos] | |
| + | |||
| + | ~ Edited by [https://artofproblemsolving.com/wiki/index.php/User:Aoum aoum] | ||
==See also== | ==See also== | ||
Revision as of 13:37, 14 February 2025
Problem
Let
![\[f(x)=\frac{(x-18)(x-72)(x-98)(x-k)}{x}.\]](http://latex.artofproblemsolving.com/0/d/b/0dbe0a5b2a8136e0b1e02478bc822fe2ba6ada41.png)
There exist exactly three positive real values of 
such that 
has a minimum at exactly two real values of 
. Find the sum of these three values of .
Solution 1 ('clunky', trial and error)
Let 
be the minimum value of the expression (changes based on the value of , however is a constant). Therefore we can say that
\begin{align*}
f(x)-n=\frac{(x-\alpha)^2(x-\beta)^2}{x}
\end{align*}
This can be done because is a constant, and for the equation to be true in all the right side is also a quartic. The roots must also both be double, or else there is an even more 'minimum' value, setting contradiction.
We expand as follows, comparing coefficients:
\begin{align*} (x-18)(x-72)(x-98)(x-k)-nx=(x-\alpha)^2(x-\beta)^2 \\ -2\alpha-2\beta=-18-72-98-k \implies \alpha+\beta=94+\frac{k}{2} \\ \alpha^2+4\alpha \beta +\beta^2=(-18\cdot -72)+(-18\cdot-98)+(-18\cdot-k)+(-72\cdot-98)+(-72\cdot-k)+(-98\cdot-k)=10116+188k \\ (\alpha^2)(\beta^2)=(-18)(-72)(-98)(-k) \implies \alpha \beta=252\sqrt{2k} \\ \end{align*}
Recall 
, so we can equate and evaluate as follows:
\begin{align} (94+\frac{k}{2})^2+504\sqrt{2k}=10116+188k \tag{1}\\ \end{align} \begin{align*} (47-\frac{k}{4})^2+126\sqrt{2k}=2529 \\ \frac{k^2}{16}-\frac{47}{2}k+126\sqrt{2k}-320=0 \\ \end{align*}
We now have a quartic with respect to 
. Keeping in mind it is much easier to guess the roots of a polynomial with integer coefficients, we set 
. Now our equation becomes
\begin{align*} 4a^2-188a+504\sqrt{a}-320=0 \\ a^2-47a+126\sqrt{a}-80=0 \\ \end{align*}
If you are lucky, you should find roots 
and 
. After this, solving the resulting quadratic gets you the remaining roots as 
and 
. Working back through our substitution for 
, we have generated values of as 
.
However, we are not finished, trying 
into the equation 
from earlier does not give us equality, thus it is an extraneous root. The sum of all then must be 
.
~ Edited by aoum
See also
| 2025 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Last Problem | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
