2025 AIME II Problems/Problem 15

Problem

Let 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 expresson (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 (1) \\ (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 subsitution 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 .

-lisztepos

See also

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