2025 AIME II Problems/Problem 15

Problem

Let $f(x)=\frac{(x-18)(x-72)(x-98)(x-k)}{x}.$ There exist exactly three positive real values of $k$ such that $f$ has a minimum at exactly two real values of $x$ . Find the sum of these three values of $k$ .

Solution 1 ('clunky', trial and error)

Let $n$ be the minimum value of the expression (changes based on the value of $k$ , however is a constant). Therefore we can say that $\begin{array}{r} f (x) - n = \frac{(x - α)^{2} (x - β)^{2}}{x} \end{array}$ This can be done because $n$ is a constant, and for the equation to be true in all $x$ 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:

$\begin{array}{r} (x - 18) (x - 72) (x - 98) (x - k) - n x = (x - α)^{2} (x - β)^{2} \\ - 2 α - 2 β = - 18 - 72 - 98 - k ⟹ α + β = 94 + \frac{k}{2} \\ α^{2} + 4 α β + β^{2} = (- 18 \cdot - 72) + (- 18 \cdot - 98) + (- 18 \cdot - k) + (- 72 \cdot - 98) + (- 72 \cdot - k) + (- 98 \cdot - k) = 10116 + 188 k \\ (α^{2}) (β^{2}) = (- 18) (- 72) (- 98) (- k) ⟹ α β = 252 \sqrt{2 k} \end{array}$

Recall $(\alpha+\beta)^2+2\alpha \beta=\alpha^2+4\alpha \beta +\beta^2$ , so we can equate and evaluate as follows:

$\begin{array}{r} (1) & (94 + \frac{k}{2})^{2} + 504 \sqrt{2 k} = 10116 + 188 k \end{array}$ $\begin{array}{r} (47 - \frac{k}{4})^{2} + 126 \sqrt{2 k} = 2529 \\ \frac{k^{2}}{16} - \frac{47}{2} k + 126 \sqrt{2 k} - 320 = 0 \end{array}$

We now have a quartic with respect to $\sqrt{k}$ . Keeping in mind it is much easier to guess the roots of a polynomial with integer coefficients, we set $a=\frac{k}{8}$ . Now our equation becomes

$\begin{array}{r} 4 a^{2} - 188 a + 504 \sqrt{a} - 320 = 0 \\ a^{2} - 47 a + 126 \sqrt{a} - 80 = 0 \end{array}$

If you are lucky, you should find roots $\sqrt{a}=1$ and $2$ . After this, solving the resulting quadratic gets you the remaining roots as $5$ and $8$ . Working back through our substitution for $a$ , we have generated values of $k$ as $(8, 32, 200, 512)$ .

However, we are not finished, trying $k=512$ into the equation $(1)$ from earlier does not give us equality, thus it is an extraneous root. The sum of all $k$ then must be $8+32+200=\boxed{240}$ .

-lisztepos

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2025 AIME II Problems/Problem 15

Problem

Solution 1 ('clunky', trial and error)

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