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 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 .

~ lisztepos

~ Edited by aoum

Solution 2 (AM-GM)

Consider the function \begin{align*} f(x) = \frac{(x - 18)(x - 98)(x - 72)(x - k)}{x^2}. \end{align*} Expanding this, we obtain \begin{align*} f(x) = \left( x + \frac{18 \times 98}{x} - (18 + 98)\right)\left( x + \frac{72k}{x} - (72 + k) \right). \end{align*}

Let \( y = x + \frac{m}{x} \) (where \( x > 0 \)). By the AM-GM inequality \( a + b \geq 2\sqrt{ab} \), we have \begin{align*} x + \frac{m}{x} \geq 2\sqrt{m}. \end{align*} Assuming \( x = \sqrt{m} \), the minimum value is \( 2\sqrt{m} \).

Let \( y_1 = x + \frac{18 \times 98}{x} - (18 + 98) \). Then,

when \(x = \sqrt{18 \times 98} = 42.\),We obtain \(y_{min} = 42-42-116 =-32\)

Let \( y_2 = x + \frac{72 \times k}{x} - (72 + k) \).

When \(x = \sqrt{72 \times k}\),We obtain \(y_{min} = 2(\sqrt{72 \times k})-(72+k)\).

Since \( y_1 = y_2 \), we have \begin{align*} 2(\sqrt{72 \times k})-(72+k) = -32, \end{align*} which yields \( k = 8 \) or \( k = 200 \).

With same method, consider the function \begin{align*} f(x) = \underbrace{\left( x + \frac{18 \times 72}{x} - 90 \right)}_{y_1} \cdot \underbrace{\left( x + \frac{98k}{x} - (98 + k) \right)}_{y_2}. \end{align*} When \( x_1 = \sqrt{18 \times 72} = 36 \), \begin{align*} y_1 = 2\times 36 - 90 = -18, \end{align*} When \( x_2 = \sqrt{98 \times k}\), \begin{align*} y_2 = 2(\sqrt{98 \times k})-(98+k), \end{align*} Thus, \(y_1 = y_2 \) gives \( 2(\sqrt{98 \times k})-(98+k) = -18 \), and the minimum value corresponds to \( k = 32 \) or \( k = 200 \).

In summary, \( k_1 = 8 \), \( k_2 = 32 \), and \( k_3 = 200 \), with their sum being .

~ Edited by dongjiu0728

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.