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 $$\begin{array}{r}f(x)-n=\frac{(x-\alpha {)}^2(x-\beta {)}^2}{x}\end{array}$$ 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:

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

Recall , so we can equate and evaluate as follows:

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

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

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

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

Note: I'm not sure what the first author meant by "or else there is an even more 'minimum' value." The most noticeable reason there are two double roots is because there are two distinct positive solutions per the conditions of the problem

~ fermat_sLastAMC

Further notes (from the author): To clarify why the equation takes the form as above, if you do not regard the context of the problem, you could say that is another form, also having two roots. However when we consider this form, (WLOG assuming ) if we take a value such that , then , therefore implying that or otherwise resulting , which is contradiction as we assumed is the least value of the function. Similarly logic proceeds if the third power is on the other binomial.

Solution 2 (AM-GM)

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

Let $y=x+\frac{m}{x}$ (where $x>0$). By the AM-GM inequality $a+b\ge 2\sqrt{ab}$, we have $$\begin{array}{r}x+\frac{m}{x}\ge 2\sqrt{m}.\end{array}$$ 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 $$\begin{array}{r}2(\sqrt{72\times k})-(72+k)=-32,\end{array}$$ which yields $k=8$ or $k=200$.

With same method, consider the function $$\begin{array}{r}f(x)=\underset{y_1}{\underbrace{(x+\frac{18\times 72}{x}-90)}}\cdot \underset{y_2}{\underbrace{(x+\frac{98k}{x}-(98+k))}}.\end{array}$$ When $x_1=\sqrt{18\times 72}=36$, $$\begin{array}{r}{y}_1=2\times 36-90=-18,\end{array}$$ When $x_2=\sqrt{98\times k}$, $$\begin{array}{r}{y}_2=2(\sqrt{98\times k})-(98+k),\end{array}$$ 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.