1983 AIME Problems/Problem 9

Problem

Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$ .

Solution 1

Let $y=x\sin{x}$ . We can rewrite the expression as $\frac{9y^2+4}{y}=9y+\frac{4}{y}$ .

Since $x>0$ , and $\sin{x}>0$ because $0< x<\pi$ , we have $y>0$ . So we can apply AM-GM:

$9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12$

The equality holds when $9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23$ .

Therefore, the minimum value is $\boxed{012}$ . This is reached when we have $x \sin{x} = \frac{2}{3}$ in the original equation (since $x\sin x$ is continuous and increasing on the interval $0 \le x \le \frac{\pi}{2}$ , and its range on that interval is from $0 \le x\sin x \le \frac{\pi}{2}$ , this value of $\frac{2}{3}$ is attainable by the Intermediate Value Theorem).

Solution 2

We can rewrite the numerator to be a perfect square by adding $-\dfrac{12x \sin x}{x \sin x}$ . Thus, we must also add back $12$ .

This results in $\dfrac{(3x \sin x-2)^2}{x \sin x}+12$ .

Thus, if $3x \sin x-2=0$ , then the minimum is obviously $12$ . We show this possible with the same methods in Solution 1; thus the answer is $\boxed{012}$ .

Solution 3 (uses calculus)

Let $y = x\sin{x}$ and rewrite the expression as $f(y) = 9y + \frac{4}{y}$ , similar to the previous solution. To minimize $f(y)$ , take the derivative of $f(y)$ and set it equal to zero.

The derivative of $f(y)$ , using the Power Rule, is

$f'(y)$ = $9 - 4y^{-2}$

$f'(y)$ is zero only when $y = \frac{2}{3}$ or $y = -\frac{2}{3}$ . It can further be verified that $\frac{2}{3}$ and $-\frac{2}{3}$ are relative minima by finding the derivatives at other points near the critical points, or by checking that the second derivative $f''(y)=8y^{-3}$ is positive. However, since $x \sin{x}$ is always positive in the given domain, $y = \frac{2}{3}$ . Therefore, $x\sin{x}$ = $\frac{2}{3}$ , and the answer is $\frac{(9)(\frac{2}{3})^2 + 4}{\frac{2}{3}} = \boxed{012}$ .

Solution 4 (also uses calculus)

As above, let $y = x\sin{x}$ . Add $\frac{12y}{y}$ to the expression and subtract $12$ , giving $f(x) = \frac{(3y+2)^2}{y} - 12$ . Taking the derivative of $f(x)$ using the Chain Rule and Quotient Rule, we have $\frac{\text{d}f(x)}{\text{d}x} = \frac{6y(3y+2)-(3y+2)^2}{y^2}$ . We find the minimum value by setting this to $0$ . Simplifying, we have $6y(3y+2) = (3y+2)^2$ and $y = \pm{\frac{2}{3}} = x\sin{x}$ . Since both $x$ and $\sin{x}$ are positive on the given interval, we can ignore the negative root. Plugging $y = \frac{2}{3}$ into our expression for $f(x)$ , we have $\frac{(3(\frac{2}{3})+2)^2}{y}-12 = \frac{16}{\left(\frac{2}{3}\right)}-12 = \boxed{012}$ .

Solution 5

Set $\frac{9x^2\sin^2 x + 4}{x\sin x}$ equal to $y$ . Then multiply by $x\sin x$ on both sides to get $9x^2\sin^2 x + 4 = y\cdot x\sin x$ . We then subtract $yx\sin x$ from both sides to get $9x^2\sin^2 x + 4 - yx\sin x = 0$ . This looks like a quadratic so set $z= x\sin x$ and use quadratic equation on $9z^2 - yz + 4 = 0$ to see that $z = \frac{y\pm\sqrt{y^2-144}}{18}$ . We know that $y$ must be an integer and as small as it can be, so $y$ = 12. We plug this back in to see that $x\sin x = \frac{2}{3}$ which we can prove works using methods from solution 1. This makes the answer $\boxed{012}$

-awesomediabrine

Solution 6

Seeing that we need to minimize, we think inequalities, and seeing squares, we think RMS-AM-GM-HM. From this inequality, we know that $\sqrt{\frac{(3x\sin x)^2+2^2}{2}} \geq \sqrt{(3x\sin x)(2)}$ , with equality holding when $3x\sin x=2$ . From this inequality, we can see the following: \begin{align*} \sqrt{\frac{(3x\sin x)^2+2^2}{2}} \geq \sqrt{(3x\sin x)(2)} \\ \frac{9x^2\sin^2x+4}{2} \geq 6x\sin x \\ \frac{9x^2\sin^2x+4}{x\sin x} \geq 12 \end{align*} We can prove that the equality condition is possible as in Solution $1$ . Thus, our answer is $\boxed{012}$ .

Video Solution

https://youtu.be/WQaL5cPDVVo

~Lucas

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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