AIME 2006 I · 第 8 题

Solution 1

Let $x$ denote the common side length of the rhombi, and let $y$ denote one of the smaller interior angles of rhombus $\mathcal{P}$. Then $x^2\sin(y)=\sqrt{2006}$. We also see that $K=x^2\sin(2y) \implies K=2x^2\sin y \cdot \cos y \implies K = 2\sqrt{2006}\cdot \cos y$. Thus $K$ can be any positive integer in the interval $(0, 2\sqrt{2006})$. Since $2\sqrt{2006} = \sqrt{8024}$ and $89^2 = 7921 < 8024 < 8100 = 90^2$, $K$ can be any integer between $1$ and $89$, inclusive, so the number of positive values for $K$ is $\boxed{089}$.

Solution 2

Call the side length of each rhombus $w$. $w$ is the width of the rhombus. Call the height h, where $wh=\sqrt{2006}$. The height of rhombus T would be 2h, and the width would be $\sqrt{w^2-h^2}$. Substitute the first equation to get $\sqrt{\frac{2006}{h^2}-h^2}$. Then the area of the rhombus would be $2h * \sqrt{\frac{2006}{h^2}-h^2}$. Combine like terms to get $2 * \sqrt{2006-h^4}$. This expression equals an integer K. $2006-h^4$ specifically must be in the form $n^2/4$. There is no restriction on h as long as it is a positive real number, so all we have to do is find all the positive possible values of $n^2$ for $2006-h^4$. Now, quick testing shows that $44^2 < 2006$ and $45^2>2006$, but we must also test $44.5^2$, because the product of two will make it an integer. $44.5^2$ is also less than $2006$, so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5 is also a valid value. (notice 0 is not valid because the height must be a positive number) That gives us $44*2+1=$ $\boxed{089}$

-jackshi2006

Solution 3

To determine the possible values of $[GIHJ],$ we must determine the maximum and minimum possible areas.

In the case where the $4$ rhombi are squares, we have $[GIHJ]=0,$ implying the minimum possible positive-integer-valued area is $1.$

Denote the length $HC=a$ and $KH=b.$ We have

$$KI=\sqrt{a^2-b^2}$$ by the Pythagorean Theorem, which implies

$$[IBCH]=a\sqrt{a^2-b^2}=\sqrt{2006}$$ and

$$[GIHJ]=2b\sqrt{a^2-b^2}.$$ The first equation yields

$$\sqrt{a^2-b^2}=\frac{\sqrt{2006}}{a}.$$ Plugging into the second, we have

$$[GIHJ]=2\sqrt{2006}\frac{b}{a}.$$ The maximal value of $\frac{b}{a}$ occurs when the height of $ABCDEF$ is minimized, which means

$$\frac{b}{a}\leq 1.$$ Plugging back up, we have

$$[GIHJ]\leq 2\sqrt{2006}=\sqrt{8024}.$$ We have

$$\lfloor \sqrt{8024} \rfloor = 89,$$ thus our answer is

$$89-1+1=\boxed{089}.$$