AIME 2015 I · 第 7 题

Solution 1

Let us find the proportion of the side length of $KLMN$ and $FJGH$. Let the side length of $KLMN=y$ and the side length of $FJGH=x$.

Now, examine $BC$. We know $BC=BJ+JC$, and triangles $\Delta BHJ$ and $\Delta JFC$ are similar to $\Delta EDC$ since they are $1-2-\sqrt{5}$ triangles. Thus, we can rewrite $BC$ in terms of the side length of $FJGH$.

$$BJ=\frac{1}{\sqrt{5}}HJ=\frac{x}{\sqrt{5}}=\frac{x\sqrt{5}}{5}, JC=\frac{\sqrt{5}}{2}JF=\frac{x\sqrt{5}}{2}\Rightarrow BC=\frac{7x\sqrt{5}}{10}$$ Now examine $AB$. We can express this length in terms of $x,y$ since $AB=AN+NH+HB$. By using similar triangles as in the first part, we have

$$AB=\frac{1}{\sqrt{5}}y+\frac{\sqrt{5}}{2}y+\frac{2}{\sqrt{5}}x$$ $$AB=BC\Rightarrow \frac{7y\sqrt{5}}{10}+\frac{2x\sqrt{5}}{5}=\frac{7x\sqrt{5}}{10}\Rightarrow \frac{7y\sqrt{5}}{10}=\frac{3x\sqrt{5}}{10}\Rightarrow 7y=3x$$ Now, it is trivial to see that $[FJGH]=\left(\frac{x}{y}\right)^2[KLMN]=\left(\frac{7}{3}\right)^2\cdot 99=\boxed{539}.$

Solution 2

We begin by denoting the length $ED$ $a$, giving us $DC = 2a$ and $EC = a\sqrt5$. Since angles $\angle DCE$ and $\angle FCJ$ are complementary, we have that $\triangle CDE \sim \triangle JFC$ (and similarly the rest of the triangles are $1-2-\sqrt5$ triangles). We let the sidelength of $FGHJ$ be $b$, giving us:

$$JC = \sqrt5 \cdot FC = \sqrt5 \cdot FJ/2 = \frac{b\sqrt 5}{2}$$ and

$$BJ = \frac{1}{\sqrt5} \cdot HJ = \frac{b}{\sqrt5}$$ Since $BC = CJ + BJ$,

$$2a = \frac{b\sqrt 5}{2} + \frac{b}{\sqrt5}$$ Solving for $b$ in terms of $a$ yields

$$b = \frac{4a\sqrt5}{7}$$ We now use the given that $[KLMN] = 99$, implying that $KL = LM = MN = NK = 3\sqrt{11}$. We also draw the perpendicular from $E$ to $ML$ and label the point of intersection $P$ as in the diagram at the top

This gives that

$$AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}$$ and

$$ME = \sqrt5 \cdot MP = \sqrt5 \cdot \frac{EP}{2} = \sqrt5 \cdot \frac{LG}{2} = \sqrt5 \cdot \frac{HG - HK - KL}{2} = \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2}$$ Since $AE$ = $AM + ME$, we get

$$2 \cdot \frac{3\sqrt{11}}{\sqrt5} + \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2} = a$$ $$\Rightarrow 12\sqrt{11} + 5(\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}) = 2\sqrt5a$$ $$\Rightarrow \frac{-21}{2}\sqrt{11} + \frac{20a\sqrt5}{7} = 2\sqrt5a$$ $$\Rightarrow -21\sqrt{11} = 2\sqrt5a\frac{14 - 20}{7}$$ $$\Rightarrow \frac{49\sqrt{11}}{4} = \sqrt5a$$ $$\Rightarrow 7\sqrt{11} = \frac{4a\sqrt{5}}{7}$$ So our final answer is $(7\sqrt{11})^2 = \boxed{539}$.

Solution 3

This is a relatively quick solution but a fakesolve. We see that with a ruler, $KL = \frac{3}{2}$ cm and $HG = \frac{7}{2}$ cm. Thus if $KL$ corresponds with an area of $99$, then $HG$ ($FGHJ$'s area) would correspond with $99*(\frac{7}{3})^2 = \boxed{539}$ - aops5234