AIME 2009 II · 第 3 题

Solution

Solution 1

From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$, and $ABE$ are also right triangles. By $AA$, $\triangle FBA \sim \triangle BCA$, and $\triangle FBA \sim \triangle ABE$, so $\triangle ABE \sim \triangle BCA$. This gives $\frac {AE}{AB}= \frac {AB}{BC}$. $AE=\frac{AD}{2}$ and $BC=AD$, so $\frac {AD}{2AB}= \frac {AB}{AD}$, or $(AD)^2=2(AB)^2$, so $AD=AB \sqrt{2}$, or $100 \sqrt{2}$, so the answer is $\boxed{141}$.

Solution 2

Let $x$ be the ratio of $BC$ to $AB$. On the coordinate plane, plot $A=(0,0)$, $B=(100,0)$, $C=(100,100x)$, and $D=(0,100x)$. Then $E=(0,50x)$. Furthermore, the slope of $\overline{AC}$ is $x$ and the slope of $\overline{BE}$ is $-x/2$. They are perpendicular, so they multiply to $-1$, that is,

$$x\cdot-\frac{x}{2}=-1,$$ which implies that $-x^2=-2$ or $x=\sqrt 2$. Therefore $AD=100\sqrt 2\approx 141.42$ so $\lfloor AD\rfloor=\boxed{141}$.

Solution 3

Similarly to Solution 2, let the positive x-axis be in the direction of ray $BC$ and let the positive y-axis be in the direction of ray $BA$. Thus, the vector $BE=(x,100)$ and the vector $AC=(2x,-100)$ are perpendicular and thus have a dot product of 0. Thus, calculating the dot product:

$$x\cdot2x+(100)\cdot(-100)=2x^2-10000=0$$ $$2x^2-10000=0\rightarrow x^2=5000$$ Substituting AD/2 for x:

$$(AD/2)^2=5000\rightarrow AD^2=20000$$ $$AD=100\sqrt2$$

Solution 4

Draw $CX$ and $EX$ to form a parallelogram $AEXC$. Since $EX \parallel AC$, $\angle BEX=90^\circ$ by the problem statement, so $\triangle BEX$ is right. Letting $AE=y$, we have $BE=\sqrt{100^2+y^2}$ and $AC=EX=\sqrt{100^2+(2y)^2}$. Since $CX=EA$, $\sqrt{100^2+y^2}^2+\sqrt{100^2+(2y)^2}=(3y)^2$. Solving this, we have

$$100^2+ 100^2 + y^2 + 4y^2 = 9y^2$$ $$2\cdot 100^2 = 4y^2$$ $$\frac{100^2}{2}=y^2$$ $$\frac{100}{\sqrt{2}}=y$$ $$\frac{100\sqrt{2}}{2}=y$$ $$100\sqrt{2}=2y=AD$$ , so the answer is $\boxed{141}$.