2000 AMC 10 Problems/Problem 16

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Problem

The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$ . Find the length of segment $AE$ .

$\mathrm{(A)}\ \frac{4\sqrt{5}}{3} \qquad\mathrm{(B)}\ \frac{5\sqrt{5}}{3} \qquad\mathrm{(C)}\ \frac{12\sqrt{5}}{7} \qquad\mat...$

Solution

Solution 1

Let $l_1$ be the line containing $A$ and $B$ and let $l_2$ be the line containing $C$ and $D$ . If we set the bottom left point at $(0,0)$ , then $A=(0,3)$ , $B=(6,0)$ , $C=(4,2)$ , and $D=(2,0)$ .

The line $l_1$ is given by the equation $y=m_1x+b_1$ . The $y$ -intercept is $A=(0,3)$ , so $b_1=3$ . We are given two points on $l_1$ , hence we can compute the slope, $m_1$ to be $\frac{0-3}{6-0}=\frac{-1}{2}$ , so $l_1$ is the line $y=\frac{-1}{2}x+3$

Similarly, $l_2$ is given by $y=m_2x+b_2$ . The slope in this case is $\frac{2-0}{4-2}=1$ , so $y=x+b_2$ . Plugging in the point $(2,0)$ gives us $b_2=-2$ , so $l_2$ is the line $y=x-2$ .

At $E$ , the intersection point, both of the equations must be true, so $\begin{align*}y=x-2, y=\frac{-1}{2}x+3 &\Rightarrow x-2=\frac{-1}{2}x+3 \\&\Rightarrow x=\frac{10}{3} \\&\Rightar...$

We have the coordinates of $A$ and $E$ , so we can use the distance formula here: $\sqrt{\left(\frac{10}{3}-0\right)^2+\left(\frac{4}{3}-3\right)^2}=\frac{5\sqrt{5}}{3}$

which is answer choice $\boxed{\text{B}}$

Solution 2

Draw the perpendiculars from $A$ and $B$ to $CD$ , respectively. As it turns out, $BC \perp CD$ . Let $F$ be the point on $CD$ for which $AF\perp CD$ .

$m\angle AFE=m\angle BCE=90^\circ$ , and $m\angle AEF=m\angle CEB$ , so by AA similarity, $\triangle AFE\sim \triangle BCE \Rightarrow \frac{AF}{AE}=\frac{BC}{BE}$

By the Pythagorean Theorem, we have $AB=\sqrt{3^2+6^2}=3\sqrt{5}$ , $AF=\sqrt{2.5^2+2.5^2}=2.5\sqrt{2}$ , and $BC=\sqrt{2^2+2^2}+2\sqrt{2}$ . Let $AE=x$ , so $BE=3\sqrt{5}-x$ , then $\frac{2.5\sqrt{2}}{x}=\frac{2\sqrt{2}}{3\sqrt{5}-x}$ $x=\frac{5\sqrt{5}}{3}$

This is answer choice $\boxed{\text{B}}$

2000 AMC 10 (Problems • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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