2009 AMC 10B Problems/Problem 12

Problem

Distinct points $A$ , $B$ , $C$ , and $D$ lie on a line, with $AB=BC=CD=1$ . Points $E$ and $F$ lie on a second line, parallel to the first, with $EF=1$ . A triangle with positive area has three of the six points as its vertices. How many possible values are there for the area of the triangle?

$\text{(A) } 3 \qquad \text{(B) } 4 \qquad \text{(C) } 5 \qquad \text{(D) } 6 \qquad \text{(E) } 7$

Solution 1

Consider the classical formula for triangle area: $\frac 12 \cdot b \cdot h$ . Each of the triangles that we can make has exactly one side lying on one of the two parallel lines. If we pick this side to be the base, the height will always be the same - it will be the distance between the two lines.

Hence each area is uniquely determined by the length of the base. And it can easily be seen, that the only possible base lengths are $1$ , $2$ , and $3$ . Therefore there are only $\boxed{(A)3}$ possible values for the area.

(To be more precise in the last step, the possible base lengths are $AB=BC=CD=EF=1$ , $AC=BD=2$ , and $AD=3$ .)

Solution 2

No matter what how we draw a triangle by selecting three non-linear points, its height will always remain the same. Therefore, we will only get different areas with different base-lengths. The possibilities are $1$ , $2$ , and $3$ units for a total of $\boxed{\text{(A) } 3}$ .

~MathFun1000

Video Solution

https://youtu.be/57B-aRqONv8

~savannahsolver

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2009 AMC 10B Problems/Problem 12

Contents

Problem

Solution 1

Solution 2

Video Solution

See Also