2015 AMC 12A Problems/Problem 12
Problem
The parabolas and intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area . What is ?
Solutions
Solution 1
Clearly, the parabolas must intersect the x-axis at the same two points. Their distance multiplied by (the distance between the y-intercepts), all divided by 2 is equal to 12, the area of the kite (half the product of the diagonals). That distance is thus 4, and so the x-intercepts are Then , and Then .
Solution 2
The parabolas must intersect the x-axis at the same two points for the kite to form. We find the x values at which they intersect by equating them and solving for x as shown below. and or . The x-values of the y-intercepts is 0, so we plug in zero in each of them and get and . The area of a kite is . The is . The is . Solving for the area , therefore .
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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All AMC 12 Problems and Solutions |