2019 AMC 12B Problems/Problem 18
Contents
Problem
Square pyramid has base , which measures cm on a side, and altitude perpendicular to the base, which measures cm. Point lies on , one third of the way from to ; point lies on , one third of the way from to ; and point lies on , two thirds of the way from to . What is the area, in square centimeters, of ?
Solution 1 (Coordinate Bash)
Let and . We can figure out that and .
Using the distance formula, , , and . Using Heron's formula or dropping an altitude from P to find the height, we can compute that the area of is .
Alternative Finish (Vectors)
Upon solving for and , we can find vectors <> and <>, take the cross product's magnitude and divide by 2. Then the cross product equals <> with magnitude , yielding .
Faster way to compute area
Once we get the coordinates of the desired triangle and , we notice that the plane defined by these three points is perpendicular to the plane defined by . To see this, consider the 'bird's eye view' looking down upon , , and projected onto : Additionally, we know that is parallel to the plane since and have the same coordinate. From this, we can conclude that the height of is equal to coordinate of coordinate of . We know that , therefore the area of .
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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