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 (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 .
Finding area with perpendicular planes
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 .
Solution (Old Fashioned Geometry)
Use Pythagorean Teorem we can quickly obtain the following parameters: EB=ED=3sqrt(5),EC=3sqrt(6),ER=sqrt(6),EP=EQ=2sqrt(5) Inside triangle EBC, using cosine law: COS(EBC)=(EB^2+EC^2-BC^2)/(2*EB*EC)=sqrt(30)/6 Now move to triangle EPR, use cosine law again PR^2=ER^2+EP^2-2*ER*EP*COS(EBC)=6, therefore PR=sqrt(6), noticing that triangle ERP is congruent to triangle ERQ, QR=PR=sqrt(6). Now look at points P, Q and triangle EDB, PQ is parallel to DB, and therefore triangle EQP is similiar to triangle EDB, we have QP/DB=EP/EB=2/3, since DB=3sqrt(2), we have PQ=2sqrt(2). Now we have the three side lengths of isosceles triangle PQR: PR=QR=sqrt(6), PQ=2sqrt(2). Suppose the midpoint of PQ is S, connect RS, it would be perpendicular bisector of PQ and act as the height of side PQ. Use Pythagorean again we have RS=sqrt(PR^2-PS^2)=2, therefore the area of triangle PQR is = 1/2*PQ*RS=2sqrt(2)
(by Zhen Qin)
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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