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 Pyhtagorean 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, PQis 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 |
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 | |
All AMC 12 Problems and Solutions |
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