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
minus the
coordinate of
. We know that
, therefore the area of
.
Solution (Old Fashioned Geometry)
Use Pythagorean Teorem we can quickly obtain the following parameters:
Inside
, using cosine law:
Now move to
, use cosine law again
, therefore
, noticing that
is congruent to
,
.
Now look at points
,
, and
,
is parallel to
, and therefore
EQP
\triangle EDB
\frac{QP}{DB}=\frac{EP}{EB}=\frac{2}{3}
DB=3\sqrt{2}
PQ=2 \sqrt{2}
\triangle PQR
PR=QR=\sqrt{6}
PQ=2 \sqrt{2}
PQ
S
RS
PQ
PQ
RS=\sqrt{PR^2-PS^2}=2
\triangle PQR
\frac{1}{2} \cdot PQ \cdot RS= \boxed{\textbf{(C) } 2 \sqrt{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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.