Difference between revisions of "2019 AMC 12B Problems/Problem 18"
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Upon solving for <math>P,Q,</math> and <math>R</math>, we can find vectors <math>\overrightarrow{PQ}=</math><<math>-2,2,0</math>> and <math>\overrightarrow{PR}=</math><<math>-1,1,2</math>>, take the cross product's magnitude and divide by 2. Then the cross product equals <<math>4,4,0</math>> with magnitude <math>4\sqrt{2}</math>, yielding <math>\boxed{\textbf{(C) }2\sqrt{2}}</math>. | Upon solving for <math>P,Q,</math> and <math>R</math>, we can find vectors <math>\overrightarrow{PQ}=</math><<math>-2,2,0</math>> and <math>\overrightarrow{PR}=</math><<math>-1,1,2</math>>, take the cross product's magnitude and divide by 2. Then the cross product equals <<math>4,4,0</math>> with magnitude <math>4\sqrt{2}</math>, yielding <math>\boxed{\textbf{(C) }2\sqrt{2}}</math>. | ||
− | === | + | ===Finding area with perpendicular planes=== |
Once we get the coordinates of the desired triangle <math>P(2, 0, 2), Q(0, 2, 2),</math> and <math>R(1, 1, 4)</math>, we notice that the plane defined by these three points is perpendicular to the plane defined by <math>ABCD</math>. To see this, consider the 'bird's eye view' looking down upon <math>P</math>, <math>Q</math>, and <math>R</math> projected onto <math>ABCD</math>: | Once we get the coordinates of the desired triangle <math>P(2, 0, 2), Q(0, 2, 2),</math> and <math>R(1, 1, 4)</math>, we notice that the plane defined by these three points is perpendicular to the plane defined by <math>ABCD</math>. To see this, consider the 'bird's eye view' looking down upon <math>P</math>, <math>Q</math>, and <math>R</math> projected onto <math>ABCD</math>: |
Revision as of 00:05, 15 February 2019
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
.
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
.
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.