Difference between revisions of "2019 AMC 12B Problems/Problem 18"
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<math>\textbf{(A) } \frac{3\sqrt2}{2} \qquad\textbf{(B) } \frac{3\sqrt3}{2} \qquad\textbf{(C) } 2\sqrt2 \qquad\textbf{(D) } 2\sqrt3 \qquad\textbf{(E) } 3\sqrt2</math> | <math>\textbf{(A) } \frac{3\sqrt2}{2} \qquad\textbf{(B) } \frac{3\sqrt3}{2} \qquad\textbf{(C) } 2\sqrt2 \qquad\textbf{(D) } 2\sqrt3 \qquad\textbf{(E) } 3\sqrt2</math> | ||
− | ==Solution | + | ==Solution (Coordinate Bash)== |
Let <math>A(0, 0, 0), B(3, 0, 0), C(3, 3, 0), D(0, 3, 0),</math> and <math>E(0, 0, 6)</math>. We can figure out that <math>P(2, 0, 2), Q(0, 2, 2),</math> and <math>R(1, 1, 4)</math>. | Let <math>A(0, 0, 0), B(3, 0, 0), C(3, 3, 0), D(0, 3, 0),</math> and <math>E(0, 0, 6)</math>. We can figure out that <math>P(2, 0, 2), Q(0, 2, 2),</math> and <math>R(1, 1, 4)</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 (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.