Difference between revisions of "2019 AMC 12B Problems/Problem 18"
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Now we have the three side lengths of isosceles <math>\triangle PQR</math>: <math>PR=QR=\sqrt{6}</math>, <math>PQ=2 \sqrt{2}</math>. Letting the midpoint of <math>PQ</math> be <math>S</math>, <math>RS</math> is the perpendicular bisector of <math>PQ</math>, and so can be used as a height of <math>\triangle PQR</math> (taking <math>PQ</math> as the base). Using the Pythagorean Theorem again, we have <math>RS=\sqrt{PR^2-PS^2}=2</math>, so the area of <math>\triangle PQR</math> is <math>\frac{1}{2} \cdot PQ \cdot RS = \frac{1}{2} \cdot 2\sqrt{2} \cdot 2 = \boxed{\textbf{(C) } 2 \sqrt{2}}</math>. | Now we have the three side lengths of isosceles <math>\triangle PQR</math>: <math>PR=QR=\sqrt{6}</math>, <math>PQ=2 \sqrt{2}</math>. Letting the midpoint of <math>PQ</math> be <math>S</math>, <math>RS</math> is the perpendicular bisector of <math>PQ</math>, and so can be used as a height of <math>\triangle PQR</math> (taking <math>PQ</math> as the base). Using the Pythagorean Theorem again, we have <math>RS=\sqrt{PR^2-PS^2}=2</math>, so the area of <math>\triangle PQR</math> is <math>\frac{1}{2} \cdot PQ \cdot RS = \frac{1}{2} \cdot 2\sqrt{2} \cdot 2 = \boxed{\textbf{(C) } 2 \sqrt{2}}</math>. | ||
+ | |||
+ | ==Solution 4 (Simplest)== | ||
+ | Follow as Solution 1 to find the coordinates <math>P \in (2,2,0)</math>, <math>Q \in (0,2,2)</math>, and <math>R \in (1,4,1)</math>. Clearly there is symmetry, so drawing the midpoint in 3-D space yields a midpoint <math>M \in (1,2,1)</math>, so the height of <math>\triangle PQR</math> in the y direction is <math>2</math> and <math>MP=MQ=\sqrt{2}</math>, so the answer is <math>\boxed{\textbf{(C)}\ 2 \sqrt{2}}</math> | ||
+ | |||
+ | ~Geometry285 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=17|num-a=19}} | {{AMC12 box|year=2019|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:09, 7 November 2021
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)
Using the given data, we can label the points and
. We can also find the points
. Similarly,
and
.
Using the distance formula, ,
, and
. Using Heron's formula, or by dropping an altitude from
to find the height, we can then find that the area of
is
.
Note: After finding the coordinates of and
, we can alternatively find the vectors
and
, then apply the formula
. In this case, the cross product equals
, which has magnitude
, giving the area as
like before.
Solution 2
As in Solution 1, let and
, and calculate the coordinates of
,
, and
as
. Now notice that the plane determined by
is perpendicular to the plane determined 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 determined by
, since
and
have the same
-coordinate. Hence the height of
is equal to the
-coordinate of
minus the
-coordinate of
, giving
. By the distance formula,
, so the area of
is
.
Solution 3 (geometry)
By the Pythagorean Theorem, we can calculate and
. Now by the Law of Cosines in
, we have
.
Similarly, by the Law of Cosines in , we have
, so
. Observe that
(by side-angle-side), so
.
Next, notice that is parallel to
, and therefore
is similiar to
. Thus we have
. Since
, this gives
.
Now we have the three side lengths of isosceles :
,
. Letting the midpoint of
be
,
is the perpendicular bisector of
, and so can be used as a height of
(taking
as the base). Using the Pythagorean Theorem again, we have
, so the area of
is
.
Solution 4 (Simplest)
Follow as Solution 1 to find the coordinates ,
, and
. Clearly there is symmetry, so drawing the midpoint in 3-D space yields a midpoint
, so the height of
in the y direction is
and
, so the answer is
~Geometry285
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.