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 (Coordinate Bash)== | + | ==Solution 1 (Coordinate Bash)== |
| − | Let <math>A(0, 0, 0), B(3, 0, 0), C( | + | 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>. |
Using the distance formula, <math>PQ = 2\sqrt{2}</math>, <math>PR = \sqrt{6}</math>, and <math>QR = \sqrt{6}</math>. Using Heron's formula or dropping an altitude from P to find the height, we can compute that the area of <math>\triangle{PQR}</math> is <math>\boxed{\textbf{(C) }2\sqrt{2}}</math>. | Using the distance formula, <math>PQ = 2\sqrt{2}</math>, <math>PR = \sqrt{6}</math>, and <math>QR = \sqrt{6}</math>. Using Heron's formula or dropping an altitude from P to find the height, we can compute that the area of <math>\triangle{PQR}</math> is <math>\boxed{\textbf{(C) }2\sqrt{2}}</math>. | ||
| − | ==Alternative Finish (Vectors)== | + | ===Alternative Finish (Vectors)=== |
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>. | ||
| + | |||
| + | ===Faster way to compute area=== | ||
| + | |||
| + | 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>: | ||
| + | <asy> | ||
| + | unitsize(40); | ||
| + | for(int i =0; i<=3; ++i) { | ||
| + | draw((0,i)--(3,i)); | ||
| + | draw((i,0)--(i,3)); | ||
| + | } | ||
| + | label("$A$", (0,0), SW); | ||
| + | label("$B$", (3,0), SE); | ||
| + | label("$C$", (3,3), NE); | ||
| + | label("$D$", (0,3), NW); | ||
| + | label("$P$", (2,0), S); | ||
| + | label("$Q$", (0,2), W); | ||
| + | label("$R$", (1,1), NE); | ||
| + | dot((2,0)); | ||
| + | dot((0,2)); | ||
| + | dot((1,1)); | ||
| + | draw((0,2)--(2,0)); | ||
| + | </asy> | ||
| + | Additionally, we know that <math>PQ</math> is parallel to the plane <math>ABCD</math> since <math>P</math> and <math>Q</math> have the same <math>z</math> coordinate. From this, we can conclude that the height of <math>\triangle PQR</math> is equal to <math>z</math> coordinate of <math>R - z</math> coordinate of <math>P = 4-2= 2</math>. We know that <math>\overline{PQ} = 2\sqrt{2}</math>, therefore the area of <math>\triangle PQR = \boxed{\textbf{(C) } 2\sqrt{2}}</math>. | ||
==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 00:04, 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 .
Faster way to compute area
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 :
![[asy] unitsize(40); for(int i =0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$A$", (0,0), SW); label("$B$", (3,0), SE); label("$C$", (3,3), NE); label("$D$", (0,3), NW); label("$P$", (2,0), S); label("$Q$", (0,2), W); label("$R$", (1,1), NE); dot((2,0)); dot((0,2)); dot((1,1)); draw((0,2)--(2,0)); [/asy]](http://latex.artofproblemsolving.com/0/7/1/071ab396144784a2cceee50ab064ae389ad1434e.png)
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. 
