Difference between revisions of "2019 AMC 12B Problems/Problem 18"

Revision as of 23:05, 14 February 2019

Problem

Square pyramid $ABCDE$ has base $ABCD$ , which measures $3$ cm on a side, and altitude $AE$ perpendicular to the base, which measures $6$ cm. Point $P$ lies on $BE$ , one third of the way from $B$ to $E$ ; point $Q$ lies on $DE$ , one third of the way from $D$ to $E$ ; and point $R$ lies on $CE$ , two thirds of the way from $C$ to $E$ . What is the area, in square centimeters, of $\triangle{PQR}$ ?

$\textbf{(A) } \frac{3\sqrt2}{2} \qquad\textbf{(B) } \frac{3\sqrt3}{2} \qquad\textbf{(C) } 2\sqrt2 \qquad\textbf{(D) } 2\sqrt3 \qquad\textbf{(E) } 3\sqrt2$

Solution (Coordinate Bash)

Let $A(0, 0, 0), B(3, 0, 0), C(3, 3, 0), D(0, 3, 0),$ and $E(0, 0, 6)$ . We can figure out that $P(2, 0, 2), Q(0, 2, 2),$ and $R(1, 1, 4)$ .

Using the distance formula, $PQ = 2\sqrt{2}$ , $PR = \sqrt{6}$ , and $QR = \sqrt{6}$ . Using Heron's formula or dropping an altitude from P to find the height, we can compute that the area of $\triangle{PQR}$ is $\boxed{\textbf{(C) }2\sqrt{2}}$ .

Alternative Finish (Vectors)

Upon solving for $P,Q,$ and $R$ , we can find vectors $\overrightarrow{PQ}=$ < $-2,2,0$ > and $\overrightarrow{PR}=$ < $-1,1,2$ >, take the cross product's magnitude and divide by 2. Then the cross product equals < $4,4,0$ > with magnitude $4\sqrt{2}$ , yielding $\boxed{\textbf{(C) }2\sqrt{2}}$ .

Finding area with perpendicular planes

Once we get the coordinates of the desired triangle $P(2, 0, 2), Q(0, 2, 2),$ and $R(1, 1, 4)$ , we notice that the plane defined by these three points is perpendicular to the plane defined by $ABCD$ . To see this, consider the 'bird's eye view' looking down upon $P$ , $Q$ , and $R$ projected onto $ABCD$ : $[asy] unitsize(40); for(int i =0; i<=3; ++i) { draw((0,i)--(3,i)); draw((i,0)--(i,3)); } label("$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 $PQ$ is parallel to the plane $ABCD$ since $P$ and $Q$ have the same $z$ coordinate. From this, we can conclude that the height of $\triangle PQR$ is equal to $z$ coordinate of $R - z$ coordinate of $P = 4-2= 2$ . We know that $\overline{PQ} = 2\sqrt{2}$ , therefore the area of $\triangle PQR = \boxed{\textbf{(C) } 2\sqrt{2}}$ .

@@ Line 4: / Line 4: @@
 <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 1 (Coordinate Bash)==
+==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>.

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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