2019 AMC 12B Problems/Problem 18

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 $\boxed{\textbf{(C) }2\sqrt{2}}$ .

Solution (Perpendicular planes)

Let We start by finding the coordinates of $P$ , $Q$ , and $R$ , getting $P=(2,0,2), Q=(0,2,2), R=(1,1,4)$ . We notice that the plane defined $\triangle PQR$ 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 $R$ minus the $z$ coordinate of $P = 4-2= 2$ . We know that $\overline{PQ} = 2\sqrt{2}$ , therefore the area of $\triangle PQR = \boxed{\textbf{(C) } 2\sqrt{2}}$ .

Solution (Old Fashioned Geometry)

Use Pythagorean Teorem we can quickly obtain the following parameters: $EB=ED=3\sqrt{5},EC=3\sqrt{6},ER= \sqrt{6},EP=EQ=2 \sqrt{5}$ Inside $\triangle EBC$ , using cosine law: $\cos{(\angle EBC)}=\frac{(EB^2+EC^2-BC^2)}{2 \cdot EB \cdot EC}=\frac{\sqrt{30}}{6}$ Now move to $\triangle EPR$ , use cosine law again $PR^2=ER^2+EP^2-2 \cdot ER \cdot EP \cdot \cos{(\angle EBC)}=6$ , therefore $PR=\sqrt{6}$ , noticing that $\triangle ERP$ is congruent to $\triangle ERQ$ , $QR=PR=\sqrt{6}$ . Now look at points $P$ , $Q$ , and $\triangle EDB$ , $PQ$ is parallel to $DB$ , and therefore $\triangle EQP$ is similiar to $\triangle EDB$ , we have $\frac{QP}{DB}=\frac{EP}{EB}=\frac{2}{3}$ , since $DB=3\sqrt{2}$ , we have $PQ=2 \sqrt{2}$ . Now we have the three side lengths of isosceles $\triangle PQR$ : $PR=QR=\sqrt{6}$ , $PQ=2 \sqrt{2}$ . Suppose the midpoint of $PQ$ is $S$ , connect $RS$ , it would be perpendicular bisector of $PQ$ and act as the height of side $PQ$ . Use Pythagorean again we have $RS=\sqrt{PR^2-PS^2}=2$ , therefore the area of $\triangle PQR$ is = $\frac{1}{2} \cdot PQ \cdot RS= \boxed{\textbf{(C) } 2 \sqrt{2}}$ .

(by Zhen Qin)

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
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