# 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 1 (coordinate bash)

Using the given data, we can label the points $A(0, 0, 0), B(3, 0, 0), C(3, 3, 0), D(0, 3, 0),$ and $E(0, 0, 6)$. We can also find the points $P = B + \frac{1}{3} \overrightarrow{BE} = (3,0,0) + \frac{1}{3}(-3, 0, 6) = (3,0,0) + (-1,0,2) = (2, 0, 2)$. Similarly, $Q = (0, 2, 2)$ and $R = (1, 1, 4)$.

Using the distance formula, $PQ = \sqrt{\left(-2\right)^2 + 2^2 + 0^2} = 2\sqrt{2}$, $PR = \sqrt{\left(-1\right)^2 + 1^2 + 2^2} = \sqrt{6}$, and $QR = \sqrt{1^2 + \left(-1\right)^2 + 2^2} = \sqrt{6}$. Using Heron's formula, or by dropping an altitude from $P$ to find the height, we can then find that the area of $\triangle{PQR}$ is $\boxed{\textbf{(C) }2\sqrt{2}}$.

Note: After finding the coordinates of $P,Q,$ and $R$, we can alternatively find the vectors $\overrightarrow{PQ}=[-2,2,0]$ and $\overrightarrow{PR}=[-1,1,2]$, then apply the formula $\text{area} = \frac{1}{2}\left|\overrightarrow{PQ} \times \overrightarrow{PR}\right|$. In this case, the cross product equals $[4,4,0]$, which has magnitude $4\sqrt{2}$, giving the area as $2\sqrt{2}$ like before.

## Solution 2

As in Solution 1, let $A=(0, 0, 0), B=(3, 0, 0), C=(3, 3, 0), D=(0, 3, 0),$ and $E=(0, 0, 6)$, and calculate the coordinates of $P$, $Q$, and $R$ as $P=(2,0,2), Q=(0,2,2), R=(1,1,4)$. Now notice that the plane determined by $\triangle PQR$ is perpendicular to the plane determined 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 determined by $ABCD$, since $P$ and $Q$ have the same $z$-coordinate. Hence the height of $\triangle PQR$ is equal to the $z$-coordinate of $R$ minus the $z$-coordinate of $P$, giving $4-2= 2$. By the distance formula, $\overline{PQ} = 2\sqrt{2}$, so the area of $\triangle PQR$ is $\frac{1}{2} \cdot 2\sqrt{2} \cdot 2 = \boxed{\textbf{(C) } 2\sqrt{2}}$.

## Solution 3 (geometry)

By the Pythagorean Theorem, we can calculate $EB=ED=3\sqrt{5},EC=3\sqrt{6},ER= \sqrt{6},$ and $EP=EQ=2 \sqrt{5}$. Now by the Law of Cosines in $\triangle BEC$, we have $\cos{\left(\angle BEC\right)}=\frac{EB^2+EC^2-BC^2}{2 \cdot EB \cdot EC}=\frac{5}{\sqrt{30}}$.

Similarly, by the Law of Cosines in $\triangle EPR$, we have $PR^2=ER^2+EP^2-2 \cdot ER \cdot EP \cdot \cos{\left(\angle BEC\right)}=6$, so $PR=\sqrt{6}$. Observe that $\triangle ERP \cong \triangle ERQ$ (by side-angle-side), so $QR=PR=\sqrt{6}$.

Next, notice that $PQ$ is parallel to $DB$, and therefore $\triangle EQP$ is similiar to $\triangle EDB$. Thus we have $\frac{QP}{DB}=\frac{EP}{EB}=\frac{2}{3}$. Since $DB=3\sqrt{2}$, this gives $PQ=2 \sqrt{2}$.

Now we have the three side lengths of isosceles $\triangle PQR$: $PR=QR=\sqrt{6}$, $PQ=2 \sqrt{2}$. Letting the midpoint of $PQ$ be $S$, $RS$ is the perpendicular bisector of $PQ$, and so can be used as a height of $\triangle PQR$ (taking $PQ$ as the base). Using the Pythagorean Theorem again, we have $RS=\sqrt{PR^2-PS^2}=2$, so the area of $\triangle PQR$ is $\frac{1}{2} \cdot PQ \cdot RS = \frac{1}{2} \cdot 2\sqrt{2} \cdot 2 = \boxed{\textbf{(C) } 2 \sqrt{2}}$.