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

Revision as of 21:31, 21 December 2020

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 $\frac{1}{2} \cdot PQ \cdot RS = \frac{1}{2} \cdot 2\sqrt{2} \cdot 2 = \boxed{\textbf{(C) } 2 \sqrt{2}}$ .

@@ Line 37: / Line 37: @@
 ==Solution 3 (geometry)==
-By the Pythagorean Theorem, we can calculate <math>EB=ED=3\sqrt{5},EC=3\sqrt{6},ER= \sqrt{6},</math> and <math>EP=EQ=2 \sqrt{5}</math>. Now by the Law of Cosines in <math>\triangle EBC</math>, we have
+By the Pythagorean Theorem, we can calculate <math>EB=ED=3\sqrt{5},EC=3\sqrt{6},ER= \sqrt{6},</math> and <math>EP=EQ=2 \sqrt{5}</math>. Now by the Law of Cosines in <math>\triangle BEC</math>, we have
-<math>\cos{\left(\angle EBC\right)}=\frac{EB^2+EC^2-BC^2}{2 \cdot EB \cdot EC}=\frac{\sqrt{30}}{6}</math>.
+<math>\cos{\left(\angle BEC\right)}=\frac{EB^2+EC^2-BC^2}{2 \cdot EB \cdot EC}=\frac{5}{\sqrt{30}}</math>.
 Similarly, by the Law of Cosines in <math>\triangle EPR</math>, we have
-<math>PR^2=ER^2+EP^2-2 \cdot ER \cdot EP \cdot \cos{\left(\angle EBC\right)}=6</math>, so <math>PR=\sqrt{6}</math>. Observe that <math>\triangle ERP \cong \triangle ERQ</math> (by ''side-angle-side''), so <math>QR=PR=\sqrt{6}</math>.
+<math>PR^2=ER^2+EP^2-2 \cdot ER \cdot EP \cdot \cos{\left(\angle BEC\right)}=6</math>, so <math>PR=\sqrt{6}</math>. Observe that <math>\triangle ERP \cong \triangle ERQ</math> (by ''side-angle-side''), so <math>QR=PR=\sqrt{6}</math>.
 Next, notice that <math>PQ</math> is parallel to <math>DB</math>, and therefore <math>\triangle EQP</math> is similiar to <math>\triangle EDB</math>. Thus we have <math>\frac{QP}{DB}=\frac{EP}{EB}=\frac{2}{3}</math>. Since <math>DB=3\sqrt{2}</math>, this gives <math>PQ=2 \sqrt{2}</math>.

2019 AMC 12B (Problems • Answer Key • Resources)
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