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

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==Problem==
 
==Problem==
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Square pyramid <math>ABCDE</math> has base <math>ABCD</math>, which measures <math>3</math> cm on a side, and altitude <math>AE</math> perpendicular to the base, which measures <math>6</math> cm. Point <math>P</math> lies on <math>BE</math>, one third of the way from <math>B</math> to <math>E</math>; point <math>Q</math> lies on <math>DE</math>, one third of the way from <math>D</math> to <math>E</math>; and point <math>R</math> lies on <math>CE</math>, two thirds of the way from <math>C</math> to <math>E</math>. What is the area, in square centimeters, of <math>\triangle{PQR}</math>?
  
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==Solution==
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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>
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==Solution 1 (coordinate bash)==
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 +
Using the given data, we can label the points <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 also find the points <math>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)</math>. Similarly, <math>Q = (0, 2, 2)</math> and <math>R = (1, 1, 4)</math>.
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Using the distance formula, <math>PQ = \sqrt{\left(-2\right)^2 + 2^2 + 0^2} = 2\sqrt{2}</math>, <math>PR = \sqrt{\left(-1\right)^2 + 1^2 + 2^2} = \sqrt{6}</math>, and <math>QR = \sqrt{1^2 + \left(-1\right)^2 + 2^2} = \sqrt{6}</math>. Using Heron's formula, or by dropping an altitude from <math>P</math> to find the height, we can then find that the area of <math>\triangle{PQR}</math> is <math>\boxed{\textbf{(C) }2\sqrt{2}}</math>.
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''Note'': After finding the coordinates of <math>P,Q,</math> and <math>R</math>, we can alternatively find the vectors <math>\overrightarrow{PQ}=[-2,2,0]</math> and <math>\overrightarrow{PR}=[-1,1,2]</math>, then apply the formula <math>\text{area} = \frac{1}{2}\left|\overrightarrow{PQ} \times \overrightarrow{PR}\right|</math>. In this case, the cross product equals <math>[4,4,0]</math>, which has magnitude <math>4\sqrt{2}</math>, giving the area as <math>2\sqrt{2}</math> like before.
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 +
==Solution 2==
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As in Solution 1, 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>, and calculate the coordinates of <math>P</math>, <math>Q</math>, and <math>R</math> as <math>P=(2,0,2), Q=(0,2,2), R=(1,1,4)</math>. Now notice that the plane determined by <math>\triangle PQR</math> is perpendicular to the plane determined 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>:
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<asy>
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unitsize(40);
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for(int i =0; i<=3; ++i) {
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draw((0,i)--(3,i));
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draw((i,0)--(i,3));
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}
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label("$A$", (0,0), SW);
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label("$B$", (3,0), SE);
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label("$C$", (3,3), NE);
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label("$D$", (0,3), NW);
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label("$P$", (2,0), S);
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label("$Q$", (0,2), W);
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label("$R$", (1,1), NE);
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dot((2,0));
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dot((0,2));
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dot((1,1));
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draw((0,2)--(2,0));
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</asy>
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Additionally, we know that <math>PQ</math> is parallel to the plane determined by <math>ABCD</math>, since <math>P</math> and <math>Q</math> have the same <math>z</math>-coordinate. Hence the height of <math>\triangle PQR</math> is equal to the <math>z</math>-coordinate of <math>R</math> minus the <math>z</math>-coordinate of <math>P</math>, giving <math>4-2= 2</math>. By the distance formula, <math>\overline{PQ} = 2\sqrt{2}</math>, so the area of <math>\triangle PQR</math> is <math>\frac{1}{2} \cdot 2\sqrt{2} \cdot 2 = \boxed{\textbf{(C) } 2\sqrt{2}}</math>.
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==Solution 3 (geometry)==
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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
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<math>\cos{\left(\angle BEC\right)}=\frac{EB^2+EC^2-BC^2}{2 \cdot EB \cdot EC}=\frac{5}{\sqrt{30}}</math>.
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Similarly, by the Law of Cosines in <math>\triangle EPR</math>, we have
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<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>.
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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>.
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Now we have the three side lengths of isosceles <math>\triangle PQR</math>: <math>PR=QR=\sqrt{6}</math>, <math>PQ=2 \sqrt{2}</math>. Letting the midpoint of <math>PQ</math> be <math>S</math>, <math>RS</math> is the perpendicular bisector of <math>PQ</math>, and so can be used as a height of <math>\triangle PQR</math> (taking <math>PQ</math> as the base). Using the Pythagorean Theorem again, we have <math>RS=\sqrt{PR^2-PS^2}=2</math>, so the area of <math>\triangle PQR</math> is <math>\frac{1}{2} \cdot PQ \cdot RS = \frac{1}{2} \cdot 2\sqrt{2} \cdot 2 = \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}}

Latest revision as of 18:30, 12 November 2022

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

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

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