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

m (→‎Solution (Old Fashioned Geometry))
m (→‎Solution (Old Fashioned Geometry))
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Now move to <math>\triangle EPR</math>, use cosine law again
 
Now move to <math>\triangle EPR</math>, use cosine law again
 
<math>PR^2=ER^2+EP^2-2 \cdot ER \cdot EP \cdot \cos{(\angle EBC)}=6</math>, therefore <math>PR=\sqrt{6}</math>, noticing that <math>\triangle ERP</math> is congruent to <math>\triangle ERQ</math>, <math>QR=PR=\sqrt{6}</math>.
 
<math>PR^2=ER^2+EP^2-2 \cdot ER \cdot EP \cdot \cos{(\angle EBC)}=6</math>, therefore <math>PR=\sqrt{6}</math>, noticing that <math>\triangle ERP</math> is congruent to <math>\triangle ERQ</math>, <math>QR=PR=\sqrt{6}</math>.
−
Now look at points <math>P</math>, <math>Q</math>, and <math>\triangle EDB</math>, <math>PQ</math> is parallel to <math>DB</math>, and therefore <math>\triangle </math>EQP<math> is similiar to </math>\triangle EDB<math>, we have </math>\frac{QP}{DB}=\frac{EP}{EB}=\frac{2}{3}<math>, since </math>DB=3\sqrt{2}<math>, we have </math>PQ=2 \sqrt{2}<math>.
+
Now look at points <math>P</math>, <math>Q</math>, and <math>\triangle EDB</math>, <math>PQ</math> is parallel to <math>DB</math>, and therefore <math>\triangle EQP</math> is similiar to <math>\triangle EDB</math>, we have <math>\frac{QP}{DB}=\frac{EP}{EB}=\frac{2}{3}</math>, since <math>DB=3\sqrt{2}</math>, we have <math>PQ=2 \sqrt{2}</math>.
−
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>. Suppose the midpoint of </math>PQ<math> is </math>S<math>, connect </math>RS<math>, it would be perpendicular bisector of </math>PQ<math> and act as the height of side </math>PQ<math>. Use Pythagorean again we have </math>RS=\sqrt{PR^2-PS^2}=2<math>, therefore the area of </math>\triangle PQR<math> is = </math>\frac{1}{2} \cdot PQ \cdot RS= \boxed{\textbf{(C) } 2 \sqrt{2}}$.
+
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>. Suppose the midpoint of <math>PQ</math> is <math>S</math>, connect <math>RS</math>, it would be perpendicular bisector of <math>PQ</math> and act as the height of side <math>PQ</math>. Use Pythagorean again we have <math>RS=\sqrt{PR^2-PS^2}=2</math>, therefore the area of <math>\triangle PQR</math> is = <math>\frac{1}{2} \cdot PQ \cdot RS= \boxed{\textbf{(C) } 2 \sqrt{2}}</math>.
  
 
(by Zhen Qin)
 
(by Zhen Qin)

Revision as of 07:41, 15 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 $R$ 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)

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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