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

m (→‎Finding area with perpendicular planes)
m (→‎Solution (Old Fashioned Geometry))
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==Solution (Old Fashioned Geometry)==
 
==Solution (Old Fashioned Geometry)==
  
−
Use Pythagorean Teorem we can quickly obtain the following parameters: EB=ED=3sqrt(5),EC=3sqrt(6),ER=sqrt(6),EP=EQ=2sqrt(5)
+
Use Pythagorean Teorem we can quickly obtain the following parameters: <math>EB=ED=3sqrt(5),EC=3sqrt(6),ER=sqrt(6),EP=EQ=2sqrt(5)</math>
−
Inside triangle EBC, using cosine law:
+
Inside triangle <math>EBC</math>, using cosine law:
−
COS(EBC)=(EB^2+EC^2-BC^2)/(2*EB*EC)=sqrt(30)/6
+
<math>\cos{(\angle EBC)}=\frac{(EB^2+EC^2-BC^2)}{2 \cdot EB \cdot EC}=\frac{\sqrt{30}}{6}</math>
−
Now move to triangle EPR, use cosine law again
+
Now move to triangle <math>EPR</math>, use cosine law again
−
PR^2=ER^2+EP^2-2*ER*EP*COS(EBC)=6, therefore PR=sqrt(6), noticing that triangle ERP is congruent to triangle ERQ, QR=PR=sqrt(6).
+
<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 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 QP/DB=EP/EB=2/3, since DB=3sqrt(2), we have PQ=2sqrt(2).
 
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 QP/DB=EP/EB=2/3, since DB=3sqrt(2), we have PQ=2sqrt(2).
 
Now we have the three side lengths of isosceles triangle PQR: PR=QR=sqrt(6), PQ=2sqrt(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 = 1/2*PQ*RS=2sqrt(2)
 
Now we have the three side lengths of isosceles triangle PQR: PR=QR=sqrt(6), PQ=2sqrt(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 = 1/2*PQ*RS=2sqrt(2)

Revision as of 08:35, 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: $EB=ED=3sqrt(5),EC=3sqrt(6),ER=sqrt(6),EP=EQ=2sqrt(5)$ 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 QP/DB=EP/EB=2/3, since DB=3sqrt(2), we have PQ=2sqrt(2). Now we have the three side lengths of isosceles triangle PQR: PR=QR=sqrt(6), PQ=2sqrt(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 = 1/2*PQ*RS=2sqrt(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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