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Difference between revisions of "2007 AMC 12B Problems/Problem 25"

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Also refer to the [https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_25#Solution| 2007 AMC 10B #25] (same problem)
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==Problem==
 
==Problem==
 
Points <math>A,B,C,D</math> and <math>E</math> are located in 3-dimensional space with <math>AB=BC=CD=DE=EA=2</math> and <math>\angle ABC=\angle CDE=\angle DEA=90^o</math>. The plane of <math>\triangle ABC</math> is parallel to <math>\overline{DE}</math>. What is the area of <math>\triangle BDE</math>?
 
Points <math>A,B,C,D</math> and <math>E</math> are located in 3-dimensional space with <math>AB=BC=CD=DE=EA=2</math> and <math>\angle ABC=\angle CDE=\angle DEA=90^o</math>. The plane of <math>\triangle ABC</math> is parallel to <math>\overline{DE}</math>. What is the area of <math>\triangle BDE</math>?

Revision as of 00:37, 25 December 2020

Also refer to the 2007 AMC 10B #25 (same problem)

Problem

Points $A,B,C,D$ and $E$ are located in 3-dimensional space with $AB=BC=CD=DE=EA=2$ and $\angle ABC=\angle CDE=\angle DEA=90^o$. The plane of $\triangle ABC$ is parallel to $\overline{DE}$. What is the area of $\triangle BDE$?

$\mathrm {(A)} \sqrt{2}\qquad \mathrm {(B)} \sqrt{3}\qquad \mathrm {(C)} 2\qquad \mathrm {(D)} \sqrt{5}\qquad \mathrm {(E)} \sqrt{6}$

Solution

Let $A=(0,0,0)$, and $B=(2,0,0)$. Since $EA=2$, we could let $C=(2,0,2)$, $D=(2,2,2)$, and $E=(2,2,0)$. Now to get back to $A$ we need another vertex $F=(0,2,0)$. Now if we look at this configuration as if it was two dimensions, we would see a square missing a side if we don't draw $FA$. Now we can bend these three sides into an equilateral triangle, and the coordinates change: $A=(0,0,0)$, $B=(2,0,0)$, $C=(2,0,2)$, $D=(1,\sqrt{3},2)$, and $E=(1,\sqrt{3},0)$. Checking for all the requirements, they are all satisfied. Now we find the area of triangle $BDE$. The side lengths of this triangle are $2, 2, 2\sqrt{2}$, which is an isosceles right triangle. Thus the area of it is $\frac{2\cdot2}{2}=2\Rightarrow \mathrm{(C)}$.

Solution 2

Similar to solution 1, we allow $A=(0,0,0)$, $B=(2,0,0)$, and $C=(0,2,0)$. This creates the isosceles right triangle on the plane of $z=0$

Now, note that $\angle CDE=\angle DEA=90^o$. This means that there exists some vector $DE$ parallel to the plane of $ABC$ that forms two right angles with $AE$ and $CD$. By definition, this is the cross product of the two vectors $AE$ and $CD$. Finding this cross product, we take the determinant of vectors

$AE=<x_1,y_1,z>$ and

$CD=<x_2,y_2,z>$ *Note that z is constant because the line is parallel to the plane*

to get $(y_1-y_2)zi+(x_1-x_2)zj+(x_1y_2-y_1x_2)k$

Because there can be no movement in the $z$ direction, the k unit vector must be zero. Also, because the i unit vector must be orthogonal and also 0. Thus, the vector of line $DE$ is simply $2tj+2k$

From this, you can figure out that line $BE=2$, and the area of $BDE=\frac{2\cdot2}{2}=2\Rightarrow \mathrm{(C)}$.

See also

2007 AMC 12B (Problems • Answer Key • Resources)
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Problem 24 		Followed by
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All AMC 12 Problems and Solutions

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