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

Revision as of 11:02, 9 February 2015

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

2007 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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 ==Solution==
-Let <math>A=(0,0,0)</math>, and <math>B=(2,0,0)</math>. Since <math>EA=2</math>, we could let <math>C=(2,0,2)</math>, <math>D=(2,2,2)</math>, and <math>E=(2,2,0)</math>. Now to get back to <math>A</math> we need another vertex <math>F=(0,2,0)</math>. 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 <math>FA</math>. Now we can bend these three sides into an equilateral triangle, and the coordinates change: <math>A=(0,0,0)</math>, <math>B=(2,0,0)</math>, <math>C=(2,0,2)</math>, <math>D=(1,\sqrt{3},2)</math>, and <math>E=(1,\sqrt{3},0)</math>. Checking for all the requirements, they are all satisfied. Now we find the area of triangle <math>BDE</math>. It is a <math>2-2-2\sqrt{2}</math> triangle, which is an isosceles right triangle. Thus the area of it is <math>\frac{2*2}{2}=2\Rightarrow \mathrn{(C)}</math>.
-Integrate that
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Difference between revisions of "2007 AMC 12B Problems/Problem 25"

Revision as of 11:02, 9 February 2015

Problem

Solution

See also