Question

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

$\sqrt{2}$

$\sqrt{3}$

$2$

$\sqrt{5}$

$\sqrt{6}$

Solution:

Set $D=(-1,0,0)$ and $E=(1,0,0),$ and let $\triangle ABC$ lie in the plane $z=k\gt0.$ Because $\angle CDE$ and $\angle DEA$ are right angles, $A$ and $C$ lie on radius- $2$ circles centered at $E$ and $D$ in the planes $x=1$ and $x=-1,$ so $A=(1,y_1,k),\ C=(-1,y_2,k)$ with $y_j=\pm\sqrt{4-k^2}.$

Since $\angle ABC=90^\circ,$ $AC=2\sqrt2,$ which forces $y_1=-y_2.$ Taking $y_1=1,$ $y_2=-1,$ gives $k=\sqrt3,$ so $A=(1,1,\sqrt3),$ $C=(-1,-1,\sqrt3),$ and $B$ is $(1,-1,\sqrt3)$ or $(-1,1,\sqrt3).$

In the first case $BE=2$ with $BE\perp DE;$ in the second $BD=2$ with $BD\perp DE.$ Either way $\triangle BDE$ has legs $2$ and $2,$ so its area is $\tfrac12(2)(2)=2.$

Thus, the correct answer is C.

2007 AMC 12B Problem 25

Solution:

Problem 25 in Other Years