Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC < AD$ and $AB = CD.$ There is a point $P$ in the plane such that $PA=1, PB=2, PC=3,$ and $PD=4.$ What is $\tfrac{BC}{AD}?$ $\textbf{(A) }\frac{1}{4}\qquad$ $\textbf{(B) }\frac{1}{3}\qquad$ $\textbf{(C) }\frac{1}{2}\qquad$ $\textbf{(D) }\frac{2}{3}\qquad$ $\textbf{(E) }\frac{3}{4}$
Solution
Place the trapezoid symmetric about the $y$-axis: $A=(-a,0)$, $D=(a,0)$, $B=(-b,h)$, $C=(b,h)$ with $0<b<a$, and let $P=(x,y)$.
$PA^2-PD^2=(x+a)^2-(x-a)^2=4ax$, and $PA^2-PD^2=1-16=-15$. So $4ax=-15$.
$PB^2-PC^2=(x+b)^2-(x-b)^2=4bx$, and $PB^2-PC^2=4-9=-5$. So $4bx=-5$.
Dividing, $\dfrac ba=\dfrac{-5}{-15}=\dfrac13$, so $\dfrac{BC}{AD}=\dfrac{2b}{2a}=\dfrac13$.
Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC}$ with $AD=22$ and $BC=10$. The diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. What is the area of the trapezoid?
$\textbf{(A) }\frac{1}{4}\qquad$ $\textbf{(B) }\frac{1}{3}\qquad$ $\textbf{(C) }\frac{1}{2}\qquad$ $\textbf{(D) }\frac{2}{3}\qquad$ $\textbf{(E) }\frac{3}{4}$ 