2022 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 23 Easy

Isosceles trapezoid ABCD has parallel sides \overline{AD} and \overline{BC}, with BC\lt 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 \frac{BC}{AD} ?

  • A.

    \dfrac{1}{4}

  • B.

    \dfrac{1}{3}

  • C.

    \dfrac{1}{2}

  • D.

    \dfrac{2}{3}

  • E.

    \dfrac{3}{4}

Answer:B

\begin{cases}m^2+n^2=1 \\ n^2+(m-a)^2=16 \\ (m-b)^2+(n-t)^2=2^2 \\ (m-c)^2+(n-t)^2=3^2 \end{cases}

So \begin{cases}a^2-2ma=15 \\ c^2-b^2+2(b-c)m=5 \end{cases}

Since ABCD is equilateral triangle, so a=c+b.

(c+b)(c-b)+2(b-c)m=\frac15

(c-b)(a-2m)=5

a(a-2m)=15

Thus, \frac{c-b}{a}=\frac 13.

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