Track / Stage 3 / 223 of 260 #223 of 1964

Problem 223

AMC 10/12, early questions

Geometry Difficulty 3.8 Multiple choice

Figure (Asymptote source)

import cse5; pathpen=black; pointpen=black; dotfactor=3; pair A=(1,2),B=(2,0),C=(0,0); D(CR(A,1.5)); D(CR(B,1.5)); D(CR(C,1.5)); D(MP("$A$",A)); D(MP("$B$",B)); D(MP("$C$",C)); pair[] BB,CC; CC=IPs(CR(A,1.5),CR(B,1.5)); BB=IPs(CR(A,1.5),CR(C,1.5)); D(BB[0]--CC[1]); MP("$B'$",BB[0],NW);MP("$C'$",CC[1],NE); //Credit to TheMaskedMagician for the diagram

Circles with centers

A ,~ B

, and

C

each have radius

r

, where

1 < r < 2

.
The distance between each pair of centers is

2

. If

B'

is the point of intersection of circle

A

and circle

C

which is outside circle

B

, and if

C'

is the point of intersection of circle

A

and circle

B

which is outside circle

C

,
then length

B'C'

equals

Pick one

Official solution

The circles can be described in the cartesian plane as being centered at $(-1,0),(1,0)$ and $(0,\sqrt{3})$ with radius $r$ by the equations
$x^2+(y-\sqrt{3})^2=r^2$
$(x+1)^2+y^2=r^2$
$(x-1)^2+y^2=r^2$ .
Solving the first 2 equations gives $x=1-\sqrt{3}\cdot y$ which when substituted back in gives $y=\frac{\sqrt{3}\pm \sqrt{r^2-1}}{2}$ .
The larger root $y=\frac{\sqrt{3}+\sqrt{r^2-1}}{2}$ is the point B' described in the question. This root corresponds to $x=-\frac{1+\sqrt{3(r^2-1)}}{2}$ .
By symmetry across the y-axis the length of the line segment $B'C'$ is $1+\sqrt{3(r^2-1)}$ which is $\boxed{D}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.