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Chords $\overline{UV},$ $\overline{WX},$ and $\overline{YZ}$ of a circle are parallel.  The distance between chords $\overline{UV}$ and $\overline{WX}$ is $1,$ and the distance between chords $\overline{WX}$ and $\overline{YZ}$ is also $1.$  If $UV = 4$ and $YZ = 2,$ then find $WX.$

 

 Jan 7, 2024
 #1
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Let the center be O    and let the midpoint of UV =  P

 

Draw OP   and OV

 

Triangle  OPV is a right triangle  with angle VPO =  90

 

Cal the distance from O to UV =  d   and call the distance from OV  = r

And PV =  2

 

So

 

OV^2  = OP^2 + PV^2

r^2  = d^2 + 4            (1)

 

Likewise, we can let M be the midpoint of YZ

Connect  OM and OZ

OM = d + 2

OZ =  r

And MZ = 1

 

So   another right triangle is  formed such that

OZ^2 = OM^2 + MZ^2

r^2  = (d + 2)^2 +  1      (2)

 

Equate (1) and (2)

 

d^2 + 4 =  (d + 2)^2 + 1

 

d^2 + 4 = d^2 + 4d + 4  + 1

 

4d = -1

 

d = -1/4

 

Impossible....so....no solution to this

 

 

cool cool cool

 Jan 7, 2024

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