+1768 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 $WX.$
+129895 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
