A regular hexagon of side length 1 is placed vertically along one of its edges on a flat surface. A particle is fired from the surface and passes through four of the vertices of the hexagon, as shown below, tracing the arc of a parabola. If the particle starts at A and ends at B, then find the length AB.

Guest May 10, 2019

#1**+2 **

Look at the graph here :

Let O = (0, 0)

By the Law of Cosines we can find the height, h, of point F above the x axis

h^2 = 1^2 + 1^2 - 2(1)(1)cos120

h^2 = 2 - 2(1/2)

h^2 = 2 - 1

h = sqrt(3).....so F = (.5, sqrt(3) )

And D = (1, sqrt(3)/2 )

And let the vertex of the parabola be (0, k)

Since the parabola passes through D and F we have that

sqrt(3) = a( .5)^2 + k

sqrt(3)/2 = a(1 )^2 + k subtract these and we have

sqrt(3)/2 = a (-3/4)

-4sqrt(3)/6 = a

-2sqrt(3)/3 = a

-2/sqrt(3) = a

So....we can find k as

sqrt (3) = -2/sqrt(3)(.5)^2 + k

sqrt(3) + 2/ [4sqrt(3) ] = k

[12 + 2 ] / [ 4sqrt(3) ] = k

7/[2sqrt(3) ] = k

To find the positive x coordinate of "B" we have

0 = -2/[sqrt(3)] (x)^2 + 7/ [2sqrt(3)]

-7/ [ 2sqrt(3) ] = - 2/ [ sqrt(3) ] (x)^2

7/4 = x^2

sqrt(7)/2 = x

So....by symmetry, AB = 2 *sqrt(7) / 2 = sqrt (7)

CPhill May 11, 2019