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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.

 May 10, 2019
 #1
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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) 

 

 

 

cool cool cool

 May 11, 2019
edited by CPhill  May 11, 2019

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