In convex quadrilateral ABCD, AB=BC=13, CD=DA=24, and \(\angle D=60^\circ\). Points X and Y are the midpoints of \(\overline{BC}\) and \(\overline{DA}\) respectively. Compute \(XY^2\)(the square of the length of XY).

Guest Dec 20, 2018

#1**+1 **

Here's the image :

Let D = (0,0)

Let C = ( 24sin 60, 24 sin 60) = (12, 12sqrt3 )

And by symmetry, angle BDC = 60/2 = 30°

We can find sin DBC as follows

sin BDC /13 = sin BCD/24

sin30/13 = sinDBC/24

1/26 =sinDBC/24

24/26 = sin DBC

12/13 = sin DBC

So

5/13 = cos BCD

And we can find BD using the Law of Cosines

DC^2 = BD^2 + BC^2 - 2(BD*BC) cosBCD

24^2 = BD^2 +13^2 - 2(BD *13)(5/13)

576 -169 = BD^2 - 10BD

407 = BD^2 - 10BD

BD^2 - 10BD - 407 = 0

The solution to this is

BD = 5 + 12sqrt3

So B = (0 , 5 + 12sqrt 3 )

So X is the midpoint of B and D =

[ 12/2, [ 5 +12sqrt(3) + 12sqrt(3)] /2 ] =

[ 6 , 2.5 + 12sqrt(3) ]

So....by symmetry... A is (-12, 12sqrt3)

So....Y is the midpoint of D and A = (-6, 6sqrt(3) )

So....the distance^2 X to Y =

[ 6 - (-6) ]^2 + [ 2.5 + 12sqrt (3) - 6sqrt(3) ]^2 =

12^2 + [ 2.5 + 6sqrt(3) ]^2 =

144 + 6.25 + 30sqrt(3) +108 ≈

310.21 units

CPhill Dec 21, 2018