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ABC is an equilateral triangle with side length 4. M is the midpoint of BC, and AM is a diagonal of square ALMN. Find the area of the region common to both ABC and ALMN.

Jul 27, 2019

#1
+3

See the following image: We are looking for the area [ AEMF]

Let B  = (0, 0)   C  = (4, 0)   A  =  (2, 2√3)

Angle EMC  = 45°

The segment MN  lies  on a line with a slope  = tan45°  = 1

So....the equation of this iine is   y = ( 1) (x - 2)  =  x - 2    (1)

And the segment  AC   lies on a line with a slope  of tan (120°)  =  -√3 ( x - 4) = -√3 x + 4√3   (2)

So......the x value of the intersection of (1) and (2)     can be found as

x - 2  =  -√3 x + 4√3

x ( 1 + √3) = 4√3 + 2

x  = [ 4√3 + 2 ] / [ 1 + √3]

So   the y value  of  this intersection is

y =    [ 4√3 + 2 ] / [ 1 + √3]  - 2   =   [ 2√3 ] / [ 1 + √3]

So  E  =   ( [ 4√3 + 2 ] / [ 1 + √3] ,  [ 2√3 ] / [ 1 + √3] )

So...the area of triangle EMC  = (1/2) ( MC) ([ 2√3 ] / [ 1 + √3]) =  (1/2)(2)[ 2√3 ] / [ 1 + √3] =

[ 2√3 ] / [ 1 + √3] units^2     (3)

And the area of  AMC  =  (1/2)(2√3)(2)  = 2√3  units^2   (4)

So   area  [ AEMF]  =   2 ( (4) - (3)  )   =    2  ( 2√3  -  [ 2√3 ] / [ 1 + √3]  )  =  2 (3√3 - 3)  units^2  ≈ 4.39 units^2   Jul 27, 2019
#2
+2

Here's one more way to solve this

Since AM    =  2√3   and is the diagonal of the square....then AN  must be  2√3 / √2  = √2 * √3  =  √6

And angle ANE  is right    and angle NAE = 15°

So angle AEN  =  75°

So....using the Law of Sines

EN / sin NAE  =  AN / sin AEN

EN / sin 15  =  √6 / sin 75

EN  =  √6 sin15 / sin 75  = √6 sin15/cos15  = √6 tan 15   = √6 [ 1 - cos 30] / sin (30)  =

√6 [ 1 - √3/2] / (1/2)  =  √6 [ 2 - √3]

So  the area of triangles AEN and ALF    = AN * EN  =  √6 * √6 [ 2 - √3]  =  6 [2 - √3]    (1)

So the area  common to the square and triangle =   [ AEMF]  =  area of square - (1)  =    6 - 6[2 -√3]  = [ 6√3 - 6]  units^2  ≈ 4.39 units^2   Jul 28, 2019