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.
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
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