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.

Question

I drew the diagram but I don't know how to find the answer? I think it has something to do with the law of sines and cosines
Help is appreciated, thanks

Answer 1

I started with AB, the vertical diagonal of the square.
I then drew square ALMN
At M, I drew the base BC of the equilateral triangle perpendicular to AM, letting BM = MC = 2
Joi AB and AC for the triangle.
Let the intersection of ML and AB be P, and
the intersection of MN and AC be Q

clearly triangles APM and AQM are congruent and form the overlap, or common area

So all we need is the area of triangle APM, then double that.
In triangle APM, angle PMA = 45°, anglePAM = 30°
then angle APM = 105°
Also it is easy to show that AM = 2√3
let's find PM
PM/sin30 = 2√3/sin 105
PM = 2√3(sin30)/sin105 = √3/sin105

area of triangle APM = (1/2)(AM)(PM)sin45
= (1/2)(2√3)(√3/sin105)(√2/2)
= (6/sin105)(√2/2)
= 3√2/(2sin105)
I got appr 2.196

better check my algebra and arithmetic, should have written it out on paper first.

If you need an "exact" answer,
change sin(105)
= sin(60+45)
= sin60cos45 + cos60sin45
= ....
then sub back into above answer.

Answer 2

Thanks a lot