數學比賽問題

Let the side of the square be L.
Notice that , after cutting the square out, triangle ARS is equilateral with side equal to L. Hence, height of triangle ARS is {sqrt(3) x L / 2 } . Also, notice that the height of original triangle ABC is { height of ARS + L }. We will then obtain the following equation:
{sqrt(3) x L / 2 } + L = sqrt(3) x 6 / 2
Hence, L = sqrt(3) x 3 / { [ sqrt(3) /2 ]+ 1 } = 12 x sqrt(3) - 18

Now, area of the sqare = L^2 = 432 - 432 x sqrt(3) +324 = 7.75 = apprx. 8

我們要首先劃一條由A垂直於BC的垂直線，而這條線與PQ線相交於M，而與BC線相交於N

Let L be the length of the square

AM=AN-L
=6cos30 - L

我們現在考慮三角形APM
因為PM=AM tan30

PM=L/2
therefore

L/2=AM tan30
L/2 = (6 cos30-L)tan30
L/2 = (3 √3-L)(1/√3)
L(1/2+1/√3) = 3
2.232051L=3
therefore L = 3/2.232051=1.34405531
PQRS的面積＝LxL　大約等於　2(答案至最接整數）

首先，三角形ABC的邊長係6。

三角形ABC的高是等於
=(6^2 - 3^2) ( )括弧入面係開方
=(27) ( )括弧入面係開方

三角ABC的面積
= 3 x (27) / 2 ( )括弧入面係開方

設正方形的邊長係 y
梯形PQCB的高 = y

梯形PQCB的面積
{[y + 6] x y } / 2
= [y^2 + 6y] / 2

三角形APQ的面積
= {y x [(27) - y]} / 2 ( )括弧入面係開方
= [(27)y - y^2] / 2 ( )括弧入面係開方

三角形ABC的面積 = 三角形APQ的面積 + 梯形PQCB的面積
3 x (27) / 2 = [(27)y - y^2] / 2 + [y^2 + 6y] / 2 ( )括弧入面係開方
3 x (27) = [(27)y - y^2] + [y^2 + 6y] ( )括弧入面係開方
3 x (27) = (27)y + 6y ( )括弧入面係開方
3 x (27) = [(27) + 6]y ( )括弧入面係開方
y = [3 x (27)] / [(27) + 6] ( )括弧入面係開方

正方形PQRS的面積 = y^2
因此
正方形PQRS的面積
={[3 x (27)] / [(27) + 6]}^2
= 1.93851278256125
= 2 (最接近的整數)

答案應該係=3x3平方[單位]