Question

area of a square whose vertices lie on the sides of an euilateral triangle

Answer 1

This is going to be a proof by contradiction, we'll assume that △LMN is equilateral and will try to contradict that.

triangle

First we gonna draw a perpendicular line to the side AB, cutting it at point L, like it's done on the picture. Now we gonna apply the Pythagorean Theorem to the right triangles, but before we gonna introduce some notations.

Let: a=AB=BC=CD=AD
a
=
A
B
=
B
C
=
C
D
=
A
D
, x=BN
x
=
B
N
and y=BL
y
=
B
L
. Now we apply the Pythagorean Theroem.

From △MCN we have:

MN2=CM2+CN2=(a2)2+(a−x)2
M
N
2
=
C
M
2
+
C
N
2
=
(
a
2
)
2
+
(
a
−
x
)
2
MN2=a24+a2−2ax+x2
M
N
2
=
a
2
4
+
a
2
−
2
a
x
+
x
2
MN2=5a24−2ax+x2
M
N
2
=
5
a
2
4
−
2
a
x
+
x
2
From △LBN we have:

LN2=BL2+BN2
L
N
2
=
B
L
2
+
B
N
2
LN2=x2+y2
L
N
2
=
x
2
+
y
2
From △LEM we have:

LM2=LE2+EM2=a2+(y−a2)2
L
M
2
=
L
E
2
+
E
M
2
=
a
2
+
(
y
−
a
2
)
2
LM2=a2+y2−ay+a24
L
M
2
=
a
2
+
y
2
−
a
y
+
a
2
4
LM2=5a24−ay+y2
L
M
2
=
5
a
2
4
−
a
y
+
y
2
Because △LMN is equilateral it means that the all sides are equal so we get:

MN2=LN2
M
N
2
=
L
N
2
5a24−2ax+x2=x2+y2
5
a
2
4
−
2
a
x
+
x
2
=
x
2
+
y
2
5a24−2ax−y2=0
5
a
2
4
−
2
a
x
−
y
2
=
0
For the other side we have:

LM2=LN2
L
M
2
=
L
N
2
5a24−ay+y2=x2+y2
5
a
2
4
−
a
y
+
y
2
=
x
2
+
y
2
5a24−ay−x2=0⟹y=5a24−x2a
5
a
2
4
−
a
y
−
x
2
=
0
⟹
y
=
5
a
2
4
−
x
2
a
Note that a≠0
a
≠
0
, because A,B,C,D
A
,
B
,
C
,
D
are distinct points. Now we substitute into the previous equation and we get:

5a24−2ax−y2=0
5
a
2
4
−
2
a
x
−
y
2
=
0
5a24−2ax−(5a24−x2a)2=0
5
a
2
4
−
2
a
x
−
(
5
a
2
4
−
x
2
a
)
2
=
0
Little playing and we can't transform this equation to:

−(4x2−8ax+5a2)(4x2+8ax+a2)16a2=0
−
(
4
x
2
−
8
a
x
+
5
a
2
)
(
4
x
2
+
8
a
x
+
a
2
)
16
a
2
=
0
Obviously the denominator can't be 0, so we need to find the root of the two quadratic equation. The first one produces 2 solution:

x1/2=(1±i2)a
x
1
/
2
=
(
1
±
i
2
)
a
But x
x
can't have complex number value. The second one produces also 2 solution and they are:

x3/4=a(−2±3‾√)2
x
3
/
4
=
a
(
−
2
±
3
)
2
But this provides that x
x
has negative value, but one side can't have a negative length.

This means that the assumption that △LMN is equilateral is wrong.