A B C above is an equilateral triangle and A D = 3 , B D = 4 and C D = 5 . Find A B .
Triangle
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Nice solution.
Amazing solution.
link text
In an equivalent triangle side S, if a point inside it is at distances X, Y, Z from the vertices, their relation is given in the link above by,
3
(
X
4
+
Y
4
+
Z
4
+
S
4
)
=
(
X
2
+
Y
2
+
Z
2
+
S
2
)
2
⟹
3
(
3
4
+
4
4
+
5
4
+
S
4
)
=
(
3
2
+
4
2
+
5
2
+
S
2
)
2
.
E
x
p
a
n
d
i
n
g
a
n
d
s
i
m
p
l
i
f
y
i
n
g
,
2
S
2
−
1
0
0
S
+
3
8
6
=
0
o
r
S
2
−
5
0
S
+
1
9
3
=
0
.
S
o
l
v
i
n
g
t
h
e
q
u
a
d
r
a
t
i
c
S
2
=
4
5
.
7
8
4
6
,
r
e
j
e
c
t
i
n
g
t
h
e
o
t
h
e
r
r
o
o
t
b
e
i
n
g
t
o
o
s
m
a
l
l
.
∴
S
=
6
.
7
6
6
4
3
2
5
Nice, I did not know the result. Thanks for sharing your solution
Refer here to get that twice the area of the equilateral triangle is 2 5 2 3 + 1 8 Hence, if the side of the required triangle is a , a 2 2 3 = 2 5 2 3 + 1 8 ⟹ a = 2 5 + 1 2 3 ≈ 6 . 7 6 6
A B C in magnitude of 6 0 ∘ in the counterclockwise direction as shown in my figure.
Rotate trianglesin ∠ D ’ D C = 5 3
∠ D ’ D C ≈ 3 6 . 8 7 ∘
By cosine rule, we have
( B C ) 2 = 4 2 + 5 2 − 2 ( 4 ) ( 5 ) [ cos ( 6 0 + 3 6 . 8 7 ) ]
B C = A B ≈ 6 . 7 6 6 units
Let generalize this problem for any Pythagorean triple with cathetus a and b , and hypotenuse c .
First moves △ A P C together △ A P B as image shows, clearly, △ A P ′ P is an equilateral triangle and △ P P ′ B is an rectangle triangle of sides a , b and c so ∠ B P A = 6 0 ° + 9 0 ° = 1 5 0 ° .
By law of cosines, the side of the equilateral triangle is s = a 2 + b 2 − 2 ( a ) ( b ) cos 1 5 0 ° .
For this triangle A B = 3 2 + 4 2 − 2 ( 3 ) ( 4 ) cos 1 5 0 ° ≈ 6 . 7 6 6 .
Similar problem: Equilateral triangle point
Any question, be free of asking in comments.
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