A geometry problem by Archit Tripathi
Let ′ J ′ be any point lying on the circum-circle of a square P Q R S let A B C be an equilateral triangle inscribed in the same circle. The radius of the circle is r , then ( J A ) 4 + ( J B ) 4 + ( J C ) 4 = m r n . Find the value of m − 2 n
The answer is 10.
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1 solution
A
B
C
i
s
a
n
e
q
u
i
l
a
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e
r
a
l
Δ
.
J
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n
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c
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r
c
u
m
c
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r
c
l
e
.
L
e
t
k
=
J
A
.
m
=
J
B
.
n
=
J
C
.
S
=
l
e
n
g
t
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d
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.
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6
0
o
a
t
i
t
s
o
p
p
o
s
i
t
e
v
e
r
t
e
x
a
n
d
a
t
J
.
C
o
s
6
0
=
2
1
,
a
n
d
C
o
s
1
2
0
=
−
2
1
.
I
n
Δ
s
A
J
B
,
B
J
C
a
n
d
A
J
C
,
a
p
p
l
y
i
n
g
C
o
s
R
u
l
e
,
k
2
+
m
2
−
2
∗
k
∗
m
∗
2
−
1
=
S
2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
(
1
)
m
2
+
n
2
−
2
∗
m
∗
n
∗
2
1
=
S
2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
(
2
)
n
2
+
k
2
−
2
∗
n
∗
k
∗
2
1
=
S
2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
(
3
)
(
3
)
−
(
2
)
,
g
i
v
e
s
:
−
k
2
−
m
2
=
n
∗
(
k
−
m
)
.
∴
(
k
−
m
)
∗
(
k
+
m
)
=
n
∗
(
k
−
m
)
⟹
(
A
)
k
=
m
O
R
(
B
)
k
=
m
b
u
t
n
=
k
+
m
.
.
.
.
.
.
.
.
.
.
.
.
.
(
4
)
(
A
)
k
=
m
F
r
o
m
(
1
)
,
3
k
2
=
S
2
.
F
r
o
m
(
3
)
,
n
2
+
3
1
S
2
−
3
n
∗
S
−
S
2
=
0
.
S
o
l
v
i
n
g
q
u
a
d
r
a
t
i
c
i
n
n
>
0
,
n
=
3
2
∗
S
.
∴
k
4
+
m
4
+
n
4
=
{
9
1
+
9
1
+
9
1
6
}
S
4
=
2
S
4
(
B
)
k
=
m
Δ
s
A
J
C
+
C
J
B
=
A
J
B
,
⟹
2
1
k
∗
n
S
i
n
6
0
+
2
1
n
∗
m
S
i
n
6
0
=
2
1
k
∗
m
S
i
n
1
2
0
+
2
1
S
2
S
i
n
6
0
.
∴
−
k
∗
m
+
m
∗
n
+
n
∗
k
=
S
2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
(
5
)
S
q
u
a
r
i
n
g
b
o
t
h
s
i
d
e
s
o
f
(
5
)
k
2
∗
m
2
+
m
2
∗
n
2
+
n
2
∗
k
2
+
2
∗
k
∗
m
∗
n
∗
(
k
+
m
−
n
)
=
S
4
.
B
u
t
b
y
(
4
)
(
k
+
m
−
n
)
=
0
.
⟹
k
2
∗
m
2
+
m
2
∗
n
2
+
n
2
∗
k
2
=
S
4
.
.
.
.
.
.
.
.
.
(
6
)
A
d
d
i
n
g
(
1
)
+
(
2
)
+
(
3
)
+
(
5
)
,
a
n
d
d
i
v
i
d
i
n
g
b
y
2
w
e
g
e
t
k
2
+
m
2
+
n
2
=
2
S
2
.
.
.
.
.
.
.
.
(
7
)
∴
k
4
+
m
4
+
n
4
=
{
k
2
+
m
2
+
n
2
}
2
−
2
∗
{
k
2
∗
m
2
+
m
2
∗
n
2
+
n
2
∗
k
2
}
F
r
o
m
(
6
)
a
n
d
(
7
)
,
k
4
+
m
4
+
n
4
=
4
S
4
−
2
S
4
=
2
S
4
.
⟹
(
J
A
)
4
+
(
J
B
)
4
+
(
J
C
)
4
=
2
S
4
=
2
(
3
∗
r
)
4
=
1
8
r
4
S
o
m
=
1
8
,
a
n
d
n
=
4
.
m
−
2
n
=
1
0
E
a
c
h
s
i
d
e
i
s
a
c
h
o
r
d
o
f
t
h
e
c
i
r
c
l
e
.
I
t
s
u
b
s
t
a
n
c
e
a
n
a
n
g
l
e
o
f
6
0
o
a
t
i
t
s
o
p
p
o
s
i
t
e
v
e
r
t
e
x
a
n
d
a
t
J
.