What is the maximal overlapping area between a circle and an equilateral triangle both with the same 1 cm 2 area ?
Submit the answer in cm 2 to 3 decimal places.
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You have done a good job obtaining the answer. However many steps could be shortened.
α
=
arcsin
⎝
⎛
π
1
1
2
×
π
1
2
−
a
2
×
π
⎠
⎞
=
arcsin
⎝
⎛
1
2
1
2
−
3
4
×
π
⎠
⎞
=
arcsin
(
1
−
3
3
π
)
I missed since I took
α
i
n
p
l
a
c
e
o
f
2
α
.
. My solution is as under.
LaTeX:
S
T
=
4
3
×
a
2
⇒
a
=
3
4
×
S
T
=
3
4
=
2
×
3
1
S
C
=
S
=
π
×
r
2
⇒
r
=
π
S
=
π
1
tan
(
3
0
)
=
2
a
∣
S
M
∣
=
a
2
×
∣
S
M
∣
⇒
∣
S
M
∣
=
2
a
×
tan
(
3
0
)
=
2
a
×
3
1
=
2
×
3
a
∣
C
M
∣
=
a
2
−
(
2
a
)
2
=
a
2
−
4
a
2
=
4
3
×
a
2
=
a
×
2
3
∣
C
S
∣
=
∣
C
M
∣
−
∣
S
M
∣
=
2
a
×
3
−
2
×
3
a
=
2
×
3
3
×
a
−
a
=
2
×
3
2
×
a
=
3
a
∣
N
S
∣
=
∣
C
S
∣
2
−
(
2
a
)
2
=
3
a
2
−
4
a
2
=
1
2
a
2
=
a
×
1
2
1
∣
N
R
∣
=
r
2
−
∣
N
S
∣
2
=
(
π
1
)
2
−
(
a
×
1
2
1
)
2
=
π
1
−
1
2
a
2
=
1
2
×
π
1
2
−
a
2
×
π
α
=
arcsin
⎝
⎛
π
1
1
2
×
π
1
2
−
a
2
×
π
⎠
⎞
=
arcsin
⎝
⎛
1
2
1
2
−
3
4
×
π
⎠
⎞
=
arcsin
(
1
−
2
7
π
)
S
A
R
C
=
2
1
×
(
π
1
)
2
×
(
1
8
0
π
×
2
×
α
−
sin
(
2
×
α
)
)
=
2
×
π
1
×
(
1
8
0
π
×
2
×
α
−
sin
(
2
×
α
)
)
=
1
8
0
×
π
π
×
α
−
2
×
π
sin
(
2
×
α
)
=
1
8
0
×
π
π
×
α
−
9
0
×
sin
(
2
×
α
)
S
=
S
C
−
3
×
S
A
R
C
=
1
−
3
×
1
8
0
×
π
π
×
arcsin
(
1
−
2
7
π
)
−
9
0
×
sin
(
2
×
arcsin
(
1
−
2
7
π
)
)
∼
0
.
8
1
7
5
2
8
6
9
4
.
.
.
Above is the same program whose latex your editor has given. But your editor has inserted many many
unnecessary { } and \left \right, making it very difficult to follow. You may go for a better editor. Since I write
directly in latex I have no idea of any editor. One thing you can do even with your editor. After every \\ add three or more spaces and a return. In this way your lines even in latex will be separated. With best wishes.
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S T = 4 3 × a 2 ⇒ a = 3 4 × S T = 3 4 = 2 × 3 1 S C = S = π × r 2 ⇒ r = π S = π 1 tan ( 3 0 ) = 2 a ∣ S M ∣ = a 2 × ∣ S M ∣ ⇒ ∣ S M ∣ = 2 a × tan ( 3 0 ) = 2 a × 3 1 = 2 × 3 a ∣ C M ∣ = a 2 − ( 2 a ) 2 = a 2 − 4 a 2 = 4 3 × a 2 = a × 2 3 ∣ C S ∣ = ∣ C M ∣ − ∣ S M ∣ = 2 a × 3 − 2 × 3 a = 2 × 3 3 × a − a = 2 × 3 2 × a = 3 a ∣ N S ∣ = ∣ C S ∣ 2 − ( 2 a ) 2 = 3 a 2 − 4 a 2 = 1 2 a 2 = a × 1 2 1 ∣ N R ∣ = r 2 − ∣ N S ∣ 2 = ( π 1 ) 2 − ( a × 1 2 1 ) 2 = π 1 − 1 2 a 2 = 1 2 × π 1 2 − a 2 × π sin ( α ) = r ∣ N R ∣ ⇒ α = arcsin ( π 1 1 2 × π 1 2 − a 2 × π ) = arcsin ( 1 2 × π 1 2 × π − a 2 × π 2 ) = arcsin ( 1 2 × π π × ( 1 2 − a 2 × π ) ) = arcsin ( 1 2 1 2 − a 2 × π ) = arcsin ( 1 2 1 2 − 4 × 3 1 × π ) = arcsin ( 1 2 × 3 1 2 × 3 − 4 × π ) = arcsin ( 3 × 3 3 × 3 − π ) = arcsin ( 2 7 2 7 − π ) = arcsin ( 1 − 2 7 π ) S A R C = 2 1 × ( π 1 ) 2 × ( 1 8 0 π × 2 × α − sin ( 2 × α ) ) = 2 × π 1 × ( 1 8 0 π × 2 × α − sin ( 2 × α ) ) = 1 8 0 × π π × α − 2 × π sin ( 2 × α ) = 1 8 0 × π π × α − 9 0 × sin ( 2 × α ) S = S C − 3 × S A R C = 1 − 3 × 1 8 0 × π π × arcsin ( 1 − 2 7 π ) − 9 0 × sin ( 2 × arcsin ( 1 − 2 7 π ) ) ∼ 0 . 8 1 7 5 2 8 6 9 4 . . .