A square is inscribed in a regular pentagon such that all four vertices touch the pentagon. If the side length of the pentagon is 1 and the side length of the square is s , find ⌊ 1 0 0 0 0 s ⌋ .
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One of the reasons I love mathematics is that it's such an unexpected platform for creativity. There are so many original ways to approach a problem. Thank you, Chew-Seong, for demonstrating this once again. By the way, I find a certain beauty in your penultimate expression of the solution, sin 3 8 ∘ + sin 1 8 ∘ sin 1 0 8 ∘ .
We place the regular pentagon
A
B
C
D
E
on the plane so that
C
D
lie on the x-axis and
a
on the positive part of y-axis. The four vertices
P
,
Q
,
R
,
S
of the square will lay on four of the pentagon’s sides. Let them be
C
D
,
A
B
,
A
E
, and
E
D
respectively. From points
Q
,
R
and
S
we drop perpendicular to the axes, thus forming three congruent right triangles
△
P
Q
F
,
△
R
Q
L
and
△
S
P
G
. Let
∠
Q
P
F
=
∠
Q
R
L
=
∠
P
S
G
=
θ
. Denote by
x
the x-coordinate of
P
/ and by
s
the side length of the square.
Then, we have
P ( x , 0 ) Q ( x + s ⋅ cos θ , s ⋅ sin θ ) R ( x + s ⋅ cos θ − s ⋅ sin θ , s ⋅ sin θ + s ⋅ cos θ ) S ( x − s ⋅ sin θ , s ⋅ cos θ ) The y-coordinate of vertex A is the height of the pentagon, thus A ( 0 , 2 5 + 2 5 ) .
The equations of line A B is
y = − ( tan 3 6 ∘ ) x + 2 5 + 2 5 ⇔ y = − 5 − 2 5 x + 2 5 + 2 5
The equations of line A E is
y = ( tan 3 6 ∘ ) x + 2 5 + 2 5 ⇔ y = 5 − 2 5 x + 2 5 + 2 5
The equations of line D E is
y = ( tan 1 0 8 ∘ ) ( x + 2 1 ) x ⇔ y = − 5 + 2 5 ( x + 2 1 )
Since Q ∈ A B , R ∈ A E and S ∈ D E , we get the system ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ s ⋅ sin θ s ⋅ sin θ + s ⋅ cos θ s ⋅ cos θ = − 5 − 2 5 ( x + s ⋅ cos θ ) + 2 5 + 2 5 = 5 − 2 5 ( x − s ⋅ sin θ + s ⋅ cos θ ) + 2 5 + 2 5 = − 5 + 2 5 ( x − s ⋅ sin θ + 2 1 ) Settting y = s ⋅ sin θ and z = s ⋅ cos θ , we come to a linear 3x3 system:
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ y y + z z = − 5 − 2 5 ( x + z ) + 2 5 + 2 5 = 5 − 2 5 ( x − y + z ) + 2 5 + 2 5 = − 5 + 2 5 ( x − y + 2 1 ) which solves to
x
≈
0
.
4
0
2
1
1
3
0
3
2
5
9
0
3
0
7
,
y
≈
1
.
0
0
8
5
9
3
0
3
2
1
3
0
2
0
,
z
≈
0
.
3
2
7
7
1
1
7
4
1
6
2
2
3
7
2
Now we have,
s
2
=
s
2
⋅
(
sin
2
θ
+
cos
2
θ
)
=
(
s
⋅
sin
θ
)
2
+
(
s
⋅
cos
θ
)
2
=
y
2
+
z
2
≈
(
1
.
0
0
8
5
9
3
0
3
2
1
3
0
2
0
)
2
+
(
0
.
3
2
7
7
1
1
7
4
1
6
2
2
3
7
2
)
2
=
1
.
1
2
4
6
5
4
8
9
0
0
5
8
7
5
Hence,
s
=
1
.
1
2
4
6
5
4
8
9
0
0
5
8
7
5
≈
1
.
0
6
0
4
9
7
4
7
2
9
1
4
8
4
.
For the answer, ⌊ 1 0 0 0 0 s ⌋ = 1 0 6 0 4 .
I think your insight regarding the three congruent triangles is ingenious. I wish I had noticed that. And a beautiful exposition. Thank you, Thanos.
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Thank you Fletcher. I'm glad you like it.
While I can't prove it, I'm pretty certain the square must be oriented symmetrically to the pentagon, i.e. a side of the square must be parallel to a side of the pentagon, as in the figure. In fact, I suspect this is the only way to inscribe a square. If someone knows, I'd be grateful to learn more.
Symmetry implies △ J B H is isosceles, and thus J B = B H . Let L B bisect ∠ A B C .
Since ∠ B = 1 0 8 ∘ , ∠ S H B = 3 6 ∘ , ∠ F H C = ( 1 8 0 ∘ − 9 0 ∘ − 3 6 ∘ ) = 5 4 ∘ .
Similarly ∠ H F C = ( 1 8 0 ∘ − 5 4 ∘ − 1 0 8 ∘ ) = 1 8 ∘ .
Let s = J H and x = J B .
Using the law of sines on △ H F C , sin 1 0 8 ∘ s = sin 1 8 ∘ 1 − x .
From the right △ J L B , s = 2 x ⋅ cos 3 6 ∘
Substituting for s , sin 1 0 8 ∘ 2 x cos 3 6 ∘ = sin 1 8 ∘ 1 − x
Let k = sin 1 0 8 ∘ cos 3 6 ∘ sin 1 8 ∘ ⟹ x = 2 k + 1 1
Evaluating , x = 2 1 ( 5 − 5 − 2 5 − 2 5 ) ≈ 0 . 6 5 5 4 2 3
Evaluating , s = 5 − 2 1 ( 5 − 5 ) ≈ 1 . 0 6 0 4 9 7
So the requested answer is 1 0 6 0 4
Your intuition about the existence of a square oriented this way is correct. We can easily use a continuity argument as the animation implies. The uniqueness of this solution is not obvious to me. That's why, in the solution I'm about to post, I use a different approach.
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Consider a general case of a quadrilateral F G H I inscribed in the unit regular pentagon A B C D E . Let F G = I F = s 1 , F G ⊥ I F , F G ⊥ G H , ∠ I F J = θ and G H = s 2 (note that s 2 > s 1 for θ = 5 ∘ as shown in the figure). Then the various angles in terms of θ are as shown. If D F = a , then E F = 1 − a and F J = 1 + 2 a sin 1 8 ∘ . By sine rule , we have:
sin ( 5 4 ∘ + θ ) 1 − a = sin 1 0 8 ∘ s 1 = sin ( 7 2 ∘ − θ ) 1 + 2 a sin 1 8 ∘ ⟹ s 1 = sin ( 7 2 ∘ − θ ) + 2 sin 1 8 ∘ sin ( 5 4 ∘ + θ ) sin 1 0 8 ∘ ( 2 sin 1 8 ∘ + 1 )
Let E G = b . By sine rule, b = sin 1 0 8 ∘ s 1 sin ( 1 8 ∘ − θ ) and
sin 1 0 8 ∘ s 2 s 2 = sin ( 3 6 ∘ + θ ) 1 − b = sin ( 3 6 ∘ + θ ) sin 1 0 8 ∘ − s 1 sin ( 1 8 ∘ − θ )
When quadrilateral F G H I is a square s 2 = s 1 = s , then
⟹ s 1 sin ( 7 2 ∘ − θ ) + 2 sin 1 8 ∘ sin ( 5 4 ∘ + θ ) sin 1 0 8 ∘ ( 2 sin 1 8 ∘ + 1 ) ( 2 sin 1 8 ∘ + 1 ) ( sin ( 3 6 ∘ + θ ) + sin ( 1 8 ∘ − θ ) ) = sin ( 3 6 ∘ + θ ) + sin ( 1 8 ∘ − θ ) sin 1 0 8 ∘ = sin ( 3 6 ∘ + θ ) + sin ( 1 8 ∘ − θ ) sin 1 0 8 ∘ = sin ( 7 2 ∘ − θ ) + 2 sin 1 8 ∘ sin ( 5 4 ∘ + θ )
Solving the equation above, we get θ = 0 ∘ . And s = sin 3 6 ∘ + sin 1 8 ∘ sin 1 0 8 ∘ ≈ 1 . 0 6 0 4 9 7 4 7 3 ⟹ ⌊ 1 0 0 0 0 s ⌋ = 1 0 6 0 4 .