Gaussian Gamble : Part 1
Soumya Dasgupta Sutirtha Bhattacharyya
March 18, 2021
Solutions
1. Claim : All ordered pairs of integers (x, y) such that x
3
+ y
3
= (x + y)
2
are (x, y) =
{(1, 0), (2, 1), (0, 1), (1, 2), (2, 2), (t, −t)} where t ∈ Z.
x
3
+ y
3
= (x + y)
2
⇒ (x + y)(x
2
− xy + y
2
) = (x + y)
2
Observe that x = −y is a trivial solution. So from now on we suppose x 6= −y then
x
2
− xy + y
2
= x + y
⇒ x
2
− (y + 1)x + (y
2
− y) = 0 (1)
Now treating (1) as a quadratic in x, let ∆ be its discriminant. Then
∆ = (y + 1)
2
− 4(y
2
− y)
= −3y
2
+ 6y + 1
Now let f (t) = −3t
2
+ 6t + 1
We get f
0
(t) = −6t + 6 = −6(t − 1)
∴ f is decreasing on (1, ∞) and f is increasing on (−∞, 1).
Now f(3) = −8 < 0 and as f is decreasing on (1, ∞) we get f (t) ≤ f (3) < 0 ∀ t ≥ 3
f(−1) = −8 < 0 and as f is increasing on (−∞, 1) we get f (t) ≤ f (−1) < 0 ∀ t ≤ −1
∴ f (t) < 0 ∀ t ∈ (−∞, 1] ∪ [3, ∞)
Now we must have ∆ ≥ 0
But then we must have y ∈ [0, 2].
Now plugging y = 0 we get x
2
= x ⇒ x = {0, 1}.
Plugging y = 1 we get x
2
= 2x ⇒ x = {0, 2}
Finally plugging y = 2 we get x
2
− 3x + 2 = 0 ⇒ (x − 1)(x − 2) = 0 ⇒ x = {1, 2}
Thus we get (x, y) = {(1, 0), (2, 1), (0, 1), (1, 2), (2, 2), (t, −t)} where t ∈ Z are the only
solution to the given diophantine equation.
1
2. Let ω
1
, ω
2
be two circles centered at A, C respectively which are tangent to each other
at K, and tangent to the line l at B and D respectively. Let ω
3
:= the circle tangent to
ω
1
, ω
2
and l.
Lemma : Let P
1
, P
2
be parabolas with vertex at B, D and focus at A, C respectively.
Then let T = P
1
∩ P
2
. Then T := center of ω
3
.
Proof
Figure 1: Circles tangent to each other
Let X = AT ∩ ω
1
and Y = BT ∩ ω
2
Let l
1
:= directrix of P
1
and l
2
:= directrix of P
2
Let AB ∩ l
1
= P and CD ∩ l
2
= Q
Finally let M ∈ l such that T M ⊥ l and let T M ∩ l
1
= R and T M ∩ l
2
= S
Now, as T ∈ P
1
⇒T R = T A
⇒T M + MR = T X + XA
⇒T M + BP = T X + AP
⇒T M = T X
Similarly T M = T Y
∴ T M = T X = T Y ⇒ T := center of ω
3
.
Hence, our lemma is true.
Now coming to the main problem.
Part (a)
Let O
n
:= center of C
n
and let L be the x-axis.
We let O
n
= (x
n−1
, y
n−1
) then P
n
= (x
n−1
, 0)
Then as C
n
, C
n+1
are tangent to each other we get
O
n
O
2
n+1
= (O
n
P
n
+ O
n+1
P
n+1
)
2
2
⇒ (x
n−1
− x
n
)
2
+ (y
n−1
− y
n
)
2
= (y
n−1
+ y
n
)
2
⇒ (x
n−1
− x
n
)
2
= 4y
n−1
y
n
Now let P
n
:= the parabola with vertex at P
n
and focus at O
n
.
Then
P
n
: 4y
n−1
y = (x − x
n−1
)
2
P
n+1
: 4y
n
y = (x − x
n
)
2
From our lemma we get P
n
∩ P
n+1
= O
n+2
= (x
n+1
, y
n+1
)
⇒ 2
√
y
n+1
y
n−1
= x
n+1
− x
n−1
and ⇒ 2
√
y
n+1
y
n
= x
n
− x
n+1
⇒ 2
√
y
n+1
(
√
y
n−1
+ y
n
) = x
n
− x
n−1
= 2
√
y
n−1
y
n
⇒
1
√
y
n+1
=
1
√
y
n
+
1
√
y
n−1
(2)
Let a
n
=
1
√
y
n
then a
n+1
= a
n
+ a
n−1
Now, x
n+1
= 2
√
y
n+1
y
n−1
+ x
n−1
=
x
n
√
y
n−1
+ x
n−1
√
y
n
√
y
n−1
+
√
y
n
⇒
x
n+1
√
y
n+1
=
x
n
√
y
n
+
x
n−1
√
y
n−1
(3)
Let b
n
= x
n
a
n
then b
n+1
= b
n
+ b
n−1
Now lim
n→∞
P
n
=
lim
n→∞
x
n−1
, 0
So we need to compute lim
n→∞
x
n
Lemma : Let f
n+1
= f
n
+ f
n−1
∀ n ∈ N then f
n
= c
1
φ
n
+ c
2
−1
φ
n
where φ =
√
5+1
2
and
c
1
, c
2
are constants such that c
1
=
1
√
5
f
1
+
f
0
φ
and c
2
=
1
√
5
(f
1
− φf
0
).
Proof
We know by method of characteristics roots that f
n
= c
1
φ
n
+ c
2
−1
φ
n
Now we get f
0
= c
1
+ c
2
and f
1
= c
1
φ −
c
2
φ
Solving this gives
c
1
=
1
√
5
f
1
+
f
0
φ
(4)
c
2
=
1
√
5
(f
1
− φf
0
) (5)
Which completes our lemma.
Now let a
n
= A
1
φ
n
+ A
2
(−φ)
−n
and b
n
= B
1
φ
n
+ B
2
(−φ)
−n
⇒ x
n
=
B
1
φ
n
+ B
2
(−φ)
−n
A
1
φ
n
+ A
2
(−φ)
−n
⇒ x
n
=
B
1
+ B
2
(−1)
n
φ
−2n
A
1
+ A
2
(−1)
n
φ
−2n
3
⇒ lim
n→∞
x
n
= lim
n→∞
B
1
+ B
2
(−1)
n
φ
−2n
A
1
+ A
2
(−1)
n
φ
−2n
=
B
1
A
1
Now B
1
=
1
√
5
x
1
√
y
1
+
x
0
φ
√
y
0
and A
1
=
1
√
5
1
√
y
1
+
1
φ
√
y
0
∴ lim
n→∞
x
n
=
φ
√
y
0
x
1
+
√
y
1
x
0
φ
√
y
0
+
√
y
1
∴ P
n
goes to a point P
∞
such that
P
1
P
∞
: P
2
P
∞
= φ
√
y
0
:
√
y
1
Part (b)
Now if P
1
P
∞
: P
2
P
∞
= 1
⇒
φ
√
y
0
√
y
1
= 1
⇒
y
1
y
0
= φ
2
4
3. Part (a)
An example of a 3-D object having 6 vertices such that there are only two possible values
for the distance between any two vertices is a regular octahedron.
Figure 2: Regular Octahedron
Part (b)
Lemma : There exist a set P
n
of n + 1 points in R
n
such that distance between any two
points is same.
Proof
We are going to proof the result using induction. The base case for n = 1 is trivial.
Suppose the result holds true for n.
Thus from the hypothesis we know that ∃ a set P
n
of n + 1 equidistant points in R
n
. Let
P
n
= {v
1
, v
2
, . . . , v
n+1
}. Let c be the centroid of these n + 1 points. Now as all the points
are equidistant from each other they are equidistant from the centroid ( let the distance
between the centroid and the points be d ) and also the angle subtended by any two
points with the centroid is constant ( let this angle be θ ), i.e (~v
i
−~c) ·(~v
j
−~c) = d
2
cos θ
∀ i 6= j and if i = j then (~v
i
−~c) · (~v
i
−~c) = ||~v
i
−~c||
2
= d
2
.
Let v ∈ R
n
then let v =< x
1
, x
2
, . . . , x
n
>
Now consider the function f : < x
1
, x
2
, . . . , x
n
> 7→ < x
1
, x
2
, . . . , x
n
, 0 >
Then if v ∈ R
n
⇒ f (v) ∈ R
n+1
Thus we can embed R
n
→ R
n+1
using the function f moreover this embedding is special
because it preserves the distances.
Thus as {v
i
}
n+1
i=1
are equidistant from each other in R
n
we get {f (v
i
)}
n+1
i=1
are equidistant
from each other in R
n+1
. And f(c) is equidistant from {f (v
i
)}
n+1
i=1
.
5
Now this embedding comes with a new direction let ~w be the unit vector in this direction
then ~w =< 0, 0, . . . , 1 > and ~w is perpendicular to all the f(v
i
).
Now consider the straight line ~r(t) = f(c) + t ~w. Now any point on ~r is equidistant from
{f(v
i
)}
n+1
i=1
. Thus we can choose a point v
n+2
on ~r such that distance between v
n+2
and
f(v
1
) is same as the distance between any of the f (v
1
), f (v
2
). Hence we get n + 2 points
in R
n+1
which are all equidistant from each other.
Now coming to the main problem.
Claim : There exist
n+1
2
points in R
n
and positive real numbers x, y such that distance
between any two points is either x or y.
Proof
By our lemma we know that there exist n + 1 points in R
n
which are equidistant from
each other. Let c be their centroid. We have already shown that c is equidistant to all
the n + 1 points. Let us shift c to the origin.
Let ~c
ij
be the reflection of c about v
i
v
j
where i < j. Then ~c
ij
= ~v
i
+ ~v
j
.
We will show that the set of points c
ij
gives us the required
n+1
2
points.
Let ||~v
i
− ~v
j
|| = l and ||~v
i
|| = d and angle between any two points and c be θ. Then we
have
||~c
ij
− ~c
kl
||
2
= ||(~v
i
− ~v
k
) + (~v
j
− ~v
l
||
2
= ||~v
i
− ~v
k
||
2
+ ||~v
j
− ~v
l
||
2
+ 2(~v
i
− ~v
k
) · (~v
j
− ~v
l
)
= 2l
2
+ 2(~v
i
· ~v
j
− ~v
k
· ~v
j
− ~v
i
· ~v
l
+ ~v
k
· ~v
l
)
= 2l
2
+ 2(d
2
cos θ − d
2
cos θ − d
2
cos θ + d
2
cos θ)
= 2l
2
∴ ||~c
ij
− ~c
kl
|| =
√
2l if i 6= k or j 6= l (6)
Now ~c
ij
= ~v
i
+ ~v
j
and ~c
ik
= ~v
i
+ ~v
k
||~c
ij
−~c
ik
||
2
= ||~v
j
− ~v
k
||
2
= l
2
∴ ||~c
ij
− ~c
ik
|| = l if j 6= k (7)
Hence the
n+1
2
points of the form c
ij
satisfies our condition as distance between any two
points is either l or
√
2l. Hence, our claim is true.
Thus for n = 63 we have a set S of
64
2
= 2016 distinct points in R
63
and two positive
real numbers x and y such that the distance between any two distinct points in S is either
x or y.
Part (c)
Generalised Statement : For any arbitrary positive integer n there exist
n+1
2
points
in R
n
and positive real numbers x and y such that distance between any of two points is
either x or y.
6
Comments
Post a Comment