Squarely integral
A square in a Cartesian coordinate system has vertices with integer coordinates.
No two vertices lie on the same horizontal or vertical line.
Moreover, the circle through the four vertices has an integer radius.
What is the smallest possible area of this square?
The answer is 50.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
Even if diameter is required to be an integer, then the square side will be 5root(2) having a diameter of 10. Consider a square of side n. The diameter is n root(2), so n = a root(2), with 'a' being integer. Since we are looking for a straight edge which is at an angle not equal to 0,45,90 with any of the axes, we can consider (0,0) as one end of the edge and (b,c) as the other with b != c. So we have b^2 + c^2 = 2*a^2. The smallest value for a which can be expressed as sum of two squares is a = 5, for b = 1, c = 7 or b = 7, c = 1.
Up voted. Your approach is beautiful and also general. I give my solution only as a different approach.
In the second line, you should have mentioned "No two vertices " instead of " No two point", I got confused for that
If a adjoining vertices (p,q) and (r,s) satisfy all constrains, other two also would. Hence we shall only consider these two.
Since the diagonal of the square sides S is S
2
. S must be integer*
2
.....condition (1).
Two diagonals can not be || to x-axes or y-axes .....condition (2),
More over the diagonal must be even so the radius is an integer........condition (3).
We need smallest square........condition (4).
For simplicity and WLOG we can assume the two adjoining vertices as (0,n) and (m,0) to keep the square in first quadrant only.
(1) S must have the form "integer *
2
." So
n
2
+
m
2
=
s
o
m
e
i
n
t
e
g
e
r
∗
2
.
(2) p
=
r and q
=
s. n>0, m>0.
(3) (r - p) and (s-q) must both be odd or both even. m and n both even or both odd.
(
4
)
W
e
s
t
a
r
t
w
i
t
h
o
d
d
n
=
1
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
m
=
1
,
s
q
u
a
r
e
i
s
(
0
,
1
)
,
(
1
,
0
)
,
(
2
,
1
)
,
(
1
,
2
)
c
o
n
d
i
t
i
o
n
(
2
)
v
i
o
l
a
t
e
d
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
m
=
3
,
S
=
1
0
c
o
n
d
i
t
i
o
n
(
1
)
v
i
o
l
a
t
e
d
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
m
=
5
,
S
=
2
6
c
o
n
d
i
t
i
o
n
(
1
)
v
i
o
l
a
t
e
d
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
m
=
7
,
S
=
5
2
.
a
l
l
c
o
n
d
i
t
i
o
n
s
O
K
.
C
h
e
c
k
e
v
e
n
n
=
2
:
−
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
m
=
2
,
s
q
u
a
r
e
i
s
(
0
,
2
)
,
(
2
,
0
)
,
(
4
,
2
)
,
(
2
,
4
)
c
o
n
d
i
t
i
o
n
(
2
)
v
i
o
l
a
t
e
d
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
m
=
4
,
S
=
2
5
c
o
n
d
i
t
i
o
n
(
1
)
v
i
o
l
a
t
e
d
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
m
=
6
,
S
=
2
1
0
c
o
n
d
i
t
i
o
n
(
1
)
v
i
o
l
a
t
e
d
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
m
=
8
,
S
=
2
1
7
c
o
n
d
i
t
i
o
n
(
1
)
a
n
d
(
4
)
v
i
o
l
a
t
e
d
.
S
o
w
e
s
t
o
p
h
e
r
e
.
∴
s
m
a
l
l
e
s
t
a
r
e
a
=
(
5
2
)
2
=
5
0
The square here is (0,1),(7,0),(8,7),(1,8) or add/subtract any one integer to all coordinates.
Let ( x , y ) be the center of the square (as well as the circumscribed circle); then there exist a , b such that the vertices are ( x + a , y + b ) ; ( x − b , y + a ) ; ( x − a , y − b ) ; ( x + b , y − a ) . It is easy to see that a , b , x , y must be integers or half-integers. The radius of the circle r is the distance between the center and each of the vertices, so that a 2 + b 2 = r 2 , for integer r . It is easy to see that a and b must be integers as well. The trivial solution a = 0 or b = 0 is excluded by the second condition, and without loss of generality we can assume a , b > 0 .
The length of each side of the square is s = ( a + b ) 2 + ( a − b ) 2 = r 2 ; and its area is s 2 = 2 r 2 .
The smallest area corresponds to the smallest Pythagorean triple, ( a , b , r ) = ( 3 , 4 , 5 ) . This corresponds to a square with sides 5 2 and area 5 0 . Taking ( x , y ) = ( 0 , 0 ) for a concrete example, we have the vertices P 1 : ( 4 , 3 ) ; P 2 : ( − 3 , 4 ) ; P 3 : ( − 4 , − 3 ) ; P 4 : ( 3 , − 4 ) .
Follow-up challenge : How would the answer change if, instead of the radius, we had required the diameter to be an integer?