Solutions

283.

(a) Determine all quadruples (a, b, c, d)
of positive integers for which the greatest common divisor
of its elements is 1,
and a + b + c = d.


(b) Of those quadruples found in (a), which also satisfy


(c) For quadruples (a, b, c, d) of positive integers,
do the conditions a + b + c = d and (1/b) + (1/c) + (1/d) = (1/a)
together imply that a/b = c/d?
Solution 1. (a) Suppose that the conditions on a, b, c, d
are satisfied. Note that b and c have symmetric roles. Since
ad = bc, if b and c were both even, then either a or d
would be even, whence both would be even (since
a + b + c = d), contradicting the fact
that the greatest common divisor of a, b, c, d is equal to 1.
Hence, at most one of b and c is even.
Suppose, if possible, b and c were both odd. Then a and
d would be odd as well. If b º c (mod 4), then
bc º 1 and b + c º 2 (mod 4), whence
ad º a(a + 2) º 3 \not º bc (mod 4).
If b º c + 2 (mod 4), it can similarly be shown that
ad \not º bc (mod 4), In either case, we get an untenable
conclusion. Hence, exactly one of b and c is even and the other
is odd.
Without loss of generality, we may suppose that a and b have
opposite parity. Let g be the greatest common divisor of a
and b, so that a = gu and b = gv for some coprime pair
(u, v) of positive integers with opposite parity.
Since d > c, it follows that
b > a and v > u. Let w = v  u.
Since

b
a

= 
a + b + c
c

= 
a + b
c

+ 1 , 

it follows that
whence
c = 
gu(u + v)
w

and d = 
gv(u + v)
w

. 

Since the greatest common divisor of u and v is 1, w has no
positive divisor in common with either u or v, save 1. Any
common divisor of w and u + v must divide 2u = (u + v)
(v
 u) and 2v = (u + v) + (v
 u); such a common divisor
equals 1. Since u and v have
opposite parity and so w is odd, w must divide g.
Since the greatest common divisor of
a, b, c, d is equal to 1, we must have that g = w. Hence
(a, b, c, d) = (u(v  u), v(v  u), u(v + u), v(v + u)) 

where u and v are coprime with opposite parity.
Interchanging, the roles of b and c leads also to
(a, b, c, d) = (u(v  u), u(v + u), v(v  u), v(v + u)) 

with u, v coprime of opposite parity.
On the other hand, any quadruples of
this type satisfy the condition.
(b)


= 
1
v(v  u)

+ 
1
u(v + u)

+ 
1
v(v + u)


 
 
= 
1
v(v  u)

+ 
1
uv

= 
u + (v  u)
uv(v  u)

= 
1
v  u

= 
1
a

. 

(c) Note that the conditions imply that d  a and b + c are
nonzero. The conditions yield that d  a = b + c and
(1/a)  (1/d) = (1/b) + (1/c). The second of these can be
rewritten
so that ad = bc. Thus, all quadruples imply the required condition.
Solution 2. (a) [M. Lipnowski] Let a/b = c/d = r/s where
the greatest common divisor of r and s is equal to 1. Then
a = hr, b = hs, c = kr, d = ks. Since the greatest common
divisor of a, b, c, d equals 1, the greatest common divisor of
h and k is 1. From a + b + c = d, we have that
(h + k)r = (k  h)s. Observe that gcd(h + k, k  h) = 1
when h and k have opposite parity and gcd(h + k, k  h) = 2
when h and k are both odd. (Why?)
Thus, when h and k have oppposite parity, r = k  h,
s = k + h and
(a, b, c, d) = (h(kh), h(k+h), k(kh), k(k+h)) 

and, when h and k are both odd, then r =
^{1}/
_{2}(k
 h),
s =
^{1}/
_{2}(k + h) and
(a, b, c, d) = ((1/2)h(kh), (1/2)h(k+h), (1/2)k(kh),(1/2)k(k+h)) . 

It can be checked that these always work. (Collate these with
the result given in Solution 1.)
(b) Since a/b = c/(a + b + c), c = a(a + b)/(b  a) and
d = (a + b) + [a(a+b)/(ba)] = b(a + b)/(b  a). Hence


= 
1
b

+ 
b  a
a + b


æ è


1
a

+ 
1
b


ö ø


 
 

(c) [M. Lipnowski]

1
b

+ 
1
c

+ 
1
a + b + c

= 
1
a



is equivalent to


= bc(a + b + c)  a(b + c)(a + b + c)  abc 
 
 
= (b + c)(bc  a^{2}  ab  ac) , 

which in turn is equivalent to
0 = bc  a^{2}  ab  ac Û bc = a(a + b + c) = ad . 


284.

Suppose that ABCDEF is a convex hexagon for which
ÐA + ÐC + ÐE = 360^{°} and

AB
BC

· 
CD
DE

· 
EF
FA

= 1 . 

Prove that

AB
BF

· 
FD
DE

· 
EC
CA

= 1 . 

Solution 1. [A. Zhang] Since the hexagon is convex, all its
angles are less than 180
^{°}. A dilation of factor
CD
/
DE
 followed by a rotation, both with
centre D, takes E to C and F to a point G so that
DDCG
~ DDEF,
ÐDEF =
ÐDCG and
DE:EF:FD = DC:CG:GD. Since DE:DC = FD:GD and
ÐEDC =
ÐFDG,
DEDC
~ DFDC and
DE:DC:CE = FD:DG:GF. Now
ÐDCG + ÐBCD = ÐDEF + ÐBCD = 360^{°} ÐFAB > 180^{°} 

so that C lies within the triangle BDG and
ÐBCG = 360
^{°}  (
ÐDCG +
ÐBCD) =
ÐFAB.
Also,

CG
CD

= 
EF
DE

= 
AF
AB

· 
BC
CD



so that CG:BC = AF:AB, with the result that
DBCG
~ DBAF, AB:BF:FA = CB:BG:GC and
ÐFBG =
ÐABC. From the equality of these angles
and AB:CB = BF:BG, we have that
DABC
~ DFBG and AB:BC:CA = FB:BG:GF. Hence

AB
BF

· 
FD
DE

· 
EC
CA

= 
CA
GF

· 
GF
CE

· 
CE
CA

= 1 

as desired.
Solution 2. [T. Yin] Lemma. Let ABCD be a convex
quadrilateral with a, b, c, d, p, q the respective lengths
of AB, BC, CD, DA, AC and BD. Then
p^{2} q^{2} = (ac + bd)^{2}  4abcd cos^{2} q 

where 2
q =
ÐA +
ÐC.
Proof of Lemma. Locate E within the quadrilateral so that
ÐEDC = ÐADB and ÐECD = ÐABD. Then
DABD ~ DECD whence ac = qx where x is the
length of EC. Now ÐADE = ÐBDC and AD:DE = BD:CD
whence DADE ~ DBDC and bd = qy with y the
length of AE.
Hence abcd = q^{2}xy and ac + bd = q(x + y). Therefore,
a^{2}c^{2} + b^{2}d^{2} + 2abcd = q^{2}(x^{2} + 2xy + y^{2}) = q^{2} (x^{2} + y^{2}) + 2abcd 

which reduces to a
^{2} c
^{2} + b
^{2} d
^{2} = q
^{2}(x
^{2} + y
^{2}).
Since ÐDEC = ÐBAD and ÐAED = ÐBCD,
ÐAEC = ÐAED + ÐDEC = ÐC + ÐA = 2q . 

By the law of cosines,
p^{2} = x^{2} + y^{2}  2xy cos2q = x^{2} + y^{2}  2xy(2 cos^{2} q 1) Þ 


a^{2} c^{2} + b^{2} d^{2} 

= p^{2} q^{2} + 4q^{2}xy cos^{2} q 2q^{2} xy 
 
 
= p^{2} q^{2} + 4abcd cos^{2} q 2abcd 

so that the desired result follows.
ª
Note that, when ÐA + ÐC = 180^{°}, then we
get Ptolemy's Theorem. Consider the hexagon of the problem with
AB  = a, BC  = b, CD  = c,
DE  = d, EF  = e, FA  = f,
BF  = g, CA  = h, CF  = m,
DF  = u and CE  = v. We are given that
ace = bdf and need to prove that auv = dgh.
From the lemma applied to ABDF, we obtain that
g^{2} h^{2} = a^{2} m^{2} + 2abfm + b^{2} f^{2}  4abfmcos^{2} a 

where 2
a =
ÐBAC +
ÐBCF. Applying the lemma to
CDEF yields that
u^{2} v^{2} = d^{2} m^{2} + 2cdem + c^{2} e^{2}  4cdem cos^{2} b 

where 2
b =
ÐFCD +
ÐDEF. Since
ÐA+
ÐC +
ÐE = 360
^{°},
a+
b = 180
^{°}
and cos
^{2} a = cos
^{2} b. Finally,

d^{2}g^{2}h^{2}  a^{2}u^{2}v^{2} 

= (a^{2}d^{2}m^{2} + 2abd^{2}fm + b^{2}d^{2}f^{2}  4abd^{2}fmcos^{2}a) 
 
 
 (a^{2}d^{2}m^{2} + 2a^{2}cdem + a^{2}c^{2}e^{2}  4a^{2}cdemcos^{2}b) 
 
 
= 2adm(bdf  ace) + (b^{2}d^{2}f^{2}  a^{2}c^{2}e^{2}) 4adm(bdf  ace)cos^{2}a = 0 , 

whence auv = dgh as required.
Solution 3. [Y. Zhao] The proof uses inversion in a circle
and directed angles. Recall that, if O is the centre of a
circle of radius r, then inversion is that involution
X« X¢ for which X¢ is on the ray from O through
X and OX ·OX¢ = r^{2}. It is not too hard to check using
similar triangles that ÐOPQ = ÐOQ¢P¢ and using the
law of cosines that P¢Q¢ = PQ ·(r^{2}/(OP ·OQ)). For this
problem, we make F the centre of the inversion. Then


= ÐFAB + ÐBCD + ÐDEF = ÐFAB + ÐBCF + ÐFCD + ÐDEF 
 
 
= ÐA¢B¢F + ÐFB¢C¢+ ÐC¢D¢F + ÐFD¢E¢ = ÐA¢B¢C¢+ ÐC¢D¢E¢ 

whence
ÐC
¢B
¢A
¢ =
ÐC
¢D
¢E
¢.
In the following, we suppress the factor r^{2}. We obtain that


= 
æ è


AB
FA ·FB

· 
FB ·FC
BC


ö ø

· 
æ è


CD
FC ·FD

· 
FD ·FE
DE


ö ø


 
 
= 
AB
FA

· 
CD
BC

· 
EF
DE

= 1 

so that A
¢B
¢: B
¢C
¢ = D
¢E
¢:C
¢D
¢. This, along with
ÐC
¢B
¢A
¢ =
ÐC
¢D
¢E
¢ implies that
DC
¢B
¢A
¢ ~ DC
¢D
¢E
¢, so that A
¢B
¢:A
¢C
¢ = D
¢E
¢:E
¢C
¢
or A
¢B
¢·E
¢C
¢ = A
¢C
¢·E
¢D
¢.
Therefore


= 
æ è


A¢B¢
FA¢·FB¢

·B¢F 
ö ø

· 
æ è


1
F¢D¢

· 
FD¢·FE¢
D¢E¢


ö ø

· 
æ è


E¢C¢
FE¢·FC¢

· 
FC¢·FA¢
C¢A¢


ö ø


 
 
= 
A¢B¢
A¢C¢

· 
E¢C¢
E¢D¢

= 1 , 

as desired.
Solution 4. [M. AbdehKolahchi] Let A, B, C, D, E, F be
points in the complex plane with
B  A = a = a (cosa+ i sina) 

C  B = b = b (cosb+ i sinb) 

D  C = c = c (cosg+ i sing) 

E  D = d = d (cosd+ i sind) 

F  E = e = e (cose+ i sine) 

A  F = f = f (cosf+ i sinf) . 

Modulo 360
^{°}, we have that
ÐA = ÐFAB º 180^{°}  (f a) 

ÐC = ÐBCD º 180^{°}  (d b) 

ÐE = ÐDEF º 180^{°}  (e g) . 

Also a + b + c + d + e + f = 0 and


= 
a c e (cosa+i sina)(cosg+ i sing)(cose+i sine)
b d f (cosb+i sinb)(cosd+ i sind)(cosf+ i sinf)


 
 
= 1 (cos(a f+ d b+ e g)) 
 
 
= cos(ÐA  180^{°} + ÐC  180^{°}+ ÐE  180^{°}) = cos(180^{°}) = 1 , 

whence ace + bdf = 0. Therefore,
0 = ad(a + b + c + d + e + f) + (ace + bdf) = a(d + e)(c + d) + d(a + f)(a + b) , 

whence

a(d+e)(c+d)
d(a+f)(a+b)

= 1Þ 
a 
a + f 

· 
d + e 
d 

· 
c + d 
a + b 

= 1 

Þ 
AB
BF

· 
FD
DE

· 
EC
CA

= 1 . 


285.

(a) Solve the following system of equations:
(1 + 4^{2x  y})(5^{1  2x + y}) = 1 + 2^{2x  y + 1} ; 

y^{2} + 4x = log_{2} (y^{2} + 2x + 1) . 



(b) Solve for real values of x:
Express your answers in a simple form.
Solution. Let u = 2x
 y. Then
(1 + 4^{u})(5^{1u}) = 1 + 2^{u+1} 

so that
5^{u1} = 
1 + 2^{2u}
1 + 2^{u+1}

= 2^{u1} + 
1  2^{u1}
1 + 2^{u+1}

. 

Thus,
5^{u  1}  2^{u  1} = 
1  2^{u1}
1 + 2^{u+1}

. 

When u > 1, the left side of this equation is positive while the
right is negative; when u < 1, the reverse is true. Hence, the
only possible solution is u = 1, which checks out.
Substituting for x leads to
y^{2} + 2y + 2 = log_{2} (y^{2} + y + 2) . 

Since
y
^{2} + y + 2 = ( y +
^{1}/
_{2} )
^{2} +
^{7}/
_{4} > 0,
the right side is defined and is in fact positive. Let
f(y) = y^{2} + 2y + 2  log_{2} (y^{2} + y + 2) . 

Then
f¢(y) = 
2y(y+1)^{2} + 4(y + 1)  (log_{2} e)(2y + 1)
y^{2} + y + 2

. 

f¢(y) = 0 Û (y+1)^{2} =  
æ è

(2  log_{2} e) + 
4  log_{2} e
2y


ö ø

. 

From the graphs of the two sides of the equation, we see that the
left side and the right side have opposite signs when y > 0
and become equal for exactly one value of y. It follows that
f¢(y) changes sign exactly once so that
f(y)
decreases and then increases. Thus,
f(y) vanishes at
most twice. Indeed,
f(
2) =
f(
1) = 0, and so
(x, y) = (0,
1), (
^{1}/
_{2},
2) are the only solutions of
the equation.
(b) The equation can be rewritten
1 = 3^{1x} 2^{2(1x)/(x+2)} 

whence
0 = (1  x)( log3 + (2/(x+2))log2) . 

Thus, either x = 1 or 0 = log
_{2} 3 + 2/(x+2).
The latter leads to
x = 2(1 + log_{3} 2) = 2(log_{3} 6) = log_{3} 36 . 


286.

Construct inside a triangle ABC a point P such that,
if X, Y, Z are the respective feet of the perpendiculars from
P to BC, CA, AB, then P is the centroid (intersection of
the medians) of triangle XYZ.
Solution 1. Let AU, BV, CW be the medians of triangle
ABC and let AL, BM, CN be their respective images in the
bisectors of angles A, B, C. Since AU, BV, CW intersect in
a common point (the centroid of
DABC). AL, BM, CN must
intersect in a common point P. This follows from the sine version
of Ceva's theorem and its converse. Let X, Y, Z be the
respective feet of the perpendiculars from P to sides BC,
AC, AB.
Let I, J, K be the respective feet of the perpendiculars from
the centroid G to the sides BC, AC and AB. The quadrilateral
PYAZ is the image of the quadrilateral GJAK under a reflection
in the angle bisector of A followed by a dilation with centre
A and factor AP/AG. Hence PY:PZ = GK:GI.
Since triangles AGB and AGC have the same area,
AB ·GK = AC ·GJÞ PY:PZ = AC:AB = b:c . 

Applying a similar argument involving PX, we find that
Let PX = ae, PY = be, PZ = ce. Then, since
ÐXPY +
ÐACB = 180
^{°},
[PXY] = 
1
2

abe^{2} sinÐXPY = e^{2} 
æ è


1
2

absinC 
ö ø

= e^{2}[ABC] . 

Similarly, [PYZ] = [PZX] = e
^{2}[ABC] = [PXY], whence P
must be the centroid of triangle XYZ.
Solution 2. [M. Lipnowski] Erect squares ARSB,
BTUC, CVWA externally on the edges of the triangle.
Suppose that RS and VW intersect at A¢, RS and
TU at B¢ and TU and UW at C¢.
We establish that AA¢, BB¢ and CC¢ are concurrent.
They are cevians in the triangle A¢B¢C¢. We have that


· 
sinÐVC¢C
sinÐUC¢C

· 
sinÐTB¢B
sinÐSB¢B


 
 
= 
(AR/AA¢)
(AW/AA¢)

· 
(VC/CC¢)
(UC/CC¢)

· 
(TB/BB¢)
(BS/BB¢)


 
 
= 
AR
AW

· 
VC
UC

· 
TB
BC

= 
c
b

· 
b
a

· 
a
c

= 1 . 

Hence AA
¢, BB
¢, CC
¢ intersect in a point P by the converse
to Ceva's Theorem. P is the desired point.
To prove that this works, we first show that PX:PY:PZ = a:b:c,
and then that [XPY] = [YPZ] = [ZPX]. Observe that, since
DPZA ~ DARA¢ and DPYA ~ DAWA¢,

PY
PZ

= 
PY(AA¢/PA)
PZ(AA¢/PA)

= 
AW
AR

= 
b
c

, 

and similarly that PX:PZ = a:c. Now
ÐXPY = 360^{°}  ÐPXC  ÐPYC  ÐXCY = 180^{°}  ÐXCY = 180^{°}  ÐACB , 

so that [XPY] =
^{1}/
_{2}PX ·PY sin
ÐXPY =
^{1}/
_{2} PX ·PY sin
ÐACB. We find that


: [YPZ] : [ZPX] = 
1
2

PX ·PY sinÐACB : 
1
2

PY ·PZ sinÐACB : 
1
2

PZ ·PX sinÐABC 
 
 
= 
1
2

ab sinC : 
1
2

bc sinA : 
1
2

ca sinB = [ABC] : [ABC] : [ABC] = 1:1:1 . 

Hence [XPY] = [YPZ] = [ZPX] =
^{1}/
_{3}[XYZ], so that the
altitudes of these triangle from P to the sides of triangle XYZ
are each onethird of the corresponding altitudes for triangle
XYZ. Hence P must be the centroid of triangle XYZ.
Comment. A. Zhang and Y. Zhao gave the same construction.
Zhang first gave an argument that P, being the centroid of
triangle XYZ is characterized by PX:PY:PZ = a:b:c. This is
a result of the characterization [XPY] = [YPZ] = [ZPX] and
the law of sines, with the argument similar to Lipnowski's.
Zhao used the fact that PX:PY:PZ = BC:CA:AB and that the
vectors [( ®)  PX], [( ®)  PY],
[( ®)  PZ] were dilated versions of
[( ®)  BC], [( ®)  CA],
[( ®)  AB] after a 90^{°} rotation, so that
[( ®)  PX] +[( ®)  PY] +[( ®)  PZ] = [( ®)  O].

287.

Let M and N be the respective midpoints of the
sides BC and AC of the triangle ABC. Prove that the centroid
of the triangle ABC lies on the circumscribed circle of the triangle
CMN if and only if
4 ·AM ·BN  = 3 ·AC ·BC  . 

Solution 1.
4 AM BN  = 3 AC BC Û12 AM GN  = 12 AN MC Û AM : MC  = AN : GN  

Û DAMC ~ DANGÛ ÐAMC = ÐANG 

Û GMGN is concyclic.
Solution 2. [A. Zhang] Since M and N are respective
midpoints of BC and AC, [ABC] = 4[NMC], so that
[ABMN] = 
3
4

[ABC] = 
3
8

AC BC sinÐACB . 

However, [ABMN] =
^{1}/
_{2}AM
BN
sin
ÐNGM (why?). Hence
4 AM BN sinÐNGM = 3 AC BC sinÐACB . 

Observe that G lies inside triangle ABC, and so lies within
the circumcircle of this triangle. Hence
ÐNGM =
ÐAGB >
ÐACB. We deduce that
4 AM BN  = 3 AC BC Û sinÐNGM = sinÐACBÛ ÐNGM + ÐACB = 180^{°} 

Û CMGN is concyclic.

288.

Suppose that a_{1} < a_{2} < ¼ < a_{n}. Prove that
a_{1} a_{2}^{4} + a_{2} a_{3}^{4} + ¼+ a_{n} a_{1}^{4} ³ a_{2} a_{1}^{4} + a_{3} a_{2}^{4} + ¼+ a_{1} a_{n}^{4} . 

Solution. The result is trivial for n = 2. To deal with
the n = 3 case, observe that, when x < y < z,
(xy^{4} + yz^{4} + zx^{4})(yx^{4} + zy^{4} + xz^{4}) = (1/2)(z  x)(y  x)(z  y)[(x + y)^{2} + (x + z)^{2} + (y + z)^{2}] ³ 0 . 

As an induction hypothesis, assume that the result holds for
the index n
³ 3. Then

(a_{1} a_{2}^{4} + a_{2} a_{3}^{4} + ¼+ a_{n} a_{n+1}^{4}+ a_{n+1} a_{1}^{4}) 

 (a_{2} a_{1}^{4} + a_{3} a_{2}^{4} + ¼+a_{n+1} a_{n}^{4} + a_{1} a_{n+1}^{4}) 
 
 
= (a_{1} a_{2}^{4} + a_{2} a_{3}^{4} + ¼+ a_{n} a_{1}^{4}) (a_{2} a_{1}^{4} + a_{3} a_{2}^{4} + ¼+ a_{1} a_{n}^{4}) 
 
 
+ (a_{1} a_{n}^{4} + a_{n} a_{n+1}^{4} + a_{n+1} a_{1}^{4}) (a_{n} a_{1}^{4} + a_{n+1} a_{n}^{4} + a_{1} a_{n+1}^{4}) ³ 0 , 

as desired.

289.

Let n(r) be the number of points with integer
coordinates on the circumference of a circle of radius r > 1
in the cartesian plane. Prove that
Solution. Let A =
pr
^{2} be the area of the circle, so that
the right side of the inequality is 6A
^{1/3}. We observe that
A > 3,
p^{2} < (22/7)
^{2} < 10 < (2.2)
^{3}.

6A^{1/3}  2p^{2/3}A^{1/3} 

= (6  2p^{2/3})A^{1/3} > (6  2 ×10^{1/3})A^{1/3} 
 
 
> (6  4.4) ×3^{1/3} > 1.6 ×1.25 = 2 , 

so that there is an even integer k for which
6 = 2 ×3^{2/3} ×3^{1/3} < 2 p^{2/3} A^{1/3} < k < 6A^{1/3} . 

In particular, 8
p^{2} A < k
^{3}.
Let P_{1}P_{2} ¼P_{k} be a regular kgon inscribed in the
circle. Locate the vertices so that none have integer coordinates.
(How?) Identify P_{k+1} = P_{1} and P_{k+2} = P_{2}, and let
v_{i} = [( ®)  ( P_{i}P_{i+1})] for 1 £ i £ k.
Observe that v_{i} has length less than
2pr/k = (2/k)(pA)^{1/2}. Then, for each i, the area
of triangle P_{i} P_{i+1} P_{i+2} is equal to

1
2

v_{i} ×v_{i+1}  = 
1
2

v_{i} v_{i+1} sin(2p/k) < 
1
2

× 
4
k^{2}

×pA × 
2p
k

= 
1
2

× 
8p^{2}
k^{3}

×A < 
1
2

. 

Suppose, if possible, that the arc joining P
_{i} and P
_{i+2}
(through P
_{i+1}) contains points U, V, W, each with integer
coordinates. Then, if
u,
v,
w are the
corresponding vectors for these points, then
(
v  u) ×(
w  u)
 must
be a positive integer, and so the area of triangle UVW must be
at least 1/2. But each of the sides of triangle UVW has length
less than the length of P
_{i}P
_{i+2} and the shortest altitude of
triangle UVW is less than the altitude of triangle P
_{i} P
_{i+1}P
_{i+2} from P
_{i+1} to side P
_{i} P
_{i+2}. Thus,

1
2

£ [UVW] £ [P_{i} P_{i+1} P_{i+2}] < 
1
2

, 

a contradiction. Hence, each arc P
_{i} P
_{i+2} has at most two points
with integer coordinates. The whole circumference of the circle
is the union of k/2 nonoverlapping such arcs, so that there
must be at most k points with integer coordinates. The result follows.