Solutions

325.

Solve for positive real values of x, y, t:
(x^{2} + y^{2})^{2} + 2tx(x^{2} + y^{2}) = t^{2}y^{2} . 

Are there infinitely many solutions for which the values of
x, y, t are all positive integers?
What is the smallest value of t for
a positive integer solution?
Solution. Considering the equation as a quadratic in
t, we find that the solution is given by
t = 
(x^{2} + y^{2})[x +  Ö

x^{2} + y^{2}

] 
y^{2}

= 
x^{2} + y^{2}

. 

where x and y are arbitrary real numbers. The choice of
sign before the radical is governed by the condition that
t > 0. Integer solutions are those obtained by selecting
(x, y) = (k(m
^{2}  n
^{2}), 2kmn) for integers m, n, k where
m and n are coprime and k is a multiple of 2n
^{2}.
Then
t = 
k(m^{2} + n^{2})^{2}
2n^{2}

. 

The smallest solution is found by taking n = 1, m = k = 2
to yield (x, y, t) = (6, 8, 25).
Comment. L. Fei gives the solution set
(x, y, t) = (2n^{2} + 2n, 2n + 1, (2n^{2} + 2n + 1)^{2}) . 

This set of solutions satisfies in particular y
^{2} = 2x + 1.
If, in the above solution, one takes (x, y) = (2mn, m
^{2}  n
^{2}),
then t = (m
^{2} + n
^{2})
^{2}/(m
 n)
^{2}; in particular, this gives
the solution (x, y, t) = (4, 3, 25).

326.

In the triangle ABC with semiperimeter
s = ^{1}/_{2}(a +b + c), points U, V, W lie on the
respective sides BC, CA, AB. Prove that
s < AU + BV + CW  < 3s . 

Give an example for which the sum in the middle is equal to 2s.
Solution. The triangle inequality yields that
AB
+
BU
 >
AU
 and
AC
+
CU
 >
AU
. Adding these
and dividing by 2 gives s >
AU
. Applying the same
inequality to BV and CW yields that
3s > AU + BV + CW  . 

Again, by the triangle inequality, AU + BU  > AB  and AU + CU  > AC . Adding these inequalities gives
2 AU  > AB + AC  BC . Adding this to analogous inequalities for BV and
CW and dividing by 2 yields that
AU + BV + CW  > s.
Comment. For the last part, most students gave a degenerate
example in which the points U, V, W coincided with certain vertices
of the triangle. A few gave more interesting examples. However, it was
necessary to make clear that the stated lengths assigned to AU,
BV and CW were indeed possible, i.e. they were at least
as great as the altitudes. The nicest example came from D.
Dziabenko: AB  = 6, BC  = 8,
CA  = 10, AU  = 8, BV  = 7
CW  = 9.

327.

Let A be a point on a circle with centre O
and let B be the midpoint of OA. Let C and D be points
on the circle on the same side of OA produced for which ÐCBO = ÐDBA. Let E be the midpoint of CD and let F be
the point on EB produced for which BF = BE.


(a) Prove that F lies on the circle.


(b) What is the range of angle EAO?
Solution 1. [Y. Zhao] When
ÐCBO =
ÐDBA = 90
^{°}, the result is obvious. Wolog, suppose that
ÐCBO =
ÐDBA < 90
^{°}. Suppose that the
circumcircle of triangle OBD meets the given circle at G.
Since OBDG is concyclic and triangle OGD is isosceles,
ÐOBC = ÐABD = 180^{°}  ÐOBD = ÐOGD = ÐODG = ÐOBG , 

so that G = C and OBDC is concyclic.
Let H lie on OA produced so that OA = AH. Since OB ·OH = OA^{2}, the inversion in the given circle with centre O interchanges
B and H, fixes C and D, and carries the circle OBDC
(which passes through the centre O of inversion) to a straight
line passing through H, D, C. Thus C, D, H are
collinear.
This means that CD always passes through the point H on OA
produced for which OA = AH. Since E is the midpoint of CD,
a chord of the circle with centre O, ÐOEH = ÐOED = 90^{°}. Hence E lies on the circle with centre A and
radius OA.
Consider the reflection in the point B (the dilation with centre
B and factor 1). This takes the circle with centre O and
radius OA to the circle with centre A and the same radius,
and also interchanges E and F. Since E is on the latter
circle, F is on the given circle.
Ad (b), E lies on the arc of the circle with centre A and
radius OA that joins O to the point R of intersection of
this circle and the given circle. Since RB ^OA,
and OA = OR = RA, ÐRAO = 60^{°}. It can be seen
that ÐEAO ranges from 0^{°} (when CD is a diameter)
to 60^{°} (when C = D = R).
Solution 2. [A. Wice] We first establish a Lemma.
Lemma. Let UZ be an angle bisector of triangle UVW
with Z on VW. Then
UZ^{2} = UV ·UW  VZ·WZ . 

Proof. By the Cosine Law,
UV^{2} = UZ^{2} + VZ^{2}  2 UZ ·VZ cosÐUZW 

and
UW^{2} = UZ^{2} + WZ^{2} + 2 UZ ·WZ cosÐUZW . 

Eliminating the cosine term yields that
UV^{2} ·WZ + UW^{2} ·VZ = (UZ^{2} + VZ ·WZ)(WZ + VZ) . 

Now,
UV : VZ = UW : WZ = (UV + UW) : (VZ + WZ) , 

so that
and
(UV + UW) ·WZ = UW ·(WZ + VZ) . 

Thus
It follows that UW ·UV = UZ
^{2} + VZ ·WZ.
ª
Let R be a point on the circle with BR ^OA, S be the
intersection of CD in OA produced, and D¢ be the reflection
of D in OA. (Wolog, ÐCBO < 90^{°}.) Since SB is
an angle bisector of triangle SCD¢, from the Lemma, we have that
BS^{2} = SC ·SD¢ CB ·D¢B = SC ·SD  BR^{2} = SC ·SD  (SR^{2}  BS^{2}) 

whence SC ·SD = SR
^{2}. Using power of a point, we deduce that
SR is tangent to the given circle and OR
^SR.
Now
(OA + AS)^{2}  OR^{2} = RS^{2} = BR^{2} + (AB + AS)^{2} = 3AB^{2} + AB^{2} + 2AB ·AS + AS^{2} 

from which 4AB ·AS = 4AB
^{2} + 2AB ·AS, whence
AS = 2AB = OA. Since OE
^CD, E lies on the circle
with diameter OS.
Consider the reflection in the point B (dilation in B with
factor 1). It interchanges E and F, interchanges O and
A, and switches the circles ADRC and OERS. Since E lies
on the latter circle, F must lie on the former circle, and the
desired result (a) follows.
Ad (b), the locus of E is that part of the circle with centre A
that lies within the circle with centre O. Angle EAO is
maximum when E coincides with R, and minimum when
D coincides with A. Since triangle ORA is
equilateral, the maximum angle is 60^{°} and the minimum angle
is 0^{°}.
Solution 3. [M. Elqars] Let the radius of the circle
be r. Let ÐCBO = ÐDBA = a,
ÐDOB = b and ÐCOF = g. By the Law of Sines,
we have that
sina: sin(g a) = OC : OB = OD : OB = sin(180^{°}  a) : sin(a b) = sina: sin(a b) , 

whence
a b =
g a. Thus 2
a =
b+
g. Therefore,
ÐDOC = 180^{°}  (b+ g) = 180^{°}  2a = ÐCBD . 

Thus,
ÐDOE = 
1
2

ÐDOC = 90^{°}  a , 

whence
ÐEDO =
a and
OE
 = rsin
a.
Observe that sin(a b) : sina = OB : OD = 1:2,
so that sin(a b) = ^{1}/_{2}sina.


= OE ^{2} + OA ^{2}  2 OE OA cosÐEOA 
 
 
= r^{2} sin^{2} a+ r^{2}  2r^{2} sinacos(90^{°}  a+ b) 
 
 
= r^{2} [1 + sin^{2} a 2sinasin(a b)] 
 
 
= r^{2} [1 + sin^{2} a sin^{2} a] = r^{2} . 

The segments EF and OA bisect each other, so they are
diagonals of a parallelogram OFAE. Hence
OF
 =
AE
 = r, as desired. As before, we see that
ÐEAO
ranges from 0
^{°} to 60
^{°}.
Solution 4. Assign coordinates: O ~ (0, 0),
B ~ (^{1}/_{2}, 0), C ~ (1, 0), and let the
slope of the lines BC and BD be respectively m and m.
Then C ~ (^{1}/_{2} + s, ms) and D ~ (^{1}/_{2} + t,mt) for some s and t. Using the fact that the coordinates of
C and D satisfy x^{2} + y^{2} = 1, we find that
C ~ 
æ è


2(m^{2} + 1)

, 
2(m^{2} + 1)


ö ø



D ~ 
æ è


2(m^{2} + 1)

, 
2(m^{2} + 1)


ö ø



E ~ 
æ è


m^{2}
2(m^{2} + 1)

, 
2(m^{2} + 1)


ö ø



F ~ 
æ è


m^{2} + 2
2(m^{2} + 1)

, 
2(m^{2} + 1)


ö ø

. 

It can be checked that the coordinates of F satisfy the equation
x
^{2} + y
^{2} = 1 and the result follows.
Solution 5. [P. Shi] Assign the coordinates O ~ (0, 1),
B ~ (0, 0), A ~ (0, 1). Taking the coordinates of
C and D to be of the form (x, y) = (r cosq,r sinq) and (x, y) = (s cosq,  s sinq)
and using the fact that both lie of the circle of equation
x^{2} + (y + 1)^{2} = 4, we find that
C ~ ((  Ö

sin^{2} q+ 3

 sinq)cosq,(  Ö

sin^{2} q+ 3

 sinq)sinq) 

D ~ ((  Ö

sin^{2} q+ 3

+ sinq)cosq,(  Ö

sin^{2} q+ 3

+ sinq)sinq) 

E ~ ((  Ö

sin^{2} q+ 3

)cosq, sin^{2} q) 

F ~ ((  Ö

sin^{2} q+ 3

)cosq, sin^{2} q) . 

It is straightforward to verify that
OF
^{2} = 4,
from which the result follows.

328.

Let \frakC be a circle with diameter AC
and centre D. Suppose that B is a point on the circle for
which BD ^AC. Let E be the midpoint of DC and
let Z be a point on the radius AD for which EZ = EB.


Prove that


(a) The length c of BZ is the length of the side of
a regular pentagon inscribed in \frakC.


(b) The length b of DZ is the length of the side of
a regular decagon (10gon) inscribed in \frakC.


(c) c^{2} = a^{2} + b^{2} where a is the length of a
regular hexagon inscribed in \frakC.


(d) (a + b):a = a:b.
Comment. We begin by reviewing the trigonetric functions of
certain angles. Since


= 2 cos^{2} 36^{°}  1 = 2 cos^{2} 144^{°}  1 
 
 
= 2(2 cos^{2} 72^{°}  1)^{2} = 8cos^{4} 72^{°}  8cos^{2} 72^{°} + 1 , 

t = cos72
^{°} is a root of the equation
0 = 8t^{4}  8t^{2}  t + 1 = (t  1)(2t + 1)(4t^{2} + 2t  1) . 

Since it must be the quadratic factor that vanishes when
t = cos72
^{°}, we find that
sin18^{°} = cos72^{°} = 
Ö5  1
4

. 

Hence cos144
^{°} =
(
Ö5 + 1)/4 and
cos36
^{°} = (
Ö5 + 1)/4.
Solution. [J. Park] Wolog, suppose that the radius of the
circle is 2.
(a) Select W on the arc of the circle joining A and B such that
BW = BZ. We have that BE  = EZ  = Ö5,
b = ZD  = Ö5  1, c^{2} = BZ^{2} +BW ^{2} = 10  2Ö5 and, by the Law of Cosines
applied to triangle BDW,
Hence BW subtends an angle of 72
^{°} at the centre of the
circle and so BW is a side of an inscribed regular pentagon.
(b) The angle at each vertex of a regular decagon is 144^{°}.
Thus, the triangle formed by the side of an inscribed regular pentagon
and two adjacent sides of an inscribed regular decagon has angles
144^{°}, 18^{°}, 18 ^{°}. Conversely, if a triangle
with these angles has its longest side equal to that of an inscribed
regular pentagon, then its two equal sides have lengths equal to
those of the sides of a regular decagon inscribed in the same circle.
Now b = Ö5  1, c^{2} = 10  2Ö5, so 4b^{2}  c^{2} = 14  2Ö{45} > 0. Thus, c < 2b, and we can construct an
isosceles triangle with sides b, b, c. By the Law of Cosines, the
cosine of the angle opposite c is equal to (2b^{2}  c^{2})/(2b^{2}) = ^{1}/_{4}(Ö5  1). This angle is equal to 144^{°},
and so b is the side length of a regular inscribed decagon.
(c) The side length of a regular inscribed hexagon is equal to
the radius of the circle. We have that
a^{2} = BD ^{2} = BZ ^{2}  ZD ^{2} = c^{2}  b^{2} . 

(d) Since b^{2} + 2b  4 = (b + 1)^{2}  5 = 0, a^{2} = 4 = (2 + b)b = (a + b)b, whence (a + b): a = a:b.

329.

Let x, y, z be positive real numbers.
Prove that
 Ö

x^{2}  xy + y^{2}

+  Ö

y^{2}  yz + z^{2}

³  Ö

x^{2} + xz + z^{2}

. 

Solution 1. Let ABC be a triangle for which
AB
 = x,
AC
 = z and
ÐBAC = 120
^{°}. Let
AD be a ray through A that bisects angle BAC and has length
y. By the law of cosines applied respectively to triangle
ABC, ABD and ACD, we find that
BC  =  Ö

x^{2} + xz + z^{2}

, 

BD  =  Ö

x^{2}  xy + y^{2}

, 

CD  =  Ö

y^{2}  yz + z^{2}

. 

Since
BD
+
CD
 ³ BC
, the
desired result follows.
Solution 2. [B. Braverman; B.H. Deng] Note that
x^{2}  xy + y^{2}, y^{2}  yz + z^{2} and z^{2} + xz + z^{2} are
always positive [why?]. By squaring, we see that the
given inequality is equivalent to
x^{2} + z^{2} + 2y^{2}  xy  yz + 2  Ö

(x^{2}  xy + y^{2})(y^{2}  yz + z^{2})

³ x^{2} + xz + z^{2} 

which reduces to
2  Ö

(x^{2}  xy + y^{2})(y^{2}  yz + z^{2})

³ xy + yz + zx  2y^{2} . 

If the right side is negative, then the inequality holds trivially.
If the right side is positive, the inequality is equivalent (by
squaring) to
4(x^{2}  xy + y^{2})(y^{2}  yz + z^{2}) ³ (xy + yz + zx  2y^{2})^{2} . 

Expanding and simplifying gives the equivalent inequality
x^{2}y^{2} + x^{2}z^{2} + y^{2}z^{2}  2x^{2}yz  2xyz^{2} + 2xy^{2}z ³ 0 

or (xy + yz
 zx)
^{2} ³ 0. Since the last always holds, the result
follows.
Comment. The above writeup proceeds by a succession of
equivalent inequalities to one that is trivial, a workingbackwards
from the result. The danger of this approach is that one may come
to a step where the reasoning is not necessarily reversible, so that
instead of a chain of equivalent statements, you get to a stage where
the logical implication is in the wrong direction. The possibility that
xy + yz + zx < 2y^{2} complicates the argument a litle and needs
to be dealt with. To be on the safe side, you could frame the
solution by starting with the observation that (xy + yz  zx)^{2} ³ 0 and deducing that
(xy + yz + zx  2y^{2})^{2} £ 4(x^{2}  xy + y^{2})(y^{2}  yz + z^{2}) . 

From this, we get that
xy + yz + zx  2y^{2} £ xy + yz + zx  2y^{2}  £  Ö

(x^{2}  xy + y^{2})(y^{2}  yz + z^{2})

, 

from which the required inequality follows by rearranging the terms
between the outside members and taking the square root.
Solution 3. [L. Fei] Let x = ay and z = by for some
positive reals a and b. Then the given inequality is
equivalent to
 Ö

a^{2}  a + 1

+  Ö

b^{2}  b + 1

³  Ö

a^{2} + ab + b^{2}



which in turn (by squaring) is equivalent to
2  a  b + 2  Ö

a^{2}  a + 1

 Ö

b^{2}  b + 1

³ ab . 
 (*) 
We have that


£ 3(ab  a  b)^{2} = 3a^{2}b^{2}  6a^{2}b  6ab^{2} + 3a^{2} + 3b^{2} + 6ab 
 
 
= 4(a^{2}b^{2}  a^{2}b + a^{2}  ab^{2} + ab  a + b^{2}  b + 1) (a^{2}b^{2} + a^{2} + b^{2} + 4 + 2a^{2}b + 2ab^{2}  2ab  4a  4b) 
 
 
= [2  Ö

a^{2}  a + 1

 Ö

b^{2}  b + 1

]^{2} (ab + a + b  2)^{2} . 

Hence
2  Ö

a^{2}  a + 1

 Ö

b^{2}  b + 1

³ ab + a + b  2  ³ ab + a + b  2 . 

Taking the inequality of the outside members and rearranging the
terms yields (*).
Comment. The student who produced this solution worked backwards
down to the obvious inequality (ab  a  b)^{2} ³ 0, However,
for a proper argument, you need to show how by logical steps you
can go in the other direction, i.e. from the obvious
inequality to the desired one. You will notice that this has
been done in the writeup above; the price that you pay is that
the evolution from (ab  a  b)^{2} ³ 0 seems somewhat artificial.
There is a place where care is needed. It is conceivable that
ab + a + b  2 is negative, so that
2Ö{a^{2}  a + 1}Ö{b^{2}  b + 1} ³ ab + a + b  2
is always true, even when 4(a^{2}  a + 1)(b^{2}  b + 1) ³ (ab + a + b  1)^{2} fails. The inequality A^{2} ³ B^{2}
is equivalent to A ³ B only if you know that A and B are
both positive.

330.

At an international conference, there are four official
languages. Any two participants can communicate in at least one of
these languages. Show that at least one of the languages is spoken
by at least 60% of the participants.
Solution 1. Let the four languages be E, F, G, I. If anyone
speaks only one language, then everyone else must speak that
language, and the result holds. Suppose there is an individual
who speaks exactly two languages, say E and F. Then everyone
else must speak at least one of E and F. If 60% of the
participants speaks a particular one of these languages, then
the result holds. Otherwise, at least 40% of the participants,
constituting set A, must speak E and
not F, and 40%, constituting set B, must speak F and not E.
Since each person in A must communicate with each person in B,
each person in either of these sets must speak G or I.
At least half the members of A must speak a particular one of
these latter languages, say G. If any of them speaks only
G (as well as E), then everyone in B must speak G and
so at least 20% + 40% = 60% of the participants speak G.
The remaining possibility is that everyone in A speaks both
G and I. At least half the members of B speaks a particular
one of these languages, say G, and so 40% + 20% = 60% of the
participants speak G. Thus, if anyone speaks only two languages,
the result holds.
Finally, suppose that every participant speaks at least three
languages. Let p% speak E, F and G (and possibly,
but not necessarily I), q% speak E, F, I but not G,
r% speak E, G, I but not F, s% speak F, G, I but not
E. Then p + q + r + s = 100 and so
(p + q + r) + (p + q + s) + (p + r + s) + (q + r + s) = 300 . 

At least one of the four summands on the left is at least 75,
Suppose it is p + r + s, say. Then at least 75% speaks G.
The result holds once again.
Solution 2. [D. Rhee] As in Solution 1, we take the languages
to be E, F, G, I, and can dispose of the case where someone speaks
exactly one of these. Let (A¼C) denote the set of people
who speaks the languages A ¼C and no other language,
and suppose that each person speaks at least two languages.
We first observe that in each of the pairs of sets,
{(EF),(GI)}, {(EG),(FI)}, {(EI),(FG)}, at least one of the
sets in each pair is empty. So either there is one language that
is spoken by everyone speaking exactly two languages, or else
there are only three languages spoken among those that speak exactly
two languages. Thus, wolog, we can take the two language sets
among the participants to be either { (EF), (EG), (EI) }
or { (EF), (FG), (EG) }.
Case 1. Everyone is in exactly one of the language groups
(EF), (EG), (EI), (EFG), (EFI), (EGI), (FGI), (EFGI) . 

If no more than 40% of the participants are in (FGI), then at
least 60% of the participants speak E. Otherwise, more than
40% of the participants are in (FGI). If no more that 40% of
the participants are in (EG)
È(EI)
È(EGI), then at least
60% of the participants speak F. The remaining case is that
more that 40% of the participants are in each of (FGI) and
(EG)
È(EI)
È(EGI). It follows that, either at least 20%
of the participants are in (EG)
È(EGI), in which case at least
60% speak G, or at least 20% of the participants are in
(EI)
È(EGI), in which case at least 60% speak I.
Case 2. Everyone is in exactly one of the language groups
(EF), (EG), (FG), (EFG), (EFI), (EGI), (FGI), (EFGI) . 

Since the three sets (EF)
È(EFI), (EG)
È(EGI) amd
(FG)
È(FGI) are disjoint, one of them must include fewer than
40% of the participants. Suppose, say, it is (EF)
È(EFI).
Then more than 60% must belong to its complement, and each of
these must speak G. The result follows.

331.

Some checkers are placed on various squares of a
2m ×2n chessboard, where m and n are odd. Any number
(including zero) of checkers are placed on each square. There are
an odd number of checkers in each row and in each column. Suppose that
the chessboard squares are coloured alternately black and white
(as usual). Prove that there are an even number of checkers on the
black squares.
Solution 1. Rearrange the rows so that the m oddnumbered
rows move into the top m positions and the m evennumbered rows
move into the bottom m positions, while all the columns remain
intact except for order of entries. Now move all the n
oddnumbered columns to the left n positions and all the n
evennumbered columns to the right n positions, while the
rows remain intact except for order of entries. The conditions of
the problem continue to hold. Now the chessboard consists of
two diagonally opposite m ×n arrays of black squares and
two diagonally opposite m ×n arrays of white squares.
Let a and b be the number of checkers in the top m ×n
arrays of black and white squares respectively, and c and d be
the number of checkers in the arrays of white and black squares
respectively. Since each row has an odd number of checkers, and
m is odd, then a + b is odd. By a similar argument, b + d
is odd. Hence
a + d = (a + b) + (b + d)  2b 

must be even. But a + d is the total number of checkers on the
black squares, and the result follows.
Solution 2. [F. Barekat] Suppose that the (1, 1) square on
the chessboard is black. The set of black squares is contained in
the union U of the oddnumbered columns along with the union
V of all the evennumbered rows. Note that U contains an odd
number of columns and V an odd number of rows.
Since each row and each column
contains an odd number of checkers, U has an odd number u of
checkers and V has an odd number v of checkers. Thus u + v
is even. Note that each white square belongs either to both of
U and V, or to neither of them. Thus, the u + v is equal to
the number of black checkers plus twice the number of checkers on
the white squares common to U and V. The result follows.