Toán Olympic quốc tế 1998 Tiếng Anh

Problem 1
A convex quadrilateral ABCD has perpendicular diagonals. The perpendicular bisectors
of AB and CD meet at a unique point P inside ABCD. Prove that ABCD is cyclic if and
only if triangles ABP and CDP have equal areas.
Solution:
Let AC and BD intersect at E. Suppose by symmetry that P is in ABE. Denote by M
and N the respective feet of perpendiculars from P to AC and BD. Without assuming that
PA  PB and PC  PD, we express the areas ABP and CDP as follows:
2ABP  2ABE  2PAE  2PBE
 AM  PNBN  PM  AM  PNPM  BN  PMPN
 AM  BN  PM  PN,
2CDP  2CDE  2PCE  2PDE
 CM  PNDN  PM  CM  PNPM  DN  PMPN
 CM  DN  PM  PN.
Therefore
2ABP  CDP  AM  BN  CM  DN.
We now assume that PA  PB and PC  PD. Suppose that ABCD is cyclic. Then the
uniqueness of P implies that it must be the circumcenter. So M and N are the midpoints
of AC and BD, respectively. Hence AM  CM and BN  DN,so(*)
implies ABP  CDP. Conversely, suppose that ABP  CDP. Then, by (*), we have
AM  BN  CM  DN.IfPA  PC, assume by symmetry that PA  PC. Then AM  CM
and also BN  DN, because PB  PD. Thus AM  BN  CM  DN, a contradiction. Hence
PA  PC, which implies that P is equidistant from A, B, C and D. We conclude then that
ABCD is cyclic.
Alternative Solution:
Let AC and BD meet at E again. Assume by symmetry that P lies in BEC and denote
ABE   and ACD  . The triangles ABP and CDP are isosceles. If M and N are the
respective midpoints of their bases AB and CD, then PM  AB and PN  CD. Note that M,
N and P are not collinear due to the uniqueness of P.
Consider the median EM to the hypotenuse of the right triangle ABE.Wehave
BEM  , AME  2 and EMP  90


 2. Likewise, CEN  , DNE  2
and ENP  90

 2. Hence MEN  90

   , and a direct computation yields
NPM  360

 EMP  MEN  ENP  90

     MEN.
It turns out that, whenever AC  BD, the quadrilateral EMPN has a pair of equal opposite
angles, the ones at E and P.
We now prove our claim. Since AB  2EM and CD  2EN,wehaveABP 

CDP

if and only if EM  PM  EN  PN,or,EM/EN  PN/PM. On account
of MEN  NPM, the latter is equivalent to EMN  PNM. This holds if and only if
EMN  PNM and ENM  PMN, and these in turn mean that EMPN is a
parallelogram. But the opposite angles of EMPN at E and P are always equal, as noted
above. So it is a parallelogram if and only if EMP  ENP;thatis,if
90

 2  90

 2. We thus obtain a condition equivalent to   ,ortoABCD being
cyclic.
Another Solution:

Choose a coordinate system so that the axes lie on the perpendicular lines AC and BD,
and so that the coordinates of A, B, C and D are 0,a, b,0, 0,c and d,0, respectively.
By assumption, the perpendicular bisectors of AB and CD have a unique common point.
Hence the linear system formed by their equations 2bx  2ay  b
2
 a
2
and 2dx  2cy  d
2
 c
2
has a unique solution, and it is given by
x
0

cb
2
 a
2
  ad
2
 c
2

2ad  bc
, y
0

db
2

 a
2
  bd
2
 c
2

2ad  bc
.
Naturally, x
0
,y
0
 are the coordinates of P. Since it is interior to ABCD, ABP and CDP
have the same orientation. Then ABP  CDP if and only if
||

||
This is equivalent to ax
0
 by
0
 ab  cx
0
 dy
0
 cd. Inserting the expressions for x
0
and y
0

, after the inevitable algebra work we obtain the equivalent condition
ac  bda  c
2
 b  d
2
  0.
Now, the choice of the coordinate system implies that a and c have different signs, as well
as b and d. Hence the second factor is nonzero, so ABP  CDP if and only if ac  bd.
The latter is equivalent to AE  CE  BE  DE, where E is the intersection point of the
diagonals. However, it is a necessary and sufficient condition for ABCD to be cyclic.
Problem 2
Let ABCD be a cyclic quadrilateral. Let E and F be variable points on the sides AB and
CD, respectively, such that AE : EB  CF : FD.LetP be the point on the segment EF
such that PE : PF  AB : CD. Prove that the ratio between the areas of triangles APD and
BPC does not depend on the choice of E and F.
Solution:
We first assume that the lines AD and BC are not parallel and meet at S.SinceABCD is
cyclic, ASB and CSD are similar. Then, since AE : EB  CF : FD, ASE and CSF are
also similar, so that
DSE  CSF.
Moreover, we have
SE
SF

SA
SC

AB
CD


PE
PF
,
which implies that ESP  FSP. Thus, ASP  BSP,andsoP is equidistant from
the lines AD and BC. Therefore,
APD : BPC  AD : BC,
which is a constant.
Next, assume that the lines AD and BC are parallel. Then ABCD is an isosceles
trapezoid with AB  CD, and we have BE  DF.LetM and N be the midpoints of AB and
CD, respectively. Then ME  NF and clearly, E and F are equidistant from the line MN.
Thus P, the midpoint of EF, lies on MN. It follows that P is equidistant from AD and BC,
and hence
APD : BPC  AD : BC.
Alternative Solution:
Let AE : EB  CF : FD  a : b,wherea  b  1. Since PE : PF  AB : CD,we
have
d

P,AD


CD
AB  CD
 d

E,AD


AB
AB  CD

 d

F,AD

,
d

P,BC


CD
AB  CD
 d

E,BC


AB
AB  CD
 d

F,BC

,
where dX,YZ stands for the distance from the point X to the line YZ. Thus, we obtain
APD 
CD
AB  CD
AED 
AB

AB  CD
AFD

a  CD
AB  CD
ABD 
b  AB
AB  CD
ACD,
BPC 
CD
AB  CD
BEC 
AB
AB  CD
BFC

b  CD
AB  CD
BAC 
a  AB
AB  CD
BDC.
Next, since A, B, C and D areconcyclic,wehavesinBAD  sinBCD and
sinABC  sinADC. Thus,
APD
BPC

a  CD ABD  b  AB ACD
b  CD BAC  a  AB BDC


a  CD  AB  AD  sinBAD  b  AB  CD  AD  sinADC
b  CD  AB  BC  sinABC  a  AB  CD  BC  sinBCD

AD
BC

a  sinBAD  b  sinADC
b  sinABC  a  sinBCD

AD
BC
.
Problem 3
Let I be the incenter of triangle ABC.LetK,L and M be the points of tangency of the
incircle of ABC with AB,BC and CA, respectively. The line t passes through B and is
parallel to KL. The lines MK and ML intersect t at the points R and S. Prove that RIS is
acute.
Solution:
Since the lines KL and RS are parallel, we have, in BKR ,
BKR  90° A/2,
KBR  90° B/2,
BRK  90° C/2.
Hence, by the law of sine,
BR 
cosA/2
cosC/2
 BK. #
Similarly, we have, in BLS,
BLS  90° C/2,

BSL  90° A/2,
LBS  90° B/2.
so that
BS 
cos

C/2

cos

A/2

 BL 
cos

C/2

cos

A/2

 BK. #
Notice now that BI  RS and IK  AB. Then, on account of (1) and (2), we obtain
IR
2
 IS
2
 RS
2



BI
2
 BR
2



BI
2
 BS
2



BR  BS

2
 2

BI
2
 BR  BS

 2

BI
2
 BK
2


 2IK
2
 0.
So,bythelawofcosine,RIS is acute.
Alternative Solution:
Let W be the midpoint of KL and Q the midpoint of KM. Then Q  AI, W  BI,
AIKM and BIKL. We first prove that AWRI and CWSI.
Since RBI  RQI  90

, the points R, B, I, Q are concyclic, which implies
BRI  BQI. Also, in the right triangles AIK and BIK,wehave
IQ  IA  KI
2
 IW  IB,
so that IQ/IW  IB/IA. Hence AIW and BIQ are similar. It follows that
BRI  BQI  AWI.SinceBR  IW, this implies that RIAW.Byasimilar
argument, we can prove that SICW. Thus, RIS  180

AWC. It remains to prove
that AWC is obtuse.
Let T be the midpoint of AC.Then2
WT  WC  WA  LC  KA.SinceLC and KA are
not collinear, we have
WT 
LC  KA
2

CM  AM
2


AC
2
.
This implies that W is inside the circle with diameter AC,andsoAWC  90°.
Therefore, RIS is acute.
Comment:
Another proof for the fact AW  RI is as follows. Since RBI  RQI  90°, RBIQ
is cyclic and its circumcircle is orthogonal to the diameter RI. Consider the inversion  with
respect to the incircle of ABC.Since takes B and Q into W and A, respectively, it takes
the circumcircle of RBIQ into the line AW.Since leaves the line RI invariant, we have
AW  RI.
Problem 4
Let M and N be points inside triangle ABC such that
MAB  NAC MBA  NBC.
Prove that
AM  AN
AB  AC

BM  BN
BA  BC

CM  CN
CA  CB
 1.
Solution:
Let K be the point on the ray BN such that BCK  BMA. Note that K is outside
ABC, because BMA  ACB.SinceMBA  CBK,wehaveABM  KBC;so,
AB
BK


BM
BC

AM
CK
. #
Then, since ABK  MBC,andAB/KB  BM/BC, we see that ABK  MBC. Hence
AB
BM

BK
BC

AK
CM
. #
Now we have CKN  MAB  NAC. Consequently, the points A, N, C and K are
concyclic. By Ptolemy’s theorem, AC  NK  AN  CK  CN  AK,or
AC

BK  BN

 AN  CK  CN  AK. #
From (1) and (2), we find CK  AM  BC/BM, AK  AB  CM/BM and BK  AB  BC/BM.
Inserting these expressions in (3), we obtain
AC 
AB  BC
BM
 BN 

AN  AM  BC
BM

CN  AB  CM
BM
,
or
AM  AN
AB  AC

BM  BN
BA  BC

CM  CN
CA  CB
 1.
Alternative Solution:
Let the complex coordinates of A, B, C, M and N be a, b, c, m and n, respectively. Since
the lines AM, BM and CM are concurrent, as well as the lines AN, BN and CN, it follows
from Ceva’s theorem that
sinBAM
sinMAC

sinCBM
sinMBA

sinACM
sinMCB
 1. #
sinBAN

sinNAC

sinCBN
sinNBA

sinACN
sinNCB
 1. #
By hypotheses, BAM  NAC and MBA  CBN. Hence BAN  MAC
and NBA  CBM. Combined with (1) and (2), these equalities imply
sinACM  sinACN  sinMCB  sinNCB.
Thus,
cos

NCM  2ACM

 cosNCM  cos

NCM  2NCB

 cosNCM.
and hence ACM  NCB.
Since BAM  NAC, MBA  CBN and ACN  MCB, the following
complex ratios are all positive real numbers:
m  a
b  a
:
c  a
n  a
,

m  b
a  b
:
c  b
n  b
and
m  c
b  c
:
a  c
n  c
.
Hence each of these equals its absolute value, and so
AM  AN
AB  AC

BM  BN
BA  BC

CM  CN
CA  CB


m  a

n  a


b  a


c  a



m  b

n  b


a  b

c  b



m  c

n  c


b  c

a  c

 1.
Comment:
Contestants who are familiar with the notion of isogonal conjugate points may skip
over the early part of the proof.
Problem 5
Let ABC be a triangle, H its orthocenter, O its circumcenter, and R its circumradius. Let

D be the reflection of A across BC, E be that of B across CA,andF that of C across AB.
Prove that D,E and F are collinear if and only if OH  2R.
Solution:
Let G be the centroid of ABC,andA

, B

and C

be the midpoints of BC, CA and AB,
respectively. Let A

B

C

be the triangle for which A, B and C are the midpoints of B

C

,
C

A

and A

B

, respectively. Then G is the centroid and H the circumcenter of A


B

C

.
Let D

, E

and F

denote the projections of O on the lines B

C

, C

A

and A

B

,
respectively.
Consider the homothety h with center G and ratio 1/2. It maps A, B, C, A

, B


and C

into
A

, B

, C

, A , B and C, respectively. Note that A

D

 BC, which implies
AD : A

D

 2:1 GA : GA

and DAG  D

A

G. We conclude that hD  D

.
Similarly, h

E


 E

and h

F

 F

. Thus, D, E and F are collinear if and only if D

, E

and F

are collinear. Now D

, E

and F

are the projections of O on the sides B

C

, C

A

and A


B

, respectively. By Simson’s theorem, they are collinear if and only if O lies on the
circumcircle of A

B

C

. Since the circumradius of A

B

C

is 2R, O lies on its
circumcircle if and only if OH  2R.
Alternative Solution:
Let the complex coordinates of A, B, C, H and O be a, b, c, h and 0, respectively.
Consequently, aa

 bb

 cc

 R
2
and h  a  b  c.SinceD is symmetric to A with
respect to line BC, the complex coordinates d and a satisfy

d  b
c  b

a  b
c  b
,or

b

 c


d 

b  c

a



bc

 b

c

 0. #
Since
b


 c

 
R
2

b  c

bc
and bc

 b

c 
R
2

b
2
 c
2

bc
,
by inserting these expressions in (1), we obtain
d 
bc  ca  ab
a

k  2bc

a
.
d


R
2

a  b  c

bc

R
2

h  2a

bc
.
where k  bc  ca  ab. Similarly, we have
e 
k  2ca
b
, e


R
2

h  2b


ca
, f 
k  2ab
c
andf


R
2

h  2c

ab
.
Since

dd

1
ee

1
ff

1

e  de

 d


f  df

 d



ba

k2ab

ab
R
2

ab

h2c

abc

ca

k2ca

ca
R
2

ac


h2b

abc

R
2

c  a

a  b

a
2
b
2
c
2



ck  2abc

h  2c


bk  2abc




h  2b


R
2

b  c

c  a

a  b

hk  4abc

a
2
b
2
c
2
and h

 R
2
k/abc, it follows that
D,EandFarecollinear
0
 hk  4abc  0
 hh


 4R
2
 OH  2R.
Problem 6
Let ABCDEF be a convex hexagon such that B  D  F  360

and
AB
BC

CD
DE

EF
FA
 1.
Prove that
BC
CA

AE
EF

FD
DB
 1.
Solution:
Let P be the point such that FEA  DEP and EFA  EDP, where the angles
are directed. Then FEA and DEP are similar. Hence
FA

EF

DP
DE
. #
EF
ED

EA
EP
. #
It follows from (1) and the given conditions that ABC  PDC and
AB
BC

DE  FA
CD  EF

DP
CD
.
Therefore, ABC and PDC are similar, which gives BCA  DCP and
CB
CD

CA
CP
. #
Since FED  AEP, (2) implies that FED and AEP are similar. Analogously, since
BCD  ACP, (3) yields the similarity of BCD and ACP. Hence

FD
EF

PA
AE
and
BC
DB

CA
PA
.
Multiplying these together, we have the desired result.
Alternative Solution:
Let the complex coordinates of A,B,C,D,E and F be a,b,c,d,e and f, respectively.
Since ABCDEF is convex, B, D and F are the arguments of the complex numbers

a  b

/

c  b

,

c  d

/

e  d


and

e  f

/

a  f

, respectively. Then the condition
B  D  F  360

implies that the product of these three complex numbers is a
positive real number. It is equal to the product of their absolute values AB/BC,CD/DE and
EF/FA.Since

AB/BC

CD/DE

EF/FA  1, we have
a  b
c  b

c  d
e  d

e  f
a  f
 1.

On the other hand, since
a  bc  de  f  c  be  da  f
 b  ca  ef  d  a  cf  eb  d,
we deduce immediately that
b  c
a  c

a  e
f  e

f  d
b  d
 1.
Taking absolute values on both sides gives
BC
CA

AE
EF

FD
DB
 1.
Comment:
Considering the arguments of the complex numbers on both sides of the equality
b  c
a  c

a  e
f  e


f  d
b  d
 1.
it shows that BDF  AEF  ACB.
Problem 7
Let ABC be a triangle such that ACB  2ABC.LetD be the point on the side BC
such that CD  2BD. The segment AD is extended to E so that AD  DE. Prove that
ECB  180

 2EBC.
Solution:
Let H be the midpoint of CD,sothatABEH is a parallelogram. Extend BC to G so that
CG  CA.TakeBD  DH  HC  a/3,CA  b,AB  c,BE  AH  x,AD  DE  y
and CE  z.Since2ABC  ACB  CAG  CGA  2CGA, ABG and CAG
are similar. Hence AB/BG  CA/AG,or
c
2
 ba  b. #
Applying Apollonius’ theorem to the triangles ACD,ABH and CDE,wehave
b
2
 y
2
 2x
2

2a
2
9

, #
x
2
 c
2
 2y
2

2a
2
9
, #
y
2
 z
2
 2c
2

2a
2
9
. #
Eliminating y from (2) and (3), we have
x
2
 c
2
 2b
2

 4x
2

2a
2
3
.
Combining this with (1), we have
x
2
 b 
2a
3
b 
a
3
. #
Eliminating y from (3) and (4), we have
x
2
 c
2
 2z
2
 4c
2

2a
2
3

.
Combining this with (1) and (5), we have z  b  2a/3. From this and (5), we obtain
x
2
 zz  a,orBE
2
 CECE  BC  CE  EP,
where P is the point on CE such that CP  BC. Then BE/CE  EP/BE. Hence BEP and
CEB are similar since BEP  CEB. It follows that
ECB  EBP  EBC 
1
2
180

ECB,
which simplifies to ECB  180

 2EBC.
Alternative Solution:
Lemma. In a triangle PQR,ifS lies on the side QR and divides it in the ratio k :1,
then

k  1

cotPSR  cotPQR  kcotPRQ.
Proof. Let PSR  , PQR   and PRS  .Thenwehave
k 
QS
SR



QSP


SRP


PQsin

  

PRsin

  


sinsin

  

sinsin

  


sinsincos  sincossin
sinsincos  sincossin

cot  cot
cot  cot

.
The result follows at once.
Proof of the main result.
Put a  BC,b  CA and c  AB.LetABC   and ACB  2.LetH be the
midpoint of CD. Then ABEH is a parallelogram. We have
cotACB 
1  tan
2

2tan

1
2
r 
1
r
,
where r  cot. Note that r  0 because  must be acute. By the lemma with k 
1
2
on
ABC,wehave
3
2
cotADC  r 
1
4
r 
1
r

,
cotADC 
1
6
3r 
1
r
.
By the lemma with k  1onACD,wehave
2cotAHC  cotADC  cotACB,or
cotAHC 
1
12
3r 
1
r

1
4
r 
1
r

1
3r
,
cotEBC  cotAHB  
1
3r
 0.

By the lemma with k  2onECB,wehave
3cotEDB  cotECB  2cotEBC,
cotECB 
3
6
3r 
1
r

2
3r

3
2
r 
1
6r
 
1
2cotEBC

1
2
cotEBC
 cot2EBC
 cot

2EBC  180



.
Since 0  ECB  180° and 90°  EBC  180°, we conclude that
ECB  2EBC  180°.
Another Solution:
Let the complex coordinates of A, B, C, D and E be a, b, c, d and e, respectively. Then
d 

2b  c

/3 and e  2d  a.SinceACB  2ABC, the ratio
a  b
c  b
2
:
b  c
a  c
is real and positive. It is equal to AB
2
 AC/BC
3
. On the other hand, a direct computation
shows that the ratio
e  c
b  c
:
c  b
e  b
2
is equal to
1


b  c

3


b  a

 2

c  a

3
2
4

b  a



c  a

3

4
27


b  a


2

c  a


b  c

3

4
27

AB
2
 AC
BC
3
,
which is a real number. Hence the arguments of e  c/b  c and c  b
2
/e  b
2
,
namely, ECB and 2EBC, differ by an integer multiple of 180°. We easily infer that
either ECB  2EBC or ECB  2EBC  180

, according as the ratio is positive or
negative. To prove that the latter holds, we have to show that AB
2
 AC/BC

3
is greater than
4/27. Choose a point F on the ray AC such that CF  CB.
Since CBF is isosceles and ACB  2ABC,wehaveCFB  ABC. Thus, ABF
and ACB are similar and AB : AF  AC : AB.SinceAF  AC  BC,
AB
2
 AC

AC  BC

.LetAC  u
2
and AC  BC  v
2
. Then AB  uv and BC  v
2
 u
2
.
From AB  AC  BC, we obtain u/v  1/2. Thus
AB
2
 AC
BC
3

u
4
v

2

v
2
 u
2

3


u/v

4

1  u
2
/v
2

3


1/2

4

1  1/4

3


4
27
.
and the conclusion follows.
Problem 8
Let ABC be a triangle such that A  90

and B  C. The tangent at A to its
circumcircle  meets the line BC at D.LetE be the reflection of A across BC, X the foot of
perpendicular from A to BE,andY the midpoint of AX. Let the line BY meet  again at Z.
Prove that the line BD is tangent to the circumcircle of triangle ADZ.
Solution:
Let G be diametrically opposite A,andH the point of intersection of AE and BD. Note
that B and G are on the same side of AE because of the condition B  C. We claim
that G, H and Z are collinear. Indeed, AEG  90

 AXB and, in addition,
AGE  ABE  ABX. Hence AGE and ABX are similar. It follows that
GAE  BAX and GA/BA  AE/AX. Moreover, AE/AX  AH/AY,sinceH and Y are
the midpoints of AE and AX, respectively. So AGH and ABY are also similar. Hence
AGH  ABY,andifGH meets  at Z

, then ABZ

 AGZ

 ABY  ABZ.
This implies that Z and Z

coincide, both of them being on the minor arc AE. This justifies

the claim.
We now prove that DAZ  ZDH.SinceG, H and Z are in a line,
AZH  AZG  90

.SinceDA and DE are tangents to ,wehaveDAZ  AEZ
and DEZ  EAZ. Hence DHZ  90

AHZ  EAZ  DEZ,sothatZHED
is a cyclic quadrilateral. It follows that ZDH  ZEA  DAZ,andDB is indeed
tangent to the circumcircle of ADZ.
Alternative Solution:
As in the solution above, we prove that G, H and Z are collinear. Next, the
circumcircle  of ADZ and BD are tangent if and only if so are their images under the
inversion  with respect to . Note that  leaves A and Z unchanged, as well as the line BD.
Also, it takes D to the midpoint H of the chord AE.Sotheimageof under the inversion is
the circumcircle 

of AHZ. It always has a point in common with BD, namely, H.
Besides, AH  BD. Hence 

and BD are tangent if and only if AH is a diameter of 

,
which is equivalent to AZH  90

. On the other hand, AZG  90

,sinceAG is a
diameter of . We infer that  is tangent to BD if and only if G, H and Z are collinear,
which was already proven.

Another Solution:
Let the rectangular coordinates of A, B and C be A

a
1
,a
2

, B

0,0

and C

c,0

,
respectively. Then the equation of the circumcircle of ABC is x
2
 y
2
 cx.SinceA lies
on the circumcircle, we have a
1
2
 a
2
2
 ca
1

. Since the equation of the tangent line of the
circumcircle at A is 2a
1
x  2a
2
y  c

x  a
1

, the coordinates of D are
D

ca
1

/

2a
1
 c

,0

or
D
a
1

a

1
2
 a
2
2

a
1
2
 a
2
2
,0 .
Since E is symmetric to A across the x-axis, its coordinates are E

a
1
,a
2

. Then, the
equation of line BE is a
2
x  a
1
y  0, and the equation of the perpendicular AX is
a
1
x  a
2

y  a
1
2
 a
2
2
. Solving these two equations, we obtain the coordinates of the point X
as follows : X

a
1

a
1
2
 a
2
2

/

a
1
2
 a
2
2

,a
2


a
1
2
 a
2
2

/

a
1
2
 a
2
2

, and the coordinates of the
midpoint of AX are Y

a
1
3
/

a
1
2
 a
2

2

,a
2
3
/

a
1
2
 a
2
2

. Thus, the equation of line BY is
a
2
3
x  a
1
3
y  0, and the coordinates of the point of intersection of line BY and the
circumcircle are
Z
a
1
5

a
1

2
 a
2
2

a
1
6
 a
2
6
,
a
1
2
a
2
3

a
1
2
 a
2
2

a
1
6
 a

2
6
.
The equation of the circle through A, D and Z is
x
2
 y
2

2a
1

a
1
2
 a
2
2

a
1
2
 a
2
2
x 
a
2

a

1
2
 a
2
2

2

a
1
2
 a
2
2

2
y 
a
1
2

a
1
2
 a
2
2

2


a
1
2
 a
2
2

2
 0.
which is obviously tangent to the line BD (the x-axis) at D.
Comment:
Minor changes in the proofs show that the result remains true if B  C. Also, one
may actually claim (see the inversive solution) that: if Z is a point on the minor arc
AC
of , then the circumcircle of ADZ is tangent to BD if and only if BZ passes through the
midpoint of AX, the perpendicular from A to BE. Again, no new ideas are involved in
achieving this. It therefore seems reasonable not to add artificial difficulty, and leave the
problem the way it was proposed. The synthetic proofs are deceivingly short; they do
require time and thinking.
Problem 9
Let a
1
,a
2
,,a
n
be positive real numbers such that a
1
 a
2

   a
n
 1. Prove that
a
1
a
2
a
n

1  a
1
 a
2
   a
n


a
1
 a
2
   a
n
1  a
1
1  a
2
1  a
n



1
n
n1
.
Solution:
Set a
n1
 1  a
1
 a
2
   a
n
. Clearly a
n1
 0; we thus obtain n  1 positive
numbers a
1
,a
2
,,a
n1
adding up to 1. With this notation, the inequality takes the form
n
n1
a
1
a

2
a
n
a
n1
 1  a
1
1  a
2
1  a
n
1  a
n1
.
Using the AM-GM inequality, we have for each i  1,2,,n  1,
1  a
i
 a
1
   a
i1
 a
i1
   a
n1
 n
n
a
1
a

i1
a
i1
a
n1
 n
n
a
1
a
2
a
n
a
n1
a
i
.
Multiplying these n  1 inequalities, we conclude that
1  a
1
1  a
2
1  a
n
1  a
n1

 n
n1

n
a
1
a
2
a
n
a
n1

n1
a
1
a
2
a
n
a
n1
 n
n1
a
1
a
2
a
n
a
n1
,

as desired. If n  2, the equality occurs if and only if a
1
 a
2
   a
n
 a
n1
;thatis,if
and only if the original numbers a
1
,a
2
,,a
n
are all equal to 1/n  1.Inthecasen  1,
however, we have equality for any a
1
 0,1.
Problem 10
Let r
1
,r
2
,,r
n
be real numbers greater than or equal to 1. Prove that
1
r
1

 1

1
r
2
 1
  
1
r
n
 1

n
n
r
1
r
2
r
n
 1
.
Solution:
The case n  1 is clear. We now prove the inequality for all n of the form n  2
k
,
k  1,2, , by induction on k.Fork  1, we have
1
r
1

 1

1
r
2
 1

2
r
1
r
2
 1

 r
1
r
2
 1 r
1
 r
2

2
r
1
 1r
2
 1 r
1

r
2
 1
 0.
To make the inductive step, it suffices to prove that if the claim holds for any n numbers,
then it also holds for any 2n numbers. Indeed, if r
1
,r
2
,,r
2n
 1, we have

i1
2n
1
r
i
 1


i1
n
1
r
i
 1


in1

2n
1
r
i
 1

n
n
r
1
r
2
r
n
 1

n
n
r
n1
r
n2
r
2n
 1

2n
2n
r
1

r
2
r
2n
 1
.
The induction is complete.
To prove the inequality for any n,letk be a positive integer such that m  2
k
 n. Then
put r
n1
 r
n2
   r
m

n
r
1
r
2
r
n
to obtain
1
r
1
 1
  

1
r
n
 1

m  n
n
r
1
r
2
r
n
 1

m
n
r
1
r
2
r
n
 1
.
This completes the proof.
Comment:
Here is another approach to this problem. Let r
i
 e

a
i
. Note that a
i
 0, since r
i
 1.
Then the inequality is equivalent to
1
e
a
1
 1

1
e
a
2
 1
  
1
e
a
n
 1

n
e
1
n


i1
n
a
i
 1
.
Consider the function
fx 
1
e
x
 1
.
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
-4 -2 2 4
x
The graph of fx 
1
e
x
1

We compute
f

x  e
x
 1
2
e
x
,
f

x  e
x
 1
3
e
x
e
x
 1.
Since f

x  0forx  0, fx is convex on 0,. Thus,
1
n

fa
1
  f a

2
    fa
n


 f
a
1
 a
2
   a
n
n
;
that is,
1
e
a
1
 1

1
e
a
2
 1
  
1
e
a

n
 1

n
e
1
n

i1
n
a
i
 1
.
Problem 11
Let x,y and z be positive real numbers such that xyz  1. Prove that
x
3
1  y1  z

y
3
1  z1  x

z
3
1  x1  y

3
4

.
Solution:
The inequality is equivalent to the following one:
x
4
 x
3
 y
4
 y
3
 z
4
 z
3

3
4
x  1y  1z  1.
In fact, a stronger inequality holds true, namely
x
4
 x
3
 y
4
 y
3
 z
4

 z
3

1
4
x  1
3
 y  1
3
 z  1
3
.
(It is indeed stronger, since u
3
 v
3
 w
3
 3uvw for any positive numbers u,v and w.) To
represent the difference between the left- and the right-hand sides, put
ft  t
4
 t
3

1
4
t  1
3
, gt  t  14t

2
 3t  1.
We have ft 
1
4
t  1gt.Also,g is a strictly increasing function on 0,,takingon
positive values for t  0. Since
x
4
 x
3
 y
4
 y
3
 z
4
 z
3

1
4
x  1
3
 y  1
3
 z  1
3

 fx  fy  fz


1
4
x  1gx 
1
4
y  1gy 
1
4
z  1gz,
it suffices to show that the last expression is nonnegative.
Assume that x  y  z; then gx   gy  gz  0. Since xyz  1, we have x  1
and z  1. Hence x  1gx  x  1gy and z  1gy  z  1gz.So,
1
4
x  1gx 
1
4
y  1gy 
1
4
z  1gz

1
4
x  1  y  1  z  1gy

1
4
x  y  z  3gy


1
4
3
3
xyz  3gy  0,
because xyz  1. This completes the proof. Clearly, the equality occurs if and only if
x  y  z  1.
Alternative Solution:
Assume x  y  z so that
1
1  y1  z

1
1  z1  x

1
1  x1  y
.
Then Chebyshev’s inequality gives that
x
3
1  y1  z

y
3
1  z1  x

z
3

1  x1  y

1
3
x
3
 y
3
 z
3

1
1  y1  z

1
1  z1  x

1
1  x1  y

1
3
x
3
 y
3
 z
3

3  x  y  z

1  x1  y1  z
.
Now, setting x  y  z/3  a for convenience, we have by the AM-GM inequality
1
3
x
3
 y
3
 z
3
  a
3
,
x  y  z  3
3
xyz  3,
1  x1  y1  z 
1  x  1  y  1  z
3
3
 1  a
3
.
It follows that
x
3
1  y1  z

y

3
1  z1  x

z
3
1  x1  y
 a
3

3  3
1  a
3
.
So, it suffices to show that
6a
3
1  a
3

3
4
;
or, 8a
3
 1  a
3
. This is true, because a  1. Clearly, the equality occurs if and only if
x  y  z  1. The proof is complete.
Comment:
None of the solutions above actually uses the condition xyz  1. They both work,

provided that x  y  z  3. Moreover, the alternative solution also shows that the
inequality still holds if the exponent 3 is replaced by any number greater than or equal to 3.
Problem 12
Let n  k  0 be integers. The numbers cn,k are defined as follows:
cn,0  cn,n  1n  0;
cn  1,k  2
k
cn,k  cn,k  1n  k  1.
Prove that c n,k  cn,n  k for all n  k  0.
Solution:
The assertion is clear for n  0:
c0,0  c0,0  0  1.
The general case follows by induction on n. Suppose
cm,j  cm,m  j for all n  m  j  0.
Then, by the given recurrence relation and the induction hypothesis, we have
cn  1,k  2
k
cn,k  cn,k  1,
cn  1,n  1  k  2
n1k
cn,n  1  k  cn,n  k
 2
n1k
cn,k  1  cn,k.
To finish the proof we show that:
2
k
 1cn,k  2
n1k
 1cn,k  1. #

Note that the induction hypothesis implies that
2
k
 1cn  1,k  2
nk
 1cn  1,k  1, #
2
k1
 1cn  1,k  1  2
nk1
 1cn  1,k  2. #
We compute
2
k
 1cn,k  2
n1k
 1cn,k  1
 2
k
 1

2
k
cn  1,k  cn  1,k  1

 2
n1k
 1

2

k1
cn  1,k  1  cn  1,k  2

 2
k
2
nk
 1cn  1,k  1  2
k
 1cn  1,k  1
 2
n1k
 1

2
k1
cn  1,k  1  cn  1,k  2

 2
n
 2
k
 2
k
 1  2
n
 2
k1
cn  1,k  1
 2

n1k
 1cn  1,k  2
 2
k1
 1cn  1,k  1  2
n1k
 1cn  1,k  2
 0
where we have used the given recurrence relation in the first step, used (2) in the second
step, and used (3) in the last step. Thus, (1) holds, which completes the proof.
Alternative Solution:
Consider the sequence of polynomials:
P
n
x 

j0
n1
x  2
j
 

k0
n
an,kx
k
, n  1,2,3,,
and notice the two recurrence relations:
P
n1

x  P
n
xx  2
n
,
P
n1
x  2
n
x  1P
n
x/2.