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

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

, 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


 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


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,

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

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

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


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

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


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

 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

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

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

 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

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

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

, #
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

.
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


.
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

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



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

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

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

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

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

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

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

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
,

 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

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

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
  

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

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

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

 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


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  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

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,

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

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