of the triangle BIC Then T lies on the internal bisector of the angle A.

Proof.

Let us draw the external bisectors of the angles B and

C

as shown in the figure below.

B

A

C

They intersect at the excenter

E,

which lies on the internal bisector of the angle

A.

Since

BE

..1

BI

and

GE

..1

GI,

the quadrilateral

BEGI

is cyclic with the center of the circumscribed circle on

IE.

This center will be also the circumcenter of BIC. The lemma is proved.

3.2.

SOLUTIONS

113

Let us prove the main statement. For

i

1,2,3

we denote by

Qi

the center of the

circle Ci and by

Ti

the circumcenter of the triangle

Ai+1IAi+2.

Clearly,

Oi

lies on the internal bisector of the angle

Ai'

By the lemma,

Ti

also lies on the same bisector. Thus the triangles

010203

and

T1T2T3

are perspective from the point

I.

By Desargues' theorem these triangles are perspective from a line. This is to say that if we denote

Qi, i

1,2,3,

to be the point of intersection of the lines

0i+10i+2

and

Ti+1 Ti+2'

then the points

Q1, Q2, Q3

are collinear. But since

Ti+l Ti+2

is the perpendicular bisector of

Ail

and

0i+10i+2

is the perpendicular bisector of

Bili

these points are exactly the circumcenter of the triangles

AlB1I, A2B2I, A3B3I,

respectively.

Remark.

A student not familiar with Desargues' theorem may proceed from the point as follows. Applying Menelaus' theorem to the triangles

10102, 10203, 10301

and to the triples of points

(T1' T2, Q3), (T2' T3, Q1), (T3' TI, Q2),

respectively, one can, observing usual agreement about the signs, write:

01 T1 IT2 02Q3

1 IT1 . 02T2 . 01Q3 -

IT3 02T2 Q3Q1 - 1

03T3 . IT2 . 02Q1 - ,

ITI 03T3 01 Q2 - 1

OlTl . IT3 . 03Q2 -

. Multiplying them all one gets

02Q3 . 03Q1 . 01 Q2

1,

01 Q3 02Q1 03Q2

which means that the points

Q1, Q2, Q3

are collinear.

Alternative solution.

This proof will be based on inversion. We take the incenter

I

to be the center of the inversion and the power of the inversion is arbitrary. Using primes to denote images of points under the inversion we have the following " dual" figure shown below.

114

B' I

3. G EOMETRY

B' 2

Indeed, the image of the circle Ci is a straight line Bi+1 Bi+2 ' with these lines forming the triangle Bi B�B� . The line AiAi+1 will be transformed into the circle

ri+2 , with the side AiAi+1 becoming the arc AiAi+1 which does not contain I. Note

that all these circles have equal radii since the distances from I to the sides of AI A2A3 were equal.

Let us note that if �I ' �2' �3 are three circles passing through the common point I and no two of them touch, then their centres are collinear if and only if there is another common point

J

f:. I through which all these three circles pass.

We will use this observation for �i being the circumcircle of AiBiI. Since the inversion takes :Ei to the line AiB�, the desired result is to show that the lines A� Bi , A� B� , A�B� are concurrent. For this, it suffices to show that the triangles A�A�A� and Bi B�B� are homothetic, which is the same to say that their corresponding sides are parallel. Since the radii of the circles

r

I ,

r

3 are equal, the triangle PI P2 P3 formed by their centre has its sides parallel to the corresponding sides of the triangle Bi B�B�. The homothety of ratio

1/2

centred at I takes the triangle A� A�A� into the triangle whose vertices are the midpoints of the triangle PIP2P3 . Therefore the corresponding sides of the triangles A� A�A� and PI P2P3 are also parallel and the result follows.

(Titu Andreescu,

IMO 1997 Shortlist)

22.

Let I be the intersection point of lines CB' and C' B and let A' be the intersection point of lines AI and BC.

We have

AB' _{+ C'C = IA"}AC' AI

from Van Aubel's theorem, therefore

::,

is constant. Hence the locus of point I is a line segment parallel to BC.

(Titu Andreescu)

3.2. SOLUTIONS

23.

Without loss of generality assume that PC = max{PA, PB, PC}. The condition in the hypothesis is

PB · PC + PA · PC = PA · PB + 1 PC

₌

PA · PB + l · 1 _{P A . 1}

₊

_{PC . 1 .}

115

From the converse of the second theorem of Ptolemy it follows that P AC B is a cyclic quadrilateral. Note that P cannot be A, B or C otherwise the denominator of the right-hand side equals

O.

Hence the locus of point P is the circumcircle of triangle ABC without the vertices A, B, C.

(Titu Andreescu,

Revista Matematidl Timi§oara (RMT), No. 1(1985), Problem C7:3)

24.

We have

a

b

2 - 4S2 +

a

c

2 - 4S2 =

a

b2 - a2b2 sin2 C+

a

c

2 - a2c2 sin2 B = ab cos C + ac cos B =

a2 + b2 - c2 a2 + c2

-b

= ab 2ab +ac 2ac

a

2 , as desired.

(Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No. 8(1971), pp. 25, Problem 1006)

25.

The relation is equivalent to

Fa + y'rb + Fc _{= - ,}1

On the other hand,

r

1 1 1 1

--- + ---+ ---

= - .

.;r;. fo vr;.;r;. .;r;.

y'rb

r

1 1 1 1 - + - + - = - ,

ra rb re r

1 1 1 1 1 1

---

---+ ---

= -

+ - + - .

Fay'rb y'rbFc FcFa

ra rb re

Then

(

1 1

)

(

1 1

)

(

1 1

)2

vr;. - .;rb

fo -Fc

vr; - .;r;.

= 0,

ra = rb = re'

It follows that the triangle is equilateral, as desired.

(Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No. 1(1974) , pp. 21, Problem 1903)

116

3.

GEOMETRY

26.

If the triangle is equilateral the conclusion is true.

To prove the converse, we assume by way of contradiction that the triangle is not equilateral and say that

b f:. c.

Then

2 b2 + c2 a2

(b + C)2 - a2

ma -

-

>

=p(p - a)

and likewise m

�

p(p - b),

m� 2:

p(p - c).

It follows that 1 1 1 1

(

1 1 1

)

- + - + - < -

--+--+ -- =

m� m

�

p p - a

p - b

p - c

On the other hand

-p2 _{+ ab + bc + ca}

82

1 1 1 1

+ a2 + b2 + c2

- + - + - = -[(p -a)2 + (p _ b)2 + (p _ C)2] =

r� r� r�

82

Since

b f:. c

then

ab + bc + ca

a2 + b2 + c2

hence

1 1 1 1 1 1

-2 ₊-2 ₊-2

< 2

+

ma mb me

r

a rb re which is a contradiction.

(Titu Andreescu,

Revista Matematidi Timi§oara

(

RMT

)

, No. 2(1977), pp. 66, Problem 3063)

27.

We know that

On the other hand,

. A . B . C 1 sm - sm - sm - < -₂ ₂ _{2 - 8 '}

A B C _{' A ' B ' C}

cos "2 cos "2 cos "2 2: sm sm sm .

' A

' B ' C 4 A B C

+

=

cos "2 cos "2 cos "2'

so inequality (1) gives

sin A

+

sin B

+

sin C 2: 4 sin A sin B sin C.

Since the circumradius is 1, we have

and relation

(2)

yields

as claimed.

a = 2sinA, b =

2 sin B,

c =

2 sin C,

a + b + c

abc,

(1)

(2)

3.2.

SOLUTIONS

117 Alternative solution.

Let

Zl, Z2, Z3

be the afixes of

p

oints A, B, C such that

Izd =

IZ21 ₌IZ31 ₌

1. We have

a

=:= BC

= IZ2 - z31,

b =

= IZ3 - zd,

c =

= IZI - z21·

U sing the identity

Z;(Z2 - Z3) ₊Z�(Z3 - zd ₊Z�(ZI - Z2) ₌(Zl - Z2)(Z2 - Z3)(Z3 - zd

and triangle inequality it follows

abc = IZI - z211z2 - z311z3 - zd ::; IZll21z2 - z31 ₊IZ21211z3 - zll ₊IZ3121z1 - z21 ₌

= IZ2 - z31 ₊IZ3 - zd ₊IZI - z21 _{= a + b + c.}

(Titu Andreescu,

Revista Matematica Timi§oara

(

RMT

)

, No. 2(1978) , pp. 49, Problem 3513; Gazeta Matematica

(

GM-B

)

, No. 11(1981), Problem 0258; No. 2(1988), pp. 78, Problem 21353)

28.

For any positive real numbers

x, y, Z

we have

Setting gives so It follows that then Hence as desired. 8

xyz ::; (x + y)(y + z)(z + x)

x = -a + b + c,

y _{= a - b + c,}

Z _{= a + b - c,}

(

-a + b + c) (a - b + c) (a + b - c) ::; abc,

pabc

2 ' 32

(Titu Andreescu,

Revista Matematica Timi§oara

(

RMT

)

, No. 2(1974), pp. 51, Problem 2028)

29.

Using the inequality

118

we obtain Hence as desired.

3.

GEOMETRY

a2

b2

C2 p2

4S2 � 3S2 ' 1 1 1 1

h�

� 3r2 '

(Titu Andreescu,

Revista Matematica Timi§oara

(

RMT

)

, No. 2(1977), pp. 66, Problem 3062)

30.

From the inequality

we obtain Then hence as claimed. 1 1 1 1 1 1 - + - + - >--+--+-- ra rb rc - 1 1 1 1 - > --+ -- +--.

r -

+ + � 9r,

(Titu A ndreescu,

Revista Matematica Timi§oara

(

RMT

)

, No. 2(1978), pp. 64, Pro blem 3277)

31.

We have + ,?

a2 (b

C)2 - a2

ma = 4 and likewise It follows that

mambmc �

- a)(p - b)(P - c) =

pS = rarbrc,

as desired.

(-Titu Andreescu,

Revista Matematica Timi§oara

(

RMT

)

, No. 1(1978), pp. 64, Problem 3276)

32.

By the AM-GM inequality, so 8

p3

� 27

abc.

Hence

(a

b

c

)

3 � 27abc,

2 abc

2p

> 27- ·

-

₄_S -

p

= 27Rr

,

3.2.

SOLUTIONS

119

as desired.

(Titu Andreescu,

Revista Matematica Timi§oara

(

RMT

)

, No. 1(1973), pp. 43, Problem 1585)

33.

We have It follows that hence as desired. 1 < -

bc

- 2 '

bc

(p - b)(P - c)

:s; 4'

(Titu Andreescu,

Revista Matematica Timi§oara

(

RMT

)

, No. 1(1984), pp. 67, Problem 5221)

34.

Let

01, O2, 03

be the centers of the three circles and

S

the area of the common region .

The three sectors with centers

01, O2, 03

which subtend the arcs

0203, 0103,0201,

respectively, cover the surface of area

S

and twice more

the surface of triangle

010203 (

which is r2 ). On the other hand, the area of these three circular sectors equals the area of a semicircle, which is

�7fT2.

Hence

therefore 1

2 r2J3

-7fT = S + 2 . -- 2 4 ' 1 S

=

2r2 (1/'

- V3).

(Dorin Andrica,

Revista Matematica Timi§oara

(

RMT

)

, No. 2(1978) , pp. 50, Problem 3522)

120 3. G EOMETRY

35.

The parallel BD through C meets AB at point E. By Stewart's formula, we obtain

AC2 · BE + CE2 . AB - CB2 · AE = AB · BE · AE Because CE = BD and BE = CD, we deduce

AC2 . CD + BD2 . AB - BC2 . (AB + CD) = AB . CD . (AB + CD)

D C

B E

Drawing the parallel to AC through D and using similar computations yields BD2 . CD + AC2 . AB - AD2 . (AB + CD) = AB · CD · (AB + CD) Subtracting the relation (2) from (1) gives

(AC2 - BD2) (AB - CD) = (AD2 - BC2 ) (AB + CD),

as desired.

(1)

(2)

(Dorin Andrica,

Gazeta Matematidi (GM-B), No. 9(1977), Problem 6852; Revista Matematidi Timi§oara (RMT), No. 1-2(1980), pp. 64, Problem 4119)

36.

Let I be the intersection point of the diagonals AC and BD. Since IA = AB · AC and IB = AB · BD

AB + CD AB + CD

the condition in the statement becomes Hence

AiB

= 90°, as desired.

(Dorin Andrica,

Revista Matematidi Timi§oara (RMT) , No. 2(1978), pp. 59, Problem 3524)

37.

Let ABCD be the trapezoid. Point I is the intersection of diagonals and M, N are the midpoints of AB and DC. In a right triangle the length of the median corresponding to the hypothenuse is half of lenght of the the hypothenuse. Hence

IM = AB and IN = CD . ₂ 2 3.2. SOLUTIONS 121 D F N C I A M E B Then AB + CD IM + IN = 2 = EF,

which is the length of the midline and hence the length of the altitude. It follows that IM and IN are also altitudes in triangles lAB and ICD therefore lAB and ICD are isosceles. Thus ABCD is isosceles, as claimed.

(Titu Andreescu,

Revista Matematidi Timi§oara (RMT), No. 1(1978), pp. 48, Pro blem 2817)

38.

From the Law of Cosines we deduce that

Note that

2AB . AC cos

IiAC

= AB2 - BC2 + AC2 2DC · AC cos

DcA

= CD2 - AD2 + AC2 2AB . DB cos

ISiiA

= AB2 - AD2 + DB2 2DC · DB cos

CiJij

= CD2 - BC2 + DB2

IiAC

DcA, ISiiA

CiJij,

so dividing relations (1) and (2), (3) and (4) yields

AB2 - BC2 + AC2 AB2 - AD2 + DB2 AB CD2 - AD2 + AC2 = CD2 - BC2 + DB2 = DC ' as desired.

(Dorin Andrica)

(1) (2) (3) (4)

39.

Let

a, b

the lengths of the bases,

c,

d be the lengths of the nonparallel sides and d1 , d2 be the lengths of the diagonals. From Euler's theorem for quadrilaterals, it follows that

Hence

122 3. GEOMETRY

and

d1 - d2

= C -

d

implies

d1d2

ab

cd.

From Ptolemy's Theorem, we deduce that the trapezoid is cyclic and so isosceles, as claimed.

(Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No. 1(1978), pp. 48, Problem 2817)

40. First solution.

Assume the opposite. Then lAC - BDI > lAB - CDI or lAC - BDI > lAD - BCI. Without loss of generality, lAC - BDI > lAB - CDI , otherwise switch B and D. We have

AC2 - 2AC . BD + BD2 > AB2 - 2AB . CD + CD2

and, from Euler's relation,

AB2 + BC2 + CD2 + AD2 = AC2 + BD2 + 4MN2, where M and N are the midpoints of AC and BD, respectively.

From (1) and (2) ,

AD2 + BC2 - 2AC · BD > 4MN2 - 2AB · CD.

(1)

(2)

(3)

Let P be the midpoint of AB. Then NP = AD/2, MP = BC/2 and since MN � INP - MPI , it follows that

4MN2 � (AD - BC)2. (4)

From

(3)

and (4) , -2AC · BD > -2AB · CD - 2AD · BC, in contradiction with Ptolemy's Theorem. We are done.

Note.

The cyclicity is essential. The inequality fails if ABCD is a parallelogram.

Second solution.

Let E be the intersection of AC and BD. Then the triangles ABE and DCE are similar, so if we let x = AE,

y

= BE,

z

= AB, then there exists

k such that kx = l)E, ky = CE, kz = CD. Now lAB - CDI = Ik -

l iz

and

lAC - BDI = I (kx + y) - (ky + z) 1 = Ik - 11 · Ix -

yl·

Since Ix -

yl

::; z by the triangle inequality, we conclude that lAB - CDI �

lAC � BD I, and similarly IBC - DAI � lAC - BDl. These two inequalities imply the

desired result.

Third solution.

Let 20', 2,8, 2,, 28 be the measures of the arcs subtended by AB, BC, CD, DA, respectively, and take the radius of the circumcircle of ABCD to be 1. Assume without loss of generality that ,8

�

8. Then a + ,8 + , + 8 = 7r, and

(by the Extended Law of Sines)

lAB - CDI = 21 sin a - sin ,1 =

I

Sin a , cos a

I

3.2. SOLUTIONS 123

and

lAC - BDI = 21 sin(a

+,6)

- sinCa + ,) 1 =

I

Sin a , cos

(

a ,

+,6) I.

Since 0

� (a

_{+ ,)/2}_::;_{(a + ,)/2 + ,8 ::; 7r/2 (by the assumption ,8}

�

_{8) and the}

cosine function is nonnegative and decreasing on [0, 7r /2], we conclude that l AB _

CD I � lAC - BDI , and similarly lAD - BCI � lAC - BDl .

(Titu Andreescu,'

USA Mathematical Olympiad, 1999)

41.

Let E be the intersection point of the diagonals. Consider AD < BC the basis of the trapezoid and AB the altitude. Since

then

AB2 = AD . BC,

AB BC

AD AB '

so the right triangles ABC and ABD are similar.

On the other hand we have

hence

It follows that

ill

= 90° so the diagonals are perpendicular, as claimed.

(Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No. 2(1972), pp. 28, Problem 1 164)

42.

Let I be the intersection point of the diagonals AF and BE of the rectangle ABF E. Notice that N I is the median of the right triangle AN F with hypothenuse AF, so Likewise, BE IM = - = IE ₂ _' AF IN = - = IA ₂ AF IQ = - = IF ₂ _' IP = - = IB BE 2 .

Since IA = IE = IF = IB, it follows that 1M = IN = IP = IQ. Hence MNPQ

124

3.

GEOMETRY

A

B

P

N

Q

D E

F

C

(Titu Andreescu)

43.

By Sturm's theorem, we know that if 0

< a1 :::; a2 :::; a3 :::; a4 < a1 + a2 + a3,

then there is a cyclic quadrilateral having side lengths

aI, a2, a3, a4·

Denote by

a, b, c, d, e, j, m

the lengths of the segments

AB, BC, CD, DA, AC,

BD, M N,

respectively. Without loss of generality assume that

b + d 2: a + c.

Let

P

be the midpoint of the side

BC.

The segments

M P

and

N P

are midlines in triangles

CAB

and

BDC,

so Then 1 1

MP

-a 2

and

NP

-c. 2

2m

2MN < 2MP + 2NP

a+c

2m < a + c < b + d.

On the other hand, if 0 is the intersection point of the diagonals, we have

b+ d

)

3.2.

SOLUTIONS

125

44.

Let

I

be the intersection point of lines

BD

and

AF.

The parallel to

BD

through

C

meets line

AF

at point

T.

First we consider

AF .l BD

and prove that

AB .l CD.

_F

F

A

)

Assume that

D

lies on the segment

CEo

Then

ArC

= 90° . Since

IfAjJ : BnC,

we obtain

Assume that

C

lies on the segment

DE.

In the right triangle

CT F, T E

is the median, so

Ere

EaT

. (3)

Because

CTIIBD,

we have

(

)

Also,

(5)

so from

(

)

, (4), (5), we obtain

Ere : Mo.

Hence

ATEC

is cyclic, then

AEC

ArC

= 90° , and

AB

CD,

as desired. Conversely, consider that

AB .l CD.

)

D

is on the segment

CE,

then

ME

AcE

. On the other hand

AcE

Aci5

ABi5

, so

ME

ABi5

, and

FBI E

is cyclic. It follows that

JiiF

liEF

90°, hence

B D .l AF,

as desired.

)

C

is on the segment

DE,

then

126

3. GEOMETRY

11oreover,

J.U515 = ACB15,

AcE = ABl.

(7)

From

(6)

and

(7)

we obtain

ME = ABl,

hence

FEB!

is cyclic. Note that

liEF = 900,

iiiF = 900

and

BD

..L AF, as desired.

(Titu Andreescu,

Revista 11atematica Timi§oara (R11T) , No.

1(1986),

pp. 106, Problem

C6:4)

45.

Let

a, b, c, d

be the side lengths of the quadrilateral and let

S

be its area. Because the quadrilateral is cyclic, we have

S2 = (p - a)(p - b)(P - c)(P - d).

1 _S

S

N �mbers

a, b, c, d

are odd, hence

is an integer.

a + b+c+d

(1)

p

is odd, then

p - a, p - b, p - c, p - d

are even and so

82

is divisible by

16,

which is false. Hence p is even.

3.2. SOLUTIONS

127 S

1 S

Without loss of generality assume that

a � b � c � d.

Since

S

is a prime number,

from relation

(1)

we obtain

p- d=p - c = l md p - a = p- b = �

Summing up these equalities yields

4p-2p = 2 + 2S

p = S

1.

Hence

a = b = 1

and

c= d = S.

The required quadrilaterals are either rectangles or kites.

(Titu Andreescu,

Revista 11 at em atidi Timi§oara (R11T) , No.

2(1977),

pp.

66,

Problem

3067)

46.

From the A11-G11 inequality it follows that

b d

(a +b+ c+d)4 1 4

a c < _-

₄

= -p . ₄

Hence

16abcd � p4,

8(ac + bd)2 -p4 � 8(a2c2 + b2d2).

The desired inequality is now obtained from Ptolemy's Theorem:

ac+ bd = ef.

(Titu Andreescu,

Revista 11atematidi Timi§oara (R11T) , No.

3(1973),

pp.

36,

Problem

1811;

Gazeta 11atematidi (G11-B), No.

8(1980),

pp.

364,

Problem

18370)

47.

Let m be the length of the segment determined by the midpoints of the

diagonals.

From Euler's Theorem for quadrilaterals we have

1

28 3. G EOMETRY

and from Ptolemy's Theorem,

ae + bd = ef.

From relations

(1)

and

(2)

we obtain

(a -

e)2 + (b

- d)2

= (e

- 1)2

+

4m2.

Since max{la - el,

Ib

- dl} �

1,

then

2 = 1 + 1

� (a -

e)2 + (b

- d)2

= (e - 1)2 +

m2

�

(e

1)2.

Hence

Ie

11

�

V2,

as desired.

(2)

(Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No.

2(1978),

pp.

51,

Problem 3527)

48.

The quadrilateral is cyclic so

S =

- a) (p -

b)(P

- e) (p - d)

Since

S = (�)

2, we have

- a) (p -

b)(P

e

)(p -

d) =

₂=

(p

- a)

+

(p -

b) + (p - e

)

+ (p

- d) 4

Note that this is the equality case in the AM-GM inequality, hence p - a

= p-b =

p

e =

p -

d.

It follows that a

= b = e =

d, so the quadrilateral is a square.

(Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No.

1(1977),

pp.

24,

Problem

2136)

4�.

Observe that

SPAB . SPCD = SPBC . SPDA,

since both are equal to

�PA .

P B· PC· PD·

sin P. The numbers

SPAB, SPCD

and

SPBC, SPDA

have the same sum and the same product, thus

SPAB = SPBC

and

SPCD = SPDA

SPAB = SPDA

and

SPBC = SPCD,

i.e.

P

is the midpoint of

AC

BD,

as desired.

(Titu Andreescu,

Korean Mathematics Competitions,

2001)

50.

(i) Let

MI

and

Mil

be the projections of point

M

onto diagonals

AC

and

BD.

3.2. SOLUTIONS

129 D

AI

We recall the Simpson's theorem: the projections of a point of the circumcircle of a triangle onto the sides of the triangle are collinear. Applying this result to triangles

is on the lines

BI CI

and

DIAl.

Since

Mil

is a point of

AC,

the conclusion follows.

(Dorin Andrica,

Revista Matematidi Timi§oara (RMT), No.

1(1979),

pp. 54, Problem

3855;

Romanian Regional Mathematical Contest " Grigore MoisH" ,

1995)

51.

We have

₁

it follows the diagonals

AC

and

BD

are perpendicular and intersect at

P.

Thus,

AB = + 322 =

40,

BC = + 322 =

y'fi3

CD =

452 = 53,

and

In document 360 Problems for Mathematical Contests (Page 59-67)