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Some problems from geometry olympiads

Collected by pupils of 10A1 Math - HSGS School

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Preface

After a month of studying and researching under the advices of our English and mathematicteachers. We have collected some nice geometric problems from around the world. All solutions arewritten by members of our class 10A1 Math. We would like to introduce them readers. Thank foryour attention.

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Problem 1 (Russia Sharygin Geometry Olympiad 2008 problem 3 grade 10). SupposeX and Y arethe common points of two circles 1 and 2. The third circle is internally tangent to 1 and 2 inP and Q respectively. Segment XY intersects in points M and N. Rays P M and P N intersect1 in points A and D; rays QM and QN intersect 2 in points B and C respectively. Prove thatAB = CD.

Proof by Gii.

Lemma 1. Let circle (O1) is tangent(O) in P. Let G, H be on the circle (O). Ray P G and P Hintersect(O1) in points E and F. Prove EF GH.

O1 O2PO

1 O2P

A1

B1

B2

A2

A1

B1

A2

B2

Proof of lemma. We have P, O1, O is collinear and triangle P O1E and P OG are isoceles. Deduce

triangle P O1E and P OG are similar P O1

P O=

P E

P G(1).

Similarity, triangle P O1F and P OH are similar. HenceP O1

P O=

P F

P H(2).

From (1) and (2), we deduceP E

P G =P F

P H EF GH.

1

2

O1

O2

OP

Q

X

Y

M

N

AB

D

C

Proof of problem. Let O1, O2 be center of1, 2,respectively. According to lemma, we have AD M N and BD M N. Deduce AD BC or AD XY and BC XY. But XY O1O2 (sinceX and Y are the common points of two circles1 and 2). Deduce AD O1O2 and BC O1O2. But A, D lie on the circle ); B, C lie onthe circle 2. Deduce A, B; C, D are two pairsreflection points with respect to the lines O1O2,

respectively. Therefor AB = CD. We are done.

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Problem 2 (All-Russian Olympiad 2009 problem 2 grade 9). Let be given a triangle ABC and itsinternal angle bisector BD (D AC). The line BD intersects the circumcircle of triangle ABCat B and E. Circle with diameter DE cuts again at F. Prove that BF is the symmedian line oftriangle ABC.

A

B C

O

D

E

M

N

F

Proof by ng Phc. Denote by O the center of circle

OE AC = {N}; OE = {E, M}.

Because BD is the angle bisector os triangle ABC. Hence OE AC or M N D = 90 (1).On the other hand, M E is a diameter of (O), so

M BE = M F E = 90 (2).

And we have that DF E = 90 (3).(1) and (2) hence BDNM is cyclic

DBN = DM N ().

(2) and (3) we have that M , D , F is collinear. Hence

F M E = DM N

F M E =

F BE

DM N = F BE ()

(Because BFEM is cyclic). From (*) and (**) we have

F BE = DBN

Or BF is the symmedian line of triangle ABC (Q.E.D).

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Problem 3 (Romania Junior Balkan Team Selection Tests 2008 problem 4 day 1) . Let ABC bea triangle, and D the midpoint of the side BC. On the sides AB and AC we consider the pointsM and N, respectively, both different from the midpoints of the sides, such that AM2 + AN2 =

BM2 + CN2 and M DN = BAC. Prove that BAC = 90.

A

B CD

E F

N

M

N'

Proof by Tun Anh. Let E, F be the midpoints of AB, AC, respectively. Without loss of generality,

we may assume that M, N are points on the segments BE,AF, respectively (since M DN = BAC =F DE).We easily have

AB = 2FD,AC= 2ED , BE D = BAC = DFC.

From the condition of the problem, we have

(AE+ EM)2 + (AF F N)2 = (BE EM)2 + (CF + F N)2.

It is equivalent toAB.EM = FN.AC ( since AE = EB,AF = F C).

It follows thatF N

EM=

AB

AC=

2F D

2ED=

F D

ED.

Let N be point on the F C such that F is the midpoint of N N. From this,

F N

F D=

F N

F D=

EM

ED.

Since

F N

F D =

EM

ED andM ED =

BAC =

N

F D, we deduce that

M ED NF D.

It implies that F DN = M DE = F DN (since M DN = EDF). We deduce that N DN isoscelesat D.From this, we have N F D = NF D = 90. Therefore, we have BAC = 90.

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Problem 4 (Romania Junior Balkan Team Selection Tests 2008 problem 3 day 1). LetABC be anacute-angled triangle. We consider the equilateral triangle AU V, where A (BC), U (AC) andV (AB) such that U V BC. We define the points B, C in the same way. Prove that AA, BB

and CC are concurrent.

A

B C

V U

H KA'

Proof by Anh Th. Drop V H BC,UK BC (H BC,K BC). We have

U V BC V H = U K.

We have BH = V H. cot B

AH = V H. cot V AB = V H. cot60

BA = BH + AH = V H(cot B + cot 60) (1).

In the other way

CK = U K. cot CAK = U K. cot 60

CA = CK + AK = U K(cot C + cot 60) (2).

From (1) and (2), we have

BA

CA=

V H(cot B + cot 60)

U K(cot C + cot 60)=

cot B + cot 60

cot C + cot 60.

Simiary, we have

CB

AB=

cot C + cot 60

cot A + cot 60

AC

BC= cot A + cot 60

cot B + cot 60Hence

BA

CA

CB

AB

AC

BC=

(cot B + cot 60)(cot C + cot 60)(cot A + cot 60)

(cot C + cot 60)(cot A + cot 60)(cot B + cot 60)= 1.

Then from Cevas theorem, we deduce AA, BB , CC are concurrent (Q.E.D)

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Problem 5 (Romania Junior Balkan Team Selection Tests 2008 problem 1 day 2). Consider theacute-angled triangle ABC, altitude AD and point E - intersection of BC with diameter from A ofcircumcircle. Let M, N be symmetric points of D with respect to the lines AC and AB respectively.

Prove that EM C = BN E.

A

BC

O

D

M

N

E

K

H

Proof by Huy Hong. Since M, N be symmetric points of D with respect of D to the lines AC andAB, respectively so that

AM = AN = AD and ANE = AME.

Denote {K} = N E AB and {H} = M E AC.

So that ANE = ADKAME = ADHi.e, 90 ADK = 90 ADH; KB D = HDC.So by the typical of symmetric of point D, we conclude that

BN E = KDB = HDC = EMC.

i.e, BN E = EM C.

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Problem 6 (China Team Selection Test 2005 Quiz 6 problem 2). In triangle ABC, BC = a,CA =b,AB = c, and a > b > c. I , O , H are the incentre, circumcentre and orthocentre of ABC respec-tively. Point D BC, E CA and AE = BD, CD + CE = AB. Let the intersection of BE andAD beK. Prove that KH IO and KH = 2IO.

a

b

c

A

BC

IK

O

G

H

E

DM

F

Proof by Khnh. Let F and M be the tangent point of the incircle of triangle ABC and AC. Hencewe have

CF = p c (1)

with p is equal to one-second the qrea of triangle ABC. But

AE+ AB = BD + DC + CE

AE+ AB = BC + CE

and AE+ AB + BC + CE = AB + BC + CA = 2p. So

AE + AB = p AE = p c (2)

From (1) and (2), we obtain that E and F are symmetric points together to the midpoint of sideAC. Similarity, D and M are symmetric points together to the midpoint of side BC. Hence, K whichis the intersection of AD and BE, is the "Nagel point" of triangle ABC. Let G be the centroid oftriangle ABC, we have

G, H , O are collinear and GH = 2GO (Euler line of triangle ABC)

G, K , I are collinear and GK = 2GI (Nagel line of triangle ABC)

So, according to Thales theorem, we have OI runs parallel with HK and 2.OI = KH (QED).

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Problem 7 (All-Russian Olympiad 2009 problem 7 grade 9). The incircle (I) of a given scalenetriangle ABC touches its sides BC, CA, AB at A1, B1, C1, respectively. Denote by (O1), (O2) theincircles of quadrilateralsBA1IC1 andCA1IB1, respectively. Prove that the internal common tangentof (O1) and (O2) different from IA1 passes through A.

A

BC

I

B1

C1

O1

O2E

GF

H

N

M

Proof by Lng An. We easily seen O1E = O1F.Similarly, O2H = O2G. Let M, N be projection of O2, O1 on AI.

It is easy to proof that F O1N M O2H becauseF O1N = HO2M

N F O1 = HM O2 = 90

C

SoO1F

O1N=

O2M

O2H

O1E

O1N=

O2M

O2G.

But EO1N = GO2M so EAO1 = IAO2 BAO1 + O2AC =1

2BAC.

Let Ax be the tangent of (O1),

O1AB = O1Ax

but BAO1 + O2AC =1

2BAC.

Deduce, Ax is the tangent of(O2). We are done.

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