501-849.pdf

Geometri Günlüğü

Geometri Günlüğü (Geometry Diary) B

A'

F

O

E R P Q

T

I

N

Problems 501-850

G

I K

C

A

D G

D

M

A

1

2

Geometri Günlüğü

Problem 501.

Problem 504.

F

E

G

E

AEFGBCDG, düzgün yedigen ABC yedigensel (heptagonal) üçgen BC=a, AC=b, AB=c Bu kenarlara ait yükseklikler ha, hb,hc

D

F

AHCDEFB regular heptagon BH∩AC={N} G,centroid of ABC Proove that:

c

B

A

hc

a

ha=hb+hc

C

B

b

G

hb

ha

H

A

F

G

Problem 505.

E AEFGBCDG, regular heptagon ABC, heptagonal triangle BC=a, AC=b, AB=c

A E c

10° a

b

20°

Proove that

c

a

A

x 30°

30°

B

archiµedes26

N

D

C

Problem 502.

F,G and N collinear

B 2

C

D

AD=a, BE=b, AB=c, EC=x

b + c + a =5 a 2 b 2 c2 2

2

Proove that:

x=a+b+c

b

archiµedes26

D

C

Problem 503.

D Problem 506. A E

30°+α x

30°

A

x

C a

b

30°-α

c

20°

B DB=DC

Proove that x=a

80°

B

60°

C

D

x=a+b+c

archiµedes26

3

Geometri Günlüğü

Problem 507.

Problem 510.

A

A

60°+a AB=AC=DC

24°

a

15°

x=30° m(∠DBC)=x=3°

D

x

C

B

D

51°

81°

x

C

B

Problem 511. A

a E

Problem 508. A

B

c b

D

x=a+b+c

D

a 20°

b c

30°

70°

d

x

40° E

x

C Proove that:

20°

archiµedes26

CD=x=a+b+c+d

archiµedes26

10°

50°

40°

B

C

Problem 512.

Problem 509.

ABC üçgeninin iç çemberinin yarıçapı r, Çevrel çember yarıçapı R Dış teğet çember yarıçapı ra m(∠CAB)=60°

A C

B

9°

75°

φ 2

D Prove that:

φ=24°

5 -1

ra

r

C A

ra=R+r

60°

R

B

4

Geometri Günlüğü

Problem 513.

Problem 516.

C

A 80°

b

a

b

D

150°

a

F

C

70°

a+b+c

a+b=2d

80°

x

c

E

D

30°

d

x=5°

B

archiµedes26

40° 60°

archimedes26

B

A

Problem 517.

Problem 514.

A

a+b

b

c

+c

162°

48° 60°

C

x

a

D

B

x=6° archimedes26

archimedes26

Problem 518. Problem 515.

A A

b 150°

60°

C

B x

Proove that:

B

x=10°

D archiµedes26

156°

48°

a+b

a

30°

b

c

80°

C a

a+b+c

x D

Proove that:

x=6°

archimedes26

5

Geometri Günlüğü

Problem 519.

Problem 522. A

A

a+c

b 150°

70° 60°

C

a

a+b

B

130°

c

100°

x

60°

C

x

a

D

D

B

x=30°

archiµedes26

x=10°

archimedes26

Problem 523.

Problem 520.

A c

A

B

48°

b

c

160°

70°

c

C

60°

x

a+b+c

a

a

+ +b

a

D

B

x=5° archimedes26

b 12°

84°

x

C

D archiµedes26

x=24°

Problem 524. A

Problem 521. A

b 48° C

b

a+

a

100°

c

100° 60° B

c

b+c

a

C

D

a+b+c

36°

x

B

x D

x=20° archimedes26

24°

x=12°

archiµedes26

6

Geometri Günlüğü

Problem 525.

Problem 528.

A

b

A c B

4α

C

c

b+c

a+

a

2α

a D

36°

b 24°

3α

x

B Proove that: archiµedes26

84°

x

x=α

D

C

a+b+c archiµedes26

x=18°

Problem 526. B

a

C

80° c x

D

A

Problem 529.

b

40°

60°

x=10°

x

A

a+2b+3c

archimedes26 (inspiration of Angel LAZO)

D 12°

63° 39°

18°

C

B Problem 527.

A

x=15°

c B

48°

a

b 12°

Problem 530. A x

60°

x D

C

a+b+c archiµedes26

x=24°

12°

6°

D

B x=42°

6°

18°

C

Geometri Günlüğü

Problem 531.

Problem 534.

A

D

x 15° 9°

D

18°

7

9°

a

C

C

B

c

b

x=54°

10° 20° 30°

30°

A archiµedes26

c=a+b

B

Problem 535.

Problem 532. A

B

33°

x

15°

D

12°

6°

A

18°

C

C

B

ABCDE düzgün beşgen

x=24° 9°

x=27° x

archiµedes26

E

D

F

Problem 533. Problem 536.

C

A 40°

1 D

100°

x

20°

67,5° C

E

B

A ve E odaklı hiperbol C noktasından geçiyor. Hiperbolle AB kenarının kesişimi D.

x=20°

A

x 6

B

D

AC⊥CB, CD=1, AB= 6 archiµedes26

archimedes26

x=60°

8

Geometri Günlüğü

Problem 537.

Problem 540. A

B

C B ve D çemberlerin kesişim Noktaları H, ABC üçgeninin diklik merkezi M,AC nin orta Noktası

D

OA=1 C, AB yayının orta Noktası N, BD nin orta Noktası M, OC nin orta noktası

B M

1

N O

M,H ve D doğrusaldır

H

MN = 1 2

C M

Düzgün 7-gen

A

Problem 541.

A

B

Problem 538. A

AO⊥OC BE⊥OC D, AO nun orta noktası

D B b

O

a

E

C m(∠ODE)=m(∠EDC)

a+b=c

O

C

c

O merkezli düzgün 15-gen

Pentagon Construction (Coxeter 1969, Wells 1991).

O merkezli düzgün 5-gen

archiµedes26

Problem 542.

D

A

Problem 539. A N a E

1 F

B

AD,BE,CF açıortay D,E ve F noktalarından geçen çember ABC üçgenini M,N,P noktalarında kesiyor.

a+b=c

b P

E

EB=EC= 2

C 1

Düzgün 7-gen

D

B

c M

D

C

Geometri Günlüğü

Problem 543.

9

Problem 546.

C

C O, çemberin merkezi K, teğet noktası

a

b

4α

3α

B

c

D

GO=OE

E

O

G D

2α

A

A B

Proove that:

K archiµedes26

1 =1 +1 c a b

Problem 547.

Problem 544. B

D

ABC heptagonal üçgen

ABCDE, bir kenar uzunluğu 1 br. olan düzgün beşgen AG=AB DG= 2

= 2r ,+C = 4 r +A = r 7 ,+B 7 7

2r 7

CE açıortay

E

E

a CE=b-a

2r 7

2r 7

r 7

b

C

Proove that:

C

2

m(∠GAB)=x=12° 1

A

G archiµedes26

x

B

A

Problem 548.

Problem 545.

C C

D

DC//AB A ve C teğet Noktaları

D

a

MA=MB

18° 18°

A

A

M

B

D

a a 2

x=9° x archiµedes26

B

10

Geometri Günlüğü

Problem 552.

Problem 549.

C

D

AC=CB

60°

D a

a 3

a

ABCDE düzgün beşgen DF=FG m(GDC)=60°

x

E

x

18° 18°

A

x=6°

C

archiµedes26

F

x=6° , y=18°

B

y archiµedes26

Problem 550.

A

D 30°

Problem 553.

D

x=60°

2

archiµedes26

B

G

42°

ABCDE düzgün beşgen DF=FG m(∠GDC)=42°

E

C

A

x=30°

x

1

54° 1

B

F

C

archiµedes26

x

A

B

G

Problem 551. A

A,G,K,D,A,E,H çemberlerin kesişim noktaları

G A

C

B

Problem 554. C 46°

E BC=AE

H

D 8°

K

x

50°

2°

B

A

D x=12°

Geometri Günlüğü

Problem 558.

Problem 555.

D

b

C

A

x=I α-45° I

90°-α

BCD eşkenar üçgen m(∠BAD)=30°

30°

3α−90°

c

D

α

11

archiµedes26

1

A

c2 = a 2 + b 2

a

x

B

2

PME, Pr. 1131

C

B

Problem 559.

C Problem 556.

C 60°

Proove that:

40°

D

A

ABC eşkenar Üçgen

a

50°

D a 3

x=20°

x

6°

archiµedes26

c

d

B

a

a 2 + b 2 = c2 + d 2

18°

archiµedes26

42°

54°

A

Problem 557.

C x=39°

d

D

b 20°

9° 27°

A

B

Problem 560.

C x

b

24°

B

A

40°

c

D

a

a 2 + b 2 = c2 + d 2 archiµedes26

10° 10°

B

12

Geometri Günlüğü

Problem 561.

Problem 564.

A

A

a D x

c

D 9°

18° 15°

10°

12°

B

C

Proove that:

x=75°

d

40°

20°

B

archiµedes26

C

b archiµedes26

Proove that:

a+b=c+d

Problem 562.

Problem 565.

A

A α 2α M, BC nin Orta Noktası m(∠MAC)=2m(∠BAM) AB⊥BD

B ve C teğet noktaları

BM=MC

C

M

M

B

C AD=2.AC

B Geometry In Figures

A.Posamentier

D

D

Problem 566. (Sangaku)

Problem 563.

A

A

B a ABCD dikdörtgen

c

12° b c

b 108°

132° a

C

x a+b+c

D

D

B x=18°

archiµedes26

a+b=c

C

Geometri Günlüğü

13

Problem 570.

Problem 567.

A

A

c 12°

b

c

d

B

b

(b + c) 2 - a 2 = 4d 2

24° 60° a

a+b+c

x

C

r

r

C

D archiµedes26

x=6°

B

D a

Problem 568.

Problem 571.

E

A

A

A, EB yayının Orta Noktası OM=MA, BC=CD

M B

x

24° c

Proove that:

O C

b

108°

132°

C

x

a+b+c

D

a

x=18° B

archiµedes26

x=12°

archiµedes26

D

O merkezli düzgün beşgen

Problem 569.

Problem 572 .

A

A

D,E,G teğet Noktaları GF⊥DE

D

E

F

6°

m(∠BFG)=m(∠GFC)

D 48° J. Warendorff

B

G

C ABC üçgeninin iç çemberi

B

x

6°

12°

x=9°

C

14

Geometri Günlüğü

Problem 573.

Problem 576.

A

Düzgün beşgen

H

c

K, ABC üçgeninin simedyan Noktası BH yükseklik M, BH nin orta Noktası N, AC nin orta Noktası

b

a

c2 = a 2 + b 2

E

B

N

K

M

A

I

M,K,N doğrusaldır

A

V.Prasolov Plane Geometry, Pr. 6.47

archiµedes26

C

B

Düzgün beşgen

C

D

Düzgün beşgen

Problem 574. ABC heptagonal üçgen

B

= 2r ,+C = 4 r +A = r 7 ,+B 7 7 E

I, ABC üçgeninin iç merkezi

Problem 577.

B

ABC heptagonal üçgen

F I

= 2r ,+C = 4 r +A = r 7 ,+B 7 7

O

EF=ED

H, ABC üçgeninin diklik merkezi O, ABC üçgeninin çevrel merkezi K, ABC üçgeninin simedyan Noktası

Bankoff and Garfunkel

K

C

OK=KH

A

D

A C

archiµedes26

H

Problem 575. B

ABC heptagonal üçgen

= 2r ,+C = 4 r +A = r 7 ,+B 7 7

r 7

r 7

E

G

2r 7 C

Problem 578.

A

G, ABC üçgensel bölgesinin ağırlık merkezi BD açıortay, CE açıortay

a

D d

24° c 48°

D,G,E doğrusaldır

x b

B 2r 7

A

D Abdilkadir Altintas (archiµedes26) “Some Collinearities In Heptagonal Triangle”, Forum Geometricorum, Vol16 (2016)

a+b=c+d archiµedes26

x=12°

C

Geometri Günlüğü

Problem 579.

15

Problem 582.

C

C

a+b=2c 30°-φ

b

a

E

1

Proove that:

A

cot A + cot B = 2cot C 2 2 2

x

30°-2φ

φ 3

B

D

AC=1, AD= 3 archiµedes26

Proove that:

C.V. Durell

c

A

x=60°

B

Problem 583.

A

Problem 580. C

D

1

A

20°

E

x

20°

5°

D

2

12°

6° B

B

AC=1, AD= 2 archiµedes26

54°

24°

C AD is angle bisector

Proove that:

archimedes26

x=25°

Problem 584. A 72°

D AB=AC

Problem 581.

Proove that:

C E

x

x=6°

E A

3θ

Mathematical Gazette Adventitous Angles

x

60°−3θ

θ AC=AD

D archiµedes26

x=30°−θ

B

48° B

24° C

16

Geometri Günlüğü

Problem 585.

Problem 588.

A

N, teğet Noktası M, BH nin orta Noktası

42° 1

A

φ

54° AB=1, DC= 3

M I B

R.Honsberger “Episodes In Nineteenth and Twentieth Century Euclidean Geometry”

C ABC üçgeninin I merkezli iç çemberi

Proove that:

φ=24°

M,I N doğrusaldır

N

C

3

D

B

H

SPEED 2001 - FOROGEOMETRAS

ABC üçgeninin dış teğet çemberi

Problem 586. Problem 589.

ABC heptagonal triangle

circumcircle of ABC

B

A

= 2r ,+C = 4 r +A = r 7 ,+B 7 7 H, orthocenter of ABC D,E tangency points

D

1

E

φ I

48°

Proove that

G

B

A

12°

HD⊥HE

C

12°

5

D

C

AB=1, DC= 5 Bankoff and Garfunkel

archiµedes26

φ=72°

E

H

Problem 587.

A

Problem 590.

ABC dik üçgen c

A

φ

AI.BI = 2 .r.c

I

E

1 84° B

r

G. Huvent

12°

12°

5

D AB=1, DC= 5 archiµedes26

B

C ABC üçgeninin I merkezli iç çemberi

φ=72°

C

Geometri Günlüğü

Problem 591.

17

Problem 594.

A

A

30° 50°

E

E

D

φ

D 15°

10°

AE = BD

20° B

3° 6°

B

C

C

φ=24°

Problem 595.

Problem 592. A

A M,N,P,R teğet noktaları MN∩AF={K}

E

E

D

K

φ

N AD=AK=AE

P

M

Geometry In Figures, A.Akopyan

33°

R B

D

C

F

18°

24°

B

C

φ=12°

Problem 596. Problem 593.

A

A E

E φ

φ

D

D 9°

B

33°

48°

12°

C

φ=6°

30°

12°

B

φ=24°

C

18

Geometri Günlüğü

Problem 600.

Problem 597. D H

C

ABC heptagonal üçgen

B

+A = r ,+B = 2r ,+C = 4 r 7 7 7

1

A

78°

6°

42°

66° X

AB//DC XH⊥DC, XH=1

I, ABC üçgeninin iç merkezi BD simedyan, CE simedyan

E

B

I

Nick's Mathematical Puzzles

D,I,E doğrusaldır

AD+DX-(BC+CX)=8

A

D

C

Abdilkadir Altintas (archiµedes26) “Some Collinearities In Heptagonal Triangle”, Forum Geometricorum, Vol16 (2016)

Problem 598. A E φ

Problem 601.

D

A

15°

B

E φ

9° 12°

D

C 30°

10° 10°

φ=18° B

C

φ=20°

Problem 599.

Problem 602.

A

A

E φ

ABC eşkenar Üçgen

D

2α

12°

B

6° 18°

φ=90°-3α C

φ=30°

E

B

α

φ D

60−2α

C

19

Geometri Günlüğü

Problem 606.

Problem 603.

A

A E

ABC eşkenar Üçgen

φ

2α

D 25°

φ=60°-4α

5° 10°

B

C

φ=30°

E

B

φ

α

30°−α

D

Problem 604.

C

Problem 607.

A

ABC eşkenar Üçgen

20°

A

D

φ=25°

10° φ

30°

20°

B

E

B

5°

φ

10°

D

C

AB=CD

φ=20° C

Problem 608.

A

Problem 605. A E φ

D

D 15°

B

5° 20°

6°

C

φ

18° B

12° AB=CD

φ=40°

φ=18°

C

20

Geometri Günlüğü

Problem 609.

Problem 612. A

A

AB=CD D

AB=CD

φ=45°

D

5°

B

φ

55°

φ=10°

10°

10°

C φ 100°

30°

C

B

Problem 610. Problem 613.

A

A AB=CD AB=CD

D

φ=18°

D

φ=40°

12°

B

20°

φ

54°

30°

C

φ

50°

B

10°

C

Problem 614.

A

Problem 611.

A AB=CD

AB=CD

φ=30°

φ=30°

D 66° B

D 36°

6°

42° 42°

φ C

B

12°

φ C

Geometri Günlüğü

Problem 615.

Problem 618.

D

A

Bankoff Çemberi AB çap, C,D,E,F teğet noktaları

AB=CD

F E

D

d

O1 A

1-r

C

φ=24°

d = 1 r^1 - r h 2

O2

r

21

12°

B

φ

Bankoff Çemberi

B

30°

18°

C

Problem 619.

A

Problem 616. A merkezli, AC yaıçaplı çember

AB çap, C,D,E,F teğet noktaları

B merkezli, BC yaıçaplı çember

D

AB=CD

d

D

E

d = 1 r^1 - r h 2

r

12°

O2

O1 A

φ=18°

F

1-r

C

B

B

φ

66°

30°

C

Schoch Doğrusu

Problem 620. D Bankoff Arşimedyen Çemberi (2)

Problem 617.

d

A

AB çap, C,D,E,F,G teğet noktaları

F G

E O1

D 36° B

12°

6° AB=BD

φ=12°

A

φ C

r

d = 1 r^1 - r h 2

O2 C

1-r

B

22

Geometri Günlüğü

Problem 621.

Problem 624.

A

C x

D

10°

A

30° m

24°

30°

B

70° B

n

D

x = ` m j2 + m n y n

φ

42°

y

G. Aydın

C

AB=CD

φ=12°

Problem 625. C

Problem 622. C x

30°

40°

10°

A

y

m

n

D

x = ` m j2 + m - 1 n y n

x y

B

A

archimedes26

100°

40°

20°

m

n

D

x = ` m j2 + m - 1 n y n

B

archimedes26

Problem 626.

Problem 623.

E C x

a

y

b A

20°

70° m

40° n

D

β B

c α

x = `m j + m - 2 n y n 2

archimedes26

A

10°

b

30°

C a 20°

D

a = 30c, b = 50c

B

Geometri Günlüğü

Problem 627.

23

Problem 630.

A

A

φ 27°

D

φ=51°

15° φ

30°

15°

B

AB=CD

E

C

φ=30°

21°

9°

B

Problem 631.

Problem 628.

A

φ

φ φ φ a

15°

C

D

42°

d

φ=72°

c

b

30° B

E

a+c =b b+d c

Çok Kıymetli Ahmet ELMAS hocamızın anısına. ALLAH rahmet eylesin. Mekanı Cennet olsun inşallah....

Problem 629.

36°

12°

B

C

D

A 80°

30°

Problem 632.

A φ

φ B

E

60°

D

D

80°

18° 12°

C

φ=70°

φ=30°

B

18° 24°

C

24

Geometri Günlüğü

Problem 633.

Problem 636.

A

Sayın Tezcan ŞAHİN hocama çok teşekkür ederim...

AB=AC M, AD nin orta Noktası C,H,M doğrusal M

A

φ AB=AC

φ=45°

BH = AH

H

D 2

A.Akopyan Geometry In Figures

27° B

Problem 634.

B

54°

C

D

B

A

D

5°

10°

C

10

Problem 637.

55°

C

E

M D A

H

O

B

O, AB çaplı çemberin merkezi H, teğet Noktası M, EH nin orta noktası EH⊥AB D,C; E merkezli çemberle O merkezli çemberin kesişim Noktaları D,M,C doğrusaldır

φ=30°

φ

A.Akopyan, “Geometry In Figures”

C

Problem 635.

A

AB=BD AC=CD

Problem 638.

A E

C

φ

m (+EAD) = m (+DAF)

B

F

B

27°

9° 12°

C

φ=24°

E D

D

Geometri Günlüğü

Problem 642.

Problem 639.

25

A

A E φ

D

9° 12°

B

a

57°

φ=60°-a C

φ=18° D 60°-2a

φ

3a

B

Problem 640.

A

E

Problem 643.

C

A

φ D

φ=48°

ABC eşkenar Üçgen T, Teğet Noktası

D

30° 66°

T E AD + AE = 1 DB EC

12° 24° B

C

E

C

B ABC üçgeninin iç çemberi

Problem 644. Problem 641.

E

G

A

c

B

x=a

3α+3β D B

AEFGBCDG, düzgün yedigen ABC yedigensel (heptagonal) üçgen BC=a, AC=b, AB=c Bu kenarlara ait yükseklikler ha, hb,hc

DC=2a AD=x AB⊥AC x

α

F

2a

β

hc

a hb

α C

A

b Bankoff and Garfunkel “Heptagonal Triangle”

ha C

2 2 2 h 2a + h 2b + h 2c = a + b + c 2

D

26

Geometri Günlüğü

Problem 648. A

Problem 645. F

AB=1 BD= 5

E

G

1

E

AEFGBCDG, O merkezli R yarıçaplı düzgün yedigen ABC, yedigensel üçgen H, ABC nin diklik merkezi

R

φ

42°

φ=42°

12°

18°

B

D

5

C

O OH = R 2

A

B

Bankoff and Garfunkel “Heptagonal Triangle”

D

C H

Problem 649. Problem 646.

F C

G

A

24°

54°

42° m

n

D

E

AEFGBCDG, O merkezli R yarıçaplı düzgün yedigen ABC, yedigensel üçgen H, ABC nin diklik merkezi I, ABC üçgeninin r yarıçaplı iç çemberi

R

B

O A

B m `m n j + n =1 2

archimedes26

I

2 2 IH 2 = R + 4r 2

r

Bankoff and Garfunkel “Heptagonal Triangle”

D

C H

Problem 647.

C

Problem 650. Sayın Kemal AYDIN hocama çok teşekkür ederim...

20°

A

80°

CE∩BD={K} M, BE nin Orta noktası N, DC nin Orta Noktası MN∩BD={P}, X; MP nin Orta Noktası Y; EK nın Orta Noktası Z, AD nin Orta Noktası

Z

D

φ

A

D

E

20° AC=BC

Y K

M

B

X

X,Y,Z Doğrusaldır

N P

archimedes26

φ=10°

B

C

Geometri Günlüğü

Problem 651.

27

Problem 654. A

A E φ

D

51°

B

φ

D

C

a

B

+c

b

66°

24°

3° 12°

12°

c

a+b

Proove that: archiµedes26

C

φ=18°

φ=30°

Problem 655.

Problem 652.

A

A

60

30

-a

-2a

E φ

D

D φ

3° 6°

B

a

51°

3a

B

C

C φ=30°

φ=24°

Problem 656.

Problem 653.

A

A

60

φ

30

-4a

E

-2

a

D

D φ

B

a

27°

3° 18°

3a

B

C φ=30°+a

φ=48°

C

28

Geometri Günlüğü

Problem 657.

Problem 660.

A

A x

a

10°

b

10°

z

y 30°

20°

D

m

B

c 10° 20°

C

d

B

C

archiµedes26

x2 + y2 + z2 = m 2 + n 2

40°

n

D

Proove that: archiµedes26

a+d=b+c

Problem 661. B c

Problem 658.

a

A φ

A

54°

24°

c=a+b

D 46°

60° b

D

8°

12°

2°

C

B φ=50°

Problem 662. C

Problem 659. C c

a

A

24°

1

60° b

D

54°

12°

d

48°

φ=18°

B

D a+b=c+d

A

φ

30°

3

B

C

Geometri Günlüğü

Problem 663.

Problem 666. D

C 36°

24°

54° DB=1 AC= 3

1

φ=48°

1 6°

φ=96°

D

A

29

42°

C

φ

A

B

3

3

24°

φ

B

Problem 667. Problem 664.

C

C 6°

A

18°

1

D

AC nin Orta Noktası

φ

12°

φ=42°

B

3

CD ve AE yükseklik M, AC nin orta Noktası

E

M

A

α+2φ=180°

α

φ D

B

Problem 665.

C 6° D A

Problem 668.

E

54°

C

1

φ φ

φ

12°

B

3

P

M

F Q

CD açıortay AF∩BE={M}∈CD AC∩DE={P} BC∩DF={Q}

φ=6° PQ//AB A

B

D

INMO 2010

30

Geometri Günlüğü

Problem 669.

Problem 672. C

C O, çevrel çember merkezi I, iç merkez AD açıortay AB=a, BC=a+1, AC=a+2 a+2

AB=AC C,A,D doğrusal I, ABC üçgeninin iç merkezi BC=AB+AI

a+1 D

O

I

OI⊥AD

m(∠C)=45°

I Indian National Math Olympiad 2006

A

B

a

A

B

D

Problem 670. B

ABC heptagonal üçgen

r 7

r 7

Problem 673. A

= 2r ,+C = 4 r +A = r 7 ,+B 7 7 BE açıortay

54°

c B

a

φ

AC=1 CD= 2 D BD= 3

3

1

BE=c-a

φ=48°

4r 7 C

66°

r 7

2

A

E

C

Problem 671.

A Problem 674.

α

C

60−α

AD=1 BC= 3

27°

9°

1

1

φ=α

φ=18°

D

30° B

D

A

φ 3

C

φ

18°

2

B

31

Geometri Günlüğü

Problem 675.

Problem 678.

A

A φ D

30°

a=b

a 30°-2a

2a

B

a

C

φ=90°+a

D

96°

6°

B

b

18° 12°

C

Problem 679. A

Problem 676.

φ

A

D

30°

φ

φ=12°

a 30°-a

2a

B

C

φ=90°-a

D 24° 24°

6°

B

30°

C

Problem 680. A

Problem 677. A

E

F

B

O

φ

m(∠BAC)=60° O, ABC üçgeninin çevrel çember merkezi H, ABC üçgeninin diklik merkezi OH ∩AB={F}, OH∩AC={E}

60°

E

Çevre (∆AFE)=AB+AC OH= I AB-AC I

H

C Iran NMO

B

D

35°

5° 20° Proove that

φ=40°

C

32

Geometri Günlüğü

Problem 681.

Problem 684.

A

D

φ

E

A

φ=18° 24°

15°

B

27°

24°

42°

m

C

B

54°

n

D

C

2 m = `m nj + n 1

Problem 685. Problem 682. B

ABC heptagonal üçgen

2r 7

AE açıortay

O, center of the circle G,D,E,F points of tangency OB⊥AC

O

ha E

G Ia

Proove that

Ia=2ha

4r 7 H

Dual Butterfly Theorem

= 2r ,+C = 4 r +A = r 7 ,+B 7 7

r 7

F

P

Bankoff and Garfunkel

D

Q

E

m(∠OBQ)=m(∠OBP)

A

C

A

B

C

Problem 686.

A

Problem 683. r4 r7

30° 7 =2 +5 r4 r7 r1

B

m

48° D

r1

A

D

O

C

B

2 m = `m nj + n 1

n

66°

C

33

Geometri Günlüğü

Problem 687.

A

a

B

A

b

m

a

105°

30°

15°

Problem 690.

n

D

C

B

a = ` m j2 + m 1 n n 2 b

a=b

D

12° 18°

6°

b

42°

C

Problem 691.

Problem 688.

A

A

D a

a=b

D 96°

b

24°

b

a

12° 12°

3°

C

B

33°

a=b

69°

39°

C

B Problem 689.

A Problem 692.

A

a

a=b

a

D c

6° 24° 12°

54° B

B

D b

6°

48° C

d

a 2 + b 2 = c2 + d 2 30°

84° b

archiµedes26

C

34

Geometri Günlüğü

Problem 693.

A

Problem 696.

A

a

c

b

a 2 + b 2 = c2 + d 2

D c

6°

d

a 2 + b 2 = c2 + d 2 30°

96°

18°

b

B

D

d

18°

a

6°

B

30° 24°

C

archiµedes26

archiµedes26

C

Problem 697. A 96°

1

Problem 694. A

AB=1 BD= 6

E

B

φ

36° D

5

φ

1 60°

B

D

6

AB=1 DC= 5

φ=30°

15°

15°

C

archiµedes26

C

φ=12°

Problem 698.

A Problem 695. A

AB=1 BD= 5

E 1

φ 78°

B

5

D

1 18°

φ=42°

12°

18°

AB=1 DC= 5

φ=18° 108°

C

B

φ D

5

C

Geometri Günlüğü

Problem 699.

35

Problem 702. A

A m(∠BAC)=60° DEF, orthic triangle of ABC J, ortho center of DEF N, midpoint of PQ M, midpoint of BC

60°

F Proove that:

18°

E

J

30°

m

B

M,N,J collinear

D

n

66°

C

Q N

P

B

M

2 m = `m nj + n 1

archimedes26

C

D

Problem 700.

Problem 703.

A

φ

D

27°

24° 39°

9°

B

φ=12°

C

Problem 704.

Problem 701.

A

A c 24°

B

12° B

m

24° D 2 m = `m nj + n 1

n

30°

C

a

12°

b φ

48°

a+b+c

C Proove that:

φ=18°

archiµedes26

D

36

Geometri Günlüğü

Problem 705.

Problem 708. A 18°

D 24°

6°

φ

12°

B

C φ=105°

Problem 709. A E

Problem 706.

φ

D 30+α

B

α 2α

C

φ=30°

Problem 710.

Problem 707.

archiµedes26

Geometri Günlüğü

Problem 711.

37

Problem 714.

Problem 715.

Problem 712.

D 15°

A

30°

φ=22,5° φ

22,5°

B

C

Problem 716. Problem 713.

ABC heptagonal triangle

B

A

= 2r ,+C = 4 r +A = r 7 ,+B 7 7

81°

B

φ

12°

D

I, incenter of ABC G, centroid of ABC AD, symmedian line

C

D I

AB + BD = AC

Proove that G D,I,G collinear

φ=6°

Proposed by Özgür Demirkapu

C

A

archiµedes26

38

Geometri Günlüğü

Problem 717.

Problem 720. ABC heptagonal triangle

B

B = 2r ,+C = 4 r +A= r 7 7 7 , Br1, first Brocard Point of ABC BL, angle bisector AZ, symmedian line

Proove that

Z

Z, Br1, L collinear

Br1

archiµedes26

C L

Problem 718.

Problem 721.

A

B

ABC heptagonal triangle

= 4r B = 2r +A= r 7 ,+C 7 7 ,

a 30°

Proove that:

Br1, First Brocard Point of ABC Br2, Second Brocard Point of ABC AN, altidue

Br2

D

c

A

Proove that

a+b=c b

18° Br2N⊥AC Br1N⊥AB

Br1 archiµedes26

48°

36°

B

C

archimedes26

C

A N

Problem 722.

Problem 719.

B

A

ABC heptagonal triangle

= 4r B = 2r +A= r 7 ,+C 7 7 ,

α 12° 6°

B

1

D

T

18°

3

Br1, First Brocard Point of ABC CT, angle bisector BQ, symmedian

C

Proove that

Proove that: T, Br1, Q collinear

Br1

C

α=36°

archiµedes26

Q A

39

Geometri Günlüğü

Problem 723.

Problem 726.

A

A Proove that:

a 30°

18°

Proove that:

AD 2 = BD.BC

D

c

b

b 2 = ac

18°

proposed by Tezcan Şahin

30° proposed by Ayhan Yanağlıbaş

48°

36°

B

66°

D

C

C

B

Problem 727.

A Problem 724.

D

φ

A

AH, angle bisector F,D,E tangency points

Proove that:

Proove that:

α

30°

D

30°

B

F

φ=3α

30°−α

C

F,D,E collinear

C

H

B

E

Incircle of ABH

Geometry In Figures

Excircle of AHC

Problem 725.

A E B

A,E,B collinear A,B,H,M concyclic EH⊥CD m(∠AEC)=m(∠BED)

Problem 728.

A D

E AB⊥AC

φ

Proove that:

Proove that:

5°

CM=MD Geometry In Figures

C

M

H

φ=50° archimedes26

30°

D

15°

B

C

40

Geometri Günlüğü

Problem 732.

Problem 729.

A

Proove that:

A 70° Proove that:

a

D 30°

10°

1 = 1 + 1 c a b

b

c

φ

archimedes26

40°

φ=60°

D

10° 20°

archimedes26

20°

C

B

40°

C

B

Problem 733. Sangaku

P

A Problem 730.

D

r

A

ABCD rectangle AD//EF

r

Proove that:

a 6° c

d

D b

18°

B

12°

18°

C

Q

E

a2 + b2 = c2 + d2

E

archimedes26

R

R =U= 1+ 5 2 r

C

B

Problem 731.

A 48°

Proove that:

Problem 734.

A m(∠A)=45° O, circumcircle center of ABC M, midpoint of AC CD⊥AB

45°

φ=30°

12° 1

archimedes26

D

O

M

D

Prove that:

M,O,D collinear

78°

B

Proove that:

φ 3

C

B

C

Geometri Günlüğü

Problem 735.

41

Problem 738.

C

A

φ

Proove that:

AD=DB CD=CB m(∠B)=2m(∠A)

φ=30°+α

3α Proove that:

φ=60°-α

D

E

Hong Kong, Mock Test, 1994

α

2α

Problem 736.

φ

α

D

B

B

A

30−2α

D

Problem 739.

A

φ

α

Proove that:

2

9° 9°

A 60°

φ=30°

66°

D 18°

B

C

10 Proove that:

3 1

C

proposed by Tezcan Şahin

archimedes26

α=9°

B

C

2

Problem 737.

A Problem 740.

A

Proove that:

φ

φ=15°

3

φ B

6° 12°

1

D

18°

Proove that: archimedes26

120° 1

B

2

φ=30°

D C

C

5 archiµedes26

42

Geometri Günlüğü

Problem 741.

Problem 744.

A

A

E

M

D

ABC right-angled triangle I, incenter of ABC M, midpoint of DE

φ Proove that:

Proove that:

I

φ=30°

MI⊥BC V.Nicula

B

C

90°-α

30° 30°-α

B

30°-α

proposed by Özgür Demirkapu

C

Problem 745.

Problem 742.

A

A

Proove that:

24°

φ

Proove that:

E D

D

18°

φ=60°-α

α=6°

β β

30°+α

α α

B

30° 30°-α

B

30°-α

C

D proposed by Özgür Demirkapu

C

Problem 746. A 54°

Problem 743.

A

24°

Proove that:

6°

a2 + b2 = c2

Prove that:

b

B

3

3

a - b = 3ab

a

2

30° 24°

120°

20°

a

C

B

b

c

FOROGEOMETRAS, 2012

C

Geometri Günlüğü

Problem 750.

Problem 747.

A

A

T

6°

B

43

φ 2

12°

5 -1

D Prove that:

I, incenter of ABC D,E,T tangency points

D

C I

φ=54°

B

Proove that:

F

C

E

D,I,E collinear

IMO Training 2007

Problem 748.

A

Problem 751.

A m(∠A)=60° I, incenter of ABC A(∆FEI)=S1 A(∆FIB)=S2 A(∆EIC)=S3

60°

9°

B

φ 2

D Prove that:

15°

5 -1

C

Proove that:

E F

φ=84°

S1 S2

1 = 1 1 S1 S2 + S3

S3

I

V. Nicula

C

B

Problem 749.

Problem 752. A

A DE, diameter I, incenter of ABC A,D,F collinear

D

c

b

Proove that:

I

10° 10°

BE=FC

D

d Proove that:

IMO Training 2007

E incircle of ABC

F

C

a

30°

B

B

10°

archiµedes26

a +b =c +d 2

2

2

2

C

44

Geometri Günlüğü

Problem 753.

Problem 756.

B

A ABC right-angled triangle D, midpoint of AD

mid-point of AD

D

φ, acute angle

30° 45°

Proove that:

Proove that:

D

2

sin z # 13

φ=24°

1 archimedes26

British M.O.-1991

φ

A

C

φ

B

Problem 754.

C

3

Problem 757.

A φ Proove that:

D

2

A

Proove that:

45°

2

φ=30°-α

1 archimedes26

D 30°

φ=36°

60°-α 3α

α

B

φ

C B

C

3

Problem 758.

D

Problem 755. A

H

2 ABC isosceles triangle AB=AC H, orthocenter of ABC O, circumcircle of ABC c, circle passing from B,O and H with center P

A Proove that:

45°

φ=60°

Proove that:

P

2

1

archimedes26

P∈AB

O B

C c circle c with center P

British M.O. 2008

φ

B 3

C

Geometri Günlüğü

Problem 759.

A

Problem 762.

D

45

A

φ, acute angle

18° 45°

Proove that: Proove that:

2

d

φ=12°

b

a2 + b2 = c2 + d2

1

archimedes26

archimedes26

φ

B

C

3

Problem 760.

c

10°

B

30° 20°

C

Problem 763.

A D

D

a

40°

c3

E

AB common tangent of c1 and c2 C,Q,P,D are intersection points c3, circle passing from A and B points

AB⊥AC

φ

Proove that:

5°

φ=50°

25°

C

Proove that:

P

D

c2

QC PC = QD PD

archimedes26

30°

2016 China Western Math. Oly.

c1

Q

C

B

A

B common tangent of c1 and c2

Problem 761.

Problem 764 .

A

C 60°

m(∠BAC)=60° DEF orthic triangle of ABC J, orthocenter of DEF S1=Area(BACJ) S2=Area(BJC)

F

N, midpoint of AC

J

Proove that:

E I

S1 $ 2 S2

S2 archiµedes26

D

D

a+2

Proove that:

B

a+1

N

S1

ABC has side lengths in arithmetic progression AB=a, BC=a+1, AC=a+2 I, incenter of ABC AD, angle bisector M, N midpoints of AB and AC

C

ID = IN = IM A M, midpoint of AB

M

a

B

Indian National Math Olympiad 2006

46

Geometri Günlüğü

Problem 765.

Problem 768.

A

ABC right-angled triangle BD, angle bisector BE, angle bisector AM=MC

A

w1

w2

C

D I1

E

I2

Proove that:

M

CA tangent to w2 DA tangent to w1 A and B intersection points of w1 and w2 I1, incenter of ABC I2, incenter of ABD E, intersection point of I1I2 and AB

Proove that:

1 = 1 + 1 AE AC AD

α

B

5 1 AB 1 3 2 BC

D α

Chinese TST 2002

Indian National Math Olympiad 2007

2α

C

B

Problem 769. Problem 766.

E,D tengent points of excircles of ABC M, mid point of AB N, midpoint of DE MN∩BC={P}

A

A incircle of ABC

D c

AC>BC>AB AM, median c, incircle of ABC D,E intersection points of AM with c AD=DE=EM

M

E N o

B

o

P

Proove that:

C

D

perimeter(AMPC)=perimeter(MPB)

E B

Tournament of Towns

C

M Proove that:

Indian National Math Olympiad 2005

AB:BC:AC=5:10:13

Problem 770. A

B ABCD square P, any point on circumcircle of ABC

Problem 767.

A Proove that:

B

6°

φ 2

D Prove that:

φ=156°

6°

5 -1

max " PA,PC , = 1 ^ PB + PD h 2

C

Buffet Contest 2015

D

C P

Geometri Günlüğü

Problem 771.

47

Problem 774.

A

A

φ

G, centroid of ABC G,E,C,D con-cyclic E

P

Proove that: AC BC $ 2

G

BP⊥PC N, midpoint of AC m(∠BAP)=m(∠PCB) BP=2.PN M, midpoint of BC

a

N Proove that:

2a

A,P,M collinear

INMO-1994 24. th Indian National Math. Olym.

C

D

B

φ

B

M

C

Problem 775. Problem 772.

B

A

F P

ABC equilatiral P, any point on incircle of ABC PD, PE, PF altidues from P

E

b

ABCD convex quadrilateral

a

C

Proove that:

Proove that: PD = PE + PF

A c

Romantics of Geometry, V. Stamatiadis

1 ^ a + b h ^ c + d h $ 6 ABCD@ 4

d Romantics of Geometry, T. Papadapoulos

B

C

D

D

incircle of ABC

Problem 776.

D Problem 773.

B

φ

G, centroid of ABC K,G,L,P collinear A,C,P collinear

AD=BC+2AB Proove that:

K

Proove that:

G

1 = 1 + 1 GK GL GP

L A centroid of ABC

C

φ=40°

B

P Problems In geometry, Prithwijit De ICFAI Business School, Kolkata, Problem 88

A

40°

20°

C

48

Geometri Günlüğü

Problem 777.

Problem 780.

A

D

O, circumcircle of ABC I, incenter of ABC AD, angle bisector b+c=2a

85°

B c φ 10°

A

100°

b

O

I

AC=CD

C

Proove that:

OI = AD

Proposed by Tezcan Şahin

B

Proove that:

C

a D

St. Petersburg Math. Contest

φ=5°

Problem 778.

A

Problem 781.

B

B

E 12°

D

ABC heptagonal triangle

= 4r B = 2r +A= r 7 ,+C 7 7 ,

18°

Br2

C

E

ABCD trapezoid, AC=DC Proove that:

O

BD and CE angle bisectors Br2, Second Brocard Point of ABC O, circumcircle center of ABC

Proove that

Proposed by Tezcan Şahin

DE = OBr2

AB + ED = 1 EC + CB

C

archiµedes26

D A

circumcircle of ABC

Problem 779.

Problem 782.

A

A

G, centroid of ABC P,G,Q collinear

18°

P Proove that:

Proove that:

G

PB . QC # 1 PA QA 4

Q

CA - CB = CA.CB 2

B

30°

C

2

B

centroid of ABC

C

31 st Spanish Math Oly., 1995 (Second Round)

Geometri Günlüğü

Problem 783.

49

Problem 786. C

C AD, angle bisector BE, angle bisector T∈ED P,Q,R; feet of altidues from T

Q R

D

T

c1, quarter circle with radius AB c3, semi circle with diameter AB P,Q,T tangency points d, common tangent of c2 and c3, d∩PB={R}, TH⊥AB

P

c2

c1

d

Q

R T

E

c3

Proove that:

Proove that:

TP = TQ + TR

φ=α φ α

SOBRE GEOMETRİAS

A

B

P

A

Problem 784.

H

C

D

27°

15°

Proove that:

B

φ

2

φ=3° 21°

135°

1

1

D

B

O

Problem 787.

A 18°

SOBRE GEOMETRIAS, A.R.M. Romero

A

archimedes26

φ

Prove that:

C

2

B

3 archiµedes26

φ=12°

Problem 788.

A Problem 785.

BD, median CE, median G. centroid T∈ED

C y ^ a - b h ^ a + ac - b h = b c 2

b

2

2

2

2

2

A

c

b x

Mathematical Reflections, J264, T.Andreescu

3α

D

z

E

Proove that:

a

T

Proove that:

ax = by + cz

G

2α

c

B

B

a

C

50

Geometri Günlüğü

Problem 789.

Problem 792.

A

A G, centroid of ABC G,E,C,D con-cyclic

180c 7

E

B

Prove that: 2 2 cot A = 2AC - AB 46 ABC@

G

AB=AC=CD,

Leonard Giugiuc-Romania and Kadir Altintas-Turkey

Problem 790.

AE = 2 AD

A ABC is triangle with sides 2a,2b,2c respectively G, centroid of ABC G,E,C,D con-cyclic

Problem 793. A

ABC rectangular triangle at B F,E,D tangency Points DG⊥FE

E E

Prove that: G

a 2 + b 2 = 2c 2

b

c

G

Leonard Giugiuc-Romania and Kadir Altintas-Turkey

a

D

B

Prove that:

F

CG⊥BG

C B

Problem 791.

C

D

Romantics of Geometry Stamatiadis Vaggelis

A

K

c

b

Problem 794.

G

A4

Q

B

AD=EC

Prove that:

C

D

B

C

E

D

C

a ABC is right angled triangle with sides a,b,c respectively G, centroid of ABC K, symmedian point of ABC

P

A5

A3

Prove that: N

Z2 N Z2 X X2Y M P2Q R

R M

Prove that: 6 A(ABC)@ = ^ 3a.GK + b - c h^ 3a.GK - b + c h 2

2

2

Leonard Giugiuc-Romania and Kadir Altintas-Turkey

2

2

A1A2A3A4A5 regular pentagon S,R,M,N,P,Q are mid-points of the segments they belong to and form a pentagon MNPQR M,N,P,Q,R are interior angles

A1

S

archiµedes26

A2

Geometri Günlüğü

Problem 795.

Problem 798.

A De Gua’s Theorem ABCD tetrahedron, three right angles at D Sa=Area(∆BDC) Sb=Area(∆ADC) Sc=Area(∆ABD) Sd=Area(∆ABC)

A

N

P F

G, centroid of ABC M,N,P points on circumcircle of ABC

E

c

D

51

b

G

Prove that: Prove that:

a + b + c = AD.AM + BE.BN + CF.CP 2

S d2 = S a2 + S b2 + S c2 C

B

a D

B

2

2

C SSM, 4140, Jack Garfunkel

M

Problem 799. Problem 796.

A φ

A

I, Incenter of ABC M, anypoint on circumcircle of ABC, on arc BC IE⊥BM, IF⊥MC

AB=AC Prove that:

I Prove that:

φ=60°

IE + IF = sin+A AM B

C F

48°

M

Problem 797.

6°

B

N

P F

I, Incenter of ABC M,N,P points on circumcircle of ABC

A

D G is point on semicircle with diameter BC, center O DC and AB are tangents GH⊥BC M, midpoint of GH B,G,D and C,G,A collinear

E

E

c

I

b

G

Prove that:

Prove that: OM⊥AD

abc = AD.AM.BE.BN.CF.CP

M B

a D

C

S.Vaggelis, Romantics of Geometry MM, Q646, Jack Garfunkel

B M

archimedes26

C

5

Problem 800.

A

D

3

Mathematics Magazine, Pr. 1809, C. Pohoata

E

H O

C

52

Geometri Günlüğü

Problem 804.

Problem 801. D

A

Commandino’s Theorem ABCD tetrahedron M,N,P,Q are centroids of BDC,ADC,ABD,ABC

P, any point on circumcircle of ABC K, symmedian point of ABC X,Y,Z interction poits of PK with sides BC,AC,AB respectively

P

Prove that:

Z

AM,BN,CP,DQ coincides at a point G

QG MG NG PG 1 GA = NB = GC = QD = 3

N P

Prove that:

K

M

B

G A

3 = 1 + 1 + 1 PX PY PZ PK

Y X

C

C Q B

Problem 802.

Problem 805.

A

B

A

a b

ABCD rectangle

d

2 2 ` a j +` c j = 1 b d

c

B

φ

15°

2

D Prove that:

27°

5 -1

φ=78°

Proposed by Borislav Mirchev

C

D

Problem 806. Problem 803.

A

A

m(∠A)=45° N, center of nine-point circle K, symmedian point of ABC

45°

70°

Prove that:

Prove that:

N

b

a 3 - b 3 = 3ab 2

A,N,K collinear

K B

Leonard Giugiuc, Romania-Kadir Altintas, Turkey

30°

a

C

B

C

C

Geometri Günlüğü

Problem 807.

Problem 810.

A

A m(∠A)=45° O, circumcircle center of ABC K, symmedian point of ABC AH⊥BC

45°

m(∠A)=60° N, center of nine-point circle I, incenterof ABC G, centroid of ABC

60°

Prove that:

Prove that:

O

I

K H B

C

B

archiµedes26

C

Problem 811.

A

m(∠A)=60° B,C,D collinear DE, Euler line of ABC m(∠EDB)=φ

A 60°

B

φ 2

D Prove that:

9°

5 -1

E

C

x=2φ y=60+φ

x

φ=105°

y

φ

archiµedes26

C

B

D

Problem 812.

A

A

d

m(∠A)=60° N, center of nine-point circle I, incenter of ABC

60°

m(∠A)=60° B,C,D collinear DE, Euler line of ABC Line d is perpendicular to DE

60° φ

Prove that:

E Euler Line of ABC

I N C

Prove that:

φ=30°

A,I,N collinear Leonard Giugiuc, Romania-Kadir Altintas, Turkey

B

Prove that: Euler Line of ABC

archiµedes26

Problem 809 .

AB - AC 6

GN⊥IN

N

Leonard Giugiuc, Romania-Kadir Altintas, Turkey

Problem 808.

GN =

G

H,K,O collinear

6°

53

D

B

C

archiµedes26

54

Geometri Günlüğü

Problem 813.

Problem 816.

A

C E

AC⊥CB

G, centroid of ABC lies on incircle of ABC

Prove that 10°

Prove that:

b

c

5^a 2 + b 2 + c 2h= 6 (ab + bc + ca)

G

Sharygin Geometry Oly., 2008

40°

10°

A

DB = 2 CE

D

B

Yiu, P., Introduction to the Geometry of the Triangle

a

B

C Incircle of ABC

Problem 817. C

Problem 814.

Ω1 First Brocard Point of ABC Ω2 Second Brocard Point of ABC O, circumcircle center of ABC φ, Brocard angle

A

Prove that

Ω2

60°

m(∠A)=60° DE = EC + DB

x

Prove that:

E

Ω1

φ

A

D

OΩ1=OΩ2, x=2φ

O

B

DE, Euler line of ABC Ruben Dario

C

B

Problem 818.

C 60° m(∠C)=60°

Problem 815.

GD and FE are mid-perpendiculars C,A,G collinear

C EB = z + 1 (z: golden ratio) AD

2α

E

D

Prove that

Prove that

α

GF is tangent to incircle of ABC

α=12° 42°

A

30°

D

E

F

archiµedes26

A

B G

Sharigyn Geometry Oly., 2011

B Incircle of ABC

Geometri Günlüğü

Problem 819.

Problem 822.

C

A

EB = z (golden ratio) AD α

α

Prove that

m(∠A)=60° N, nine-point center of ABC

60°

α=12° 42°

A

30°

D

55

Prove that:

archiµedes26

E

B

N

3 (BN 2 + CN 2) = 2BC 2

C

B

Leonard Giugiuc, Romania-Kadir Altintas, Turkey

Problem 820.

Problem 823.

B

C

P, any point in Triangle ABC G, centroid of ABC G1,G2,G3 are centroids of CPB, CPA and APB respectively

ABC heptagonal triangle

= 4r B = 2r +A= r 7 ,+C 7 7 ,

P

Prove that

Ω1 First Brocard Point of ABC CP, symmedian BF, median I, incenter of ABC CP∩BF=X

X I

G2

Prove that

F

PM = 3 MG

M G

X, I, Ω1 collinear

C

G1

P

Ω1

AG1,BG2,CG3 concurrent at a point M P,M,G collinear

A

Leonard Giugiuc-Romania, Kadir Altintas-Turkey

G3 B

A

Leonard Giugiuc, Romania-Kadir Altintas, Turkey

Problem 821. A

AH⊥BC M, midpoint of BC T, tangency point T, midpoint of MH p: perimeter of ABC

Problem 824.

A CE, median Ω, Brocard Point of ABC BD, symmedian

D

Prove that E

P

CE, BD and AΩ concurrent

3.BC=p Sharygin Geometry Oly., 2009

B

M

T H

Ω

C incircle of ABC

Prove that

B

archiµedes26

C

56

Geometri Günlüğü

Problem 826.

Problem 829.

A

B

ABC heptagonal Triangle

m(∠B)=75°,m(∠C)=45°

Ω2

DEF orthic triangle of ABC DE∩CF={P}

E

= 4r B = 2r +A= r 7 ,+C 7 7 ,

Ω2, Brocard Point of ABC G, centroid of ABC DEF, peadal triangle of ABC from G

F

Prove that:

G

Prove that:

D,E,F,

D

AP is symmedian from A

F P

A

E

Leonard Giugiuc, Romania-Kadir Altintas, Turkey

archiµedes26

75°

B

C

Ω2, are con-cyclic.

45°

D

C

Problem 827. A

Problem 830.

60°

A

m(∠A)=60°

ABCD square M, any point in the plane of ABCD P,Q,R,S intersection points oc circle with MA,MB,MC,MD

S

DEF orthic triangle of ABC d; Euler line of DEF

Prove that:

P

Prove that:

E d // BC

D

B

M

Q

PQRS is harmonic PQ.RS=PS.QR

Euler Line of DEF

F

Reference. http://users.math.uoc.gr/~pamfilos/eGallery/problems/HarmonicQuad.html

R

d

C archiµedes26

B

C

D

Problem 828. A DEF orthic triangle of ABC P, midpoint of BC Q, midpoint of FE N, ninepoint circle center

E

Problem 831.

A I, incenter of ABC G, centroid of ABC b+c=2a

Prove that:

Q

Prove that:

P,N,Q collinear AQ, symmedian from A

F

c

b

Leonard Giugiuc, Romania-Kadir Altintas, Turkey

N

IG//BC

I

G Reference. Gheorghe Duca

B

D

P

C

B

a

C

Geometri Günlüğü

Problem 832.

Problem 835.

B

C H, orthocenter of ABC DEF orthic triangle of ABC H’, orthocenter of DEF M, midpoint of HH’ G’, centroid of DEF O, circumcirle center of ABC

K, symmedian point of ABC DEF, pedal triangle of ABC from K ma,mb,mc median lengths of ABC

F

mb

Proove that

D

E

Prove that

ma = mb = mc DF DE EF K

O

ma

G’

Honsberger, 1995

M H’

D

O,G’,M collinear

OG' = 2 G'M

H

mc C

57

A

E

A

Problem 833.

C

B

F

Problem 836. m(∠A)=α° m(∠B)=120°-α H, orthocenter of ABC

A

DEH orthic triangle of ABC J, orthocenter of DEF

E

ABCD harmonic quadrangle ( AB.CD=AD.BC ) P, intersection of diagonals

Prove that

H

Prove that:

D

φ

m(∠CHJ)=φ=30°+α

J

B

P

AP is symmedian line of ABD

D

archiµedes26

120°−α

α° A

B

F

Reference. "Some Properties of the Harmonic Quadrilateral" Ion Patraşcu, Florentin Smarandache

C

Problem 834.

A ABC is triangle with sides D,E are midpoints G, centroid of ABC K, symmedian point of ABC G,E,C,D con-cyclic

E

Problem 837.

A

Miguel Ochoa's van Schooten Like Theorem ABC equilateral triangle D tangency point E,F are intersection points with sides AC and BC Prove that:

F

K G

AD=DF+DE

Prove that:

E

K, lies on this circle Leonard Giugiuc-Romania and Kadir Altintas-Turkey

B

D

C

B

D

C

circle passing from A, tangent to BC at D

58

Geometri Günlüğü

Problem 838.

A

m(∠A)=60°

60°

DE, nagel line of ABC (I: incenter, G; centroid and Sp; Spieker center)

Problem 841. B 2φ

CH, altidue BD, median Sp, Spieker point of ABC

H

Prove that:

Prove that:

D

BD=EC

E I G Sp

Nagel line of ABC

H,Sp,D are collinear.

Sp

archiµedes26

φ D

C

B

A

Refrerence. Kadir Altintas (archimedes26)

C

Problem 842. Problem 839.

A

m(∠A)=60°

A 60°

D

DE, nagel line of ABC Sp:Spieker center of ABC Prove that:

D Sp

Nagel line of ABC

R

Q

P

S

Sp, is midpoint of DE

E

Extended Butterfly Theorem

T

Prove that:

1 - 1 = 1 - 1 PR RT RS QR

archiµedes26

B

C

Problem 840.

B

C

Problem 843 . A

A

m(∠A)=60°

I, Incenter of ABC T, tangency point M, midpoint of BC

O: Circumcircle center of ABC N: Nine-point center of ABC Sp: Spieker center of ABC

60°

Prove that:

Prove that:

O N

AI - TK = 1 IK KM

I

NSp=OSp

Sp

Carlos Hugo Olivera Diaz

archiµedes26

B

B

C

T

K

M

C

Geometri Günlüğü

Problem 844.

Problem 847.

A

59

ABC heptagonal Triangle

B = 2r ,+C = 47r +A= r 7 7 , AM median, AA’,BB‘,CC’ angle bisectors A'B'C' angle bisector triangle of ABC I’, incenter of A’B’C’

B

ABCD circumscribed quadrilateral O; center of Incircle Prove that:

C’ Prove that:

O B

OA.OC + OB.OD =

M A’

AB.BC.CD.DA

I’ I’, lies on median AM

Darij Grinberg

D

A C

C

B’

Kadir Altintas (archiµedes26),Turkey

Problem 848.

Problem 845.

A ABCD circumscribed quadrilateral O; center of Incircle P; intersection of diagonals AC and BD Hx,Hy,Hv,Hz orthocenters of AOB, BOC, COD,DOA

A

H I

Prove that:

O B Hx

Hv

D

ABC is triangle with sides a,b,c BD,CE cevians with intersection point J. F is any point on ED.G,H and I are feet of altidues. If J has barycentric coordinates

c E

F J

FG = FH + FI a n - 1 b n - 1 cn - 1

^ a ,b n,c n h n

Leonard Giugiuc, Romania, Kadir Altintas, Turkey

Darij Grinberg

B

G

C

Problem 849.

Problem 846 .

C

a

A

A E

ABCD harmonic quadrangle ( AB.CD=AD.BC ) P, intersection of diagonals h1,h2,h3,h4 are altidues from P

φ D

h4

Prove that:

h1

α 2α

P

B

30-α

B

Prove that:

D

Hx,Hy,Hv,Hz and P are collinear

P Hy Hz

^ a n,b n,c n h

b

h2 h3

C

φ=30°+α

archiµedes26

D

h1 .h3 = h2 .h4

Reference. "Some Properties of the Harmonic Quadrilateral" Ion Patraşcu, Florentin Smarandache

C