INTERESTING GEOMETRIC PROBLEMS Текст научной статьи по специальности «Науки о Земле и смежные экологические науки»

INTERESTING GEOMETRIC PROBLEMS

1Ibragimov A., 2Pulatov O., 3Urozova Z.

1Professor 2Assistant 3Student

1,2,3Uzbekistan-Finland Pedagogical Institute https://doi.org/10.5281/zenodo.13120774

Abstract. In this article, we present to readers some information about some interesting geometric problems and their solution methods.

Keywords: circle, diameter, triangle, center.

Issue 1.

In a circle with center O, CD is perpendicular to the diameter AB. AE- watar OC bisects the radius. Prove it; The chord DE passes through the middle of the chord BC. Proof. Let the circle be ro. COnw = F AF = BC = BD ^ AD = BF ^ zAED = zFCB ^ (M

CO the middle of and DE n BC = N) ^ MNEC - cycle ^ zMEC = zMNC zMEC = zAEC = ZABC = ZMNC => MN || OB => CN = NB

Issue 2.

ABCDE-pentagon ro is inscribed outside the circle. Prove that if BC intersects the circle ro at point K and AB=BC=CD zEKB = 90°

Proof. LA = KC = KM = CP and BK = BL = DP = DN ^ zKCP = zLAM and zKBL = ZPDN ^ KP = LM, KL = PN KM = PL and PL = KN ^ KM = KN.

Point O is the center of the circle. MO = OK = NO and MK = KN ^ zMOK = zNOK ^ The straight line KO is the bisector of zMON, and so is the straight line EO. ^ E, O, K - collinear OK 1 BC ^ EK 1 BC ^ zEKB = 90° ^A

Issue 3.

a, b, c are the sides of a triangle. If cA4=aA4+bA4, prove it; this triangle has an acute angle. Proof. We use the theorem of cosines. c2 = a2 + b2 — 2ab • cos a ^ c4 = a4 + b4 + 2a2b2 — 4a3b cos a — 4ab3 cos a + 4a2b2(cos a)2 ^ 2a2b2 + 4a2b2 cos p = 4a3b cos a + 4ab3 cos a ^ ab + 2ab cos p = 2a2 cos a + 2b2 cos a ^ ab = 2 cos a • c2 ^ cos a > 0 ^ a < 90°. a2 = b2 + c2 — 2bc cos p ^ a4 =b4 + c4 + 4b2c2(cos p)2 — 4b3c cos p — 4bc3 cos p + 2b2c2 ^ 2b4 + 4b2c2(cos p)2 + 2b2c2 — 4b3ccos p — 4bc3 cos p ^ b3 + bc2 = 2bc cos p (b + c — c cos p) 1 > cos p ^ c — c cos p > 0 ^ cos p > 0 ^ p <

90° just like that 9 < 90° ^ a, p, y e (0; n) ^ triangle with acute angles ^A

c

Issue 4. zA=90° in AABC. The center of BC is point M, point D is taken from AC such that AD=MA. The second point of intersection of circles inscribed outside AAMC and ABDC is point P. Prove that straight line CP bisects zBCA.

Proof; Let P' be the center of the arc AM. ^If P C bisects zBCA ^BCDP' is cyclic.AD=MB, AP =P M. zBCA=2a zCAM=2a and zP'AM=a ^ zDAP =3 a^zP'MB= 3a and AD=MB, APWM^AADP^AMPB^zADP^zPBM^CDPB- cyclic =>P =P=> A

C

P (P) B

Issue 5. BE and CF are altitudes in AABC. Point P is taken from inside BC. Point Q is the continuation of BC on side B. If BQ2 = BP2 = BF • AB and QFHPE=S, prove; Points A, E, F, C lie on 1 circle. Proof.

A C

BE nCF = H => AH 1 BP AH n BP = R. QPEA - we prove that it is cyclic. AQAB = ABQF = p. QPEA -To prove cyclicity CE • CA = CP • CQ it is enough to prove that. CE • CA = CR • CB ^ CP • CQ = CR • CB .

CP • (CB + BQ) = CP(CB + BP) = CR • CB & CP(CB + BP) = (CP + PR) • CB & CP • BP = PR • BC.&(BC - BP) • BP = PR • BC = (BP - BR) • BC & -BP2 = -BR • BC.

flP • AC = flF • fl^flP2 = flF • SA This is given in the condition.-cyclic.z^ÇF = x.^zPFC = z^F5 = fi + x ^ 4FSF^,F,F,S points lie on the same circle.^ ▲

Issue 6.

Let H be the point of intersection of the altitudes of acute-angled triangle ABC. The circle with its center in the middle of the side BC and passing through the point H intersects the straight line BC at the points A1 and A2. Similarly, the circle with its center in the middle of the side CA and passing through the point H intersects the straight line CA at points B1 and B2, and the circle AB with its center in the middle of the side AB and passing through the point H intersects the straight line at points C1 and C2. Prove that points A1 ,A2 ,B1 ,B2 ,C1 ,C2 lie on one circle.

Proof. The radius of the circle passing through the point H with its center in the middle of BC is R1 and its center is O1. Similarly, let O2,R2 and O3,R3.

AC = 2a = 2ô and = 2c.^flC2 = c - P3 and = a - P1 ,flC1 = c + P3 = a + Rv 0103 II 4C ^ 0103 1 fltf. fltf n 4C = fl . = c and = a.

03H = P3 and = P1 According to Carnot's theorem, the following is appropriate;

- 03H2 = - 01H2 ^ c2 - P32 = a2 - P12 ^ (c - P3) • (c + P3) = (a - P1) • (a + P1) ^ flC2 • flC1 = • ^ ^1^2C1C2 - cyclic. Likewise CA2 • CA1 = CB2 • CB1 and AC1 • AC2 = AB1 • AB2 ^ A1A2B1B2 and B1B2C1C2 cyclic.^A1A2B1B2C1C2 hexagonal cyclic A1, A2, B1, B2, C1, C2 points lie on the same circle.

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