ILM-FAN VA INNOVATSIYA
ILMIY-AMALIY KONFERENSIYASI
in-academy.uz/index.php/si
28
SOME NON-STANDART PROBLEMS OF ANALYTIC GEOMETRY
Ahmatov Rustam Axmatovich
student of the Shakhrisabz State Pedagogical Institute, rustamaxmatov708@gmail.com
Xasanova Sitora Laziz qizi
student of the Shakhrisabz State Pedagogical Institute, sitora2204A@gmail.com
https://doi.org/10.5281/zenodo.12204507
Annotation
In this article, the solutions of some problems that belong to the analytical geometry part of
geometry, are not included in the school geometry course and the higher geometry course,
and are considered non-standard. That is, the solution to the problematic questions related to
rectangles with a relatively high level of complexity was found analytically. The considered
issues serve to develop the geometric worldview of students who want to master analytical
geometry in depth.
Keywords.
non-standart problems, analytic geometry, rectangles, bissektor, parallelogram,
The quadrilateral, fashion similarly, rhombus.
Introduction.
In this work, the problems related to the internal properties of a square, a
square and a parallelogram were solved. The goal of finding a
solution to the problems presented in the article in an
analytical way is set, and for this, the task of using the
necessary fundamental properties is assigned. The solved
problems are completely non-standard and are relevant for
the development of geometric imagination for students of
mathematics in higher education.
Difenation.
A triangle is a simple closed curve or polygon
which is created by three line-segments. In geometry, any
three points, specifically non-collinear,form a unique triangle
and separately, a unique plane [1:179].
The SAS Similarity Theorem.
Given a correspondence be-tween two triangles. If two pairs of
corresponding sides are proportional, and the included angles are congruent, then the
correspondence is a similarity[1:189].
The ASA Similarity Theorem.
If two angles and the included side of one triangle are
congruent to two angles and the included side of a second triangle, then the two triangles are
congruent[5:220].
Theorem-1
.
If a square is drawn externally on each side of a parallelogram, prove that:
(a)
The quadrilateral, determined by the centers of these squares, is itself a square;
(b)
The diagonals of the newly formed square are concurrent with the diagonals of the
original parallelogram.
Proof:
(a)
ABCD
is parallelogram. Points
,
,
P Q R
and
S
are the centres of the four squares,
,
,
DAIJ DCLK
and
,
CBFE
respectively (fig. S-1).
PA
DR
and
AQ
QD
(each is one-half a
diagonal). Also,
ADC
is supplementary to
DAB
, and
IAH
is supplementary to
DAB
(since
AID
HAB
right angle). Therefore,
ADC
IAH
.
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Since,
0
45 ,
m RDC
m QDA
m HAP
m QAI
RDQ
QAP
. Thus,
RDQ
PAQ
(s.a.s), and
.
QR
QP
In a fashion similarly, it may be proved that
QP
PS
and
PS
RS
.
Therefore,
PQRS
is a rhombus.
Since,
RDQ
PAQ
,
;
DQR
AQP
therefore,
PQR
DQA
(by addition). Since,
DQA
- right angle, (S-1)
and
PQRS
a is square.
(b)
To prove that the diagonals of
PQRS
are concurrent with the diagonals of
Parallelograms
ABCD
, we must prove that a diagonal of the square and a diagonal of the
parallelogram bisect each other. In other words, we prove that the diagonals of the square and
the diagonals of the parallelogram all share the same midpoint, (i.e., point
O
).
BAC
ACD
, and
0
45 ;
m PAB
m RCD
therefor,
PAC
RCA
. Since,
AOP
COR
and
,
AP
CR
AOP
COR
(s.a.a).
Thus,
,
AO
CO
and
PO
RO
. Since,
DB
passes through the midpoint of
AC
, and similarly
QS
passes through the midpoint of
PR
, and since
AC
and
PR
share the same midpoint (i.e.,
O
), we have shown that
,
,
,
AC PR DB
and
QS
are concurrent (i.e., all pass through point
O
).
Theorem-3
.
In right
△ 𝐴𝐵𝐶
, with right angle at
𝐶, 𝐵𝐷 = 𝐵𝐶, 𝐴𝐸 = 𝐴𝐶,
𝐸𝐹
̅̅̅̅ ⊥ 𝐵𝐶
̅̅̅̅,
and
𝐷𝐺
̅̅̅̅ ⊥ 𝐴𝐶
̅̅̅̅
.
Prove
that
𝐷𝐸 = 𝐸𝐹 + 𝐷𝐺.
(S-2)
Proof.
Draw
𝐶𝑃
̅̅̅̅ ⊥ 𝐴𝐵
̅̅̅̅
, also draw
𝐶𝐸
̅̅̅̅
and
𝐶𝐷
̅̅̅̅
(Fig. S-4).
𝑚∠3 + 𝑚∠1 + 𝑚∠2 = 90
°
𝑚∠3 + 𝑚∠1 = 𝑚∠4 (#5)
By
substitution,
𝑚∠4 + 𝑚∠2 = 90
°
but in right
△ 𝐶𝑃𝐸
,
𝑚∠4 + 𝑚∠1 = 90
°
.
Thus,
∠1 ≅ ∠2
(both are complementary to
∠4
), and right
△ 𝐶𝑃𝐸 ≅ 𝑟𝑖𝑔ℎ𝑡 △ 𝐶𝐹𝐸
, and
𝑃𝐸 = 𝐸𝐹
.
Similarly,
𝑚∠9 + 𝑚∠7 + 𝑚∠6 = 90
°
𝑚∠9 + 𝑚∠7 = 𝑚∠5
.
By substitution,
𝑚∠5 + 𝑚∠6 = 90
°
.
However, in right
△ 𝐶𝑃𝐷
,
𝑚∠5 + 𝑚∠7 = 90
°
Thus,
∠6 ≅ ∠7
(both are complementary to
∠
5), and right
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30
△ 𝐶𝑃𝐷 ≅ 𝑟𝑖𝑔ℎ𝑡 △ 𝐶𝐺𝐷
and
𝐷𝑃 = 𝐷𝐺.
Since
𝐷𝐸 = 𝐷𝑃 + 𝑃𝐸,
we get
𝐷𝐸 = 𝐷𝐺 + 𝐸𝐹.
Theorem-4
.
Prove that the sum of the measures of the perpendiculars from any point on a side of a rectangle
to the diagonals is constant.
(S-3)
Proof.
Let
P
be any point on side
AB
̅̅̅̅
of rectangle
ABCD
(Fig. S-5).
PG
̅̅̅̅
and
PF
̅̅̅̅
are perpendiculars to the diagonals. Draw
𝐴𝐽
̅̅̅
perpendicular to
𝐷𝐵
̅̅̅̅,
and then
𝑃𝐻
̅̅̅̅
perpendicular to
𝐴𝐽
̅̅̅
. Since
𝑃𝐻𝐽𝐹
is a rectangle (a quadrilateral with three right angles),we get
𝑃𝐹 = 𝐻𝐽.
Since
𝑃𝐻
̅̅̅̅̅
and
𝐵𝐷
̅̅̅̅
are both perpendicular to
𝐴𝐽
̅̅̅, 𝑃𝐻
̅̅̅̅
is parallel to
𝐵𝐷
̅̅̅̅
. Thus,
∠𝐴𝑃𝐻 ≅
∠𝐴𝐵𝐷
. Since,
𝐴𝐸 = 𝐸𝐵
,
∠𝐶𝐴𝐵 ≅ ∠𝐴𝐵𝐷
. Thus, by transitivity,
∠𝐸𝐴𝑃 ≅ ∠𝐴𝑃𝐻;
also in
△ 𝐴𝑃𝐾,
𝐴𝐾 = 𝑃𝐾
. Since
∠𝐴𝐾𝐻 ≅ ∠𝑃𝐾𝐺 ,
right
△ 𝐴𝐻𝐾 ≅ 𝑟𝑖𝑔ℎ𝑡 △ 𝑃𝐺𝐾
(S.A.A.). Hence,
𝐴𝐻 = 𝑃𝐺
and, by addition,
𝑃𝐹 + 𝑃𝐺 = 𝐻𝐽 + 𝐴𝐻 = 𝐴𝐽
, a constant.
References:
1.
Edwin E. Moise “Elementary Geometry from an Advanced Standpoint” ., Emeritus, Quens
College of the City Univertsity of New York, 2014;
2.
C.A Hart “Plane and Solid Geometry”., Published by Forgotten books, 2013;
3.
Sultonov, Sherzod. "KOMPAKT TO ‘PLAMDA QUVISH MASALASI."
Академические
исследования в современной науке
2.18 (2023): 9-10;
4.
H.S.Hall “Text-Book of Euclid’s elements”., Cambridge Syndicate on Geometry, 2010;
5.
Sultonov, Sherzod, Abdurasul Amirov, and Ibrohimxon Umarov. "MATEMATIKANI
QANDAY O’RGATISH BO’YICHA 15 TA STRATEGIYA."
Conference Proceedings: Fostering Your
Research Spirit
. 2024.
