KOMATH

Advanced Geometric Problem & Solution (Twice a week update) Now 4563 problems posted.

HOME
PROBLEM
- TRIANGLE
  - EQUILATERAL
    - AREA
    - ANGLE
    - BISECTOR
    - COLLINEARITY
    - CONCURRENCY
    - CENTER
    - CONSTRUCTION
    - CIRCLE
    - DISTANCE
    - LOCUS
    - MEDIANS
    - PERIMETER
    - PERPENDICULAR
    - SIMILAR
    - OTHERS
  - ISOSCELES
    - AREA
    - ANGLE
    - BISECTOR
    - COLLINEARITY
    - CONCURRENCY
    - CENTER
    - CONSTRUCTION
    - CIRCLE
    - DISTANCE
    - LOCUS
    - MEDIANS
    - PERIMETER
    - PERPENDICULAR
    - SIMILAR
    - OTHERS
  - RIGHT
    - AREA
    - ANGLE
    - BISECTOR
    - COLLINEARITY
    - CONCURRENCY
    - CENTER
    - CONSTRUCTION
    - CIRCLE
    - DISTANCE
    - LOCUS
    - MEDIANS
    - PERIMETER
    - PERPENDICULAR
    - SIMILAR
    - OTHERS
  - GENERAL
    - AREA
    - ANGLE
    - BISECTOR
    - COLLINEARITY
    - CONCURRENCY
    - CENTER
    - CONSTRUCTION
    - CIRCLE
    - DISTANCE
    - LOCUS
    - MEDIANS
    - PERIMETER
    - PERPENDICULAR
    - SIMILAR
    - OTHERS
- QUADRILATERALS
  - GENERAL
    - AREA
    - ANGLE
    - BISECTOR
    - CIRCLE
    - CYCLIC
    - COLLINEARITY
    - CONCURRENCY
    - CONSTRUCTION
    - DISTANCE
    - LOCUS
    - SIMILAR
    - OTHERS
  - PARALLELOGRAM
    - AREA
    - ANGLE
    - BISECTOR
    - CIRCLE
    - CYCLIC
    - COLLINEARITY
    - CONCURRENCY
    - CONSTRUCTION
    - DISTANCE
    - LOCUS
    - SIMILAR
    - OTHERS
  - RECTANGLE
    - AREA
    - ANGLE
    - BISECTOR
    - CIRCLE
    - CYCLIC
    - COLLINEARITY
    - CONCURRENCY
    - CONSTRUCTION
    - DISTANCE
    - LOCUS
    - SIMILAR
    - OTHERS
  - SQUARE
    - AREA
    - ANGLE
    - BISECTOR
    - CIRCLE
    - CYCLIC
    - COLLINEARITY
    - CONCURRENCY
    - CONSTRUCTION
    - DISTANCE
    - LOCUS
    - SIMILAR
    - OTHERS
  - TRAPEZOID
    - AREA
    - ANGLE
    - BISECTOR
    - CIRCLE
    - CYCLIC
    - COLLINEARITY
    - CONCURRENCY
    - CONSTRUCTION
    - DISTANCE
    - LOCUS
    - SIMILAR
    - OTHERS
- CIRCLES
  - AREA
  - BISECTOR
  - CHORD
  - CYCLIC
  - COLLINEARITY
  - CONCURRENCY
  - CONSTRUCTION
  - INVERSION
  - LOCUS
  - TANGENT
  - OTHERS
- POLYGON
  - HEXAGON
    - ANGLE
    - AREA
    - CONSTURCTION
    - CIRCLE
    - DISTANCE
    - PERIMETER
    - VECTOR
    - OTHERS
  - PENTAGON
    - ANGLE
    - AREA
    - CONSTURCTION
    - CIRCLE
    - DISTANCE
    - PERIMETER
    - VECTOR
    - OTHERS
  - OTHERS
    - ANGLE
    - AREA
    - CONSTURCTION
    - CIRCLE
    - DISTANCE
    - PERIMETER
    - VECTOR
    - OTHERS
- SOLID
  - BOX
    - AREA
    - VOLUME
    - OTHERS
    - LOCUS
    - DISTANCE
  - TETRAHEDRON
    - AREA
    - VOLUME
    - OTHERS
    - LOCUS
    - DISTANCE
  - GENERAL
    - AREA
    - VOLUME
    - OTHERS
    - LOCUS
    - DISTANCE
- CONSTRUCTIONS
  - CIRCLE
    - BISECT
    - PARALLEL
    - PERPENDICULAR
    - OTHERS
  - LINE
    - BISECT
    - PARALLEL
    - PERPENDICULAR
    - OTHERS
  - QUADRILATERAL
    - BISECT
    - PARALLEL
    - PERPENDICULAR
    - OTHERS
  - TRIANGLE
    - BISECT
    - PARALLEL
    - PERPENDICULAR
    - OTHERS
  - OTHERS
    - BISECT
    - PARALLEL
    - PERPENDICULAR
    - OTHERS
- TRIGONOMETRY
  - CIRCLE
    - EVALUATION
    - INEQUALITY
    - RELATION
    - OTHERS
  - QUADRILATERAL
    - EVALUATION
    - INEQUALITY
    - RELATION
    - OTHERS
  - TRIANGLE
    - EVALUATION
    - INEQUALITY
    - RELATION
    - OTHERS
  - GENERAL
    - EVALUATION
    - INEQUALITY
    - RELATION
    - OTHERS
SOLUTION
SOURCE
LINK
COPYRIGHT

Qua-2021-652, hint

.

작성자: komath

라벨: quadrilaterals, solution

No comments:

Total Pageviews

Ads

Weekly

Qua-2021-656, area, square

.
(no title)

<Cir-2024-769>, tangent <Dutch M.O 2004>
(no title)

<Tri-2024-2115>,parallel, general <Dutch M. O 1972>
(no title)

<Tri-2024-2111>, orthocener, parallel general <Mexico national O. 2017>
(no title)

<Cir-2023-754>, hint B,M,X,C are concyclic <MBC=<XMC=<DAC Similarly <XDC=<XNC=<BEC Hence triangles XMN and YAE are...
(no title)

<Tri-2025-2183>, isosceles, angle <Romania TST 2021>
(no title)

<Tri-2024-2123>, hint Let the intersection of PB with MC be R. Line R meets BC at Q Then by Ceva's theorem CP/PA x AM/MC x BQ/QC=1...
(no title)

<Tri-2023-2062>, hint
(no title)

<Cir-2024-760>, angle <Mexico national O. 2016>
(no title)

<Cir-2024-759>, general, concurrent <Mexico National O 2014>

Powered by Blogger.

Qua-2021-652, hint

No comments:

Post a Comment