KOMATH

Advanced Geometric Problem & Solution (Twice a week update) Now 4708 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

Tri-2017-1431, bisector, right

.

작성자: komath

라벨: bisector, problem, right, triangle

No comments:

Total Pageviews

Ads

Weekly

(no title)

<Solid-2025-261 >, hint Let P on AB, Q on BC, R on,CD, S on DA such that PQRS is a rhombus. For PS//QR, PQ//SR, AP/PB=x/(1-x), ...
Trigo-2007-69, tan

.
(no title)

<Tri-2025-2225>, hint On line BC, MD extended intersect line BC at E. Draw a line , parallel to MDE, cutting line BCE at F. Triangles ...
(no title)

<Tri-2025-2270>, general, tangent <Russia Regional Olympiad 2003>
Tri-2019-1677, hint

.
(no title)

<Tri-2025-2189>, hint
(no title)

<Solid-2025-258>, hint Since MN is perpendicular to AB, CD, Triangle NBA , MDC are isosceles , and so NA=NB, MC=MD. Triangles CBD , CA...
(no title)

<Tri-2025-2217>, hint Consider Simson Line. Suppose M on arc AC and let N'. P', Q' be the projections of M onto lines AB, ...
(no title)

<Cir-2024-760>, hint Let M be the midpoint of PQ and O' S O" cuts O" at N . Quadrilaterals SMPO' and SO"QM are...
(no title)

<Tri-2025-2259>, hint <A'BC+<ABC+<ABC'=3<ABC=180 (A'.B.C' are collinear) , <ABC=60 C'B=2A'B --...

Powered by Blogger.

Tri-2017-1431, bisector, right

No comments:

Post a Comment