Interesting combinatoric problem on rectangles
by jaydenkaka, Apr 23, 2025, 2:22 PM
Define act <Castle> as following:
For rectangle with dimensions i * j, doing <Castle> means to change its dimensions to (i+p) * (j+q) where p,q is a natural number smaller than 3.
Define 1*1 rectangle as "C0" rectangle, and define "Cn" ("n" is a natural number) as a rectangle that can be created with "n" <Castle>s.
Plus, there is a constraint for "Cn" rectangle. The constraint is that "Cn" rectangle's area must be bigger than n^2 and be same or smaller than (n+1)^2. (n^2 < Area =< (n+1)^2)
Let all "C20" rectangle's area's sum be A, and let all "C20" rectangles perimeter's sum be B.
What is A-B?
For rectangle with dimensions i * j, doing <Castle> means to change its dimensions to (i+p) * (j+q) where p,q is a natural number smaller than 3.
Define 1*1 rectangle as "C0" rectangle, and define "Cn" ("n" is a natural number) as a rectangle that can be created with "n" <Castle>s.
Plus, there is a constraint for "Cn" rectangle. The constraint is that "Cn" rectangle's area must be bigger than n^2 and be same or smaller than (n+1)^2. (n^2 < Area =< (n+1)^2)
Let all "C20" rectangle's area's sum be A, and let all "C20" rectangles perimeter's sum be B.
What is A-B?
This post has been edited 1 time. Last edited by jaydenkaka, an hour ago
(help urgent) Classic Geo Problem / Angle Chasing?
by orangesyrup, Apr 23, 2025, 1:51 PM
In the given figure, ABC is an isosceles triangle with AB = AC and ∠BAC = 78°. Point D is chosen inside the triangle such that AD=DC. Find the measure of angle X (∠BDC).
ps: see the attachment for figure
ps: see the attachment for figure
Collect ...
by luutrongphuc, Apr 21, 2025, 12:59 PM
Rectangular line segments in russia
by egxa, Apr 18, 2025, 10:00 AM
Several line segments parallel to the sides of a rectangular sheet of paper were drawn on it. These segments divided the sheet into several rectangles, inside of which there are no drawn lines. Petya wants to draw one diagonal in each of the rectangles, dividing it into two triangles, and color each triangle either black or white. Is it always possible to do this in such a way that no two triangles of the same color share a segment of their boundary?
A game optimization on a graph
by Assassino9931, Apr 8, 2025, 1:59 PM
Let
be
given points in the plane, and let
be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points
, i.e., she connects some of the points with edges so that from any point you can reach any other point by moving along the edges.Then, Alice assigns to each vertex
a non-negative real number
, for
, such that
. Bob then selects a sequence of distinct vertices
such that
and
are connected by an edge for every
. (Note that the length
is not fixed and the first selected vertex always has to be
.) Bob wins if
otherwise, Alice wins. Depending on
, determine the largest possible value of
for which Bobby has a winning strategy.














![\[
\frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r;
\]](http://latex.artofproblemsolving.com/1/7/9/1795db8e4a509dd465f6ff462093ae75b04de2b4.png)


hard problem
by Cobedangiu, Apr 2, 2025, 6:11 PM
A Hard Geometry
by utkarshgupta, Apr 16, 2016, 5:54 AM
It took me so long to solve. Only if I knew a few projective theorems.
Problem : (RMM 2013 Problem 3)
Let
be a quadrilateral inscribed in a circle
. The lines
and
meet at
, the lines
and
meet at
, and the diagonals
and
meet at
. Let
be the midpoint of the segment
, and let
be the common point of the segment
and the circle
. Prove that the circumcircle of the triangle
and
are tangent to one another.
Solution :
Let
be the center of
.
Let
and
.
Let
be the reflection of
in 
Lemma :
Proof :
Well known
Obviously,
is a straight line (radical axis are concurrent).
Let
Then, since
is the polar od
, 

Thus
and
are straight lines..
We know that
and
are cyclic quadrilaterals.

Hence
It is well known that
Since
lies on the polar of
,

Also since
is a straight line;
is the midpoint of
and 
Since
is the reflection of
about
,
is the reflection of
about 
Hence
That is
Using the above results,
is a cyclic pentagon.
Now invert with radius
and centre
.
Obviously
The cyclic pentagon implies,
and 
Thus one of
maps to
.
Since
are tangent to
;
and
are tangent to
.
QED.
Problem : (RMM 2013 Problem 3)
Let


















Solution :
Let


Let


Let



Lemma :

Proof :
Well known
Obviously,

Let

Then, since




Thus


We know that



Hence

It is well known that

Since



Also since




Since






Hence

That is

Using the above results,

Now invert with radius


Obviously

The cyclic pentagon implies,


Thus one of


Since





QED.
Problem 1
by SpectralS, Jul 10, 2012, 5:24 PM
Given triangle
the point
is the centre of the excircle opposite the vertex
This excircle is tangent to the side
at
, and to the lines
and
at
and
, respectively. The lines
and
meet at
, and the lines
and
meet at
Let
be the point of intersection of the lines
and
, and let
be the point of intersection of the lines
and
Prove that
is the midpoint of 
(The excircle of
opposite the vertex
is the circle that is tangent to the line segment
, to the ray
beyond
, and to the ray
beyond
.)
Proposed by Evangelos Psychas, Greece























(The excircle of







Proposed by Evangelos Psychas, Greece
Factor of P(x)
by Brut3Forc3, Apr 4, 2010, 2:45 AM
If
, and
are all polynomials such that
prove that
is a factor of
.


![\[ P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x),\]](http://latex.artofproblemsolving.com/1/6/a/16a71abc110ece558d427a07d7be2d1f72148024.png)


Composite sum
by rohitsingh0812, Jun 3, 2006, 5:39 AM
Let
,
,
,
,
,
be positive integers and let
.
Suppose that the number
divides
and
. Prove that
is composite.







Suppose that the number




Stay insane,Coz it's your will, labour and pain,which takes you to the top of the mountain.
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