AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
with no parallel sides is inscribed in a circle 
. Circles 
are inscribed in triangles 
, respectively. Common external tangents are drawn between 
and 
, and 
, and 
, and and , not containing any sides of quadrilateral . A quadrilateral whose consecutive sides lie on these four lines is inscribed in a circle 
. Prove that the lines joining the centers of and , and , and the centers of and all intersect at one point.
, 
, 
, 
are integers such that 
.
and 
of 
such that for each integer 
with 
, we have![\[ n\mid a_i - b_i - c_i
\]](http://latex.artofproblemsolving.com/1/a/e/1ae8d54f068b646ac7a1e43d75e7f26510cfdc95.png)
be an odd prime and 
be integers so that the integers 
are divisible by . divides each of .
be real numbers such that 
Prove that 
Where 
be a triangle with circumcircle 
. Let 
and 
respectively denote the midpoints of the arcs 
and 
that do not contain the third vertex. Let 
denote the midpoint of arc 
(the arc 
including 
). Let 
be the incenter of . Let 
be the circle that is tangent to and internally tangent to at , and let 
be the circle that is tangent to and internally tangent to at . Show that the line 
, and the lines through the intersections of and , meet on .
friends, 
people are seated around a circular table in a fixed order. Ali places apples for serving directly in front of himself and wants to distribute them among everyone. Since Ali and his friends dislike eating alone and won't start unless everyone receives an apple at the same time, in each step, each person who has at least one apple passes one apple to the first person to their right who doesn't have an apple (in the clockwise direction). such that after some number of steps, the situation reaches a point where each person has exactly one apple.
be an arbitrary triangle and 
is the circumcenter of 
.Points 
lie on 
,respectively such that the reflection of WRT 
is tangent to circumcircle of 
.Prove that the circumcircle of triangle 
is tangent to circumcircle of triangle 
.
be an odd prime number. Let 
be the sequence defined as follows: 
, and for 
, 
is the smallest integer greater than 
such that in 
there are no arithmetic progressions of length . We say that a positive integer is a ghost if it doesn’t appear in 
.?
with 
, prove that
and 
Thus letting 
and 
be the incenters of 
and 
we get:![$[ABC]=[BI_1D]+[BI_2D]+[ABI_1]+[CBI_2]+[AI_1D]+[CI_2D]=$](http://latex.artofproblemsolving.com/7/6/e/76ef8ad40ba83c845804fca81805bdf87942cd07.png)
![$BD=\frac{[ABC]}{r'}-s=\frac{[ABC]}{r'}-\frac{[ABC]}{r}=[ABC] \cdot \left (\frac{1}{r'}-\frac{1}{r} \right) \ (\star).$](http://latex.artofproblemsolving.com/9/9/1/991a958934ce50c0cfcf5af2ed60d49f9eb2c411.png)
![$[ABC]=BD^2.$](http://latex.artofproblemsolving.com/b/f/1/bf1cb8278ad9ef8cd8f9d49ac7f1e73e486b987f.png)
Thus combining with 
gives
is a point on the side 
of 
such that the inradii of and 
are equal to 
If 
is the inradius of then we have the relation:![\[ \frac{1}{\varrho}=\frac{1}{r}+\cot \frac{A}{2} \cdot \frac{1}{AD}\]](http://latex.artofproblemsolving.com/a/c/0/ac04ea87ce75e4f2559c6600bf6af2cacb7627c1.png)
![$[ABC]=r \cdot s=\varrho \cdot (AD+s) \ (\star)$](http://latex.artofproblemsolving.com/e/c/9/ec9b2c0cbfb3fa86aac701569f3511e0fd8fb157.png)
But from the infamous inradii problem, we have 
![$\frac{s-a}{s}=\left (1-\frac{a}{s+AD} \right)^2 \Longrightarrow AD= \frac{a \sqrt{s}}{\sqrt{s}-\sqrt{s-a}}-s=\sqrt{s(s-a)}=\sqrt{[ABC] \cot \frac{A}{2}}.$](http://latex.artofproblemsolving.com/c/e/b/ceb76ca27f85fbf93ae9a0a9c386b37408fad6a2.png)
yields![$\frac{1}{\varrho}-\frac{1}{r}=\frac{AD}{[ABC]}=\frac{AD}{AD^2} \cdot \cot \frac{A}{2}=\cot \frac{A}{2} \cdot \frac{1}{AD} \Longrightarrow \frac{1}{\varrho}=\frac{1}{r}+\cot \frac{A}{2} \cdot \frac{1}{AD}.$](http://latex.artofproblemsolving.com/a/9/2/a92f9a225f4db80def0ee7ef8da1ccde707d56a0.png)
and 
. Let be the point on and let 
. Let ![$[ABD]=\Delta_1$](http://latex.artofproblemsolving.com/f/d/5/fd5f156a48d91e1bd4a6aaea34614e824ffac242.png)
and ![$[CBD]=\Delta_2$](http://latex.artofproblemsolving.com/6/0/4/6046104eb4acc506fc6f617d1158fa1d6942e92f.png)
and let 
and 
.
which is equivalent to
and so computing this gives that 
. and similarly we get that 
.
and that 
.![$[ABC]=\frac{x}{2}$](http://latex.artofproblemsolving.com/e/a/6/ea69e3976050f4eda564d5aea6c53784dabfd20f.png)
and so we have that ![$\Delta_1(1+k)=[ABC]$](http://latex.artofproblemsolving.com/b/5/1/b51c006661191f6611def2561baeda9fa505b35b.png)
and so 
.
and so 
and hence, we have that
.
and thus it is enough to prove 
(as
). And I fooled around a bit more with this, to add up to about 2 pages in total. How much, according to you, should that fetch me?
and thus it is enough to prove (as). And I fooled around a bit more with this, to add up to about 2 pages in total. How much, according to you, should that fetch me?
it shall suffice to proving 
" and then not proving is worth anything reasonable, at least not at INMO and certainly not at TST, not sure for RMO.
on 
and use pyth to get 
,that is,
.
by using formula 
and ratios of areas of triangle with same height.We get
.
Now compare this two equations for 
, we get by componendo(or whatever you call that}
And done 
![$[ABC]=rs$](http://latex.artofproblemsolving.com/5/f/e/5fe68c691b13681950a6542efe4b5ddd0bc41511.png)
where is inradii.![$\frac{1}{r'}=\frac{s+BD}{[ABC]}$](http://latex.artofproblemsolving.com/8/a/a/8aa907826a7db024cb894ced0e7934027b5324c7.png)
where 
is semi-perimeter of triangle 
is equivalent to prove that![$[ABC]=BD^2$](http://latex.artofproblemsolving.com/c/2/d/c2d2ee820fa0ad778a39215b8a0ab8908bea342b.png)
have same inradii. than again using 
to get ![$2BD^2=AB.AC=2[ABC]$](http://latex.artofproblemsolving.com/9/9/6/9963858c9f21c7ed5e1bbc67d9c932f3fb552d5e.png)
as desired.
) since i was thinking up creative ideas for the others. However, in the last 2 hours reality struck and thankfully i was able to solve enough problems to pass(hopefully)...
and . Let be the point on and let . Let and and let and . which is equivalent to and so computing this gives that . and similarly we get that . and that . and so we have that and so . and so and hence, we have that.
can be anything!), and the similarity ratio would cancel out in the end, so it doesn't matter.
be a triangle with inradius , and the point on such that inradii of 
have a common inradius 
. If is the semiperimeter of and 
, then 
, and the semiperimeters of 
and 
are 
. 
. Then 
.
, so the claim follows. onto be 
. Then 
.
.
. But this is the ratio of the lengths of the tangents from to the incircle and the excircle, so the ratio must equal the distance of from the "topmost" tangent of the incircle and , which equals 
. , , and multiplying the two relations together, along with an application of the lemma to yields Claim 2.
.
.
, wherein we have used 
and 
and 
. 
generally, and also all the other lengths that are fruitful in this solution. Claim 2 can also be generalized, and for unequal radii 
instead of , we have 
.
be a right-angle triangle with 
. Let be a point on such that the inradii of the triangles and 
are equal. If this common value is 
and if is the inradius of triangle , prove that![\[ \cfrac{1}{r'}=\cfrac{1}{r}+\cfrac{1}{BD}. \]](http://latex.artofproblemsolving.com/9/7/4/9748a154243144380b12b4445d15f43cd2525e12.png)
and so 
Also, 
Then the condition gives![$$\frac{[ABD]}{[BDC]}=\frac{s_{ABD}}{s_{BDC}} \Leftrightarrow \frac{x}{b-x}=\frac{c+x+m}{a+b-x+m}$$](http://latex.artofproblemsolving.com/0/f/3/0f3bb79878e219b14f475fceb17a71cf8fb167ca.png)
This simplifies to
Now, ![$$r'=\frac{[ABD]}{s_{ABD}}=\frac{[BDC]}{s_{BDC}}=\frac{[ABD]+[BDC]}{s_{ABD}+s_{BDC}}=\frac{ac}{a+b+c+2m}$$](http://latex.artofproblemsolving.com/a/a/d/aad32a9f8955df3dd4d267b44aa024cf5e9752b1.png)
Thus, it suffices to show
By Stewart's theorem
This is not much computation, and the end expression is really neat:
Hence, 
as desired. 
where 
denoting the area of triangle.![\[r'=\frac{\Delta_{CBD}}{s_{CBD}}=\frac{\Delta_{ABD}}{s_{ABD}}=\frac{\Delta_{CBD}+\Delta_{ABD}=\Delta_{ABC}}{s_{CBD}+s_{ABD}}\]](http://latex.artofproblemsolving.com/8/5/d/85d273bcb56d479df0a1ffe28cba802ab36352e5.png)
So we have![\[\frac{1}{r'}=\frac{s_{ABC}+BD}{\Delta_{ABC}}=\frac{1}{r}+\frac{BD}{\Delta_{ABC}}\]](http://latex.artofproblemsolving.com/d/4/8/d488d9159d8d00af427f17cd5f548cff33068d09.png)
![\[\frac{\Delta_{CBD}}{\Delta_{ABD}}=\frac{s_{CBD}}{s_{ABD}}=\frac{CD}{AD}\]](http://latex.artofproblemsolving.com/d/c/7/dc77c029d2081ed38671e92fe8e3099d9c380276.png)
Therefore,![\[\frac{s_{CBD}}{CD}=\frac{s_{ABD}}{AD}\]](http://latex.artofproblemsolving.com/9/f/c/9fc3d80c5f9639db905e9859868239043c617310.png)
![\[\frac{BC+BD}{CD}=\frac{AB+BD}{AD}\]](http://latex.artofproblemsolving.com/d/e/d/ded9fb14b17befa5953185b192989dbefd7ad131.png)
After some intense bashing we will get the result ![\[\frac{1}{r'}=\frac{1}{r}+\frac{1}{BD}\]](http://latex.artofproblemsolving.com/5/f/f/5ff665c4c41479ee4d201ae1c69bb6107aff5797.png)
and 
Prove that
Prove that 
