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.
be an acute triangle with 
, and let 
be the projection from 
onto 
. Let 
be a point on the extension of 
past such that 
. Let 
be on the perpendicular bisector of 
such that and 
are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across 
and 
be 
and 
, respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and 
, respectively. Let 
and 
intersect 
and 
at 
and 
. Let 
intersect at 
. Prove that 
.
with incircle 
. Let 
be the point on such that the circumcircle of 
is tangent to and let the -excircle be tangent to at 
. Prove that the tangent from to and the tangent from to (distinct from ) meet on the line parallel to and passing through .
by 
, 
for every positive integer 
. Prove that 
leave distinct remainders when divided by 
.
of distinct positive integers such that there exists a positive integer 
for which the number of divisors of 
and of 
are equal.
and 
be two sequences of integers such that 
and 
for all 
. Suppose that 
divides 
for an infinity of integers . Prove that there exist an integer 
such that 
for all
are coprime. Consider the sequence 
given by 
. Suppose that there exist positive integers 
such that 
is even and 
is an integer. Prove that the numerator in its simplest form of 
is an odd integer greater than 
.
be an integer and let 
be distinct subsets of 
. Show that there exists 
such that the n subsets 
are also disjoint.
satisfies 
. If 
, find the maximum possible value of 
.
such that for all irrational numbers 
and 
,
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 
.
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
.
s
where 
