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and 
Prove that
Let and 
Prove that
Let and 
Prove that
of positive integers such that ![\[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]](http://latex.artofproblemsolving.com/3/5/3/353bbb15bac205053c26208af8cac7c32a296f55.png)
Proposed by Gabriel Chicas Reyes, El Salvador
, where O is the circumcenter and I is the incenter of triangle ABC. In other words, 
. Since 
, this becomes 
. Thus we have
.
. It worked, but definitely not as nicely.
. It worked, but definitely not as nicely.
. It worked, but definitely not as nicely.
That's also what I did in the contest... Though it is not nice at all, it took me not much time to solve this problem.
, 
and 
. Let the incenter of the triangle be 
and let the point at which the incircle is tangent to 
be 
. Now note that![\[a=2R\cdot \sin{A} = 2R \cdot \sin{\frac{A}{2}} \cdot \cos{\frac{A}{2}}\]](http://latex.artofproblemsolving.com/d/4/a/d4ace78944ea3ec255b06277a4651857f5dd47bd.png)
,![\[AD=r\cdot \frac{\cos{\frac{A}{2}}}{\sin{\frac{A}{2}}}.\]](http://latex.artofproblemsolving.com/4/0/8/408741e1b5feef9662c66de2d07834f57db55d47.png)
yields that![\[\frac{b+c-a}{a}=\frac{2AD}{a}=\frac{r}{R\cdot \sin^2{\frac{A}{2}}}.\]](http://latex.artofproblemsolving.com/5/a/e/5ae2b0a6e75022f8d355af5a3c7a9a3d13448535.png)
, it follows that![\[\frac{MN^2}{a^2}=\frac{(b-a)^2 + (c-a)^2 -2(c-a)(b-a) \cos{A}}{a^2}\]](http://latex.artofproblemsolving.com/9/b/6/9b6a23683ec87c98f0ec9e76557815d6a8617162.png)
![\[\frac{MN^2}{a^2}=\frac{(b^2 + c^2 - 2bc\cos{A})-a(1-\cos{A})(b+c-a)}{a^2}\]](http://latex.artofproblemsolving.com/0/c/d/0cd57cbdbde37636c004aa22bd89378964a7b57d.png)
![\[\frac{MN^2}{a^2}=1-\frac{(1-\cos{A})(b+c-a)}{a}\]](http://latex.artofproblemsolving.com/a/e/8/ae88a71de241504906fe6f2da4cbde769ee875e5.png)
![\[\frac{MN^2}{a^2}=1-\frac{2\sin^2{\frac{A}{2}}(b+c-a)}{a}\]](http://latex.artofproblemsolving.com/0/4/0/0406eb1b23b797ccd19e33d80e2127fff88386cf.png)
![\[\frac{MN^2}{a^2}=1-\frac{2r}{R}.\]](http://latex.artofproblemsolving.com/b/4/0/b4079b520784c3247f5ad7061072485e0f902237.png)
![\[\frac{MN}{BC}=\sqrt{1-\frac{2r}{R}}.\]](http://latex.artofproblemsolving.com/3/1/6/316d2ed9d6e27548867f188108db869b41e35359.png)
of 
be 
.
with rotation angle 
carries 
and 
.
, 
becomes equal to 
. Therefore the ratio becomes equal to 
. in terms of 
and 
. Then apply cosine rule on triangles 
and 
to get the value of in terms of . Now just expand both the sides and they become equal.
. By the problem conditions, 
and 
. Then, 
, and thus
From here, it is clear that
Because ![$$[ABC]=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}=\frac{rs}{2},$$](http://latex.artofproblemsolving.com/f/0/e/f0e4f05d668a0fa0e18f467ed307c9146a17e749.png)
it follows that 
Similarly, because ![$$[ABC]=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}=\frac{abc}{4R},$$](http://latex.artofproblemsolving.com/5/5/a/55a14a94973ae78d1426a294b7be98b409bcfe8e.png)
we can rewrite as 
As a result, 
and 
.
and 
are on sides 
,
.
denote the circumradius and the inradius of the triangle 
in terms of 
at 
,
at 
,
at 
.
,
,
,
is 
![$\frac{1}{abc} \cdot [3abc-(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)+a^3+b^3+c^3]$](http://latex.artofproblemsolving.com/0/7/2/07209efe2e89337fda28d19fedcb93190d7081d0.png)
.
and 
for ease of notation, we have that:
.
and 
,
,
Notice 
and 
so 
Hence, 
and 
Notice ![$$\sqrt{s(s-a)(s-b)(s-c)}=[ABC]=\frac{abc}{4R}=sr$$](http://latex.artofproblemsolving.com/3/4/c/34c77bbea6fd7a3eace9eb63ebe4fb3ed8184b20.png)
so 
Therefore, 