[/quote]
,
be reals such that 
and 
Prove that
Let 
Prove that
Prove that
be an acute-angled triangle 
with 
and incenter 
.
be the mixtilinear circle at vertex 
,
and also tangent to lines 
and 
.
passes through points 
and 
and is tangent to 
,
and 
.
be real numbers such that 
.
is the interval ![$\left[-\frac{2}{3}, \frac{2}{3}\right]$](http://latex.artofproblemsolving.com/4/f/2/4f22df2b391cf34b9b4f86634575e3e0ad23a386.png)
.
,
,
be positive integers such that 
.
grid as follows:
and a constant 
,
and 
are also friendly polynomials. Prove that 
.
for which the sequence of real numbers with 
for all 
is periodic from the beginning.
and circumscribed about the incircle 
,
.
,
,
at 
,
,
,
,
,
,
,
,
(distinct from 
)
intersects 
(distinct from 
be the midpoint of the arc 
of 
and 
at 
and 
,
at 
.
.
be the orthocenter of acute triangle 
to 
intersects line 
.
![\[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\]](http://latex.artofproblemsolving.com/6/b/0/6b010978a55642b7f7aabb2df2f7cf45c9db937a.png)
is an integer strictly between 
and 
.
.
denotes the greatest integer that is less than or equal to 
.
Y by vvluo, megarnie, Adventure10, Mango247
In 
, the sides have integers lengths and 
. Circle 
has its center at the incenter of . An excircle of is a circle in the exterior of that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to 
is internally tangent to , and the other two excircles are both externally tangent to . Find the minimum possible value of the perimeter of .
, the sides have integers lengths and 
. Circle 
has its center at the incenter of . An excircle of is a circle in the exterior of that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to 
is internally tangent to , and the other two excircles are both externally tangent to . Find the minimum possible value of the perimeter of .This post has been edited 2 times. Last edited by Vfire, Mar 14, 2019, 4:59 PM
Define 
to be the incenter, inradius, 
-
Using ![$r=\frac{[ABC]}{s}, r_A=\frac{[ABC]}{s-a}, r_B=\frac{[ABC]}{s-b},$](http://latex.artofproblemsolving.com/3/f/4/3f49d658fd8ca7381911a016e3ae9767698927f3.png)
and dividing by 
This gives us side lengths of 
or something
and 
to be the tangency points of 
and 
,
sending 
sending 
is another homothety of negative scale sending ![\[\angle BT_AI = 90^\circ - \dfrac{\angle T_BIT_A}2 = 90^\circ - \dfrac{\angle AIC}2 = 45^\circ - \dfrac{B}{4}.\]](http://latex.artofproblemsolving.com/e/5/e/e5e0482a4e36465cbc3c9fd745e95b3f29b9bf0f.png)
Now note that 
is a median of 
(where 
yields the trigonometric equation ![\[\cot\left(45^\circ - \dfrac{B}{4}\right) = 2\tan\left(90^\circ - \dfrac{B}{2}\right) = \dfrac{2}{\tan\frac{B}{2}}.\]](http://latex.artofproblemsolving.com/a/9/5/a958e90c6fc4160123d71ace3ec750a21eae8b79.png)
Let 
.![\[\cot(45^\circ - \beta) = \tan(45^\circ + \beta) = \dfrac{1+\tan \beta}{1-\tan \beta}\]](http://latex.artofproblemsolving.com/e/9/5/e957c367b0695910c6ed2c2aa85c0976737c5ab7.png)
and ![\[\dfrac{2}{\tan 2\beta} = \dfrac{1-\tan^2\beta}{\tan\beta}.\]](http://latex.artofproblemsolving.com/e/a/9/ea9fff16dfc20fd311eee639305802336d756b83.png)
Dividing out by 
on both sides (since we know it can not be zero) and solving yields 
.
,
,
,
.
be the incenter and 
be the inradius, ![\[r+2r_A=II_B-r_B.\]](http://latex.artofproblemsolving.com/7/b/d/7bd6f1574f75052389642a8eeb4ba02207a3a890.png)
Using the fact that 
and 
,![\[II_B=r+2r_A+r_B=r\left(1+\frac{2a+4b}{2b-a}+\frac{a+2b}{a}\right)=r\cdot\frac{2b\left(3a+2b\right)}{a\left(2b-a\right)}.\]](http://latex.artofproblemsolving.com/9/2/f/92f7a3b754b4671c7e252147b1f77279fa82a204.png)
But you can use your favorite computing methods to compute ![\[BI=\frac{a}{a+2b}\sqrt{b\left(a+2b\right)}\]](http://latex.artofproblemsolving.com/e/c/c/eccdf00b343f816c067fd64fd34a7bd0d6d9a7f1.png)
from which it follows that ![\[BI_B=\frac{ab}{BI}=\sqrt{b\left(a+2b\right)}\]](http://latex.artofproblemsolving.com/f/1/8/f18e464b054ebc179aa96d6646032d84961ea13a.png)
so ![\[II_B=\frac{2b}{a+2b}\sqrt{b\left(a+2b\right)}\]](http://latex.artofproblemsolving.com/f/6/1/f6110e6c8534a47f8829b1e984dd792310e3ab61.png)
while ![\[r=\frac{K}{s}=\frac{a\sqrt{\left(a+2b\right)\left(2b-a\right)}}{2\left(a+2b\right)}\]](http://latex.artofproblemsolving.com/f/b/2/fb2691b3176157d0af522f5d4451c348728b5013.png)
so ![\[\frac{2b}{a+2b}\sqrt{b\left(a+2b\right)}=\frac{a\sqrt{\left(a+2b\right)\left(2b-a\right)}}{2\left(a+2b\right)}\cdot\frac{2b\left(3a+2b\right)}{a\left(2b-a\right)}.\]](http://latex.artofproblemsolving.com/8/6/4/8646c2ba038e188c5b68fb3223a08844693bed5f.png)
After mass cancellation, we are left with ![\[\sqrt{b}=\frac{3a+2b}{2\sqrt{2b-a}}\]](http://latex.artofproblemsolving.com/a/7/8/a78dac21ab09a9f2a02bc0c07289ddcd559de8e6.png)
which simplifies to 
.
and thus 
to give a perimeter of 
as desired.
.
,
is the inradius and 
is the exradius. The hypotenuse is the radius of 
.
,
,
.
and 
.
,
is the semiperimeter of the triangle. (The first two can be derived using similar triangles).
and you get 
![[asy]
size(8cm);
defaultpen(fontsize(8pt));
pair A, B, C, I, IA, IB, IC;
A=(0, 4sqrt(5));
B=(-1, 0);
C=(1, 0);
I=incenter(A, B, C);
IA=2*circumcenter(I,B,C)-I;
IB=2*circumcenter(I,C,A)-I;
IC=2*circumcenter(I,A,B)-I;
draw(B -- A -- C); draw(IB -- IC); draw(incircle(A, B, C));
draw(foot(IB, B, C) -- foot(IC, B, C));
draw(circle(IA, length(IA-foot(IA, B, C))));
draw(arc(IB, IB-(4sqrt(5), 0), IB-(0, 4sqrt(5))));
draw(arc(IC, IC-(0, 4sqrt(5)), IC+(4sqrt(5), 0)));
draw(circle(I, 2/sqrt(5)+sqrt(5)));
dot("$A$", A, N);
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$I$", I, N);
dot("$I_A$", IA, S);
dot("$I_B$", IB, NE);
dot("$I_C$", IC, NW);
[/asy]](http://latex.artofproblemsolving.com/b/8/2/b82d92f26c512b3da74847a3e9f22e104c19d43e.png)
and 
,
.
,![$K=[ABC]$](http://latex.artofproblemsolving.com/e/8/b/e8b7f4b298b7acdef16cf10905ad88615ca235ce.png)
,
.
denotes the 
and 
denotes the semiperimeter, 
Then, if 
,
.
.
It follows that
whence 
.
,
.
,
,
.
be the three excircles, and let 
be the tangency points of 
be the center of 
.
,
passes through the insimilicenter, 
.
across 
subtends equal angles to 
and 
and 
.
,
so we find that 
.
;
or 
.
or 
.
.
in it or notice that 
are collinear. Honestly, it's pretty clean and were it not for the 15 silly mistakes I made while doing this it's fairly quick as well:![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
and let 
be the semiperimeter, and 
denote the 
-
denote the exradii. Finally, let 
denote the tangency point of 
,
,
.![$[ABC]=r_A(s-a)$](http://latex.artofproblemsolving.com/7/5/0/7502cdd4c82ba1b174eea9f7a78940955c78b457.png)
,
and 
.
.
.
.
and 
.
.
,
.
.
Applying the fact that 
We can shift both square terms to one side, use difference of squares, and cancel out an 
(since 
fails to satisfy the triangle inequality) to arrive at:
From here, expanding and grouping miraculously cancels out lots of the terms, and we are left with:
Since 
we have 
it turns out that we don't have to do any other work to deal with the integer condition. Then the perimeter is simply 
.
,
are the inradius and exradii, and 
is the 
-
be the radius of 
.
.
and 
.
.
and the area is 
,![\[ r = \frac{[ABC]}{s} = \frac{x^2\cos\phi\sin\phi}{x(1 +\cos\phi)} = \frac{x\cos\phi\sin\phi}{1 + \cos\phi}.\]](http://latex.artofproblemsolving.com/b/5/1/b517deed6c21075ae0168cb5ca838706aedb0d26.png)
Also,![\[ r_A = \frac{[ABC]}{s - 2x\cos\phi} = \frac{x\cos\phi\sin\phi}{1 - \cos\phi},\]](http://latex.artofproblemsolving.com/7/1/8/7186546ac2bc852e742e6bb15e1efc62d2ff1768.png)
and![\[ r_B = \frac{[ABC]}{s - x} = x\sin\phi.\]](http://latex.artofproblemsolving.com/4/5/9/4598625b0fdfa0a84a8b7ef84803a10408a2a8e3.png)
To compute 
where 
and 
.![\[ TR = TC + CR = TC + CK = (s - x) + (s - 2x\cos\phi) = x.\]](http://latex.artofproblemsolving.com/6/6/e/66e8907d54a16f9888bb28bbb5036e1873e10ccf.png)
Note that 
,![\[ IJ_B^2 = (J_BR - IT)^2+ (TR)^2 = (r_B -r)^2 +x^2 = \frac{x^2\sin^2\phi}{(1 +\cos\phi)^2} +x^2,\]](http://latex.artofproblemsolving.com/5/1/7/5173813846614c1183be4cf38d1c0a947ca0f6b7.png)
so 
.
to obtain![\[ \frac{x\cos\phi\sin\phi}{1 +\cos\phi} + \frac{2x\cos\phi\sin\phi}{1 -\cos\phi} + x\sin\phi = x\sqrt{\frac{2}{1 + \cos\phi}},\]](http://latex.artofproblemsolving.com/6/d/4/6d40dbe4bb41d25ae4a7c354d210fe54d4d92974.png)
![\[ \cos\phi\sin\phi\left(\frac{3 + \cos\phi}{1 - \cos^2\phi}\right) + \sin\phi = \sqrt{\frac{2}{1 + \cos\phi}},\]](http://latex.artofproblemsolving.com/3/1/d/31d35aef7838129f44f8e96fd65bd88aaf7d88d3.png)
![\[ \frac{3\cos\phi + 1}{\sin\phi} = \sqrt{\frac{2}{1 +\cos\phi}}\implies (3\cos\phi +1)^2(1 + \cos\phi) = 2(1 -\cos^2\phi).\]](http://latex.artofproblemsolving.com/3/7/7/3777ccaef1e94a9636195efb0b0a17aa9f63adcd.png)
Obviously 
doesn't work, so divide both sides by 
and rearrange to get 
.
.
,
.
,
.
.
plugging in the answer doesn't even make sense lmao
where 
is the excircle radius and 
.
to 
using standard excircle and incircle tangency distance lemmas. Rearranging, 
. From here we use that 
,
,
which makes it into 
Using 
,
is apparent.
