[/quote]
,
with right angle 
,
is drawn. An arbitrary circle passing through points 
meets the segments 
,
and 
for the second time at points 
,
and 
respectively. Segments 
and 
meet at point 
.
or the sum of areas of triangles 
and 
?
and 
Prove that
be reals such that 
and 
Prove that
is divisibly by 
.
and 
be two infinite sequences of integers. Prove that there exists an infinite sequence of integers 
such that for any positive integer 
,![\[
\sum_{k|n} k \cdot z_k^{\frac{n}{k}} = \left( \sum_{k|n} k \cdot x_k^{\frac{n}{k}} \right) \cdot \left( \sum_{k|n} k \cdot y_k^{\frac{n}{k}} \right).
\]](http://latex.artofproblemsolving.com/2/5/9/259ee32c89befe7a3ee2797e0d5b1df1c3066cf6.png)
such that for all reals 
Proposed by Aritra12, India
and 
imply that either 
or 
.
implies that 
.
also implies that 
implies that for any 
,
or 
.
,
is injective. 
and 
imply that 
for all 
.
.
such that 
,
gives 
for all such 
.
then gives 
and thus 
.
which means that 
.
for all 
for some 
,
which is impossible. Thus 
in this case.
let 
be the number whose 
-
-
.
.
and 
.
Prove 
Find maximum 
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 
is the midpoint of 
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 
.
,
is the incircle tangency point and 
is the excircle tangency point, both on 
,
.
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 
denotes the tangency point between the 
,
.
.
It follows that
whence 
.
,
.
,
,
.
.
be the three excircles, and let 
be the tangency points of 
be the incenter 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 
,
,
.![$[ABC]=r_A(s-a)$](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 
-
.
.
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 
.
be the intersection of the B-excircle 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 
we get that the distance is 
using standard excircle and incircle tangency distance lemmas. Rearranging, 
. From here we use that 
,
,
which makes it into 
Using 
,
is apparent.
