[/quote]
,
,
and 
lie on sides 
is tangent to 
at 
.
intersects the circumcircle of 
again at 
,
intersects the circumcircle of 
again at 
.
,
and 
are positive integers satisfying:
on a plane, such that the angle between the two consecutive rays is 
.
is located on the first ray. The projection of 
.
,
onto the seventh ray is 
.
.
for 
and suppose 
.
,![\[ \left( a_1^k + \frac{1}{a_1^k} \right) \left( a_2^k + \frac{1}{a_2^k} \right) \dots \left( a_n^k + \frac{1}{a_n^k} \right) \ge \left( n^k + \frac{1}{n^k} \right)^n. \]](http://latex.artofproblemsolving.com/5/a/d/5ad4ea89463088b023d252d18d9f6eb411ec9f64.png)
with 
has at least two cycles in it.
be the final value obtained after repeatedly summing the digits of 
,
is 
and the digit sum of 
)
be positive real numbers such that 
.
is a nonconstant polynomial with integer coefficients with the following property:
and 
are both odd.
Y by Davi-8191, OlympusHero, samrocksnature, HWenslawski, megarnie, Adventure10, Mango247
Let 
be a triangle with 
. Points 
and 
lie on sides 
and 
, respectively, such that 
and 
. Segments 
and 
meet at 
. Determine whether or not it is possible for segments , , 
, 
, 
, 
to all have integer lengths.
be a triangle with 
. Points 
and 
lie on sides 
and 
, respectively, such that 
and 
. Segments 
and 
meet at 
. Determine whether or not it is possible for segments , , 
, 
, 
, 
to all have integer lengths.
.
.
,
.
.
,
.
and 
.
is rational, so 
is rational. We are given that 
is an integer, so 
is rational, so 
is rational. Therefore 
and 
are rational. By similar reasoning using 
,
is rational. However, if we use the sine addition/subtraction formula, we get that 
must be rational, which implies 
is rational, which is clearly false. We have a contradiction, so at least one of those lengths must be non-integral.
are relatively prime. Every right triangle has legs that can be expressed as 
and 
and a hypotenuse of 
for some 
:![\[BI^2+CI^2-2\cdot BI\cdot CI\cos{135^\circ} = BC^2 = AB^2 + AC^2,\]](http://latex.artofproblemsolving.com/4/0/9/409c4d82b56eaf7521fc709dec78ba6240f98962.png)
and we're done.
and 
are rational using more similar triangles.
,
,
,
are rational.
is rational, contradiction.
is a rational thing that I forgot times 
.
is rational; hence BC = a must be. From there, triangles ABD and ACE produce 
and 
as right triangles, with 
.
and 
as squares of rational numbers.
or 
.
and 
cannot both be rational.
.
.
now gives 
a contradiction.
,
.
.
.
bisects 
,
.
,![\[AI^2+AB^2-2AB\cdot AI\cos{45}=BI^2.\]](http://latex.artofproblemsolving.com/f/5/0/f509c93c449d42af647b6e98e8cd1fe9f2d5e413.png)
,
is irrational so 
where 
(
)
and 
are as usual length of the sides 
and thus if both 
has to be rational and so 
has to be rational and so 
has to be rational and as 
in the form 
where 
are coprime numbers with different parity .As , 
has to be a perfect square dividing the thing by 
will give us another perfect squarewriting in terms of 
we get 
is perfect square and so 
is a perfect square but as we have 
must be a non-integer. 
.![\[\angle BIC = \angle BAC+ \angle ACE + \angle ABD = 135^\circ.\]](http://latex.artofproblemsolving.com/b/b/1/bb1b240a3523caf4a2b2e892c8b9e60ceefe49c7.png)
![\[AB^2+AC^2=BC^2 = BI^2+CI^2+BI \cdot CI \cdot \sqrt{2},\]](http://latex.artofproblemsolving.com/2/c/e/2cea88d419014061133fd214cfa18ef80a1761b5.png)
![[asy]
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dot((2,2),linewidth(4pt) + dotstyle);
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[/asy]](http://latex.artofproblemsolving.com/9/a/7/9a7261659c8db0bb4186e960bdfe738b2d76dc55.png)
Law of Cosines on 
gives 
If all of these segments have integer length, the left-hand side would be irrational, while the right-hand side is rational. Therefore, it is impossible for all of these segments to have integer length.
,
.
.
and 
by AA similarity. We are then able to derive the following equalities:
Thus, 
,
is rational. Analogously, 
,
,
,![$[ABC] = \frac{bc}2$](http://latex.artofproblemsolving.com/5/d/8/5d81af12c311a4126a5272d944df341a443345e6.png)
,
we have the inradius 
is rational. However, 
,
= 
= 
= 
= 
- 
= 
,
It is well-known that 
Therefore by the Sine Law and Half Angle formula 
Therefore if 
we have the square root of an integer minus an irrational, which clearly cannot be an integer.
are integers. Let 
Then for positive integers 
with 
WLOG 
Then 
The first one yields 
but the second one gives 
a contradiction. Hence the answer is no.
