[/quote]
,
Find the maximum positive integer 
such that for every subset 
with n elements, there always exist two elements a, b in T such that:
and 
Prove that
and 
Prove that
,
,
,
are integers such that 
.
and 
of 
such that for each integer 
with 
,![\[ n\mid a_i - b_i - c_i
\]](http://latex.artofproblemsolving.com/1/a/e/1ae8d54f068b646ac7a1e43d75e7f26510cfdc95.png)
be an odd prime and 
be integers so that the integers 
are divisible by 
Prove that 
be real numbers such that 
Prove that 
Where 
be a triangle with circumcircle 
.
and 
respectively denote the midpoints of the arcs 
and 
that do not contain the third vertex. Let 
denote the midpoint of arc 
(the arc 
including 
)
be the incenter of 
be the circle that is tangent to 
be the circle that is tangent to 
,
friends, 
people are seated around a circular table in a fixed order. Ali places 
is the circumcenter of 
.
lie on 
,
is tangent to circumcircle of 
.
is tangent to circumcircle of triangle 
.
Y by Adventure10
One base of a trapezoid is 100 units longer than the other base. The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio 
Let 
be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trapezoid into two regions of equal area. Find the greatest integer that does not exceed 
Let 
be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trapezoid into two regions of equal area. Find the greatest integer that does not exceed 
,
.
.
,
.
as the height of 
as the height of the trapezoid with 
as base, and the length of the parallel segment as 
.
.
and 
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,
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,
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,
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and 
.
,
and 
on 
,
.![$\frac{[ABFE]}{[CDFE]}=\frac{AB+25}{AB+75}=\frac23$](http://latex.artofproblemsolving.com/0/1/b/01b532d4b4a9f998197e484bbcf063cdb89de4eb.png)
,
and 
.
,
.
.
,
.
,
.
and 
be the segment that divides the trapezoid into equal areas. 
is on 
,
.![$[EAB]: [ABQP]: [QPCD]=9: 20: 20$](http://latex.artofproblemsolving.com/7/a/8/7a8b057a72537b62a302223f72aa4e1a54901043.png)
,
.
,
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.
?
?
and 
be the lengths of the bases. We have that the midsegment has length 
,![\[\frac 23 = \frac{a+a+50}{a+100+a+50}\implies a = 75.\]](http://latex.artofproblemsolving.com/4/d/2/4d26a10dc1bc3478e24e4ec82ac0888dc792ae1c.png)
Now, it is well-known that the area of the segment dividing the trapezoid into equal segments has length 
(proof: intersect the legs at a point and use similar triangles) so the answer is![\[\left\lfloor\frac{75^2+175^2}{200}\right\rfloor = \left\lfloor\frac{15^2+35^2}{8}\right\rfloor = \left\lfloor\frac{225+1225}{8}\right\rfloor = \left\lfloor\frac{1450}{8}\right\rfloor=181.\]](http://latex.artofproblemsolving.com/4/c/d/4cd2fd04322c33d30802569b2a1c291a2e574034.png)
,
and 
on 
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