AoPS Academy
be positive real numbers. Prove the inequality![\[
\frac{x_1 + 3x_2}{x_2 + x_3}
\;+\;
\frac{x_2 + 3x_3}{x_3 + x_4}
\;+\;
\frac{x_3 + 3x_4}{x_4 + x_1}
\;+\;
\frac{x_4 + 3x_1}{x_1 + x_2}
\;\ge\;8.
\]](http://latex.artofproblemsolving.com/5/c/0/5c062a37f6ff120232ae87eb0093a4ce73ee0eb1.png)
be a function satisfying the following conditions for all 
:![\[
\begin{cases}
f(n+1) > f(n) \\
f(f(n)) = 2n + 2
\end{cases}
\]](http://latex.artofproblemsolving.com/0/4/7/047e28c9d899d7de443ae40ac2c756fda851fafb.png)
Find the value of 
.
is the foot of the altitude from 
of triangle 
. On the lines 
and 
points 
and 
are marked such that the circumcircles of triangles 
and 
are tangent, call this circles 
and 
respectively. Tangent lines to circles and at and intersect at 
.
.
be an acute, scalene triangle, and let 
, 
, and 
be the midpoints of 
, 
, and 
, respectively. Let the perpendicular bisectors of and 
intersect ray 
in points 
and 
respectively, and let lines 
and 
intersect in point 
, inside of triangle . Prove that points 
, , , and all lie on one circle.
be a parallelogram with 
. A circle tangent to sides 
, 
, and 
intersects diagonal 
at points 
and 
with 
, as shown. Suppose that 
, 
, and 
. Then the area of can be expressed in the form 
, where 
and 
are positive integers, and is not divisible by the square of any prime. Find 
.
and 
intersect at points and . Line 
is tangent to and at and 
, respectively, with line closer to point than to . Circle 
passes through and intersecting again at 
and intersecting again at 
. The three points 
, , 
are collinear, 
, 
, and 
. Find 
.
, we know that lines and 
intersect at 
, lines and 
intersect at 
, and lines 
and 
intersect at 
. Suppose that the circumcircle of 
intersects line 
at and , and the circumcircle of 
intersects line at and , where 
and 
are collinear in that order. Prove that if lines 
and 
intersect at 
, then 
.
and 
are written on a blackboard, where 
and 
are relatively prime positive integers. At any point, Evan may pick two of the numbers 
and 
written on the board and write either their arithmetic mean 
or their harmonic mean 
on the board as well. Find all pairs 
such that Evan can write 1 on the board in finitely many steps.![\[ \dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots \]](http://latex.artofproblemsolving.com/a/c/c/acc097f524dc9fe9bd85e8eb659ea4a75aec475d.png)
in the plane with the following property: For any three distinct integers 
and 
, points 
, 
, and 
are collinear if and only if 
.
, is a point on such that 
is the internal bisector of 
. Now Suppose 
=
and 
. Prove that 
.
. then 
and 
, and 
, we have , 
[symbols have their usual meanings]
, so , 
. so , it leads to 
. so , 
and so , 
in copy and missed 
while typing.
. i award myself the 1st position in foolishness competition 2012 
, but 
; keeping in mind that 
. This one, combined with 
leads, indeed, to , a.s.o.
is the internal angle BISECTOR THEOREM.
and 
we have, 
and 
and then plugging in the given criteria we have 
and thus 
(as they are angles of a triangle). Hence, 
and 
, which implies that 
. Now, by sine rule on 
we have 
and 
. Now again by sine rule on , we have 
(
, since 
).
.
.
be the incenter of 
. Observe that 
is isosceles with 
since![\[\angle BDI = \angle DAC + \angle DCA = \angle IAB + \angle IBA = \angle BID.\]](http://latex.artofproblemsolving.com/3/c/c/3cc8e6c99cb5d794f1c9434ac15faa4ec245d790.png)
Call 
. From here, we have two finishes.
since 
. By SAS criteria, we have that 
. Now observe![\[\angle DEC = \angle BEA = \angle EBD + \angle ECB = 2\angle EBC = \angle ABC\]](http://latex.artofproblemsolving.com/1/e/1/1e1344a169a3435a148a03e4330f3c1f37f2291e.png)
which means 
is cyclic. This finally means 
implying 
is isosceles with 
. If you call 
, then 
. Finally, by angle sum in , we have![\[\theta + 2\theta + 2\theta = 180^\circ \iff \angle A = 2\theta = 72^\circ.\]](http://latex.artofproblemsolving.com/f/2/f/f2f947720aa130b9cf6488fdcce2affc3df1283a.png)
and now we're done. 
![\[\frac{AB}{AC} = \frac{BD}{DC} = \frac{BI}{BA}.\]](http://latex.artofproblemsolving.com/b/2/3/b2300387451e179bed6ecf4569b6835799652849.png)
Let 
, 
, then applying Sine Rule in and 
in light of above equation, we get![\[\frac{\sin(x)}{\sin(2x)} = \frac{\sin(y)}{\sin(x+y)}.\]](http://latex.artofproblemsolving.com/e/c/a/eca0546dc1c4bfc14bd3006e20276ea41f1a8bd8.png)
Expand with double angle and compound angle formulae to get
Since 
are angle of triangle, we can conclude that 
. An analogous calculation as above would give 
. This finishes the solution. 
from 
Similarly, 
from 
in the above equation we get, 
So, 
.
.![\[\sum ab +\sum ab^2 \le 6\]](http://latex.artofproblemsolving.com/9/f/6/9f632ab1e6df5029bcd4416eb06623123ad2cd8c.png)
is the angle-bisector of 
.
.
.
.
.