[/quote]
,
by 
and the number of distinct prime divisors of 
.
be a positive integer. Prove that there exist infinitely many positive integers 
such that 
and 
does not divide 
for any positive integers 
satisfying 
.
such that for any 
:
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
such that 
for all 
and let 
be the 
and 
be the intersection points of 
and 
be the projections of 
intersects the tangent line at 
at a point 
.
.
and 
: Prove that :
board and made 
moves so that it visited all squares and returned to the starting square. At every moment, the distance from the center of the square the king was on to the center of the board was calculated. A move is called 
if this distance becomes smaller after the move. Find the maximum possible number of pleasant moves. (The chess king moves to a square adjacent either by side or by corner.)
,
,
,
such that 
and 
for 
.
,
be an odd prime. Prove that:![\[\displaystyle\sum_{k=1}^{p-1}k^{2p-1} \equiv \frac{p(p+1)}{2} \pmod{p^2}\]](http://latex.artofproblemsolving.com/a/5/5/a55e75d4d9769ceec3a3ce359aac676c7c7875fd.png)
,
,
,
'![\[n=a_1+a_2+\dots+a_m,\]](http://latex.artofproblemsolving.com/6/c/2/6c2a979a5ff6c40bf2d7152c8c32088fc36f848b.png)
where 
,
,
,
.
be a function such that
for all 
.
.
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
.
and 
is a square of an integer for all nonnegative integers 
.
is defined as follows: 
if 
and 
if 
.
Y by pifinity, Adventure10, megarnie, Mango247, ohiorizzler1434, ItsBesi, buddyram, cubres, and 2 other users
Let 
be a triangle. A circle passing through 
and 
intersects segments 
and 
at 
and 
, respectively. Lines 
and 
intersect at 
, while lines 
and 
intersect at 
. Prove that 
if and only if 
.
be a triangle. A circle passing through 
and 
intersects segments 
and 
at 
and 
, respectively. Lines 
and 
intersect at 
, while lines 
and 
intersect at 
. Prove that 
if and only if 
.This post has been edited 1 time. Last edited by MithsApprentice, Sep 27, 2005, 10:00 PM
(we work with non-directed segments here). Hence, 
.
.
.
from 
from 
.
because 
is a parallelogram and its diagonals must intersect at their midpoints.
and 
therefore triangles 
and 
are similar and 
hence the circumcircle of triangle 
is tangent to line 
therefore 
hence 
hence the circumcircle of triangle 
tangent to line 
)
,
is 
,
because they are vertical and 
by cyclic quadrilateral ABED. By the transitive property of equality, it follows that 
so quadrilateral FBCD' is cyclic, and by Power of a Point, 
or 
.
and then the result follows).
respectively.
while lines 
.
not paralel to 
,
is intersection of 
.
,
,
,
.
and by SAS similarity, 
.
.
.
and it follows that 
.
.
.
which by SAS implies that 
.
and therefore 
.
and that 
.
so 
where the last step follows from the Tangent-Secant Theorem and its converse.
,
at 
and since quadrilateral 
is cyclic, we have 
so 
so 
thus 
.![\[\frac{FA}{AB}\cdot\frac{BE}{EC}\cdot\frac{CM}{MF}=1\]](http://latex.artofproblemsolving.com/3/4/b/34bcd642bbfb537f88570b0de1022c149783d04e.png)
,
as desired.
SO 
so 
at point 
by Power of a Point.
.
is a harmonic division. Because 
,
.
,
.
at 
,
, , we have 
as 3 concurrent cevians. hence , by using ceva's theorem , we get that 
is similar to 
.
.
, and 
.
.
be the intersection of 
and 
.
be the midpoint of 
.
and 
and 
Since 
and 
,
therefore, 
and, by Thales' Theorem 
.
y 
.
and 
.
.
implies 
is tangent to the circumcircle of 
.
.
is a cyclic quadrilateral. Thus 
.
,
.
.
.
y 
,
.
,
as desired.
and cevians 
,
Also, 
Since 
,
tangent to 
as desired.
similar to 
is cyclic, 
is similar to 
and 
and 
(cyclic quadrilateral)
with concurrent cevians 
,
if and only if 
.
as desired.
,
so it is equivalent to prove:
Now observe that we have (angles are directed):
Noting that 
and 
,
which is a direct application of our lemma. 
where the last step follows by Ceva. 
,![\[
\measuredangle DBC = \measuredangle DBE = \measuredangle DAE = \measuredangle DCF
\]](http://latex.artofproblemsolving.com/7/4/0/740458124c6e44010c8fec1d9413a9ce30132427.png)
which is just the same as 
being tangent to line 
where the last follows from Ceva's as I showed before. That angle chasing of <MCD=<CBD could then be done this way, or we could FIRST angle chase and see this. Anyways, I saw this after looking at it for a bit longer. This is a much simpler solution.
,
,
.
iff 
![\[MF=MC\Longleftrightarrow\frac{BA}{AF}=\frac{BE}{EC}\Longleftrightarrow AE\parallel FC\Longleftrightarrow\angle FCB=\angle AEB=\angle ADB=\angle CDM\]](http://latex.artofproblemsolving.com/a/f/5/af5e02e5780c805192bd734cbebbf1778f565c19.png)
![\[\Longleftrightarrow MC\text{ is tangent to }(DCB)\text{ at }C\Longleftrightarrow MC^2=MD\cdot MB\]](http://latex.artofproblemsolving.com/4/1/e/41e527f2f4ade2f2eca5d4efa356cfecafc3456f.png)
.![\[\angle MBC = \angle DBE = \angle DAE = \angle MCD,\]](http://latex.artofproblemsolving.com/c/9/1/c91790f7b59573e561af93c1adc4161be58888d6.png)
.
such that 
is a parallelogram. Because of the properties of parallelograms and Power of a Point, the required condition is equivalent to 
being cyclic, which is true as 
.
as to denote directed angles.
as the ratio follows.
with concurrency at ![\[\frac{FA}{AB} \cdot \frac{BE}{EC} \cdot \frac{CM}{MF} = 1\]](http://latex.artofproblemsolving.com/5/9/a/59a9932ad795a3e17dc6de0126bfc8894b742374.png)
![\[\frac{FA}{AB} = \frac{EC}{BE}\]](http://latex.artofproblemsolving.com/3/b/1/3b1a75381ec401446f20343a0fd13cd69acf23f0.png)
Therefore, 
so 
.![\[\angle MCD = \angle EAD = \angle EBD\]](http://latex.artofproblemsolving.com/d/a/4/da4f9c0187be5fbdab9c6e98c922402c41bfc802.png)
Therefore, 
.
follows and the ratio ![\[\frac{FA}{AB} = \frac{CE}{EB}\]](http://latex.artofproblemsolving.com/2/7/9/27933d2874bf4f00a8f1e84954ab93a80211ae04.png)
follows. Furthermore, by Ceva's, we get ![\[\frac{CM}{MF} = 1\]](http://latex.artofproblemsolving.com/3/3/0/3303148147f9929519c45281c293e50386833e26.png)
as desired.
Meanwhile, 
But since 
so the two results are equivalent. QED
. First assume 
.
,
.
,
,
we notice that 
.
is tangent to 
,
