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that verfy the following equation :
be a positive integer. We say that a positive integer 
is -good if 
is divisible by 
for all positive integers 
with 
. Suppose is a positive integer such that is -good, but 
is not -good. Prove that 
is prime.
let 
and 
be the circumcenter and the orthocenter. The point 
is the reflection of 
with respect to 
. Assume that is not on the same side of 
as . Points 
lie on 
respectively such that 
. Let 
be the intersection point of 
. Prove that 
a grid of 9*9, we choose a little square in of this grid three times, we can choose three times the same.
be the incenter of triangle 
. Let 
, 
, and 
be the orthocenters of triangles 
, 
, and 
, respectively. Prove that the lines through , , and , parallel to 
, 
, and 
, respectively, are concurrent.
of acute triangle is 
. The ray drawn from through the circumcenter intersects at 
. Let the midpoint of 
be 
. Define 
, 
, 
, 
, 
, 
similarly. Prove that 
, 
, 
are concurrent.
+1 w
be an acute triangle with orthocenter and circumcircle 
. Suppose that 
are the antipodes of 
in . Let 
be the reflection of the circle with center and radius 
with respect to 
; define 
similarly. Show that 
concur.
be the circumcenter of 
, and 
the orthocenter of 
, which is the reflection of over . We claim that the reflections of 
, and 
over 
respectively all lie on 
. This suffices because if 
and 
is the reflection of over 
, then by Simson Line (some lemma in EGMO Chapter 4), the reflections of over 
will lie on 
.
, 
such that 
, and 
such that 
. Then 
is an isosceles trapezoid. Since is the perpendicular bisector of 
, it's also the perpendicular bisector of 
, and so indeed the reflection of over lies on , so we are done. 
,
.
is the inverse of function 
and thus, they might intersect that 
axis (if they do, we are done). Thus: 
is the only solution.
.
Let 
.
