AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
on the whiteboard. In each move, she selects two distinct numbers, 
and 
, erases them, and replaces them with 
. She repeats this process until only one number, 
, remains. What are all the possible values of ?
, 
is a perfect square.
, so that the number 
is a prime.
is a cyclic quadrilateral with 
. Let 
and 
be the feet of perpendiculars from 
and 
to 
respectively. Prove that 
.
be triangle, inscribed in parabola. Tangents in points 
forms triangle 
. Prove that 
. (
is area of triangle 
).
with no parallel sides is inscribed in a circle 
. Circles 
are inscribed in triangles 
, respectively. Common external tangents are drawn between 
and 
, and 
, and 
, and and , not containing any sides of quadrilateral . A quadrilateral whose consecutive sides lie on these four lines is inscribed in a circle 
. Prove that the lines joining the centers of and , and , and the centers of and all intersect at one point. be a quadrilateral inscribed in a circle 
such that 
. Let 
and 
be the midpoints of 
and 
respectively. The line 
meets again at . Prove that the tangent at to , the line and the line 
are concurrent.
be an acute-angled triangle with circumcircle 
. A circle is internally tangent to at and also tangent to 
at 
. Let and 
intersect at 
and 
respectively. Let and be points on line such that 
is the midpoint of 
and is the midpoint of 
. Lines 
and 
meet at 
and intersect again at 
and 
respectively. The ray 
meets the circumcircle of triangle 
again at 
.
.
be a triangle and 
its circumcentre. A line tangent to the circumcircle of the triangle 
intersects sides at and at . Let 
be the image of under 
. Prove that the circumcircle of the triangle 
is tangent to the circumcircle of triangle .
be a semicircle with diameter and midpoint . Let be a point on different from and .
touches in a point , the segment 
in a point , and additionally the segment . The circle 
touches in a point and additionally the segments 
and .
and 
are perpendicular.
be an arbitrary triangle and is the circumcenter of 
.Points 
lie on 
,respectively such that the reflection of WRT 
is tangent to circumcircle of 
.Prove that the circumcircle of triangle 
is tangent to circumcircle of triangle .
be a triangle with circumcircle and let 
be the -excircle. Let 
and 
be the intersection points of and . Let and be the projections of onto the tangent lines to at and respectively. The tangent line at to the circumcircle of the triangle 
intersects the tangent line at to the circumcircle of the triangle 
at a point 
. Prove that 
.
. Let the perpendicular line from to meets 
at points 
, respectively, and the foot from to is 
. 
intersects line at 
, 
intersects line at 
, and lines 
intersect at .
is tangent to 
.
of convex polygons (in the plane). It turns out that every point of the plane lies in the same number of elements of 
as elements of 
. Prove that 
. or .
be the set of real numbers. Find all functions 
that satisfy the following condition. Here, 
is the function obtained by composing 
times, that is, 
, 
has no meaning here, so replace it with a given integer 
. Let 
be the assertion
From 
, we can easily observe that 
.
is bijective.
, we have that 
for all reals . This implies that is surjective. Now, assume that 
for some 
. Since 
, we have that 
and 
, implying with 
that 
. Hence, is injective, so we can conclude that is bijective.
.
be a nonzero real. With induction, we will prove that 
for any 
. The inductive hypothesis holds when 
. Assume that 
for any 
, for 
. From 
, we have that
Since 
and 
, and , we have that should hold. Therefore, from , we can obtain that 
, or, .
, we have that 
or 
must hold. In both cases, we have that . Hence, if a function 
is defined as 
, we have that 
is the inverse of , and so bijective. Also, 
always holds, and 
. Now, we can simplify as follows.
for any 
. such that 
. When 
for some , from injectivity, we can deduce that 
, so 
for all positive integer . From , we have that 
also holds for all positive integer . From 
, we have that
The first one cannot hold from the reasons mentioned above. Therefore, for any positive integer , we have that 
. This implies that the sequence(orbit)
is an arithmetic progression. Hence, when , 
for any positive integer , so 
, a contradiction. Therefore, must hold for any .
was the inverse of , we can conclude that for all reals , and the identity function clearly satisfies the problem condition.
or 
and in which 
,
, so 
holds.![$a_{1}\xrightarrow[]{f}a_{2}\xrightarrow[]{f}a_{3}\cdots a_{101}\xrightarrow[]{f}a_{1}$](http://latex.artofproblemsolving.com/6/5/c/65c7b115f603c715e785d050e482537fff3cb482.png)
![$2a_{1}\xrightarrow[]{f}a_{1}+a_{2}\xrightarrow[]{f}2a_{2}\cdots 2a_{101}\xrightarrow[]{f}a_{101}+a_{1}\xrightarrow[]{f}2a_{1}$](http://latex.artofproblemsolving.com/5/4/7/547bb8b8dd2fbc71e0f92c1fed25a0cc9398cfd0.png)
where c is a constant.
or 
, so by the claim that is bijective,
or 
, so 
or 
for all 
or 
, 
, and 
, so by bijective, 
, contradiction
by bijective, 
, 
