AoPS Academy
. Initially, the numbers 
are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers 
and 
are on the board, then 
can be erased as 
, or 
as 
, or 
as 
.) is it possible to end with only one number remaining on the board?
. Points 
and 
lie on segments 
and 
, respectively. Lines 
and 
intersect at point 
. The circumcircles of triangles 
and 
intersect at a second point 
. The circumcircles of triangles and 
intersect at a second point 
. Point 
lies on segment 
such that 
. Prove that triangles 
and 
are similar.
?
be a trapezoid such that 
lie on a circle with center 
, and side 
is parallel to side 
. Diagonals 
and 
intersect at point 
, and 
. It is given that 
. It is also known that the difference in length between and can be expressed in the form 
, where 
and 
are positive integers and is square-free. Compute the value of 
.
and height 
Another sphere of radius 
is internally tangent to the lateral surface of the cone, but does not intersect the larger inscribed sphere. A plane is tangent to both of these spheres, and passes through the inside of the cone. The intersection of the plane and the cone forms an ellipse. Find the maximum area of this ellipse.
, 
in 
.
and 
are homothetic.
.
are cyclic.
.
touches 
.
be the point such that 
and lies on circle. Then, 
must be collinear. Applying Pascal's Theorem on 
, we get that 
are collinear. We know that 
, and 
. So, lines 
are concurrent. But we know that 
concur at 
. Hence, lies on tangent to 
at circle. 
exists. Let 
meet the circumcircle at 
. Define 
as the intersection of the tangent from and the line parallel to thru 
(which is just 
). We then just want to prove 
, which is equivalent to proving that 
are concurrent. By Desargues theorem, this is equivalent to proving that 
are collinear. We can simplify these as 
. The second is just the point at infinity along , so we just want to show that the line formed by 
and 
is parallel to . We just use complex numbers, clearly 
, then 
. Then 
. Thus we need to verify 
is self conjugating. This is obvious.
. Note that since 
, we have
On the other hand, let point be on the segment such that 
. Thus 
implying 
is a cyclic quadrilateral. And since 
, we get that 
From which the result follows that is tangent to 
at .
.
by parallel lines, and ![\[\frac{CM}{BC} = \frac{CM}{DL} \cdot \frac{DL}{BC} =\frac{KC}{KD} \cdot \frac{AD}{AC} = \frac{AC}{BD} \cdot \frac{BC}{AD} \cdot \frac{AD}{AC} = \frac{BC}{BD}. \ \blacksquare\]](http://latex.artofproblemsolving.com/1/e/f/1efb37b2ed7d419bfe2747fda35568683deaab46.png)
, implying desired tangency.
and see that 
and so 
, , collinear. Redefine as 
. Now just use Pascal on 
.
meet 
at .
we have that 
Thus, 
so is on 
proves that 
are collinear which finishes.
and 
Prove that
Prove that
and 
Prove that![$$ a+b+kc^3\geq\sqrt[4]{\frac{4k} {27}}$$](http://latex.artofproblemsolving.com/5/6/1/561a990e5885863c77d454ffa3f1973e1bbf60b4.png)
![$$ a+b+kc^4\geq\frac{5} {8}\sqrt[5]{\frac{k} {2}}$$](http://latex.artofproblemsolving.com/4/b/1/4b14911672688250c909571d788aaf10724462eb.png)
Where 
Prove that
is inscribed in circle 
R
intersect at 
is chosen on the diagonal 
is chosen on the segment 
so that 
Prove that line 
and Pascal for 
.
.
,
,
are collinear. Now let the tangent of 
are collinear, which means 
,
.
,
and you are done.
.
is an isosceles trapezoid so 
gives that the intersection of the tangent at