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.
have incenter 
and intouch triangle 
. Let the circumcircle of intersect 
at 
and have center 
. Let 
be the midpoint of arc 
on the circumcircle. Suppose quadrilateral 
is cyclic such that 
lies on 
. Show that 
.
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.
and 
. Let 
be non-negative real numbers, such that
holds for all 
and 
. Denote
Prove that
?
be nonnegative real numbers such that 
. Find min
and of the same colour such that is the midpoint of 
. a triangle with circumcircle 
and isogonal conjugates and 
. Take 
and 
points on and respectively such that 
. If 
cut again at 
. Prove that 
and 
cut at and are external just read it by directed angles be a set of integers with the following properties:
.
and 
, then 
.
, 
is composite, then all positive divisors of are in . contains all positive integers.
such that 
is an integer for every positive integer 
and 
be real numbers with 
and 
such that 
Find the least possible value of 
and foci at 
. Another ellipse is centered at 
with semi-major axis 
and foci 
. We claim that minimizing 
means that these ellipses are tangent - indeed, assume they are not. Then, we can reduce the major axis of the first extremely slightly, by some 
, such that the circles still intersect, but decreased, a contradiction.
and 
, with foci 
and 
are tangent at 
. is either 
or 
.![[asy]
import geometry;
ellipse ellipse(point A, point B, point C) {
return ellipse(A, B, (abs(A-C)+abs(B-C))/2);
}
size(9cm);
point E1=(-4,0),E2=(4,0),F2=(10,12),F1=(10,5.6),T=(6,4);
draw(ellipse(E1,E2,T));
draw(ellipse(F1,F2,T));
draw(E1--F1);
draw(E2--F2);
label("$E_1$",E1,S);
label("$E_2$",E2,S);
label("$F_1$",F1,E);
label("$F_2$",F2,E);
label("$T$",T,E);
dot(E1);
dot(E2);
dot(F1);
dot(F2);
dot(T);
[/asy]](http://latex.artofproblemsolving.com/f/d/6/fd6a014a5e43bc44c63e9142cc8a9d576b2e0c34.png)
, and we show that if 
(given that lies on both line segments; otherwise look at ), then the ellipses are tangent. Clearly, there is at most one ellipse with foci at 
and 
that is tangent to a given ellipse, so it suffices to prove that is the one. on both ellipses. This means that![\[E_1Q+E_2Q=E_1P+E_2P\]](http://latex.artofproblemsolving.com/b/9/b/b9b5fad352f09fefbd6e36610bd6ff0ce7042de0.png)
![\[F_1Q+F_2Q=F_1P+F_2P\]](http://latex.artofproblemsolving.com/a/7/c/a7c463802afb25c362fed3dae081c82364ae0dff.png)
or upon addition![\[E_1Q+E_2Q+F_1Q+F_2Q=E_1P+E_2P+E_1P+E_2Q\tag{1}\]](http://latex.artofproblemsolving.com/f/d/6/fd672be93623e38dd5b75904cbbd559c80f66e38.png)
However, by the triangle inequality in 
and 
, we get![\[E_1Q+F_1Q\geq E_1F_1=E_1P+PF_1\]](http://latex.artofproblemsolving.com/d/9/2/d92c781baa8ccc1e6722f6396d097fd896727870.png)
![\[E_2Q+F_2Q\geq E_2F_2=E_2P+PF_2\]](http://latex.artofproblemsolving.com/4/d/d/4ddab415a5047b636c69ac7ea7f99e3d3060a1ef.png)
with both equalities impossible, as that means 
, impossible. Thus, we know that![\[E_1Q+E_2Q+F_1Q+F_2Q>E_1P+E_2P+E_1P+E_2Q\]](http://latex.artofproblemsolving.com/0/1/1/0112a38638bd43e79ddd0af815ff6bb358797c29.png)
a contradiction to (1).
with 
, and 
with 
. We can compute this point to be 
. Thus, the semi-major axis, and , are just half of the sums of the distances from the foci. In particular,
Thus, we get the minimum value of 
.
and 
and the second ellipse has foci 
and 
. Now consider the point 
. We have that 
and since 
and 
, we have that 
, where equality occurs when is the intersection of 
and 
. Then 
, and the answer is 
.
and the foci of the second ellipse are 
. By the definition of an ellipse, we have:
So
Equality can occur, just draw a picture or solve for the intersection of the lines if you aren't convinced yet.
and 
are the foci of the first ellipse and and are the foci of the second ellipse, then the lines 
and 
intersect at the tangency point. This can be proven intuitively since AM-GM equality gives that 
minimizes any sum in general. Therefore we want them to be as small as possible while not too large, giving the assertion. and 
which gives the tangency point 
by basic ellipse properties and coordinates. We then plug this point back into each equation to solve for and respectively using the substitution 
and 
. Miraculously, solving each quadratic gives 
and 
, or 
and 
, which means 
. 
is centered at the origin with a major axis of 
and minor axis 
, while 
is centered at 
with a major axis of 
and minor axis of 
. Now, let 
denote the foci of and 
denote the foci of , respectively. Let be the intersection of and . Spamming the definition of an ellipse gives that 
. This problem is boiled down to finding the minimum value of 
.
and are collinear, and 
are collinear. The rest is just simple computation:
is that of an ellipse centered at 
with foci at 
and 
. Similarly, the equation 
is that of an ellipse centered at with foci at 
and 
. Note that from the definition of an ellipse is the sum the distances from any point on the ellipse to the two foci (
is defined similarly). We know that 
lies on the two ellipses. Thus, we want to minimize the sum of the distances from to the two pairs of foci, which have coordinates , , , 
. In other words, we want to minimize the sum of the distances from to those four points. is the intersection of the segments connecting and , and (provable using triangle inequality). We can easily solve for 
, so 
(I skipped over some distance formula computation).
is the sum of 
from the four foci. The four foci are literally just 
. But minimizing 
is minimizing 
and 
and in fact these minima can happen at the same time at 
. Basic system of equations extracts 
and 
, and the foci for the second ellipse are 
and 
. For a pair 
to exist, the graphs must intersect. Let them intersect at 
. Then 
and 
. 
. To minimize this expression, we must have circles with centers at the 4 foci points intersect at the point . We must minimize the sum of all the radii of these four circles. Consider the circles 
and 
at centers and respectively. Clearly 
. Now consider circles 
and 
at centers and respectively. Similarly, 
. Both of these inequalities are equal when the two circles are tangent, and the pairs of circles are tangent at the same point, giving 
.
.
and semimajor axis 
and semimajor axis 
,
and 
.
.
The lines between foci intersect at (14,15/2) and so that is the minimum => 16+7 = 23