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such that:
such that:
+1 w
such that for any 
,
of the acute-angled triangle 
is tangent to its side 
at a point 
. Let 
be an altitude of triangle , and let 
be the midpoint of the segment . If 
is the common point of the circle and the line 
(distinct from ), then prove that the incircle and the circumcircle of triangle 
are tangent to each other at the point .
![\[ a \le cosA + \sqrt3senA \Leftrightarrow \frac {a}{2} \le cos(A - 60).
\]](http://latex.artofproblemsolving.com/d/f/0/df0c4d38eafde75321bce1b83244b8dab2dfba6c.png)
degrees... Maybe we should think about the midpoint of 
, because of the 
. We can also note that two circles intersecting give us isosceles triangles. Midpoints and isosceles triangles make us think about perpendicular bisectors. We also have a lot of unitary lengths... These things indicate equilateral triangles. In fact, if we call 
a point of interscection of the circles 
and 
, we will have the equilateral triangle 
. Now, the extreme case must be when three of the circles concur. This gives us three unitary lengths... We also need to use that the triangle is acute-angled. All this makes us...
of triangle 
. If it is covered by a circle centered in , 
or (hence covered by them all), all the triangle is covered. It happens because the triangle is acute-angled, such that is inside and, being 
its medial triangle ( opposite to , to and 
to ), 
, 
and 
, then the circles intersect outside the triangle and cover it all. Obviously, be covered implies 
be covered. So the quadrilateral is covered 
is covered by the circle centered in, let´s say, . Let 
the circumradius of . Now, we can prove that 
. First, if we consider as a horizontal axis, the horizontal distance from to is 
, that is less than the horizontal distance from 
to , once 
. Of course, the perpendicular bisector from 
, while goes from to , approaches vertically the straight line through and (does somebody know how to put the 
over in LaTeX?). Thus, is also nearer to that to vertically, from which follows that . Here we conclude that , and then the parallelogram, is (totally) covered if and only if 
, we can ignore the factor 
, obtaining what we wanted. Else, we have 
and this also finishes the problem.
such that 
.
has 
,
has all angles acute. Prove that circles radius 
and center 
cover the parallelogram if and only![\[a\le\cos A+\sqrt3\sin A.\]](http://latex.artofproblemsolving.com/e/8/9/e89d04a29c72c36367dc99c67f9958a96339a1e8.png)