[/quote]
,
for Girls, PRISMS Online Math Meet for Girls, is happening this spring! 
groups for an activity. In the afternoon, the math teacher divided the same students into 
groups for another activity. A student is considered special if the group they belonged to in the afternoon is smaller than the group they belonged to in the morning. Find the minimum number of special students that can exist in the school.
with sum 
are such that the equation 
has a positive root 
smaller than 
.
such that the inequality 
holds.
be inscribed in the circumcircle 
and circumscribed about the incircle 
,
.
,
,
at 
,
,
,
,
,
,
,
,
(distinct from 
)
intersects 
(distinct from 
be the midpoint of the arc 
of 
and 
at 
and 
,
at 
.
.
varies on its side 
.
and 
are the midpoints of the arcs 
and 
(not containing 
and 
,
passes through a fixed point, independent of the choice of 
,
,
.
that does not contain 
and 
denote the incenters of triangles 
and 
,
passes through a fixed point.
,
,
,
and 
such that 
is a parallelogram and 
is a cyclic quadrilateral. Let 
be a line passing through 
at 
and intersects line 
at 
.
.
.
is divisibly by 
.
and 
is entered, the new password is 
,
be a circle and 
a number from a set 
is written --- in the first row numbers 
,
and so on. Exactly 
of them have been chosen, no two from the same row or the same column. Let us denote by 
a number chosen from row number 
.![\[ \frac{1^2}{a_1}+\frac{2^2}{a_2}+\ldots +\frac{n^2}{a_n}\geq \frac{n+2}{2}-\frac{1}{n^2+1}\]](http://latex.artofproblemsolving.com/a/d/5/ad5ae6861da79a08d32093f63827016868856a3a.png)
lying outside the circle 
.
be the midpoints of segments 
and 
respectively. The circumcircle of triangle 
intersects 
)
passes through the centroid of triangle 
.
Y by
Two straight pipes (circular cylinders), with radii 
and 
, lie parallel and in contact on a flat floor. The figure below shows a head-on view. What is the sum of the possible radii of a third parallel pipe lying on the same floor and in contact with both?
![[asy]
size(6cm);
draw(circle((0,1),1), linewidth(1.2));
draw((-1,0)--(1.25,0), linewidth(1.2));
draw(circle((1,1/4),1/4), linewidth(1.2));
[/asy]](//latex.artofproblemsolving.com/4/0/9/4091eff69445abb83b260b3ccbdfa4e00b8f9182.png)
and 
, lie parallel and in contact on a flat floor. The figure below shows a head-on view. What is the sum of the possible radii of a third parallel pipe lying on the same floor and in contact with both?![[asy]
size(6cm);
draw(circle((0,1),1), linewidth(1.2));
draw((-1,0)--(1.25,0), linewidth(1.2));
draw(circle((1,1/4),1/4), linewidth(1.2));
[/asy]](http://latex.artofproblemsolving.com/4/0/9/4091eff69445abb83b260b3ccbdfa4e00b8f9182.png)
This post has been edited 5 times. Last edited by mathboy282, Nov 13, 2024, 6:40 PM
by Descartes Theorem.
is the desired radius let 
,
,
.
,
.
will be the smaller radius, given that it is an answer choice, along with the fact that (C) and (D) differ by 
![[asy]
size(6cm);
draw(circle((0,1),1), linewidth(1.2));
draw((-1,0)--(1.25,0), linewidth(1.2));
draw(circle((1,1/4),1/4), linewidth(1.2));
[/asy]](http://latex.artofproblemsolving.com/d/f/b/dfb2481a375b7f929011e76a6fc8c8fb09be835f.png)
![[asy]
size(6cm);
draw(circle((0,1),1), linewidth(1.2));
draw(circle((2,1),1), linewidth(1.2));
draw(circle((0.666 , 0.111),0.111), linewidth(1.2));
draw((-1,0)--(3.25,0), linewidth(1.2));
draw(circle((1,1/4),1/4), linewidth(1.2));
[/asy]](http://latex.artofproblemsolving.com/6/1/d/61d0f8545672d748fb9dd0113ed3684f20787fb2.png)
.
so 
or 
.
where 
is the smallest circle of the 
circles of radii 
and a third circle of radius 
that is externally tangent to the first two circles as well as one of their common external tangents, then we have 
Applying this formula with 
as well as 
we get 
and 
so the answer is 
and let the center of the second circle be 
.![\[
x^2 + (y-1)^2 = (y + 1)^2
\]](http://latex.artofproblemsolving.com/9/e/e/9ee71a190ba030ed6c6509b5eacdf50bbf0f27e4.png)
and the second circle consists of![\[
(x-1)^2 + (y-\frac14)^2 = (y + \frac{1}{4})^2
\]](http://latex.artofproblemsolving.com/4/c/6/4c61ce37a9b9b100ac8dd668ddcfc810365d0575.png)
Then we can solve this as a quadratic in 
to get 
and thus 
are the radii in a configuration like this, then 
Proof: 2015 AMC 12A #25.
giving 
