![\[222{,}222-22{,}222-2{,}222-222-22-2?\]](http://latex.artofproblemsolving.com/1/5/6/156c3662fe4c874a80e54887f76c302c13b9d14c.png)
are multiples of 
or 
but not 
?
,![\[a_n = \frac {(n + 9)!}{(n - 1)!}.\]](http://latex.artofproblemsolving.com/f/d/5/fd5a5ee0252b440fd5fcd9694ac0b00de61b83b2.png)
denote the smallest positive integer for which the rightmost nonzero digit of 
is odd. The rightmost nonzero digit of 
be the largest real solution to the equation ![\[\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}= x^2-11x-4.\]](http://latex.artofproblemsolving.com/e/8/7/e87fdc9fe63d29a46e57dfb92c2d0448441b4b55.png)
There are positive integers 
such that 
.
.
Y by samrocksnature, megarnie, HWenslawski, son7, Mango247
Let 
be the incircle of a fixed equilateral triangle 
. Let 
be a variable line that is tangent to and meets the interior of segments 
and 
at points 
and 
, respectively. A point 
is chosen such that 
and 
. Find all possible locations of the point , over all choices of .
Proposed by Titu Andreescu and Waldemar Pompe
be the incircle of a fixed equilateral triangle 
. Let 
be a variable line that is tangent to and meets the interior of segments 
and 
at points 
and 
, respectively. A point 
is chosen such that 
and 
. Find all possible locations of the point , over all choices of .Proposed by Titu Andreescu and Waldemar Pompe
This post has been edited 2 times. Last edited by djmathman, Jun 22, 2020, 5:29 AM
.
and 
by the Pythagorean Theorem, or the reflection of that point over 
and 
instead of 
and 
.
by perpendicularity lemma. Then proceed to length chase.
.
.
Which is trivially true 
too.
However, 
because 
.
,
.
sends 
to 
,
and the arc made by the homothety we mentioned before.
be the center of the equilateral triangle 
and 
intersect each other at 
(this is the circumcircle of 
.
lies on ![[asy]
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.
.
and 
are congruent. Because 
and there have a right angle. Therefore 
and hence done. 
,
and 
are congruent. Since the length 
also does not change. Since the line 
,
,
be the tangency point of 
,
intersect 
.
is cyclic. Indeed, since 
and 
it follows that 
,
,
where 
.
.
by the midpoint property so let us show 
.
so by SAS it is sufficient to show 
.
and 
.
.![\[ \angle QIL = \angle EIL + \angle EIQ = \angle TIB + \angle TIQ = \angle QIB \]](http://latex.artofproblemsolving.com/d/6/b/d6bc862f70eefbe3dfc0a40330bf84d8d9c482c1.png)
Proving the claim. Using the fact that 
we have 
.
so 
.
.
is the arc 
on 
.
be tangent to the incircle at 
such that 
and 
where 
is the sidelength of 
.
and 
.
be the tangency points on 
respectively. Then![\[PA^2 = PX^2 + AX^2 = PT^2 + (\frac{\sqrt3}{2} s)^2 = PT^2 + TR^2 = PR^2.\]](http://latex.artofproblemsolving.com/3/5/d/35d8e0f2e8129b7a99ef002d1cc516bdd61b632b.png)
A similar argument shows that 
.
.
,
in the direction opposite 
in the direction of 
arc 
.
be the contact points, 
the circumradius, 
and 
so 
and 
and the second part follows as 
so 
Also, 
or 
Notice 
on 
and so by homothety, 
degree arcs bounded by 
and 
That is, 
of 
or minor arc 
of 
where 
and 
I would appreciate it if anyone could look over my writeup and give me some feedback.
be the circle with center 
,
be the circle with center 
.
as the tangency points of 
,
,
.
Therefore,
as desired.
with respect to 
collinear. By the Pythagorean Theorem, we have
But as 
(inradius), we have 
.
,
and 
denote the images of 
,
circular arcs 
and 
centered at 
be the contact triangle, with 
It is clear that 
are collinear.
and 
Notice that 
Thus 
Since 
it suffices to show that 
By Law of Cosines we have 
so 
Similarly 
so we finish.
Letting 
so we finish.
are on the same side of 
If 
which is the aforementioned image of 
with ratio 
satisfies the given conditions. Set 
and 
the incircle touchpoint on 
.
and 
,
is the midpoint of 
.
.
and 
have radical axis 
.
is a point with 
and hence 
fixed.
be the point on minor arc 
,
,
,
.
and 
,
,![\[\triangle PMA \cong \triangle PXR_1 \cong \triangle PXR_2, \quad \triangle QNB \cong \triangle QXR_1 \cong \triangle QXR_2,\]](http://latex.artofproblemsolving.com/f/5/3/f53ed83a13d2c44456b3497030d811ea85d7af76.png)
,
on ![[asy]
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,
meets 
and from
and its 
homothety centered at 
be the intersection of 
