[/quote]
,
such that 
divides 
for any positive integers 
and 
.
be an acute scalene triangle inscribed in a circle 
.
intersects 
at 
and 
.
is the midpoint of 
.
be the altitude in 
,
intersects 
.
be the midpoint of 
be the reflection of 
with respect to 
and 
are similar.
and 
Prove that 
and 
.
Prove 
Find maximum 
that is not isosceles at 
,
be its incircle, which is tangent to 
at 
,
and 
intersect the line passing through 
at 
and 
,
and perpendicular to 
intersect 
and 
at 
and 
,
are collinear and that 
are concurrent.
and 
,
.
such that 
.
corresponds to 
)
be non-negative real numbers such that 
Prove that
,
is 
tastic if there exists a number 
such that 
is short, and such that 
is not short for any 
.
be the set of 
What is the maximum number of elements in 
lie on 
respectively satisfying 
.
.
is the second intersection of 
with 
,
is lying on 
at 
and 
respectively, and 
is the circumcenter of 
.
.
Y by narutomath96, Davi-8191, valsidalv007, A-Thought-Of-God, samrocksnature, mathematicsy, donotoven, jhu08, Adventure10, mathmax12, Mango247, Rounak_iitr, Funcshun840, drago.7437, MS_asdfgzxcvb
The incircle 
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 .
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 .Attachments:
This post has been edited 1 time. Last edited by orl, Oct 25, 2004, 12:16 AM
,
be the antipode of 
on 
.
respectively.
is parallel to 
and is cut into two equal parts by 
,
.
is harmonic, so the tangents to 
meet on 
.
,
is also harmonic, meaning that the tangents to 
.
are concurrent. If 
,
,
,
lies on an Apollonius circle, so 
is the bisector of 
.
,
,
.
and 
,
,
,
,
,
tangent to the incircle of triangle 
,
are collinear.
and 
are equal.
.
.
where the ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
,
because 
.
passing through 
with 
Let 
denote the antipode of 
the tangency point of the A-excircle 
with 
According to 
bisects 
internally and externally. Let 
be the image of 
is isosceles with apex 
which implies that 
due to 
But since 
we deduce that 
the incenter of 
,
the reflection of 
.
,
.
,
,
.![\[
\left( \frac{BK\cdot KC}{IK^2} \right) \left( \frac{DK}{DL} \right) = 1.
\]](http://latex.artofproblemsolving.com/9/d/9/9d98936b3fbd69064d41688a52c2327e18628077.png)
,
be the semiperimeter of 
be the 
.
be the 
.
with diameter 
intersect 
and 
.
and 
are tangent to 
is tangent to 
.
.
.
and 
bisect the exterior angles of the triangle at 
and 
is cyclic and, by power of a point with respect to 
.
is tangent to 
(the case 
is trivial.) Let 
,
,
.
be its radius, let 
be the point where 
be the 
be its radius, and let 
be its tangency point with 
be its antipode in 
.
both line on 
.
is harmonic, we must have 
(this can also be verified by computing the lengths directly, given 
,
is tangent to 
,
is tangent to the circumcircle of 
,
.
.
.
,
,![\[ \frac{ZB}{ZB + a} = \frac{KB}{KC} = \frac{s-b}{s-c} \implies ZK = ZB + s-b = \frac{(s-b)(a+b-c)}{b-c}. \]](http://latex.artofproblemsolving.com/8/0/3/8039e28c54322f1966761564baf74279598b0cb1.png)
![\[ PK \cdot KJ = \frac{ZK}{2} \cdot KJ = \frac{(s-b)(s-c)}{b-c} \cdot (b-c) = (s-b)(s-c) = r r_A, \]](http://latex.artofproblemsolving.com/9/3/c/93ce7ed92a48a57052c6be9e5d472cc83a820efc.png)
,
,
to 
and 
.
the radius of the 
is cyclic; it suffices to prove that 
.![\[ I_AK \left( I_AK + KN \right) = II_A^2 -r^2 \implies KN \cdot KI_A = II_A^2 - r^2 - I_AK^2 \]](http://latex.artofproblemsolving.com/a/5/f/a5ffcbd1d3bd663872427cb3a186430f4942086f.png)
Now we need only simplify the right-hand side using the Pythagorean Theorem; it is 
.
,
of 
(via 
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passes through the 
If 
then 
where 
are concyclic by fact 5. Thus 
where 
so 
are concyclic. But then 
done.
passes through 
and 
the inverses of 
the perpendicular bisector of 
and 
the perpendicular bisector of 
.
sends 
,
.
and 
fixing 
be the antipode of 
where 
are the 
touchpoints of 
are collinear.
from 
.
.
is harmonic quadrilateral. So, tangents at 
meet on line 
.
,
is harmonic quadrilateral. Hence, 
,
,
.
again at 
which means 
.
,
where 
be the intersection of the perpendicular bisector of 
be the midpoint of 
and 
are both perpendicular to 
(say).
,
are concyclic as desired.
meet the perpendicular bisector of 
.
and 
and 
.
and 
.
is the perpendicular bisector of 
.
is right angled at 
.
So 
is indeed tangent to the circle 
![[asy]
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,
the pole of 
and 
tangent to 
,
which will finish. First notice that 
passes through 
,
,
.
.
,
has circumcentre 
.
,
be the intouch triangle, and let 
.
be the 
be the antipode of 
and 
.
,
.
is the polar of 
.
.
,
,
,
,
.
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is the midpoint of the arc 
in 
.
and 
is the antipode of 
thus the homothety from 
maps 
,
.
.
.
because 
,
,
then the conclusion is obvious since 
.
be the midpoint of arc 
not containing 
be the tangent to 
.
.
-
and ![\[\operatorname{slope}(KI_A) = \frac{I_A K}{-KK'} = \frac{\operatorname{dist}(I_A, \ell) + M_A L}{-KK'} = -\frac{\operatorname{dist}(I, \ell) + M_A L}{2KL} = -\frac{IK + 2M_A L}{2(BL - BK)} = -\frac{r + 2BL \tan \angle M_ABC}{2(\frac a2 - (s - b))} = -\frac{r + a \tan(\frac{\alpha}{2})}{b - c}.\]](http://latex.artofproblemsolving.com/5/6/3/5639babadf4f523e0cf09fc1c0fffd7d7a23c509.png)
Since 
,![\[-\frac{r}{TK} \cdot \frac{r + a \tan(\frac{\alpha}{2})}{b - c} = -1 \implies TK = \frac{r(r + a \tan(\frac{\alpha}{2}))}{b - c}.\]](http://latex.artofproblemsolving.com/2/4/6/24655c5410b77c0160fe648c5ff44125c330aa6c.png)
Let 
.
Now we can calculate
But 
,
,
the 
E
the intouch points of 
be the reflections of 
and 
are cyclic.
.![\[\measuredangle I_aXP = \measuredangle CPF + \measuredangle (BC,AI) = 90 + \measuredangle IBA + \measuredangle ACB + \measuredangle IAC = 90 + \measuredangle ICB = \measuredangle PCI_a\]](http://latex.artofproblemsolving.com/6/8/c/68c18efcc8adde91c2f3214c6880cba1333db50c.png)
which implies that points 
lines 
and 
are parallel. Thus, 
.![\[AX \cdot AI_a = \frac{AX\cdot AC \cdot AB}{AI} = AC \cdot AF = AC \cdot AE\]](http://latex.artofproblemsolving.com/e/6/5/e657e1fab432982cf1b46cc5c1df3b048d3a3034.png)
which implies that points 
.![\[\measuredangle ENF = \measuredangle I_aNF+ \measuredangle ENI_a = \measuredangle NI_aB + \measuredangle EI_aN = \measuredangle EI_aF = \measuredangle EDF\]](http://latex.artofproblemsolving.com/a/2/5/a255bb13a88e8aa135a9e138a78eaa704f03eea1.png)
which implies that ![\[\text{Pow}_{(CEI_a)}(D)=DP \cdot DC = 2DB \cdot DC = DQ \cdot DB = \text{Pow}_{(BFI_a)}(D)\]](http://latex.artofproblemsolving.com/d/6/9/d69015c11a257a53a74780bc12a61022f912eb55.png)
which indicates that 
and 
and hence points 
(overrides a point definition in the problem). It is clear that ![\[DB \cdot DC = \frac{DP \cdot DC}{2} = \frac{DN\cdot DI_a}{2}= DM \cdot DN\]](http://latex.artofproblemsolving.com/a/7/c/a7c9ffe02ef232bf93ff79519e9ecf6bcf2e4535.png)
which implies that 
a
which is exactly what we shall do next.
which by the Linearity of Power of Point, we know is linear. Let 
denote the 
the midpoint of segment 
Thus,![\[-MN \cdot MI_a-MB^2=-2MB^2\]](http://latex.artofproblemsolving.com/0/7/7/077dc2479cc4a32e84e91329606b3ec8cfec8fff.png)
so![\[MB^2 = MN\cdot MI_a = MN \cdot MD\]](http://latex.artofproblemsolving.com/5/1/d/51da341d8a50ab8837e6bc46055ad9ceb79e0024.png)
as desired.
incircle.
as medial triangle. If 
then if 
then prove that 
is tangent to 
(the Why point).
where 
is centroid.
is tangent to 
.
homothety at 
and so ![\[\measuredangle AY_AT=\measuredangle AY_AA'=\measuredangle ODA=\measuredangle AOM=\measuredangle AOT\]](http://latex.artofproblemsolving.com/2/c/4/2c44f4e78fcf7b549ae01041652fe15b84a44ae7.png)
And done.
