such that the reflection of each of these four points about 
and 
Show that
Let 
Show that
such that 
and 
has real roots 
where 
.
has real roots 
where 
.
(s non zero), then 
be positive reals such that 
.
Given 
non-zero reals 
such that their sum is 
.
such that 
,
is a root of 
and solve the equation.
or 
Soo... can u guys help? thx!
.
is a polynomial such that 
is divisible by 
,
.
and 
be 2 distinct positive integers.
be in an integer. Prove why ?
and 
.
,
can be expressed as 
where 
is a positive integer, find 
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Let 
be a triangle with 
, and let 
be the foot of the altitude from 
. Let 
be a point in the interior of the segment 
. Let 
be the point on the segment 
such that 
. Similarly, let 
be the point on the segment 
such that 
. Let 
be the point of intersection of 
and 
.
Show that
.
Proposed by Josef Tkadlec, Czech Republic
be a triangle with 
, and let 
be the foot of the altitude from 
. Let 
be a point in the interior of the segment 
. Let 
be the point on the segment 
such that 
. Similarly, let 
be the point on the segment 
such that 
. Let 
be the point of intersection of 
and 
.Show that
.Proposed by Josef Tkadlec, Czech Republic
This post has been edited 3 times. Last edited by Eternica, Jun 19, 2024, 10:03 AM
and to 
intersect at a point 
which lies on 
which is also the intersection of 
and 
So, by some angle chasing, it follows that 
meaning that 
is tangent to 
So, by power of 
because 
is also tangent to 
Therefore 
and 
leading to 
and 
intersect at 
.
are on line 
such that 
and 
.
is a point on 
.
cut 
don't have same side with 
cut 
don't have same side with 
cuts 
at 
intersect 
at 
.
.
and so, 
is antiparallel to 
which means, 
.
.
.
.
but by similar reasonings, 
.
is cyclic. Since in circles 
and 
as diameters, we also have 
.
to 
concur at a point, 
.
,
.
intersect 
.
and so, 
is tangent to circumcircle of 
.
.
.
.
.
,
.
.
and hence, 
is tangent to cirumcircle of 
.
is cyclic. In circles 
and 
and 
as diameters and so, 
.
passing through 
concur at 
.
.
,
coincide with 
respectively. In that case, it can be seen that we get the IMO problem.
and the lines 
and 
(that trick excludes calculation of the point 
be the circle with center 
through 
be the circle with center 
through 
and 
.
be the second intersection of the line 
be the second intersection of 
,
is cyclic.
is tangent to 
,
.
,
.
,
)
and the circle
)
lies on the midpoint of 
.
,
implying 
by the power of 
.
we get 
implying 
is cyclic.
.
.
.
.
[using (1),(2),(3)]
intersects 
and the tangent in 
,
is a harmonic quadrilateral. This yields 
.
to be the point of intersection of 
.
and the power of a point now yields 
.
and now we have congruent triangles 
and 
,
.
the circle centered at 
similarly. Suppose 
,
similarly. Observe that 
of 
,
cyclic with circumcircle 
.
is harmonic 
is in the polar 
Similarly 
are tangent to 
So 
are tangent to 
of 
.
and 
intersect at 
are similar. So, 
.
is cyclic, so 
.
at 
as 
.
.
and circle 
and are tangents to them. So, if we draw a circle with centre 
and as 
and 
,
,
is the reflection of 
be the circles centered at 
,
to be the second intersections of 
with 
,
implies that 
is the midpoint of both 
and 
.
and 
,![\[C_D'K'{}^2=C_D'K'\cdot C_D'K_2'=C_D'L'\cdot C_D'L_2'=C_D'L'{}^2.\]](http://latex.artofproblemsolving.com/f/e/3/fe3310a72d4aaf34fc70b655d74abc8235c43615.png)
So 
,
where 
is the radius of inversion. Rearranging gives the desired. 
and 
as tangents to some circle. Call 
the circle through 
similarly. Let 
.
and 
.
,![\[XK'\cdot XK=XC\cdot XC'=XL\cdot XL',\]](http://latex.artofproblemsolving.com/e/4/1/e41621686f20a3b236db233b999a8d299ae3e074.png)
so 
is cyclic. Then, 
,
.
,
.
be the circle centered at 
the circle centered at ![\[AC=AL=AS \text{ and } BC=BK=BS\]](http://latex.artofproblemsolving.com/3/5/8/358c6359e87e342d22a9b973ce932fc95a3fb958.png)
we also have that 
and 
.
and 
.![\[XL\cdot XL' = XC \cdot XS = XK \cdot XK'\]](http://latex.artofproblemsolving.com/7/f/e/7fe0f06fba3d048af55c5bb9ced7d4d3c7b952a8.png)
we also have that 
is cyclic (let its circumcircle be ![\[BL \cdot BL' = \text{Pow}_{\omega_L}(B)= BA^2-AC^2 = BC^2\]](http://latex.artofproblemsolving.com/6/f/9/6f916adecc4711d1aaf2d6d79f6e388d2ab87b74.png)
Thus, ![\[BK^2=BC^2=BL \cdot BL'\]](http://latex.artofproblemsolving.com/f/7/9/f79d92c5b5260706bad4935fedf0284c9e89c8b0.png)
Thus, ![\[MK=ML\]](http://latex.artofproblemsolving.com/b/2/c/b2c728c2a4fe1049a1ad13f2307d6bd3e548ab7d.png)
which was the required conclusion.
.
.![$[X, K, A]$](http://latex.artofproblemsolving.com/8/2/c/82cd2f7bd5106902d21382df94be100f48c61558.png)
are collinear in that order as are ![$[X, L, B]$](http://latex.artofproblemsolving.com/e/2/1/e21335b424cd8534e1fcb1e7470319ba68fbfc36.png)
.
,
)
using [4-6]. Similarly, 
so [12] is true and we are done.
and 
be the other intersections of 
with 
with 
,
are orthogonal, so by definition 
is tangent to 
is tangent to 
intersect the circle centered at 
intersect the circle centered at 
and 
intersect on 
,
and 
.
.
and 
,
,
.![\[Pow_{\omega}(A)=AC^2=AD\cdot AB=Pow_{(BDK)}(A),\]](http://latex.artofproblemsolving.com/5/7/c/57c1ff79ede43ebaa23363cd9871993f86fa5229.png)
so 
is the radical axis of 
.
is the radical axis of 
and ![\[AKX\cap BLX=X\]](http://latex.artofproblemsolving.com/6/3/4/634e3dc481d3536bcad6ed2e5c00d08a269dd9a9.png)
.
.![\[\angle TLA=\angle TDA=\angle TKB=90.\]](http://latex.artofproblemsolving.com/f/0/5/f0547822ebcae611faa6931ffd09186835124dd2.png)
Therefore:![\[Pow_{\omega}(T)=TK^2=Pow_{\gamma}(T)=TL^2,\]](http://latex.artofproblemsolving.com/a/e/1/ae1a40959417a4b52b0387a92734fa2b90daa1fa.png)
so 
,
,
,
and 
be the inverses of 
is cyclic. Let its circumcircle be ![\[AK \cdot AK^* = AC^2 = AL^2.\]](http://latex.artofproblemsolving.com/d/3/5/d352d39e1c95fac80b04d5793122b0e0884efd77.png)
be the second intersection of line 
with 
likewise. Let 
and 
be the second intersections of line 
with 
sends 
is cyclic, so![\[\angle ALL_1 = \angle AL_1L = \angle AA_1L\]](http://latex.artofproblemsolving.com/d/b/9/db92f9667e5c67c7234decc691ad3366153ceb14.png)
Similarly, 
.
is cyclic, so 
,
.
is isosceles as desired.
on 
such that 
.
,
is tangent to 
is tangent to 
,
.
.
.
,
and 
.
,
is cyclic. Note that since 
,
.
,
.
.
since 
,
is shared, so then 
and 
be circles Centered at 
and 
be the intersections of 
with circles 
is cyclic.
.
and 
is cyclic. Also 
so 
respectively.
we have that 
and 
intersect on the radical axis of 
is cyclic.
so 
and 
,
.
and 
.
and suppose 
.
,
Hence 
so![\[
k=\frac{2a^2+\sqrt{4a^4+4a^2(b^2+2t)}}{2}=a^2+a\sqrt{c^2+2t}.
\]](http://latex.artofproblemsolving.com/b/1/d/b1dae0a0d64eef0567ac167c792edeb19c4972fe.png)
Similarly, 
.![\[
M=\left(a^2+a\sqrt{c^2+2t}:b^2+b\sqrt{c^2+2t}:t\right)=\left(\frac{a^2+a\sqrt{c^2+2t}}{\alpha},\,\frac{b^2+b\sqrt{c^2+2t}}{\alpha},\,\frac{t}{\alpha}\right),
\]](http://latex.artofproblemsolving.com/d/6/7/d6717601ad090872c76bc5ae50b1d891d83e7d63.png)
where 
.
so![\[
\overrightarrow{KM}=\left(\left(a^2+a\sqrt{c+2t}\right)\left(\frac{1}{\alpha}-\frac{1}{\beta}\right),\,b^2\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)+\frac{b\sqrt{c+2t}}{\alpha},\,t\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)\right).
\]](http://latex.artofproblemsolving.com/f/8/3/f83ba546b885fb5f86d4161a4f83a3a42b441edd.png)
Noting that 
,![\[
\overrightarrow{KM}=\frac{b\sqrt{c^2+2t}}{\alpha\beta}\left(-\left(a^2+a\sqrt{c+2t}\right),\,\beta-b^2,\,-t\right)
\]](http://latex.artofproblemsolving.com/8/0/5/8054b25e6f8f1ccdcb1ac078d2bb7823c5c10c95.png)
Note that 
is independent of 
which is independent of 
