be a triangle. 
is the midpoint of segment 
,
,
are selected on sides 
,
respectively such that 
and 
intersect at a point 
.
intersects line 
.
,
and the tangent to 
concur.
such that 
for all 
people playing a game at Cheesy's math casino, where 
be a positive integer. A subset of length 
from the set of integers from 
to 
,
for 
.
.
Let 
.
Let 
.
Let 
with edges 
.
three perpendiculars to the three lateral faces.
and radius 
.
.
exist that satisfy 
for all real 
and the domain of f is strictly within the set of 
,
?
with edge 
cm.
and tangent to the plane 
.
)
Y by Circumcircle, albgeo, vinhhop, trying_to_solve_br, Adventure10
Let 
be an acute scalene triangle. Let 
and 
be two distinct interior points of the segment 
such that 
. Suppose that:
and 
are the feet of the perpendiculars from from 
to the lines 
and 
respectively.
and 
are the feet of the perpendiculars from 
to the lines and respectively.
Prove that
and 
intersect on the line .
be an acute scalene triangle. Let 
and 
be two distinct interior points of the segment 
such that 
. Suppose that:
and 
are the feet of the perpendiculars from from 
to the lines 
and 
respectively.
and 
are the feet of the perpendiculars from 
to the lines and respectively.Prove that
and 
intersect on the line .
and then reflect everything across the bisector of angle 
.
.
for every point 
be the antipodal of 
in 
.
are just the intersection points of 
and 
Observe that 
because 
.
is cyclic. We need to prove that 
,
,
have a common point, different from 
Notice that 
,
So, we have to prove that 
,
and 
are corresponding points in the similar triangles 
and 
,
,
.
are concyclic.We will show 
cyclic and then the result follows from radical axis theorm
be the feet of 
concyclic and 
concyclic. It is clear from the assumption that 
and 
,
,
is concyclic. Furthermore, let 
and 
,
.
,
.
,
,
are concurrent , thus 
lies on 
,
.
and 
,
be the reflection of 
in 
.
,
are isogonal lines wrt 
Note that 
so by isogonality lemma we are done.
![$[XY]$](http://latex.artofproblemsolving.com/b/d/5/bd5db5e85aa6daea3eebecaea5d26721edd15203.png)
,
and 
these intersections. We have to prove that 
,
or after applying Menelaus' Theorem in 
,
which rewrites as :
which is true by Steiner's Theorem. Done.
and 
are cyclic, from which we can obtain that 
is also cyclic and notice that 
and 
.
,
and then 
;
,
.
.
and 
to happen.
and 
;
we infer that 
and 
concur at a point, say 
.
,
and 
Then, if we let 
be the points diametrically opposite 
in 
then by Pascal on 
we get that 
and 
are collinear, where 
and 
are all collinear, done!
,
,
and 
are cyclics.We use Power of point wrt to 
and 
, 
, 
. We will prove that points 
be the reflexion of 
is cyclic because 
and 
. 
because it's reflexion over the midpoint of 
and 
are isogonal in 
. 
and 
. 
. 
. If points 
or 
. Now using trig Ceva two times in quadrilateral 
, with point 
points 
and 
.
problem is equivalent to this:
.
(on shorter arc 
.
and 
at 
to 
lies on 
and 
.
or 
analogously 
(noticing that we just need to exchange 
with 
in intersection formula to get 
from 
by substituting 
.
and 
And now the only real computation is finding 
.Now we just bash, but first we do denominator because it seems easier. 
By symmetry (again we can just substitute 
Finishing we get 
Now we see something quite nice, we see the term 
which will after conjugation become 
,
)
and we wish to show that it lies on unit circle we need to have 
then 
with 
having modulus 
)
or 
and again by symmetry 
and finaly 
and we see 
which finishes problem.
because 
,
;
and 
; (Thales)
,
is a projection. Also, note that 
simply lie on the circle with 
as diameter and 
lie on the circle with 
as diameter. Thus, by perspectivity from 
is a projective map. Similarly, 
is also a projective map. This gives that the degree of each of the points 
is two. Now, when 
,
(since, in this case, 
)
(as 
and 
)
.
.
are concurrent for 
positions of 
,
,
,
.
and 
,
only coincide.
and 
follows similarly as above.
,
themselves coincide.
( I'd recommend 
)
)
and 
.
.
.
we get that cross ratios are equal and hence 
concur. Done
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(11cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -13.269930488359341, xmax = 14.253973384432696, ymin = -8.985982801873275, ymax = 9.190517353080942; /* image dimensions */
pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen ttffqq = rgb(0.2,1,0); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen ffqqff = rgb(1,0,1); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902);
/* draw figures */
draw((-4.061937705258801,5.8663828220423495)--(-7.84,-0.84), linewidth(0.8) + dbwrru);
draw((-7.84,-0.84)--(5.3250975403631085,-0.9374264798248775), linewidth(0.8) + dbwrru);
draw((5.3250975403631085,-0.9374264798248775)--(-4.061937705258801,5.8663828220423495), linewidth(0.8) + dbwrru);
draw((-4.061937705258801,5.8663828220423495)--(-5.92822452908305,-0.8541478294236663), linewidth(0.8) + ttffqq);
draw((-5.92822452908305,-0.8541478294236663)--(-6.0614518340739725,-1.3339016246998978), linewidth(0.8) + ttffqq);
draw((-7.84,-0.84)--(-2.329254657245816,3.8153118361890987), linewidth(0.8));
draw((-7.84,-0.84)--(-6.0614518340739725,-1.3339016246998978), linewidth(0.8) + wvvxds);
draw((3.201320105536144,-2.7315297480481187)--(-4.061937705258801,5.8663828220423495), linewidth(0.8) + ttffqq);
draw((-5.144005987567766,1.9698365005421956)--(5.3250975403631085,-0.9374264798248775), linewidth(0.8));
draw((5.3250975403631085,-0.9374264798248775)--(3.201320105536144,-2.7315297480481187), linewidth(0.8) + wrwrwr);
draw((-5.144005987567766,1.9698365005421956)--(-7.84,-0.84), linewidth(0.8) + wvvxds);
draw((-2.329254657245816,3.8153118361890987)--(5.3250975403631085,-0.9374264798248775), linewidth(0.8) + wvvxds);
draw((-2.329254657245816,3.8153118361890987)--(-9.411875854628114,-0.8283675505807218), linewidth(0.8) + dtsfsf);
draw((-6.0614518340739725,-1.3339016246998978)--(2.15484385900213,-3.6155613905528963), linewidth(0.8) + wvvxds);
draw((-4.061937705258801,5.8663828220423495)--(-4.669746318639019,1.83813491072802), linewidth(0.8) + ffqqff);
draw((-4.061937705258801,5.8663828220423495)--(2.15484385900213,-3.6155613905528963), linewidth(0.8) + ffqqff);
draw((3.201320105536144,-2.7315297480481187)--(2.15484385900213,-3.6155613905528963), linewidth(0.8) + wrwrwr);
draw((-5.144005987567766,1.9698365005421956)--(-2.983516530558368,4.221555943296035), linewidth(0.8) + wvvxds);
draw((-2.983516530558368,4.221555943296035)--(-2.329254657245816,3.8153118361890987), linewidth(0.8) + wvvxds);
draw((-7.84,-0.84)--(-9.411875854628114,-0.8283675505807218), linewidth(0.8) + dbwrru);
draw((-9.411875854628114,-0.8283675505807218)--(3.201320105536144,-2.7315297480481187), linewidth(0.8) + dtsfsf);
draw(circle((-4.365842011948909,3.8522588663851853), 2.036922466952745), linewidth(0.8) + sexdts);
draw(circle((-0.9535469231283354,1.1254107157447264), 5.669118958632441), linewidth(0.8) + sexdts);
/* dots and labels */
dot((-4.061937705258801,5.8663828220423495),dotstyle);
label("$A$", (-3.994017429931375,6.044928339047566), NE * labelscalefactor);
dot((-7.84,-0.84),dotstyle);
label("$B$", (-7.765149713800933,-0.6573209806258183), NE * labelscalefactor);
dot((5.3250975403631085,-0.9374264798248775),dotstyle);
label("$C$", (5.389131617611365,-0.7645569697405925), NE * labelscalefactor);
dot((-5.92822452908305,-0.8541478294236663),dotstyle);
label("$P$", (-5.85277457458746,-0.6751936454782806), NE * labelscalefactor);
dot((-6.0614518340739725,-1.3339016246998978),linewidth(4pt) + dotstyle);
label("$X$", (-5.99575589340716,-1.193500926199689), NE * labelscalefactor);
dot((-2.329254657245816,3.8153118361890987),linewidth(4pt) + dotstyle);
label("$Y$", (-2.260368939242526,3.95382655130947), NE * labelscalefactor);
dot((1.662820552255459,-0.9103243065214853),linewidth(4pt) + dotstyle);
label("$Q$", (1.743107987709043,-0.7645569697405925), NE * labelscalefactor);
dot((3.201320105536144,-2.7315297480481187),linewidth(4pt) + dotstyle);
label("$M$", (3.280157165020806,-2.587568784691753), NE * labelscalefactor);
dot((-5.144005987567766,1.9698365005421956),linewidth(4pt) + dotstyle);
label("$N$", (-5.066377321079116,2.112942071505847), NE * labelscalefactor);
dot((-9.411875854628114,-0.8283675505807218),linewidth(4pt) + dotstyle);
label("$G$", (-9.33794422081762,-0.6930663103307431), NE * labelscalefactor);
dot((-2.983516530558368,4.221555943296035),linewidth(4pt) + dotstyle);
label("$H$", (-2.903784873931171,4.364897842916104), NE * labelscalefactor);
dot((2.15484385900213,-3.6155613905528963),linewidth(4pt) + dotstyle);
label("$I$", (2.2256699387255265,-3.481202027314871), NE * labelscalefactor);
dot((-4.669746318639019,1.83813491072802),linewidth(4pt) + dotstyle);
label("$K$", (-4.601688034915095,1.9878334175386105), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/6/a/1/6a1c788737be029214a20f59b58f28d69e7ac035.png)
and 
are perspective with respect to a line or that if 
then 
are collinear.Let 
Firstly observe that since 
we have that 
is cyclic. Similiarly since 
we have 
is cyclic .Note that by isogonal line lemma we have that 
are isogonal. So we just need to prove 
as isogonal. Now note that 
since 
are isogonal and also 
so by 
similiarity we have
. Also note that as 
since 
so by 
Now (1), (2) imply that 
by 
similiarity 
and since 
and 
are cyclic we have 
are isogonal w.r.t 
.
are isogonal w.r.t 
. We are done. 
.
.
were collinear, then we would have 
,
Let 
and 
.
,
.
,
.
.
and 
.
Thus, plugging this is, our equation becomes
However, by the ratio lemma, 
and 
.
Since all of our steps are reversible, we have proven that the original expression is true. Thus, 
and 
come from?
is cyclic. Let 
and, consider the Miquel point 
.
, let 
denote the inverse of some point 
, and similarly 
and so 
lie on the circle with diameter 
, inverting back we're done. 
it must be 
or 
the midpoints of 
and 
so 
since 
so 
it's obvious that 
we have 
so 
suppose that 
are outside 
, done
is cyclic. Let 
and 
are cyclic, we can see that 
.
.
and 
and 
are the point 
,
and animating 
linearly (the condition has degree 4, which is now easy).
meet at 
is cyclic.
and 
.
so 
and 
are cyclic so 
and 
meet at 
so 
.
.
which means 
Thus, properties of anti-parallel lines and Phantom Points imply 
are collinear, which clearly finishes. 
and 
.
so points 
are concyclic by pop. Let 
.
.
.
and 
,
lies on the 
denote the perpendicular bisector of segment 
,
.
,
,
.
.
passes through the midpoint of 
passes through 
,
denote the polar of 
denote the pole of 
.
all pass through 
with 
being equally inclined with respect to 
(basically 
)![\[ K \in AX, AX \perp KB \Leftrightarrow P_{AX} \in \ell_K, AP_{AX} \perp AP_{KB} \]](http://latex.artofproblemsolving.com/c/e/1/ce102ad11ceb832020bda718796f5b83bd88d4b9.png)
which can be rephrased to 
and 
.
is the point on 
such that 
.
are collinear.
![\[ \measuredangle(\ell_B) + \measuredangle(\ell_C) = \measuredangle(\ell_X) + \measuredangle(\ell_Y) = \measuredangle(\overline{AP_{BK}P_{CT}}) + \measuredangle(\overline{AP_{CL}P_{BS}}) \]](http://latex.artofproblemsolving.com/9/a/1/9a10f8d64fab274427cd2380616d3a194e315111.png)
implies 
is cyclic (and allows us to delete 
and 
from the diagram). Let 
and 
meet the circle again at 
respectively. Then 
and 
are diameters, so they intersect at 
,
collinear. Similarly, 
collinear, so the result follows.
.
,
.![\[\frac{XK}{KA}\times\frac{AL}{LY}\times \frac{YD}{DX}=\frac{\tan \alpha}{\tan \beta}\times \frac{\tan \theta}{\tan \alpha}\times \frac{\tan \beta}{\tan \theta}=1\]](http://latex.artofproblemsolving.com/3/0/0/30010182c53b8f686faf51731a9f3e621a91c0dd.png)
from the right triangles, and isogonality.
are midpoint of 
and 
.
.
lie on a circle. We also have 
.
or 
.
.
and 
is 
we have 
.
then 
so 
is cyclic, we get 
and notice that 
,
.
Let 
,
Similarly, 
And by 
,
,
,
are concurrent, as desired.
.
,
.
and 
for 
and 
is collinear.
and 
.
,
and 
are collinear. If we denote 
by the parallelogram isogonolaty lemma 
and 
are isogonal as we wanted.
p
.
so 
concur. Further, note that 
are concyclic 
is the Miquel point of cyclic quadrilateral 
meet 
in 
meet 
in 
meet 
in 
,
are concyclic 
are concyclic 
are concyclic 
is the Miquel point of 
.
are concyclic with center 
and 
.
and 
are also concyclic.
so 
,
,
,
Let 
.
we have
and 
and 
as we wanted.
