intersect at the point 
.
,
is the altitude of the triangle, the point 
is marked. Points 
and 
are the feet of perpendiculars from point 
and 
,
and 
intersect at 
.
.
be real numbers such that 
.
and 
.
Prove that
,
be a point on arc 
of circle 
.
and 
.
and 
be the reflection of 
and 
,
intersects 
,
intersects 
.
lies on 
Prove that
Let 
Prove that![$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$](http://latex.artofproblemsolving.com/f/5/e/f5e68204507aa9c3e71c28844bbee3a804546b94.png)
Prove that
Let 
Prove that
Let 
Prove that![$$ a+b^2+c\geq \frac{135 \sqrt[3]{90}-2}{2}$$](http://latex.artofproblemsolving.com/8/6/6/866623c65f54d6147a43b1b241c2cba96863b624.png)
such that 
for all 
such that 
.
be a positive integer such that the sum of its positive divisors is at least 
.
distinct prime factors.
![\[{\ \{\frac{n}{10^k}} \} > \frac{n}{10^{10}} \]](http://latex.artofproblemsolving.com/0/c/a/0ca11fa76313b65fd76a3cceee90955960e44ef6.png)
.
is given. Find all functions 
such that for all 
we have
be an acute triangle, 
its orthocenter and 
to 
and 
to 
and 
and 
are the reflections of 
are the intersections of 
and 
respectively with the circumcircle of 
and 
lies on 
.
Y by Adventure10, Mango247, Rounak_iitr
Let 
be a quadrilateral and let 
be a line. Let intersect the lines 
at points 
respectively. Given that these six points on are in the order 
, show that the circles with diameter 
are coaxal.
be a quadrilateral and let 
be a line. Let intersect the lines 
at points 
respectively. Given that these six points on are in the order 
, show that the circles with diameter 
are coaxal.This post has been edited 1 time. Last edited by fattypiggy123, Mar 13, 2017, 1:34 AM
Reason: Typo
Reason: Typo
are pairs of an involution on 
has the same power 
to all three circles by harmonics, so the perpendicular bisector of 
must exist, the center of inversion is still the midpoint of these two points, so this is another way to patch the original solution. One still must take a certain amount of care to work out the details, though, when working in the complex projective line.
,
,
,
be four circles. Define 
to be the exsimilicenter of 
for 
.
are coaxal.
belong to a line 
;
similarly. If these lines are pairwise distinct, the claim follows from 
;
centered at 
,
are the points 
,
are the exsimilicenters of 
.
must be swapped under some inversion with center 
on 
Since 
However, we know that the centers of the circles are collinear, so 
must be coaxial; clearly, the former is impossible, so we're done. 
are pairs swapped under some involution. If this is a reflection, the problem is done. Otherwise, suppose that the involution is an inversion, say with center 
meaning that 
and 
Now if you draw the line perpendicular to 
circles, as desired. Note: 
(lengths are signed).
there exists a unique 
such that 
.
respectively, thus 
-
and any other conic through 
(which can be expressed as 
for some reals 
)
is the same as that of 
and 
,
)
we have (in signed distances) 
,
:
which is perpendicular to 
then 
so 
which is absurd. 
![[asy]
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
/*
A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings.
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pair A = (-112.56662,71.45286);
pair B = (-140,-80);
pair C = (98.17049,-82.53832);
pair D = (14.16640,154.49649);
pair X = (-180.85092,-305.52787);
pair Y = (-72.06909,-80.72397);
pair Zp = (118.67646,313.46266);
pair Z = (-28.24650,9.83792);
pair Xp = (25.82900,121.58806);
pair Yp = (54.56188,180.96621);
import graph;
size(12.1cm);
pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps);
draw(A--B, linewidth(0.5));
draw(B--C, linewidth(0.5));
draw(C--D, linewidth(0.5));
draw(D--A, linewidth(0.5));
draw(B--Zp, linewidth(0.5));
draw(Zp--X, linewidth(0.5) + red);
draw(X--B, linewidth(0.5));
draw(D--Yp, linewidth(0.5));
draw(A--C, linewidth(0.5));
draw(circle((-77.51096,-91.96990), 237.24703), linewidth(0.5) + blue);
draw(circle((-8.75360,50.12111), 145.35917), linewidth(0.5) + blue);
draw(circle((45.21497,161.65029), 168.65226), linewidth(0.5) + blue);
dot("$A$", A, NW);
dot("$B$", B, W);
dot("$C$", C, E);
dot("$D$", D, NW);
dot("$X$", X, SW);
dot("$Y$", Y, 2*dir(270));
dot("$Z'$", Zp, NE);
dot("$Z$", Z, 2*dir(270));
dot("$X'$", Xp, NE);
dot("$Y'$", Yp, 2*E);
[/asy]](http://latex.artofproblemsolving.com/f/6/9/f69c388758f51ac2bcc44af87538ced9b1a3a703.png)
,
and 
.
and 
share the same midpoint and thus the order of the points would be 
or 
.
with 
as given in the problem statement.
.
and 
too. Now consider the radical axis of 
)
share a common radical axis. Similarly, 
also share the common radical axis which means that the circles 
are coaxial and we are done. 
but this is just inversion for some point 
,
and hence the circles are coaxial.
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(6cm);
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 = -7.699762129321672, xmax = 6.559203545586037, ymin = -7.052415271250787, ymax = 6.240149911824218; /* image dimensions */
pen qqwwzz = rgb(0,0.4,0.6); pen ccqqqq = rgb(0.8,0,0);
/* draw figures */
draw((-4.757561676220963,-2.701927853701796)--(-2.9394930360882117,4.271331028839409), linewidth(0.7) + blue);
draw((-2.9394930360882117,4.271331028839409)--(-1.2281172214791287,-0.413771068210891), linewidth(0.7) + blue);
draw((-1.2281172214791287,-0.413771068210891)--(2.5671300893294786,0.03546708777647148), linewidth(0.7) + blue);
draw((2.5671300893294786,0.03546708777647148)--(-4.757561676220963,-2.701927853701796), linewidth(0.7) + blue);
draw(shift((-1.1277580349207321,0.9301856222504602))*xscale(2.627452391198322)*yscale(2.627452391198322)*arc((0,0),1,-23.479199827160983,156.52080017283902), linewidth(0.7) + qqwwzz);
draw(shift((-1.1277580349207321,0.9301856222504602))*xscale(2.6274523911983216)*yscale(2.6274523911983216)*arc((0,0),1,156.52080017283905,336.520800172839), linewidth(0.7) + qqwwzz);
draw(shift((-0.07117818042485069,0.4712276449589007))*xscale(1.918189669306153)*yscale(1.918189669306153)*arc((0,0),1,-23.479199827160983,156.52080017283902), linewidth(0.7) + qqwwzz);
draw(shift((-0.07117818042485069,0.4712276449589007))*xscale(1.918189669306153)*yscale(1.918189669306153)*arc((0,0),1,156.52080017283905,336.520800172839), linewidth(0.7) + qqwwzz);
draw(shift((2.3708766145451667,-0.5895540123530806))*xscale(2.526602380171547)*yscale(2.526602380171547)*arc((0,0),1,-23.479199827160983,156.52080017283902), linewidth(0.7) + qqwwzz);
draw(shift((2.3708766145451667,-0.5895540123530806))*xscale(2.5266023801715476)*yscale(2.5266023801715476)*arc((0,0),1,156.52080017283902,336.520800172839), linewidth(0.7) + qqwwzz);
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/* dots and labels */
dot((-4.757561676220963,-2.701927853701796),dotstyle);
label("$A$", (-5.019368325982243,-3.3321114106731107), NE * labelscalefactor);
dot((-2.9394930360882117,4.271331028839409),dotstyle);
label("$B$", (-3.0318654377917826,4.507922624592221), NE * labelscalefactor);
dot((-1.2281172214791287,-0.413771068210891),dotstyle);
label("$C$", (-1.1537663783090537,-0.9799290943265048), NE * labelscalefactor);
dot((2.5671300893294786,0.03546708777647148),dotstyle);
label("$D$", (2.6388996835589813,0.22293933512359834), NE * labelscalefactor);
dot((-3.5376699074458338,1.9770050119621412),linewidth(4pt) + dotstyle);
label("$X$", (-4.2617979818075026,1.9551666223555948), NE * labelscalefactor);
dot((-1.8305508995277082,1.2354653322718627),linewidth(4pt) + dotstyle);
label("$Y$", (-2.5113367224837943,0.770532164350039), NE * labelscalefactor);
dot((0.05346485343287957,0.4170851071664104),linewidth(4pt) + dotstyle);
label("$Z$", (-0.35089794848817917,0.6240867069036397), NE * labelscalefactor);
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label("$X'$", (0.7890121101994477,0.3317694779008617), NE * labelscalefactor);
dot((1.6881945386780068,-0.29301004235406136),linewidth(4pt) + dotstyle);
label("$Y'$", (1.5454350824458382,-0.9799290943265048), NE * labelscalefactor);
dot((4.688288375657454,-1.5961931318725715),linewidth(4pt) + dotstyle);
label("$Z'$", (4.936380086421348,-1.6004578093311348), NE * labelscalefactor);
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label("$P$", (0.21435516650176835,0.423513021013619), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/5/1/5/515d1883395ec90a80745949df3363426325ac56.png)
,
and 
be the circles 
,
and 
.
,
are reciprocal pairs for a unique involution a unique involution along 
Then 
and 
,
are pairs of points swapped by an involution. Also, the involution can be defined as inversion from a point 
.
by inversion. Now this means that 
and 
and 
and 
are swapped by some involution. Since involutions are inversions/reflections, the center 
so 
and 
the circles with diameters 
and 
and 
.![\[PX\cdot PX' = PY \cdot PY' = PZ \cdot PZ'\]](http://latex.artofproblemsolving.com/9/7/b/97bd37e348024f222164cc354229b83f0c613191.png)
Thus, the pairwise radical axes of 
, 
and 
where 
is moving. 
is 
To prove the three functions are linearly dependent, we only need
However we may write 
where 
thus the above equality is 
with degree 3. Therefore we only need to plug in 4 special 
and using 
it follows that the radius of inversion squared is 
Thus, 
which implies that they are coaxial.
