AoPS Academy
with 
. Suppose that the inequality 
takes place for all choiches of the 
signs. Show that there exists a line 
through 
such that all points 
are all on one side of .
angle 
. Point 
is the midpoint of 
, and 
is the foot of altitude from point 
. Points 
and 
are marked such that 
is equilateral, and 
and this points lie in the same half-plane with respect to , not in the same as .
and 
are tangent.
and 
are given in space, such that 
. Let , 
and 
be the midpoints of 
, 
and 
. Prove that:
is a right triangle;
and 
are perpendicular. with circumcircle 
, and two arbitrary points 
on . Let , 
, 
be points on lines , 
, 
, respectively, such that 
, 
, and 
concur at a point . Let 
be a point on line such that 
, 
, , are concyclic. Similarly, let 
be a point on line such that , , , are concyclic, and let 
be a point on line such that , , , are concyclic. Prove that 
, 
, 
concur at a single point.
is the foot of the altitude from of triangle . On the lines and 
points and are marked such that the circumcircles of triangles 
and 
are tangent, call this circles 
and 
respectively. Tangent lines to circles and at and intersect at 
.
.
is given.
is the interior and the incircle of meets 
at 
. and meet at . Let 
be a tangent from D to the circumcircle of 
, and define 
and 
on and , respectively.
meet at one point.
two not intersecting circles 
and 
are drawn such that is tangent to and at points and respectively, and is tangent to 
and at and 
respectively. It is known that 
. and .
, we have 
, 
, and 
centimeter. Find the length of diagonal .
is given and let be its orthocenter. Let 
be the circle through 
, 
and , and let be the circle with diameter 
. Let 
be the other intersection point of and , and let 
be the reflection of over 
. and intersect again at 
, and line and intersect again at 
. Show that the circle through 
passes through the midpoint of segment .
and intersect each other at 
,and lies on . Let be a point on such that 
. Line 
intersects at (other than ). The bisector of 
intersects at 
( and are on the same side of the line 
). Let 
be a point on such that 
( and are on the same side of the line ). Prove that 
.
be a triangle with orthocentre . The tangent lines from to the circle with diameter touch this circle at and . Prove that 
and are collinear.
be a triangle with orthocenter . The tangent lines from to the circle 
with diameter ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
touch it on and . Prove that 
.
of 
, 
, 
for which 
and 
. Observe that 
, i.e. 
what is a characterization of 
- harmonical division, i.e. is harmonical conjugate of w.r.t. 
. In conclusion, belongs to polar of w.r.t. , i.e. . separates , . Denote 
so that 
and 
so that 
. Observe that 
and 
. Thus 
, i.e. 
, 
, 
are concurrently .
be a cyclic hexagon. Prove that 
.
as the circle of reciprocation.
are collinear its the same as proving that their polars are concurrent. Because both belong to the circle, their polars are the tangents to the circle that meet at . passes through . be the perpendicular to 
( is the midpoint of ). If is the inverse of , then is the polar of and we are done.
.
~
we have :
. 
the altitudes of 
, and the midpoint of . Because 
, the points 
are concyclic. Let the circumcircle of 
. and radius 
. Because 
, we obtain that is the image of under this inversion, and 
remains invariant. For the other hand, is transformed into the line containing . Because 
, we can conclude that 
.
and are the intersections of the circles 
and 
, and being the midpoints of 
and 
respectively.
are the feet of the altitudes on 
and respectively and these altitudes concur at , then we know that 
, i.e. has equal power w.r.t. to both circles, consequently it belongs to their radical axis .
are switched.
are switched.
from Virgil's lemma, we can also assume that it's wrong i.e there is three intersection point, if I remembre well it's we get that 
with 
, 
and 
which obviously wrong, and then deduct the desired result.
and 
(X is on the same side with A from BC)
is collinear and perpendicular to 
and 
is harmonic
is harmonic
is collinear since 
is collinear
.Let me inform you that this problem is from China 1997 and i have three solutions but two of them have already posted so i'll post the third one:
the altitudes of , and the midpoint of .Since 
then the points are concyclic.
and 
be a triangle and its altitude .The tangent lines from to the circle with diametre touch this circle on and .If 
then prove that is the orthocenter of 
be the midpoint of .Extend 
to meet 
at 
where and are on the same side of .Now is the pole of wrt the circle 
.Again note that 
is the polar of .So polar of passes through 
lies on .Hence proved.
, , . Let the foot of the perpendicular from to be . By http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2173258&sid=fd7168f527cac736647f7961645922f0#p2173258, 
so the polar of with respect to the semicircle with diameter passes through , and therefore . So , , are collinear.
and .
lie on the circle 
with diameter .
cut at . 
is the polar of with respect to 
is also the polar of . It follows that 
are collinear.
be the midpoint of , 
be the foot of onto , 
be the foot of onto . We just want to show that is on the radical axis of the circles with diameters and (which is ). But this is obvious, since 
. is implied to be acute...
be the intersections of the -altitude with the circle.
which is obvious.
be the feet of perpendiculars from respectively 
as the center and radius 
. to themselves and it takes to feet of perpendicular from to (
)thus
and are concyclic hence the conclusion
is on the cicrcle such as the is perpendicular to and analogus the point F.
and 
and we have from theorem of Hire that the intersection point of BE and CF is on the polar of the point A which is PQ so P,Q and H are collinear.
be circumcircle of and 
be a circle so that 
and exists 
such that 
Let 
and the symmetric of 
w.r.t. 
The tangent lines from to the circle touch it on and 
Prove that 
(see PP10 from here).
. Then let 
be points on the unit circle, so 
. Then if 
is the orthic triangle, 
Now with the chord intersection formula, 
It remains to compute 
so 
are collinear, as required.
be a triangle with orthocenter . The tangent lines from to the circle with diameter touch it on and . Prove that .
denote the mid of segment 
Clearly, 
are concyclic in that order. Also, 
..Therefore 
must lie on the radical axes of the circles 
and 
which clearly is the claimed line the 
.
be a cyclic hexagon. Prove that . lies outside triangle in given problem.
wrt 
both pass through , so by La Hire it suffices to prove that the polar of does as well. Let the reflection of over the midpoint of be 
, and 
; we can see that
by orthocenter reflection, and since is the -queue point, we know that 
, which implies that 
is the polar of .
and radius . Then the problem is equivalent to showing that lies on the radical axis of this circle and the circle with diameter . We can do this by showing has equal powers with respect to both circles.
or 
. Notice that we have 
, and 
, giving 
. Then let the foot from to be . We can then write 
. We then want to show 
, which we can reduce to 
. Then by Law of Cosines, we have 
where is the foot from to . We can then write 
. By Power of a Point on 
, we have 
. We then just need to show 
. Multiply everything by two to get the equivalent 
and rearranging gives the equivalent 
, but 
, so we are done by LOC.
be the altitude from to 
and the circumcenter of 
. Also denote by 
the angle 
modulo 
. Then we make the following claim:
is cyclic.![\[\measuredangle APO = \measuredangle AQO = \measuredangle ADO = 90^{\circ}. \blacksquare\]](http://latex.artofproblemsolving.com/6/7/b/67b7b13bcc6f410eb5ce219358d90f2ee5ea159e.png)
and 
, that is the radical center of all of our present circles so that ![\[AP^2 = AQ^2 = AH \cdot AD \implies \triangle AHP \sim \triangle APD; \phantom{c} \triangle AHQ \sim \triangle AQD.\]](http://latex.artofproblemsolving.com/d/2/a/d2a48806b7ca4c79586873d9f66fa476062228a5.png)
so that are collinear, as desired. 
be the orthic triangle, be the midpoint of .
. Also, 
clearly lie on the circle with diameter . with radius 
, then and go to themselves, and goes to .
are concyclic, inverting back gives that collinear as desired. 
, 
, 
as the feet of the 
, 
, 
altitudes respectively and as the midpoint of 
.
and 
, thus allowing the circle with diameter to be the unit circle. and also lie on this circle. We set 
and 
as free variables.![\[A = \frac{2pq}{p+q}\]](http://latex.artofproblemsolving.com/1/5/6/1562090eb0402f0dba760e24a5a3f1d4dbee5e73.png)
We shall find , notice that it is the intersection of 
with the unit circle and is thus given by the formula![\[e = \frac{1-\frac{2pq}{p+q}}{\frac{2}{p+q}-1} = \frac{p+q-2pq}{2-p-q}\]](http://latex.artofproblemsolving.com/c/c/1/cc174555499b0bf36ff4183491aa8bbcec8775eb.png)
We similarly find![\[f = \frac{\frac{2pq}{p+q} + 1}{\frac{2}{p+q}+1} = \frac{2pq+p+q}{2+p+q}\]](http://latex.artofproblemsolving.com/c/e/a/cea3ff7bb2a70e507b7615b8cffd18ffcef05ab0.png)
To find we intersect 
and 
, we thus obtain![\[H = \frac{\frac{2pq-p-q}{2-p-q} \cdot (1 + \frac{2pq+p+q}{2+p+q}) - \frac{2pq+p+q}{2+p+q} \cdot (\frac{p+q-2pq}{2-p-q}-1)}{\frac{2pq-p-q}{2-p-q} - \frac{2pq+p+q}{2+p+q}}\]](http://latex.artofproblemsolving.com/7/3/a/73a4cdd6264e3f76a1c1cec9811c7029c31ce8cf.png)
Simplifying, we have![\[H = \frac{2(2pq-p-q)(pq+p+q+1) + 2(2pq+p+q)(pq-p-q+1)}{(2+p+q)(2pq-p-q) + (p+q-2)(2pq+p+q)}\]](http://latex.artofproblemsolving.com/7/c/6/7c6fa2016e4da7193377c76d0a73dc48d812990e.png)
Further simplifying, we obtain![\[H = 2 \cdot \frac{4pq(pq+1) - 2{(p+q)}^2}{4(p+q)(pq-1)} = \frac{2pq(pq+1)-{(p+q)}^2}{(p+q)(pq-1)}\]](http://latex.artofproblemsolving.com/2/5/1/251b49a39a403ab54910673e8b64ebe1f7ddd4f8.png)
Now we shall prove the colinearity condition, notice that![\[\frac{H-p}{H-q} = \frac{\frac{2pq(pq+1)-{(p+q)}^2}{(p+q)(pq-1)} - p}{\frac{2pq(pq+1)-{(p+q)}^2}{(p+q)(pq-1)} - q} = \frac{(2p^2q^2 - p^2 - q^2) - p(p+q)(pq-1)}{(2p^2q^2 - p^2 - q^2) - q(p+q)(pq-1)}\]](http://latex.artofproblemsolving.com/e/9/6/e96c6cc47757686914ce938a899f673f017dd322.png)
From which![\[\frac{H-p}{H-q} = \frac{q(q - p)(p^2-1)}{p(p-q)(q^2-1)} = -\frac{q(p^2-1)}{p(q^2-1)}\]](http://latex.artofproblemsolving.com/a/0/8/a0832234be68bc35b78b0f3ea61f30201004f1cc.png)
We can simply take the conjugate to obtain![\[\overline{\left(\frac{H-p}{H-q}\right)} = -\frac{\left(\frac{1}{q}\right) \cdot (1 - \frac{1}{p^2})}{\left(\frac{1}{p}\right) \cdot (1 - \frac{1}{q^2})} = -\frac{q(p^2-1)}{p(q^2-1)}\]](http://latex.artofproblemsolving.com/5/8/c/58c541677097ccd5d62e99e011862659446d81d5.png)
Thus 
.
. Then from Brokard's theorem, triangle 
is self-polar, leading to the required result.
.
then: 
and 
and since 
is cyclic, we have 
the equality becomes then: 
since 
thus: 
then 
from which follow the collinarity of 
are concyclic where 
is the midpoint of 
and 
circles with diamters 
and 
and 
then by radical axis theorem it follows immediatly that 
and hence the points 
.
by the definition of radical axis.
and 
,
after rearranging and multiplying by 4.
,
and 
we have
,
.
,
.
,
and 
.
.
,
.
so the result follows.
,
.
,
,
.
.
and 
,
.
,
,
with respect to 
,
,
,
be the inversion operator, ![\[\Psi(AD \cap PQ) = \Psi(AD) \cap \Psi(PQ) = AD \cap (APQ) = D \implies AD \cap PQ = H,\]](http://latex.artofproblemsolving.com/2/1/d/21df81fe7ef438f3a9ba23868558df7af227a6a7.png)
also.
is a cyclic quadrilateral since
Thus notice that the power of 
,
is 
.
is Humpty, since 
,