AoPS Academy
is the circumcenter of 
, and 
are points on segments 
and 
respectively with 
. Let 
be a point such that 
and 
.
intersect and at points 
and 
respectively. Let the line passing through and perpendicular to 
intersect and at points 
and 
respectively. Prove that points 
, and are concyclic.
defined by 
and![\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/9/9/4/994aa754cc1288ce4f28a95a0276e64282fb5f66.png)
a) Prove that there exist infinitely many positive integers 
such that 
.![\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]](http://latex.artofproblemsolving.com/f/f/1/ff174e7431cbfbc17c650d109651241286756a1a.png)
and 
be altitudes in an acute triangle 
which meet at 
. Suppose that 
meets the circumcircle of at 
and 
such that lies on the shorter arc of 
and lies on the shorter arc of 
. Let 
and 
meet at 
. Show that the circumcircles of 
and 
and the line 
concur.
with the midpoint 
of 
. Draw the perpendicular 
from the orthocenter 
of to 
.
.
. EF meets BC at T.
(1)
(2)
be the altitudes. It is easy to see 
are concyclic and 
are concyclic.
so 
are conyclic. Then must be the Miquel point of 
which implies 
are concyclic. Then 
(
is isosceles) and so 
is similar to 
which implies .
、B
、C
(where 
). Then M is 
and we should prove 
.
, and H is 
.
, and line HP is 
.
.
and hence 
.
,
,
be the feet of three altitudes of the triangle . Then, 
is cyclic. is the orthocenter of the triangle made by 
, 
, 
is the polar of 
w.r.t. () and then 
, which is the property of harmonic division.
the antipode of on the circle 
, 
; obviously, 
are collinear, hence by power of w.r.t. 
: 
, but 
, so 
; from power of w.r.t. 
we get 
, so it's done.
with the midpoint of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. Draw the perpendicular from the orthocenter to . Show that 
.
, 
, 
be the projections of to the sides of and 
. Are well-known that
and 
is a harmonical division. In conclusion, 
, i.e. 
.
and , thus the points are concyclic. It is well known that 
is tangent to the circle 
that passes throught the points 
(just denote 
the midpoint of 
, thus is the center of and show that 
, it is an easy angle chase).
.
with the orthocenter and the midpoint of. Denote the projection of on . Show that 
.
. Using the power of w.r.t. the circumcircle of the quadrilateral 
![$m_a^2-bc\cdot \cos A=\frac 14\cdot\left[2\left(b^2+c^2\right)-a^2-2\cdot \left(b^2+c^2-a^2\right)\right]=\frac {a^2}{4}=MB^2$](http://latex.artofproblemsolving.com/c/b/f/cbfe99d1a20a9a2c70155143c5e35db74ecac4d0.png)
.
and 
be the foot of altitudes from and 
,respectively. Let 
. It is obviously that 
where 
is the orthocenter of 
and 
,because 
is right-angled. From sine law for 
and 
, we get:
is tangent in to the circumcircle of 
. From here follows that : 
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm);
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draw((-2.468863291036393,-4.078599961106974)--(12.802714838158101,-4.587652565413459), linewidth(0.8) + blue);
draw((12.802714838158101,-4.587652565413459)--(1.2217680901856096,5.147978491948061), linewidth(0.8) + blue);
draw((1.053897946824669,0.1118741911198956)--(2.983266984406037,0.9146988913215361), linewidth(2));
draw((1.2217680901856096,5.147978491948061)--(5.166925773560854,-4.333126263260216), linewidth(0.8) + blue);
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draw((2.983266984406037,0.9146988913215361)--(-2.468863291036393,-4.078599961106974), linewidth(0.8) + blue);
draw((2.983266984406037,0.9146988913215361)--(12.802714838158101,-4.587652565413459), linewidth(0.8) + blue);
draw(circle((13.210755203392592,7.653558391621359), 12.248009741761717), linewidth(0.8) + linetype("4 4") + red);
draw(circle((1.137833018505139,2.629926341533978), 2.5194506799028615), linewidth(0.8) + linetype("4 4") + red);
draw((-0.5379741022876767,0.7486230107648315)--(1.053897946824669,0.1118741911198956), linewidth(0.8) + blue);
draw((1.053897946824669,0.1118741911198956)--(12.802714838158101,-4.587652565413459), linewidth(0.8) + blue);
draw((1.2217680901856096,5.147978491948061)--(1.053897946824669,0.1118741911198956), linewidth(0.8) + blue);
draw((2.983266984406037,0.9146988913215361)--(-0.5379741022876767,0.7486230107648315), linewidth(0.8) + blue);
draw((1.053897946824669,0.1118741911198956)--(-2.468863291036393,-4.078599961106974), linewidth(0.8) + blue);
draw(circle((5.0829907018803855,-6.851178413674303), 8.044730573858503), linewidth(0.8) + linetype("4 4") + red);
/* dots and labels */
dot((1.2217680901856096,5.147978491948061),dotstyle);
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dot((-2.468863291036393,-4.078599961106974),dotstyle);
label("$B$", (-2.3683137143186044,-3.853771759954426), NE * labelscalefactor);
dot((12.802714838158101,-4.587652565413459),dotstyle);
label("$C$", (12.902522808965257,-4.35445492465226), NE * labelscalefactor);
dot((5.166925773560854,-4.333126263260216),linewidth(4pt) + dotstyle);
label("$M$", (5.255725384489285,-4.14962999363951), NE * labelscalefactor);
dot((1.053897946824669,0.1118741911198956),linewidth(4pt) + dotstyle);
label("$H$", (1.1364684385662163,0.2882435116367458), NE * labelscalefactor);
dot((2.983266984406037,0.9146988913215361),linewidth(4pt) + dotstyle);
label("$P$", (3.0709261203532927,1.1075432356877468), NE * labelscalefactor);
dot((-0.5379741022876767,0.7486230107648315),linewidth(4pt) + dotstyle);
label("$H'$", (-0.45661435819961127,0.9254766303430799), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/3/2/9/3294730feafa33fa8152e4ef8b33c527b9965d05.png)
is the Humpty point of .
is a cyclic quadrilateral.
and let 
be the intersection of 
with .
is cyclic, since 
.
Thus we have that 
is tangent to 
.
, which implies that 
is tangent to 
.
then we have is the antipode of on the circle and 
is 
is cyclic 
with the midpoint of . Draw the perpendicular from the orthocenter of to ..
is the A-Humpty point of 
By 
be the feet of the 
and 
altitudes. Then as 
, we get 
to be concyclic. Now from the Three Tangents Lemma in EGMO Chapter 1 (Alternatively a simple angle chase), combined with Power of a Point, we get 
. Furthermore, since 
is right-angled at 
, we get 
, so 
. 
and 
)![[asy]
import olympiad;
pair B=(0,0), C=(21,0), A=(5,12), D=(5,0), M=(21/2,0);
pair H=orthocenter(A, B, C), E=foot(B, A, C), F=foot(C, B, A);
pair P=foot(H, A, M);
draw(A--B--C--cycle);
draw(M--A--D);
draw(C--F);
draw(H--P);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$M$", M, S);
label("$F$", F, NW);
label("$P$", P-(0.5, 0), NE);
label("$H$", H, SW);
draw(circumcircle(H, D, M));
draw(rightanglemark(H, P, M, 20));
draw(rightanglemark(H, D, M, 20));
draw(rightanglemark(A, F, C, 20));
[/asy]](http://latex.artofproblemsolving.com/f/4/5/f45f69dffb86d47b900d380b6a12e8220498dfe5.png)
be the foot of the altitude from to and let be the foot of the altitude from 
to 
Observe that 
so 
is cyclic. Now, by power of a point, 
As such, since 
Also, 
Thus, 
so 
so 
as desired. 
aand 
Prove that 
are altitudes of triangle,than 
lies on a circle.So 
,
lies on a circle.
,
,
.
.
,
are cyclic, and the radical axis 
concur at 
is cyclic because 
,
.
is harmonic.
be the midpoint of 
as 
which implies 
(where 
circumcentre of 
c
,
is also well known).
