AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
and 
play a game, given an integer 
, writes down 
first, then every player sees the last number written and if it is 
then in his turn he writes 
or 
, but his number cannot be bigger than . The player who writes wins. For which values of does win?
be the circumcircle of a scalene triangle 
. Let 
be a circle internally tangent to in . Tangents from touch in 
and 
, such that lies in the interior of 
. Similarly, tangents from 
touch in 
and 
, such that lies in the interior of .
and 
concur on the bisector of 
.
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
is the largest positive integer that divides both 
and 
. Integers and are coprime if 
.
, 
and 
are the midpoints of sides 
and 
respectively. Point lies inside the triangle. Let 
. Prove that 
.
, 
, and 
of a hexagon 
inscribed in a circle 
intersect at a point 
, and the acute angle between any two of them is 
. Let 
be the radius of the circle tangent to segments 
and 
and internally tangent to ; the radii 
, 
, 
, 
, and 
are defined similarly. Prove that![\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]](http://latex.artofproblemsolving.com/0/d/6/0d66d6685bbc97c7553e9ed3ba14ea0dc3328b49.png)
, let 
be the circumcentre, 
be the orthocentre, and 
be the centroid.
satisfy the following condition:
is constant.
. Find .
such that numbers 
are both perfect squares.
such that ![\[\left(g(m)+n\right)\left(g(n)+m\right)\]](http://latex.artofproblemsolving.com/5/a/c/5ac9fce315e8a524a3a01b345cfe76bc76056d68.png)
is a perfect square for all 
be an acute triangle inscribed in a circle of center . If the altitudes 
intersect at and the circumcenter of 
is 
, prove that 
is a parallelogram.
is a perfect square number.
is good if there exists a permutation 
of 
such that 
is perfect square for all 
. Determine all the good numbers in the set 
.
of positive integers such that 
and 
are both powers of 
.
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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pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
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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);
draw(circle((-2.292829886339947,1.2024021797863635), 5.2839352164545925), linewidth(0.8) + linetype("4 4") + red);
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);
label("$A$", (1.3185350439108823,5.386108461287419), NE * labelscalefactor);
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. 
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).
