AoPS Academy
be a sequence of positive real numbers such that for every 
, we have:![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
, the following holds:![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
, 
, 
, 
, where and 
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
, such that for any positive integer 
and non-negative reals 
, the following inequality always holds:
Here 
and 
denote the median and arithmetic mean of 
, respectively.
be an isosceles triangle with 
, and 
an arbitrary point on side 
. The internal angle bisector of 
meets the circumcircle of 
again at 
, and the internal angle bisector of 
meets it again at 
. Show that lines 
, 
and 
are concurrent.
, 
and 
are points on sides 
and , respectively, such that 
. If 
and 
, find the area of .
Find 
.
Taking the square root of the second equation, we get
has a right angle at 
, with 
and 
. If 
and 
are points
and 
, respectively, such that 
, find the perimeter of quadrilateral
.
, 
, 
are positive integers such that 
. Find all possible values of 
.
to rearrange the terms and place the higher-degree terms at the front of the expression in order to have a more recognizable form.
![\begin{align*}
a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 &=a^4-2a^2(b^2+c^2)+b^4-2b^2c^2+c^4 \\
&= a^4-2a^2(b^2+c^2)+(b^2+c^2)^2 - 4b^2c^2 \\
&= [a^2 - (b^2+c^2)]^2-(2bc)^2 \\
&= (a^2-b^2-c^2+2bc)(a^2-b^2-c^2-2bc)\\
&=[a^2-(b-c)^2][a^2-(b+c)^2] \\
&= (a+b-c)(a-b+c)(a+b+c)(a-b-c) = -63.
\end{align*}](http://latex.artofproblemsolving.com/0/c/4/0c430b7562ec7571fdf3e492939c354e0c3e7e20.png)
, we get
.
, 
, or 
, we get something of the form![\[
a^2(a+2b)(a-2b)=-63.
\]](http://latex.artofproblemsolving.com/1/0/0/1006c2fabc10f934c25b6e6ae822a59ce2dd2712.png)
and 
are the only square factors of 
, we have 
.
are 
and 
, including all their permutations.
are all distinct positive integers, there are no solutions. are 
.
. We have
is an imaginary number. Thus, N must be equal to +2. We then have
and 
, find the value of 
.
. We have
.
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);
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 */
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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);
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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);
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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. 
Prove that
side length of 
area of 
is a right angle, and 
,
.
is a right angle, 
,
.
and 
both contain a right angle and 
through AA postulate. Because they are similar, 
,
,
,
.
.
,
.
and 
Prove that
.
,
is a right angle, using pythagorean theorem, we can get 
.
,
.
from point 
be the intersection of said altitude and 
.
both have a right angle and 
,
through AA Postulate. 
,
,
.
(CD = DE, as stated in the problem), 
,
(they were formed by an altitude, and are therefore both right angles), we can show that 
through HL Theorem. 
corresponds with 
,
.
,
and 
.
.
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).
