[/quote]
,
such that 
,
lie on a line in this order and satisfy 
.
and 
be the midpoints of 
and 
,
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
,
and 
and 
and 
be all the positive integers smaller than 
such that
for all 
.
satisfied 
Find minimum and maximum of 
of positive integers is called central if for every positive integer 
,
terms of the sequence is equal to 
,
,
,
of positive integers such that for every central sequence 
,
,
,
.
and 
and 
.![\[
\frac{1}{x} + \frac{1}{y} + 2 \sqrt{\frac{2}{x^2 + y^2}} + 4 \geq 4 \left( \sqrt{\frac{2}{x^2 + 1}} + \sqrt{\frac{2}{y^2 + 1}} \right).
\]](http://latex.artofproblemsolving.com/1/b/8/1b8ab2e343a3043271f5a392d32b12aca748f1b9.png)
be all the positive integers smaller than ![\[gcd(N, c_i + c_{i + 1}) \neq 1\]](http://latex.artofproblemsolving.com/1/0/7/1075290d442e8c1f54237aad8ad07126f455f1f1.png)
for all 
of positive integers is called central if for every positive integer 
of positive integers such that for every central sequence 
,
.
,
,
are concurrent and the point of concurrence is 
,
are on sides 
such that 
is a cyclic quadrilateral. Let 
and 
.
.
be the midpoints of 
.
.
be non-negative real variables with 
Prove that
Equality holds iff 
or 
where 
?
be a convex quadrilateral such that diagonals 
and 
intersect at right angles, and let 
be their intersection. Prove that the reflections of 
are concyclic.
Y by Omeredip, centslordm, Adventure10, megarnie, Mango247, and 1 other user
Let 
be the center of the square inscribed in acute triangle 
with two vertices of the square on side 
. Thus one of the two remaining vertices of the square is on side 
and the other is on 
. Points 
are defined in a similar way for inscribed squares with two vertices on sides and , respectively. Prove that lines 
are concurrent.
be the center of the square inscribed in acute triangle 
with two vertices of the square on side 
. Thus one of the two remaining vertices of the square is on side 
and the other is on 
. Points 
are defined in a similar way for inscribed squares with two vertices on sides and , respectively. Prove that lines 
are concurrent.Attachments:
This post has been edited 3 times. Last edited by orl, Oct 25, 2004, 12:03 AM
coincides with the line AX. Similarly, the line 
coincides with BY, and the line 
coincides with CZ. Hence, instead of proving that the lines 
maps C to T and Z to A. Hence, the segment TA is the image of the segment CZ, so that
maps T to X and A to Y. Hence, the segment XY is the image of the segment TA, so that
be the intersections of 
with 
respectively. The homothety which takes the square with one side on 
as its center, so the center of the large square also lies on 
(the right angles are at 
)
are concurrent. this is, of course, Vecten's point. Here's a proof of the concurrence:
be the midpoint of 
.
)
and ratio 
,
)
.
,
is the image of 
through a rotation of angle 
.
.
have vertices 
and 
lying on 
and 
respectively, let 
meet 
at 
and 
at 
,![$ \frac {BX}{XC} = \frac {MS}{SN} = \frac {[AMA_1]}{[ANA_1]} = \frac {\frac {1}{2}AM \cdot MA_1 \cdot sin AMA_1}{\frac {1}{2} AN \cdot NA_1 \cdot sin ANA_1} = \frac {AM}{AN} \cdot \frac {sin(B + 45)}{sin(C + 45)} = \frac {AB}{AC} \cdot \frac {sin(B + 45)}{sin(C + 45)}$](http://latex.artofproblemsolving.com/3/9/3/39310413754f4cdb6c670a2d68a7718274f41359.png)
,
are concurrent. 
,
be 
,
,
.
.
is at 
is at 
.
and 
and show it lies on 
.
and of 
.
,![$[(m+n)(m+n+1)-(1-m+n)(m-n-2)]x=(m+n)(m+n+1)$](http://latex.artofproblemsolving.com/1/c/1/1c135a3d15f808d36ff92390ee39469600bb5f09.png)
.
and moving the 
over, we must show:
over this becomes
and subbing back in 
,
![$(1-m+n)[(2m-1)-(m-n-2)]=(m+n+1)[(2n+1)-(m+n)]$](http://latex.artofproblemsolving.com/9/0/6/9069086d910d0351430e5a3a8b7a59a1ccc33f9d.png)
,
.
.
,
is the side of the square. But 
,
.
as the measures of angles 
respectively.
and 
be the vertices of the square with center 
,
and 
.
,
and 
.
and 
,![\[ \frac{A_1D}{\sin\angle{A_1AB}}=\frac{AA_1}{\sin\angle{ADA_1}}\text{ and }\frac{A_1E}{\sin\angle{A_1AC}}=\frac{AA_1}{\sin\angle{AEA_1}}.\]](http://latex.artofproblemsolving.com/c/6/8/c6895e0e5e873928ecfcdac7bd8965dc20816a9f.png)
,![\[ \frac{\sin\angle{A_1AB}}{\sin\angle{A_1AC}}=\frac{\sin(B+45)}{\sin(C+45)}.\]](http://latex.artofproblemsolving.com/3/5/e/35ec64c44cd9b72056a085f992161ebfee693be5.png)
![\[ \frac{\sin\angle{B_1BC}}{\sin\angle{B_1BA}}=\frac{\sin(C+45)}{\sin(A+45)}\text{ and }\frac{\sin\angle{C_1CA}}{\sin\angle{C_1CB}}=\frac{\sin(A+45)}{\sin(B+45)}\]](http://latex.artofproblemsolving.com/0/8/0/0808ced55e049451748d8b4b635cff885a169e13.png)
![\[ \frac{\sin\angle{A_1AB}}{\sin\angle{A_1AC}}\frac{\sin\angle{B_1BC}}{\sin\angle{B_1BA}}\frac{\sin\angle{C_1CA}}{\sin\angle{C_1CB}}=\frac{\sin(B+45)}{\sin(C+45)}\frac{\sin(C+45)}{\sin(A+45)}\frac{\sin(A+45)}{\sin(B+45)}=1\]](http://latex.artofproblemsolving.com/8/0/5/80536ef8337aa39d5402a14acb37d6090bf0c67a.png)
concurent, where 
are the centers of the squares erected outwardly onto the sides of the triangle.
.
have the same base, hence the ratio of their areas is 
,
;
we get 
and other two similar relations for 
which, multiplied side by side will give the required 
.
respectively.Let 
(
.
and 
.
and 
respectively we get
.
.
and
.
,
.
in 
.
.
and 
,
and 
.
as 
and 
as 
.
and 
respectively. Now divide the first equation by the second to get 
.
and use Ceva's Theorem to prove concurrency. By ratio lemma, 
from which we found earlier. Doing the same with 
lies on 
,
,
are collinear.![\[
\frac{\sin BAA_2}{\sin CAA_2} = \frac{\sin BAA_2 \slash BA_2}{\sin CAA_2 \slash CA_2} = \frac{\sin ABA_2 \slash AA_2}{\sin ACA_2 \slash AA_2} = \frac{\sin(B + 45^\circ)}{\sin(C + 45^\circ)}
\]](http://latex.artofproblemsolving.com/e/0/7/e07485faf0ae8d3463e082426af39779467d81d7.png)
,
.![\[ 1=\prod \frac{\sin \angle BA_2A}{\sin \angle AA_2C} = \prod \frac{\sin \angle ABA_2 \cdot \frac{BA_2}{AA_2}}{\sin \angle ACA_2 \cdot \frac{CA_2}{AA_2}} = \prod \frac{\sin (45+B)}{\sin (45+C)} = 1,\]](http://latex.artofproblemsolving.com/4/8/d/48db2ea96a2b68736961732aa9b22a4d5bf01e32.png)
and we are done.
.
,
in the usual way. Then 
concur.
times the base.![[asy]
unitsize(1cm);
path rightangle(pair A, pair B, pair C) { pair u = unit(A-B)/5; pair v = unit(C-B)/5; return (B+u -- B+u+v -- B+v); }
real r = 2; real alpha = 1.45; pair A = r*dir(120); pair B = r*dir(210); pair C = r*dir(330); pair X = (B+C)/2; pair Y = (C+A)/2; pair Z = (A+B)/2; pair A2 = X + rotate(90) * (B-C)/2 * alpha; pair B2 = Y + rotate(90) * (C-A)/2 * alpha; pair C2 = Z + rotate(90) * (A-B)/2 * alpha;
filldraw(B -- A2 -- C -- cycle, royalblue*.2+white, blue+royalblue+.7); filldraw(C -- B2 -- A -- cycle, royalblue*.2+white, blue+royalblue+.7); filldraw(A -- C2 -- B -- cycle, royalblue*.2+white, blue+royalblue+.7); draw(X -- A2 ^^ Y -- B2 ^^ Z -- C2, black+.7); filldraw(A -- B -- C -- cycle, pink*.3+white, black+1); draw(A -- A2 ^^ B -- B2 ^^ C -- C2, purple+magenta+1); draw(rightangle(A2, X, B)); draw(rightangle(B2, Y, C)); draw(rightangle(C2, Z, A));
label(A, "$A$", N); label(B, "$B$", SW); label(C, "$C$", SE); label(A2, "$D$", S); label(B2, "$E$", NE); label(C2, "$F$", WNW); [/asy]](http://latex.artofproblemsolving.com/1/d/3/1d3b674e7aa70152dc0314cc47cbc73322b325d4.png)
![\[ d = \tfrac12 (b+c) + \tfrac\alpha2 (b-c) i, \quad e = \tfrac12 (c+a) + \tfrac\alpha2 (c-a) i, \quad f = \tfrac12 (a+b) + \tfrac\alpha2 (a-b) i. \]](http://latex.artofproblemsolving.com/1/4/0/140d151037b368274204e41ec7f0bffb585e501a.png)
If we set 
then we have ![\[ d = \omega b + \overline \omega c, \quad e = \omega c + \overline \omega a, \quad f = \omega a + \overline \omega b. \]](http://latex.artofproblemsolving.com/7/3/0/7307cace0ea40171f54610913d99a00e838b7943.png)
Since the condition 
is encoded in complex numbers as ![\[ (p - u) (\overline p - \overline v) = (p - v) (\overline p - \overline u) \iff (u - v) \overline p - (\overline u - \overline v) p - (u \overline v - v \overline u) = 0, \]](http://latex.artofproblemsolving.com/1/0/b/10bf1f22f75ee945b9e17098052a4293675278ed.png)
the system of equations in 
and 
we have to solve exists iff ![\[ \det \begin{bmatrix} a - \omega b - \overline \omega c & \overline a - \overline \omega \overline b - \omega \overline c & a (\overline \omega \overline b + \omega \overline c) - \overline a (\omega b + \overline \omega c) \\ b - \omega c - \overline \omega a & \overline b - \overline \omega \overline c - \omega \overline a & b (\overline \omega \overline c + \omega \overline a) - \overline b (\omega c + \overline \omega a) \\ c - \omega a - \overline \omega b & \overline c - \overline \omega \overline a - \omega \overline b & c (\overline \omega \overline a + \omega \overline b) - \overline c (\omega a + \overline \omega b) \end{bmatrix} \overset{?}= 0. \]](http://latex.artofproblemsolving.com/4/a/0/4a0b42ce3d1b2140626e1a9b6d4da9b9715fa5ea.png)
Since 
,
.
case corresponds to the outer Vecten point. 
are collinear. Similarly, 
and 
are collinear, so 
,
,
concur at the outer Vecten point. 
be the point on 
that are also on the square. Now we have 
.
where 
is the sidelength of the square with center 
and have one side covering 
and 
.![\[\frac{\sin{BAA_1'}}{BA_1'} = \frac{\sin{A_1'BA}}{AA_1'} = \frac{\sin{(45^{\circ}+B)}}{AA_1'}\]](http://latex.artofproblemsolving.com/4/5/9/459e5c0acc80409d67cc46f3651828ca3371f1df.png)
![\[\frac{\sin{CAA_1'}}{A_1'C} = \frac{\sin{A'CA}}{AA_1'} = \frac{\sin{(45^{\circ}+C)}}{AA'_1}\]](http://latex.artofproblemsolving.com/5/7/6/5768c7c803c8db865e1a70b63c37dd0af530db21.png)
Therefore, ![\[\frac{\sin{BAA_2}}{\sin{CAA_2}} = \frac{\sin{(45^{\circ}+B)}}{\sin{(45^{\circ}+C)}}\]](http://latex.artofproblemsolving.com/1/2/d/12d5c29ec7e5ef36d63a3d545b25df48cef514fc.png)
Similar results follow for ![\[\frac{\sin{BAA_2}}{\sin{A_2AC}}\cdot \frac{\sin{CBB_2}}{\sin{B_2BA}} \cdot \frac{\sin{ACC_2}}{\sin{C_2CB}} = \frac{\sin{(45^{\circ}+B)}}{\sin{(45^{\circ}+C)}} \cdot \frac{\sin{(45^{\circ}+C)}}{\sin{(45^{\circ}+A)}}\cdot \frac{\sin{(45^{\circ}+A)}}{\sin{(45^{\circ}+B)}} = 1\]](http://latex.artofproblemsolving.com/3/b/f/3bf667ed1917437875156f751a0909043cfc151c.png)
as desired.
intersect at 
at 
respectively. Let 
.
Hence, 
is the center of this square. Then by the Law of Sines 
and a similar equation holds for 
Multiplying these equations, we finish by the Trignometric form of Ceva's Theorem. QED
.
and 
so by law of sines we get![\[
\frac{\sin BAA_1}{\sin CAA_1}=\frac{\sin PAA_1}{\sin QAA_1} = \frac{\sin APA_1}{\sin AQA_1} = \frac{\sin (B+45)}{\sin (C+45)}
\]](http://latex.artofproblemsolving.com/7/3/2/732196c4af1eb32ef4aefade7a17e1ff06e062f2.png)
and applying Ceva's theorem we get the desired result and we're done.
