be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
.
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
,
,
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
interesting if 
of 
and 
is also a divisor of 
,
and 
.
is prime, 
with 
and 
.
be points on the line segments 
is a parallelogram. Prove that 
.
Y by malcolm, esi, Viswanath, steppewolf, pog, Ya_pank, HWenslawski, Adventure10, jhu08, megarnie, ImSh95, Mango247, Zhaom, Rounak_iitr, and 1 other user
Let 
be an acute-angled triangle with 
. Let 
be the orthocenter of triangle 
, and let 
be the midpoint of the side 
. Let 
be a point on the side 
and 
a point on the side 
such that 
and the points , , are on the same line. Prove that the line 
is perpendicular to the common chord of the circumscribed circles of triangle and triangle 
.
be an acute-angled triangle with 
. Let 
be the orthocenter of triangle 
, and let 
be the midpoint of the side 
. Let 
be a point on the side 
and 
a point on the side 
such that 
and the points , , are on the same line. Prove that the line 
is perpendicular to the common chord of the circumscribed circles of triangle and triangle 
.This post has been edited 4 times. Last edited by BogG, May 26, 2006, 2:18 PM
and the orthocentre ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
;
between 
and the bisector of the angle 
;
and 
so that 
and 
.
of the triangle 
.
and 
.
Denote the point 
,
.
and 
,
; the middlepoint 
of ![$[AH]$](http://latex.artofproblemsolving.com/0/3/b/03b8986ebe750b377f987f87b41a1dbc4c128e17.png)
; 
.
.
(the point ![$[HA']$](http://latex.artofproblemsolving.com/b/d/d/bdd6284b56db3edb70c48d53b2cef3bbf7dfdb34.png)
)
and 
.
are the middlepoints of the segments 
,
,
respectively and 
and 
.
.
without the point 
.
is an angle bisector of 
and why 
are collinear?
are collinear?
be a circumcircle of the acute triangle 
.
.
,
so that 
and 
.
- the ray without the point 
be the circumcenters of 
.
.
and 
,
.
is a parallelogram. Let 
be the midpoint of 
.
is a parallelogram 
.
is the midpoint of 
,
.
be the intersection of 
and the triangle 
.
.
.
,
.
.
.
.
.
.
.
)
so 
so 
the circumcenter of 
intersects again the circumcircle of 
then it will intersect it precisely at 
.
cuts 
.
and they are all together beautifully.
and 
are similar for 
and 
.
hence, using 3, triangles 
and 
are similar hence 
actually bisects 
and we're set because if 
is the point where it touches again the small circle the 
are diametrically opposite and hence 
is right which means 
.
being a parallelogram. Since 
,
.
,
.
meets circle 
,
,
,![\[\frac{AF}{AO'}=\frac{A'O}{A'O'}\]](http://latex.artofproblemsolving.com/8/8/4/884e0459c67210ccfc0b3908cef15f438f64d9cf.png)
Let 
,
,
,![\[AA'=R \left( \sin{\left( B+\frac{A}{2}\right)}+\sin{\left( C+\frac{A}{2}\right)}\right) =2R \sin{\frac{\pi}{2}}\cos{\frac{B-C}{2}}= 2R \cos{\frac{B-C}{2}}\]](http://latex.artofproblemsolving.com/8/8/2/8826f8bfec2ae7189cc9d0c3fba8257dbdbbe111.png)
Next we calculate 
.![\[\frac{AB}{AD}\tan{B}+\frac{AC}{AE}\tan{C}=\tan{A}+\tan{B}+\tan{C}\]](http://latex.artofproblemsolving.com/6/3/e/63eb21dc514dc59785ae534e2ca9f94cf533cefd.png)
Since in any triangle 
.![\[\frac{AB}{AD}\tan{B}+\frac{AC}{AE}\tan{C}=\tan{A}\tan{B}\tan{C}\]](http://latex.artofproblemsolving.com/6/3/7/637e8e15b8aaf4c3ae6808e6e664515767347e56.png)
Hence we have ![\[AD=\frac{AB \tan{B}+AC \tan{C}}{\tan{A}\tan{B}\tan{C}}=2R \cdot \frac{\frac{\sin{C}\sin{B}}{\cos{B}}+\frac{\sin{C}\sin{B}}{\cos{C}}}{\tan{A}\frac{\sin{C}\sin{B}}{\cos{B}\cos{C}}}=2R\frac{\cos{B}+\cos{C}}{\tan{A}}\]](http://latex.artofproblemsolving.com/0/0/b/00b5262d111cff2df2cc37873cb28ffa760a614e.png)
Then use the fact that 
,![\[r=\frac{\cos{B}+\cos{C}}{\tan{A}\cdot \cos{\frac{A}{2}}}\cdot R = \frac{2 \cos{\frac{B+C}{2}}\cos{\frac{B-C}{2}}}{\tan{A}\cdot \cos{\frac{A}{2}}}\]](http://latex.artofproblemsolving.com/e/2/6/e2668e2d85de015c2aaddf6e8c5e0c78a0787bc8.png)
Now pluge them into our original equality, we get: ![\[\frac{AF}{R}=\frac{r}{AA'-r}= \frac{1}{\frac{AA'}{r}-1}\]](http://latex.artofproblemsolving.com/a/a/b/aab9373f5b710d362e81de1d1c8eb4683d7a4276.png)
Pluge 
,
and 
,![\[\frac{AA'}{r}=\tan{A}\cdot \cot{\frac{A}{2}}=\frac{2\sin{\frac{A}{2}}\cos^{2}{\frac{A}{2}}}{\cos{A}\sin{\frac{A}{2}}}=\frac{2\cos^{2}{\frac{A}{2}}}{\cos{A}}\]](http://latex.artofproblemsolving.com/8/d/9/8d9863a4a428fa97d4f90d4f5a6f113bd8de2806.png)
Hence: ![\[\frac{R}{AF}=\frac{2\cos^{2}{\frac{A}{2}}}{\cos{A}}-1=\frac{1}{\cos{A}}\]](http://latex.artofproblemsolving.com/5/1/5/515752eb0bc30af6fc3d79586c0f512a5d3803e0.png)
Hence 
,
,
is a parallelogram, hence 
be the circumcenter, 
be the radius and 
be the circumcircle of 
and let 
be the radius and 
be the circumcircle of 
.
the intersection of 
with 
the intersection of the line 
with 
be the intersection of the line 
with 
be the intersection of 
with 
.
,
are collinear and that 
intersect in 
and that the angle 
,
and 
.
.
and 
be the reflections of 
and 
,
and thus 
and 
and thus 
.
is a parallelogramm and it follows that 
intersects 
at its middle, i.e. 
,
.
,
or equivalently 
.
,
.
is the more difficult part. We have 
and 
.
.
.
and therefore 
.
.
.
.
.
,
.
and 
is the diameter of 
.
,
and 
.
is cyclic and coaxial to 
.
and 
,
which means that it suffices to prove 
.
and 
.
we're done.
.
and 
denote the feet of the altitudes from 
and 
.
to 
Let 
.
so 
Hence, we also have 
Thus, by Angle Bisector Theorem, 
where the second equality is from PoP on 
.
,
.
are collinear (since they all lie on the line from 
which we call 
are coaxial as they all have 
,
,
,
we are then done.
First we chase some angles: 
so HD is an angle bisector. Now the problem is evident, because 
from similarity of triangles BHJ and CHI by AA (JBH=ICH by cyclic quad BCIJ), as desired. 
is harmonic. However, noting that 
,![\[\frac{XB}{XC} = \frac{BD}{CE} = \frac{HB}{HC} = \frac{H'B}{H'C}\]](http://latex.artofproblemsolving.com/e/6/e/e6e32725d98f049eedaf9905d5c2444e97a9ad74.png)
and we are done.
,
be altitudes of 
be the 
,
is cyclic. By the (forgotten) Coaxiality lemma, it's equivalent to show that
However, because 
it's easy to see that 
and 
so 
and we are done.
,
with corresponding angle bisectors 
and 
,
.
to corresponding points 
,
,
,
,![[asy]
size(200);
pair A, B, C, H, D, E, X, Y;
A = dir(110);
B = dir(200);
C = dir(340);
H = A+B+C;
X = foot(B, A, C);
Y = foot(C, A, B);
D = extension(A, B, H, incenter(H,B,Y));
E = extension(A, C, H, incenter(H,C,X));
draw(A--B--C--cycle^^D--E^^B--X^^C--Y);
draw(circumcircle(A, D, E)^^circumcircle(A, B, C));
draw(circumcircle(A, X, Y), dashed);
dot("$A$", A, dir(90));
dot("$B$", B, dir(225));
dot("$C$", C, dir(315));
dot("$D$", D, dir(200));
dot("$E$", E, dir(30));
dot("$H$", H, dir(270));
dot("$X$", X, dir(0));
dot("$Y$", Y, dir(210));
[/asy]](http://latex.artofproblemsolving.com/e/a/8/ea81bf1d03688539268a729269bcef60a385e69e.png)
= 
= 
= 
passes through 
to 
therefore It is enough to show that 
= 
which is trivial by 
is similiar to 
and 
is similiar to 
, hence spiral at 
to 
.
are collinear.
and 
.
,
are concyclic.
,
and 
,
,![\[\frac{DY}{DB} = \frac{EX}{EC} \implies \frac{\operatorname{pow}(D,(AH))}{\operatorname{pow}(D,(ABC))} = \frac{\operatorname{pow}(E,(AH))}{\operatorname{pow}(E,(ABC))}.\]](http://latex.artofproblemsolving.com/2/3/e/23ee430ef40d33352344b84f9d19b3613fe47a7f.png)
,
lies on 
and is perpendicular
.
to 
centred at 
which
to 
,
so this suffices.
which we call 
to 
to 
to 
and 
![\[\angle YHD=90^{\circ}-\angle ADH=\frac12\angle A.\]](http://latex.artofproblemsolving.com/c/f/3/cf308fd862f0df421b0d162ead14164c6d636a1c.png)
Similarly, 
.
,
.
,
.
(by our ratio equivalence earlier), we get![\[\angle DPE=\angle BPC=\angle BAC=\angle DAE,\]](http://latex.artofproblemsolving.com/a/5/0/a50098c1aaa5ccc5fc37319c7fea05132d31de22.png)
so indeed 
(since 
and the 
q
are the altitudes from 
then 
and similarily 
bisects 
.![\[\frac{Pow(E,(AHQ))}{Pow(E,(ABCQ))}=\frac{EH_C.EA}{EA.EB}=\frac{HH_C}{HB}=\frac{HH_B}{HC}=\frac{Pow(F,(AHQ))}{Pow(E,(ABCQ))}\]](http://latex.artofproblemsolving.com/9/b/d/9bd0a0760546f26e8ff633b16a91bc4cbe373234.png)
Now, coaxiality lemma finishes it as desired
where 
be foot of altitudes of ![\[\frac{\text{Pow}(D,(AH))}{\text{Pow}(D,(ABC))}=\frac{\text{Pow}(E,(AH))}{\text{Pow}(E,(ABC))} \iff \frac{DZ}{DB}=\frac{EY}{EC}\]](http://latex.artofproblemsolving.com/2/f/1/2f1af1f2c7465e7529da4cea537b95afa89d8d86.png)
But see that this is trivial because 
and done.
