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 an acute triangle with orthocenter 
and circumcenter 
.
be points on the line segments 
and 
respectively such that 
is a parallelogram. Prove that 
.
Y by Davi-8191, nguyendangkhoa17112003, FaThEr-SqUiRrEl, centslordm, jhu08, megarnie, Adventure10, Mango247, and 1 other user
Let 
be an acute triangle. Denote by 
the foot of the perpendicular line drawn from the point 
to the side 
, by 
the midpoint of , and by 
the orthocenter of . Let 
be the point of intersection of the circumcircle 
of the triangle and the half line 
, and 
be the point of intersection (other than 
) of the line 
and the circle . Prove that 
must hold.
(Here we denote
the length of the line segment .)
be an acute triangle. Denote by 
the foot of the perpendicular line drawn from the point 
to the side 
, by 
the midpoint of , and by 
the orthocenter of . Let 
be the point of intersection of the circumcircle 
of the triangle and the half line 
, and 
be the point of intersection (other than 
) of the line 
and the circle . Prove that 
must hold.(Here we denote
the length of the line segment .)This post has been edited 5 times. Last edited by syk0526, Apr 4, 2012, 6:48 AM
where 
is diameter of cirumcircle. So, 
is symmedian...
![\[\frac{BF}{FC} = \frac{\sin \angle BCF}{\sin \angle CBF} = \frac{\sin \angle BED}{\sin \angle CED} = \frac{BD \cdot CE}{CD \cdot BE} \]](http://latex.artofproblemsolving.com/0/5/8/058914d5140ef2513929894b2f5c40d0db67a281.png)
.
is the second intersection of 
with the circumcircle of 
is harmonic and therefore ![\[ \frac{CE}{BE} = \frac {BE'}{CE} = \frac{AE \cdot cos C}{ AE \cdot cos B} \]](http://latex.artofproblemsolving.com/7/1/4/714815e9d827eaaea4a6206079f2b53085b19332.png)
and therefore we get ![\[\frac{BF}{FC} = \frac{ c \cos B}{b \cos C} \cdot \frac{AE \cdot cos C}{ AE \cdot cos B} = \frac{AB}{AC} \]](http://latex.artofproblemsolving.com/5/d/b/5dbdf87fbe8893f5cba5b92b38c906d17545de07.png)
and we are done.
beyond 
such that 
.
is the antipode of 
on 
then 
are collinear. Therefore 
so 
is cyclic. Therefore 
.
.
.
.
which implies 
.
so we have 
.
and the results follows.
to intersect 
again 
.
is a diameter, so 
.
.
.
,
:
.
the second intersection of line 
with 
such that: 
.
is a parallelogram, so: 
,
.
is the midpoint of ![$[HX]$](http://latex.artofproblemsolving.com/a/c/0/ac08620a7aa1175fa436ee7e317d3081c0e6fe85.png)
,
: 
.
.
,
.
.
.
And this can be proved using metric relations...
hits 
and 
by well-known results. Then remark that 
,
because that 
,
.
so that approach would be hopeless. However, remark that if we extend 
to a point 
,
and then since 
is cyclic we have 
,
.
and that 
in the case of 
.
.
.
.
then we would be done because 
.
we have 
.
when 
so it suffices to show 
.
and therefore we are done.
,
is a parallelogram, 
collinear, hence 
,
.
intersects circle 
,
.
intersects the circle 
,
we get that the arcs 
are congruent, hence 
and hence AF is a symmedian. It follows that ABFC is harmonic and therefore 
,
,
.
is parallel to 
is harmonic.
is harnomic.
is harmonic.
.
cut the circumcircle again at 
.
and it is easy to see well known facts that 
(see the diagram) and 
,
,
are collinear.
,
,
(*)
(or 
)
,
(**)
,
is the 
.
,
is the 
is cyclic.![[asy]
size(7cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A, B, C); pair T = 2*B*C/(B+C); pair M = midpoint(B--C); pair H = orthocenter(A, B, C); draw(A--B--C--cycle, red); pair O = origin;
pair L = 2*M-H; pair E = -L+2*foot(O, M, H); pair F = -A+2*foot(O, A, T);
filldraw(unitcircle, invisible, red); draw(E--M--A--F, red); draw(A--E, orange); draw(A--D, red); draw(E--F, red); draw(circumcircle(A, D, M), dashed+brown);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$H$", H, dir(H)); dot(L, dir(L)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F));
/* Source generated by TSQ */ [/asy]](http://latex.artofproblemsolving.com/f/f/7/ff7635f1238d214e47a19ffe118a48847cfbffdd.png)
.
.
all lie on a line. But 
and 
are isogonal, which implies 
.
is parallel to 
,
cyclic
which implies 
follows from reflecting the orthocenter.![\[ \measuredangle MAC = \measuredangle MAE + \measuredangle EAC = \measuredangle MDE + \measuredangle EFC = \measuredangle CDF + \measuredangle DFC = \measuredangle DCF = \measuredangle BAF \]](http://latex.artofproblemsolving.com/5/1/b/51bf365ff6a98efb7cf3b9b2b2856b557fec12cc.png)
gives the result.
.
,
concur at a point, say 
by projecting from 
on minor arc 
,![\[\angle BAF = \angle MAC.\]](http://latex.artofproblemsolving.com/c/e/e/cee0650280b0c98f83010de47941e31aa566daea.png)
is a parallelogram using angle chase.
.![\[\angle BAD = 90 - \angle ABC = 90 - \angle AA'C = \angle A'AC.~\Box\]](http://latex.artofproblemsolving.com/e/8/f/e8f03c4584bcf20743067bd8837bb08c99a9d5c6.png)
and 
.
and 
.
![[asy]
size(225); defaultpen(linewidth(0.4)+fontsize(10));
pair A, B, C, H, D, M, E, F, A1;
A = dir(120);
B = dir(210);
C = dir(330);
H = orthocenter(A, B, C);
D = foot(A, B, C);
M = .5B + .5C;
E = IP(circumcircle(A, B, C), H--5H-4M);
F = IP(circumcircle(A, B, C), D--3D-2E);
A1 = -A;
draw(E--A--B--C--A--F--E--A1--A--D--A--M);
draw(circumcircle(A, B, C));
markscalefactor = .02;
filldraw(anglemark(B, A, D)^^anglemark(A1, A, C), cyan);
markscalefactor = .015;
filldraw(anglemark(A, D, E)^^anglemark(A, M, E), blue);
filldraw(anglemark(A, F, E)^^anglemark(A, A1, E), green);
markscalefactor = .008;
draw(rightanglemark(A1, E, A)^^rightanglemark(C, D, A));
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$H$", H, NE);
label("$D$", D, S);
label("$M$", M, S);
label("$E$", E, W);
label("$F$", F, S);
label("$A'$", A1, SE);
dot(A);
dot(B);
dot(C);
dot(H);
dot(D);
dot(M);
dot(E);
dot(F);
dot(A1);
[/asy]](http://latex.artofproblemsolving.com/6/a/d/6ad02389a050512fe4f384a34a4f143890776ea0.png)
![[asy]
unitsize(4cm);
pair A = dir(110); pair B = dir(205); pair C = dir(335); pair D=foot(A, B, C); pair BB=foot(B, C, A); pair CC=foot(C, A, B); pair H=A+B+C; pair M=(B+C)/2; pair G=IP(Line(M, H, 10),unitcircle,1); pair T=extension(A, G, B, C); pair F=IP(Line(D, G, 10), unitcircle, 0); pair I=IP(Line(D, A, 10), unitcircle, 0);
draw(A--B--C--cycle, heavygreen); draw(A--T--BB, red); draw(T--C, red); draw(G--F, heavygreen); draw(A--I, heavygreen); draw(G--M, heavygreen); draw(circumcircle(A, G, H), red); draw(circumcircle(A, B, C), heavygreen); draw(circumcircle(B, CC, C), red);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$B'$", BB, NE); dot("$C'$", CC, W); dot("$H$", H, NE); dot("$M$", M, dir(M)); dot("$G$", G, dir(G)); dot("$T$", T, dir(T)); dot("$F$", F, SE); dot("$$", I, SW);
[/asy]](http://latex.artofproblemsolving.com/5/e/6/5e6d3ed4ef72d7f69ce0809b924c1e29b2a61a35.png)
be the feet of the altitudes in 
and 
passes through the 
.
,
,
,
.
which implies that 
that 
,
be the feet of the altitudes from 
in 
.
,
concur at 
.
,
Thus 
denote the intersection between 
respectively. It is well-known that 
.
,
is cyclic. This means that![\[\angle MAD = \angle MED = \angle FAA'.\]](http://latex.artofproblemsolving.com/a/c/0/ac036952ad6da70e90ae75495f8d1f26180ae6c0.png)
and 
are isogonal with respect to 
.
.![\[\angle BAM = \angle BAD + \angle MAD = \angle CAA' + \angle FAA' = \angle CAF,\]](http://latex.artofproblemsolving.com/4/0/a/40a08c46245eb25fa7f4cb1d083bc286ee028c9b.png)
and 
,
,
.
is a diameter in 
,
which implies that 
is cyclic. However, we know that 
is the 
,
,
.
is a parallelogram, so 
,
,
,
and the desired condition follows. 
by angle chasing in cyclic quadrilaterals. However, we know that 
by angle-angle similarity, implying that 
and 
,
.
,
,
which is also the desired conclusion. 
lie on a line and that 
.
And we are done.
through 
