AoPS Academy
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
. If is an integer, then 
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 
, and that, for every such pair 
, the "next" such pair can be calculated as
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?
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
, 
and 
. Find 
and 
interesting if is composite and for every divisor 
of at least one of 
and 
is also a divisor of with 
and 
are points on a semicircle. The tangent at meets the extended diameter of the semicircle at 
, and the tangent at meets it at 
, so that and are on opposite sides of the center. The lines 
and 
meet at 
. 
is the foot of the perpendicular from to 
. Show that 
bisects angle 
and are points on a semicircle. The tangent at meets the extended diameter of the semicircle at , and the tangent at meets it at , so that and are on opposite sides of the center. The lines and meet at . is the foot of the perpendicular from to . Show that bisects angle 
and 
are points on a semicircle. The tangent at meets the extended diameter of the semicircle at 
, and the tangent at meets it at 
, so that and are on opposite sides of the center. The lines 
and 
meet at 
. 
is the foot of the perpendicular from to 
.
bisects angle 
; thus, the point C is the foot of the perpendicular from the point U to the line BG. Similarly, the point D is the foot of the perpendicular from the point U to the line AG.
quadrilaterals 
are cyclic 
- harmonical division.
and be on the semicircle so that 
are collinear, in that order. Let the tangents to the semicircle at and intersect at 
, and let 
be the intersection of 
and 
(possibly the point at infinity.) By Pascal's theorem on 
, we get that , , and are collinear. 
and 
, so is the orthocenter of 
. Hence, , , and are collinear, so , , and are collinear. Since 
is an altitude, Blanchet's theorem tells us that 
, so our proof is complete.
, let 
, 
, 
, 
, and 
. Then , 
, and 
each lie on the polar of .
, we find that , , and are collinear. By Pascal's theorem on 
, we find that , 
, and are collinear. The polars of and are and 
. Since lies on the polars of and , and lie on the polar of . Hence, , , , and all lie on the polar of . 
be on the semicircle so that 
are collinear, in that order. Let hit at 
, and let and intersect the semicircle at 
and 
, respectively. By Pascals' theorem on 
, we find that , , and are collinear. By the lemma, the polar of passes through .
hit 
at , and let 
hit 
at . 
and 
, so is the orthocenter of 
, whence 
. Furthermore, the polar of also passes through , and , so by the lemma, , , , and are collinear. It follows that is an altitude of 
, so by Blanchet's theorem .
, let , , , , and . Then , , and each lie on the polar of ., we find that , , and are collinear. By Pascal's theorem on , we find that , , and are collinear. The polars of and are and . Since lies on the polars of and , and lie on the polar of . Hence, , , , and all lie on the polar of . 
LEMMA: 
be the center of the circle. Let the tangents meet at . Let H be the foot of the perpendicular from to . The key is that and intersect on 
.
and 
are similar, so 
. 
and 
are similar, so 
. But 
, so 
. Also 
(equal tangents), so 
. Hence by Ceva's theorem applied to the triangle 
, the lines , and 
are concurrent. So 
. 
, so , and lie on the circle diameter 
. So 
. 
and 
intersect at 
, 
is the center of the circle. connect 
, 
, ., 
, 
are conliear. in 
, use ceva theorem, try to prove: 
.
, 
, suppose the radium of circle is 
, 
, 
, 
. so we can find , , are conliear.
, so 
are concylic, so 
.
, so 
are concylic, so 
. it is easy to konw 
. prove is done.
![[asy]
size(10cm); pointpen = black;
pair C = Drawing("C", dir(130), dir(130));
pair D = Drawing("D", dir(60), dir(60));
pair B = Drawing("B", -1.55572, dir(180));
pair A = Drawing("A", 2.0000000, dir(0));
draw(unitcircle);
pair Q = Drawing(" ", extension(B,C,A,D), dir(0));
draw(B--Q);
draw(A--Q);
draw(A--B);
pair O = Drawing(" ", 0.0, dir(0));
pair E = Drawing(" ", extension(A,C,B,D), dir(0));
pair F = Drawing(" ", extension(Q,E,A,B), dir(0));
label("$O$", O, SE);
label("$F$", F, SW);
label("$E$", E, SE);
label("$Q$", Q, N);
draw(Q--F);
draw(C--A);
draw(B--D);
draw(circumcircle(C,Q,D), green+dashed);
draw(C--F, orange+dashed);
draw(C--O, red+dashed);
draw(D--F, orange+dashed);
draw(D--O, red+dashed);
draw(D--C, red+dashed);
[/asy]](http://latex.artofproblemsolving.com/f/f/0/ff0eccf5e1ca866c0b026543cca46615a6dbabf7.png)
, the circle be 
, and the center of .
are collinear.
, let be the unit circle, and let be the 
axis. Let 
where 
. It suffices to show that 
.
and 
. Then line 
has equation![\[y=\tfrac{c_2}{c_1-\tfrac{1}{d_1}}(x-\tfrac{1}{d_1}).\]](http://latex.artofproblemsolving.com/5/7/d/57d92d4c259fb99e8a7713f9f883dbeccbff1b74.png)
Similarly, 
has equation![\[y=\tfrac{d_2}{1-\tfrac{1}{c_1}}(x-\tfrac{1}{c_1}).\]](http://latex.artofproblemsolving.com/e/d/e/ede646f62bd31fcdccc11b88ec35f1387f82a14a.png)
Since 
, we have
Now, line 
has equation![\[y=\tfrac{c_2}{c_1-\tfrac{1}{c_1}}(x-\tfrac{1}{c_1})\]](http://latex.artofproblemsolving.com/3/9/5/395a35d1b2cc36dc3e1af145766025c010f6f333.png)
and similarly line 
has equation![\[y=\tfrac{d_2}{d_1-\tfrac{1}{d_1}}(x-\tfrac{1}{d_1}).\]](http://latex.artofproblemsolving.com/f/8/0/f801fd7e164a9e00aa28764fe0056339885b7ff1.png)
Since 
, we have
as desired. 
is cyclic. Obviously 
is cyclic since 
Then since![\[\angle OFQ=\angle OFE =\tfrac{\pi}{2}=\angle OCQ=\angle ODQ,\]](http://latex.artofproblemsolving.com/3/f/f/3ff162e8c99c5601ff500fdea19134351ba5c266.png)
lies on the same circle.![\[\angle CFE=\angle EFD\iff\angle CFQ=\angle QFD\iff\angle QDC=\angle QCD,\]](http://latex.artofproblemsolving.com/b/f/b/bfbbacee8cef630cccae6ddeddb2c84c976529ec.png)
which is obvious by equal tangents. 
are collinear where 
. It suffices to prove that 
bisects 
. Let 
and let 
. Then 
and 
from this two facts we have that 
bisects thus bisects q.e.d
and 
so we obtain E,F,O,D,c cyclic.
and we have 
so 
bisects 
and we can obtain by cyclicity that 
as desired.
, Let 
be the orthic triangle WRT . Let tangents at 
to 
intersect 
at . Let 
meet at . Let be the foot from to .Prove, 
bisects 
be the orthocenter of and 
be the midpoint of . Then, lies on 
. Let the perpendiculars from 
to intersect 
at 
. Let 
intersect again at 
concur, suppose say at , then, Apply Pascal on 
passes through . By La Hire's Theorem are polars of 
WRT . Now since, 
are collinear, their polars are concurrent 
lies on 
and since is the incenter of 
() bisects 
()
such that 
is cyclic.
and 
Thus we have the following:
Because 
and because of the ratios above.
and are concurrent cevians,thus 
is cyclic
Thus we have shown,because 
, that PF is an angle-bisector of the angle 
be the intersection of and 
. Let 
be the foot of the perpendicular from onto . Let be the center of the given semicircle.
,
Since 
,
Multiplying, we see that 
. We also know that 
by power of a point. Therefore, by Ceva’s theorem, we see that 
, and 
are concurrent. Furthermore, we know that 
is cyclic because 
. Now, we proceed with angle-chasing. Let 
. Then, 
. Then, 
. Since is the center of the semicircle, 
as well, so 
.Thus, 
bisects , which as we showed beforehand is the same as saying that bisects .
.
and 
- reflections of 
about .
we see that 
and so 
we see that 
which implies the result.
concur. and intersect at and let be the foot of the altitude from to 
Also let the semicircle have center 
Now note![\[\triangle PAQ\sim \triangle OAD\]](http://latex.artofproblemsolving.com/4/b/8/4b88b87ff7ac65c9b2834561b8bd73a82a0fd566.png)
![\[\triangle PBQ\sim \triangle OBC\]](http://latex.artofproblemsolving.com/5/c/f/5cf4eb286f909c4ccc7627f1f193525291079e50.png)
so 
Since 
concur, is actually 
and 
concur.![\[\angle OCP=\angle ODP=\angle OFP=90^{\circ},\]](http://latex.artofproblemsolving.com/c/3/7/c371dfa6ed91ddb5b98f88df27d89d2009c8e876.png)
so 
is cyclic. Thus![\[\angle COP=\angle DOP\]](http://latex.artofproblemsolving.com/4/f/9/4f9d4f5dfdb83e83965f86131020e59b03259bc5.png)
![\[\angle CFP=\angle DFP.\]](http://latex.artofproblemsolving.com/7/1/9/7191e85a661b959ff3a241224b94977c0192624c.png)
![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
.
-
,
for which 
,
;
.
.
and 
.
is harmonically, i.e. 
.
and the triangle 
.
.
.
of the point 
,
.
are concyclic, we have: 
is inscribed. If 
,
,
.
and 
.
and 
are concurrent, where 
.
and 
be the foot of perpendicular from 
to 
.
are concurrent. However, 
,
are collinear.
and 
,
.
are collinear, and we are done. 
,
,
.
,
is on the polar of 
'
is the polar of X,so 
,
are collinear.
and 
,
.
is cyclic? Thanks. =)
,
and 
the reflections of 
and 
are tangents to the circle. Consider the quadrilateral 
,
and 
,
and therefore they are perpendicular, let 
we conclude that 
and 
and in 
to conclude that 
,
concur at 
and 
are congruent, thus 
.
and the problem is solved.