AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
satisfying the following equation:
.
and 
be two sequences defined by:![\[
a_{n+1} = \frac{1 + a_n + a_n b_n}{b_n} \quad \text{and} \quad b_{n+1} = \frac{1 + b_n + a_n b_n}{a_n}
\]](http://latex.artofproblemsolving.com/a/a/0/aa0a5c5acedc990eb345fb50a8395db05307a4e7.png)
for all 
, with initial values 
and 
.![\[
a_{2024} < 5.
\]](http://latex.artofproblemsolving.com/f/c/a/fca6e6630f33429b5a7ea867d639a2e29ff881f2.png)
children, no two of the same height, lined up. Let us call a pair of different children 
good if between them there is no child whose height is greater than the height of one of 
and 
, but less than the height of the other. What is the greatest number of good pairs that could be formed? (Here, and 
are considered the same pair.)
satisfy 
prove in equality.
![\[\sum_{cyc}(x+y)\sqrt{(z+x)(z+y)} \geq 4(xy+yz+zx),\]](http://latex.artofproblemsolving.com/0/4/7/047adb616a6be9f9ed52f2385fa592fb5464e420.png)
and 
.
and 
be elements of the set 
such that 
, 
is a divisor of 
, and 
is a divisor of 
. Determine the largest value that can take.
such that the following equation holds:![\[
7^a + 1 = 2b^2
\]](http://latex.artofproblemsolving.com/e/0/8/e08d3fc7feefd7b4acc4a0f9c7e8bcfa328b5791.png)
have the property that 
is a perfect square for infinitely many positive integers 
?
and 
respectively be the incentre, circumcentre and circumcircle of triangle 
. Points 
are chosen on , such that 
, and point 
is the foot of the altitude from 
to 
. If 
are similarly defined, prove that lines 
and 
concurr on 
.
that is not isosceles at 
, let 
be its incircle, which is tangent to 
at 
, respectively. The lines 
and 
intersect the line passing through and parallel to 
at 
and 
, respectively. The lines passing through 
and perpendicular to 
intersect 
and 
at 
and 
, respectively.
are collinear and that 
are concurrent.
positive integer numbers such that the least common multiple (LCM) of all these numbers is not a perfect square. Mykhailo consecutively hides one of these numbers and writes down the LCM of the remaining 
numbers that are not hidden. What is the maximum number of the written numbers that can be perfect squares?
be a triangle with orthocentre 
. The tangent lines from 
to the circle with diameter 
touch this circle at 
and 
. Prove that 
and are collinear.
be a triangle with orthocenter . The tangent lines from to the circle 
with diameter ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
touch it on and . Prove that 
.
of 
, 
, 
for which 
and 
. Observe that 
, i.e. 
what is a characterization of 
- harmonical division, i.e. is harmonical conjugate of w.r.t. 
. In conclusion, belongs to polar of w.r.t. , i.e. .
separates , 
. Denote 
so that 
and 
so that 
. Observe that 
and 
. Thus 
, i.e. 
, 
, 
are concurrently .
be a cyclic hexagon. Prove that 
.
as the circle of reciprocation.
are collinear its the same as proving that their polars are concurrent. Because both belong to the circle, their polars are the tangents to the circle that meet at . passes through .
be the perpendicular to 
(
is the midpoint of ). If 
is the inverse of , then is the polar of and we are done.
.
~
we have :
. 
the altitudes of 
, and 
the midpoint of . Because 
, the points 
are concyclic. Let 
the circumcircle of 
. and radius 
. Because 
, we obtain that is the image of 
under this inversion, and 
remains invariant. For the other hand, is transformed into the line containing . Because 
, we can conclude that 
.
and are the intersections of the circles 
and 
, and 
being the midpoints of 
and 
respectively.
are the feet of the altitudes on 
and respectively and these altitudes concur at , then we know that 
, i.e. has equal power w.r.t. to both circles, consequently it belongs to their radical axis .
are switched.
are switched.
from Virgil's lemma, we can also assume that it's wrong i.e there is three intersection point, if I remembre well it's we get that 
with 
, 
and 
which obviously wrong, and then deduct the desired result.
and 
(X is on the same side with A from BC)
is collinear and perpendicular to 
and 
is harmonic
is harmonic
is collinear since 
is collinear
.Let me inform you that this problem is from China 1997 and i have three solutions but two of them have already posted so i'll post the third one:
the altitudes of , and the midpoint of .Since 
then the points are concyclic.
and 
be a triangle and its altitude 
.The tangent lines from to the circle with diametre touch this circle on and .If 
then prove that is the orthocenter of 
be the midpoint of .Extend 
to meet 
at 
where and are on the same side of .Now is the pole of wrt the circle 
.Again note that 
is the polar of .So polar of passes through 
lies on .Hence proved.
, , 
. Let the foot of the perpendicular from to be 
. By http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2173258&sid=fd7168f527cac736647f7961645922f0#p2173258, 
so the polar of with respect to the semicircle with diameter passes through , and therefore . So , , are collinear.
and .
lie on the circle 
with diameter .
cut at 
. 
is the polar of with respect to 
is also the polar of . It follows that 
are collinear.
be the midpoint of , 
be the foot of onto , 
be the foot of 
onto 
. We just want to show that is on the radical axis of the circles with diameters and (which is ). But this is obvious, since 
. is implied to be acute...
be the intersections of the -altitude with the circle.
which is obvious.
be the feet of perpendiculars from respectively 
as the center and radius 
. to themselves and it takes to feet of perpendicular from to (
)thus
and are concyclic hence the conclusion
is on the cicrcle such as the is perpendicular to and analogus the point F.
and 
and we have from theorem of Hire that the intersection point of BE and CF is on the polar of the point A which is PQ so P,Q and H are collinear.
be circumcircle of and 
be a circle so that 
and exists 
such that 
Let 
and the symmetric of 
w.r.t. 
The tangent lines from to the circle touch it on and 
Prove that 
(see PP10 from here).
. Then let 
be points on the unit circle, so 
. Then if 
is the orthic triangle, 
Now with the chord intersection formula, 
It remains to compute 
so 
are collinear, as required.
be a triangle with orthocenter . The tangent lines from to the circle with diameter touch it on and . Prove that . denote the mid of segment 
Clearly, 
are concyclic in that order. Also, 
..Therefore 
must lie on the radical axes of the circles 
and 
which clearly is the claimed line the 
.
be a cyclic hexagon. Prove that . lies outside triangle in given problem.
wrt 
both pass through , so by La Hire it suffices to prove that the polar of does as well. Let the reflection of over the midpoint of be 
, and 
; we can see that
by orthocenter reflection, and since is the -queue point, we know that 
, which implies that 
is the polar of .
and radius . Then the problem is equivalent to showing that lies on the radical axis of this circle and the circle with diameter . We can do this by showing has equal powers with respect to both circles.
or 
. Notice that we have 
, and 
, giving 
. Then let the foot from to be . We can then write 
. We then want to show 
, which we can reduce to 
. Then by Law of Cosines, we have 
where is the foot from to . We can then write 
. By Power of a Point on 
, we have 
. We then just need to show 
. Multiply everything by two to get the equivalent 
and rearranging gives the equivalent 
, but 
, so we are done by LOC.
be the altitude from to 
and the circumcenter of 
. Also denote by 
the angle 
modulo 
. Then we make the following claim:
is cyclic.![\[\measuredangle APO = \measuredangle AQO = \measuredangle ADO = 90^{\circ}. \blacksquare\]](http://latex.artofproblemsolving.com/6/7/b/67b7b13bcc6f410eb5ce219358d90f2ee5ea159e.png)
and 
, that is the radical center of all of our present circles so that ![\[AP^2 = AQ^2 = AH \cdot AD \implies \triangle AHP \sim \triangle APD; \phantom{c} \triangle AHQ \sim \triangle AQD.\]](http://latex.artofproblemsolving.com/d/2/a/d2a48806b7ca4c79586873d9f66fa476062228a5.png)
so that are collinear, as desired. 
be the orthic triangle, be the midpoint of .
. Also, 
clearly lie on the circle with diameter . with radius 
, then and go to themselves, and goes to .
are concyclic, inverting back gives that collinear as desired. 
, 
, 
as the feet of the , 
, 
altitudes respectively and as the midpoint of .
and 
, thus allowing the circle with diameter to be the unit circle. and also lie on this circle. We set and as free variables.![\[A = \frac{2pq}{p+q}\]](http://latex.artofproblemsolving.com/1/5/6/1562090eb0402f0dba760e24a5a3f1d4dbee5e73.png)
We shall find , notice that it is the intersection of 
with the unit circle and is thus given by the formula![\[e = \frac{1-\frac{2pq}{p+q}}{\frac{2}{p+q}-1} = \frac{p+q-2pq}{2-p-q}\]](http://latex.artofproblemsolving.com/c/c/1/cc174555499b0bf36ff4183491aa8bbcec8775eb.png)
We similarly find![\[f = \frac{\frac{2pq}{p+q} + 1}{\frac{2}{p+q}+1} = \frac{2pq+p+q}{2+p+q}\]](http://latex.artofproblemsolving.com/c/e/a/cea3ff7bb2a70e507b7615b8cffd18ffcef05ab0.png)
To find we intersect 
and 
, we thus obtain![\[H = \frac{\frac{2pq-p-q}{2-p-q} \cdot (1 + \frac{2pq+p+q}{2+p+q}) - \frac{2pq+p+q}{2+p+q} \cdot (\frac{p+q-2pq}{2-p-q}-1)}{\frac{2pq-p-q}{2-p-q} - \frac{2pq+p+q}{2+p+q}}\]](http://latex.artofproblemsolving.com/7/3/a/73a4cdd6264e3f76a1c1cec9811c7029c31ce8cf.png)
Simplifying, we have![\[H = \frac{2(2pq-p-q)(pq+p+q+1) + 2(2pq+p+q)(pq-p-q+1)}{(2+p+q)(2pq-p-q) + (p+q-2)(2pq+p+q)}\]](http://latex.artofproblemsolving.com/7/c/6/7c6fa2016e4da7193377c76d0a73dc48d812990e.png)
Further simplifying, we obtain![\[H = 2 \cdot \frac{4pq(pq+1) - 2{(p+q)}^2}{4(p+q)(pq-1)} = \frac{2pq(pq+1)-{(p+q)}^2}{(p+q)(pq-1)}\]](http://latex.artofproblemsolving.com/2/5/1/251b49a39a403ab54910673e8b64ebe1f7ddd4f8.png)
Now we shall prove the colinearity condition, notice that![\[\frac{H-p}{H-q} = \frac{\frac{2pq(pq+1)-{(p+q)}^2}{(p+q)(pq-1)} - p}{\frac{2pq(pq+1)-{(p+q)}^2}{(p+q)(pq-1)} - q} = \frac{(2p^2q^2 - p^2 - q^2) - p(p+q)(pq-1)}{(2p^2q^2 - p^2 - q^2) - q(p+q)(pq-1)}\]](http://latex.artofproblemsolving.com/e/9/6/e96c6cc47757686914ce938a899f673f017dd322.png)
From which![\[\frac{H-p}{H-q} = \frac{q(q - p)(p^2-1)}{p(p-q)(q^2-1)} = -\frac{q(p^2-1)}{p(q^2-1)}\]](http://latex.artofproblemsolving.com/a/0/8/a0832234be68bc35b78b0f3ea61f30201004f1cc.png)
We can simply take the conjugate to obtain![\[\overline{\left(\frac{H-p}{H-q}\right)} = -\frac{\left(\frac{1}{q}\right) \cdot (1 - \frac{1}{p^2})}{\left(\frac{1}{p}\right) \cdot (1 - \frac{1}{q^2})} = -\frac{q(p^2-1)}{p(q^2-1)}\]](http://latex.artofproblemsolving.com/5/8/c/58c541677097ccd5d62e99e011862659446d81d5.png)
Thus 
.
. Then from Brokard's theorem, triangle 
is self-polar, leading to the required result.
and 
be tangency points of the incircle of 
and 
.
and 
,
.
and 
be feet of perpendiculars from 
and 
respectively. Prove that 
,
and 
are concurrent.
satisfying: 
.
min
.
then: 
and 
and since 
is cyclic, we have 
the equality becomes then: 
since 
thus: 
then 
from which follow the collinarity of 
are concyclic where 
is the midpoint of 
and 
circles with diamters 
and 
and 
then by radical axis theorem it follows immediatly that 
and hence the points 
.
by the definition of radical axis.
and 
,
after rearranging and multiplying by 4.
,
and 
we have
,
.
,
.
,
and 
.
.
,
.
so the result follows.
,
.
,
,
.
.
and 
,
.
,
,
with respect to 
,
,
,
be the inversion operator, ![\[\Psi(AD \cap PQ) = \Psi(AD) \cap \Psi(PQ) = AD \cap (APQ) = D \implies AD \cap PQ = H,\]](http://latex.artofproblemsolving.com/2/1/d/21df81fe7ef438f3a9ba23868558df7af227a6a7.png)
also.
is a cyclic quadrilateral since
Thus notice that the power of 
,
is 
.
is Humpty, since 
,