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 
.
and 
be points on the line segments 
and 
respectively such that 
is a parallelogram. Prove that 
.
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A convex quadrilateral 
satisfies 
. Point 
lies inside so that ![\[\angle{XAB} = \angle{XCD}\quad\,\,\text{and}\quad\,\,\angle{XBC} = \angle{XDA}.\]](//latex.artofproblemsolving.com/6/d/7/6d798ef2bf8136054abb232961c3b646e7f320a4.png)
Prove that 
.
Proposed by Tomasz Ciesla, Poland
satisfies 
. Point 
lies inside so that ![\[\angle{XAB} = \angle{XCD}\quad\,\,\text{and}\quad\,\,\angle{XBC} = \angle{XDA}.\]](http://latex.artofproblemsolving.com/6/d/7/6d798ef2bf8136054abb232961c3b646e7f320a4.png)
Prove that 
.Proposed by Tomasz Ciesla, Poland
This post has been edited 3 times. Last edited by djmathman, Jun 16, 2020, 4:03 AM
meet 
be 
.
,
are concyclic. Define 
similarly.
,
,
and 
are sets of concyclic points. Note that due to inversion under the 
A
subtends equal angles at 
,
.
both lie on the same side of line 
.
so trivially 
.
with 
so 
.
,
is an isosceles trapezoid. If 
;
.
by similarity, we want to show 
.
so 
.
.
be the circle centered at point 
;
.
are inverses in 
,
both lie on 
,
respectively, it's enough to show 
.
,
,
.
and 
.
are quadrilaterals such that they share the same angles and 
,
,
.
and the rays 
intersect at 
.
and let 
and 
.
.![\[ac=bd\]](http://latex.artofproblemsolving.com/8/f/1/8f15eda50639fe2dbe035965a32c45d4d5dead8b.png)
and![\[a(c+xt) = (b+yt)(d+zt).\]](http://latex.artofproblemsolving.com/f/b/b/fbbc59edce3125e2097c54939010463c6dd94973.png)
This gives us that![\[t=\frac{ax-bz-yd}{yz}\]](http://latex.artofproblemsolving.com/0/4/7/047c622d8decae0ceaea9588f279443d73dcb871.png)
and so 
.
.![\[\frac{ax}{yd} = \frac{AB\cdot CD}{YC\cdot DA} < \frac{AB\cdot CD}{BC\cdot DA} =1,\]](http://latex.artofproblemsolving.com/d/c/4/dc46e7c8eb99e8221d0bbd22630c7344c7faee6e.png)
which is a contradiction.
be the images of 
under this inversion. Then we have![\[A_0B_0\cdot C_0D_0 = \frac{r^4}{XA\cdot XB\cdot XC\cdot XD}(AB\cdot CD) = \frac{r^4}{XA\cdot XB\cdot XC\cdot XD}(BC\cdot DA) = B_0C_0\cdot D_0A_0. \]](http://latex.artofproblemsolving.com/a/c/2/ac2271ff6b87d601d7b9aadeccc2b66d63983228.png)
Moreover, we have 
.
.
.
,
such that 
and etc. It is clear that 
is the isogonal conjugate of 
.
spirals similarly to 
.
.
is equivalent to 
,
are collinear. Suppose otherwise, so WLOG assume 
.
.
to intersect circumcircle of 
at 
.
.
to intersect circumcircle of 
at 
.
and 
.
,
is minimal iff 
is maximal. But 
.
is maximal when 
is maximal when 
.
becomes cyclic and harmonic, and 
.
,
.
,
and we are done.
and vary 
is maximal when 
,
,
.
,
,
are on a circle.
,
,
,
,
,
,
,
,
be the circle passes through 
are collinear. By 
.
.
be the image of 
,
.
.
,
.
under inversion w.r.t. 
under the inversion w.r.t. 
.
gives 
.
to 
.
.
so 
.
of quadrilaterals 
such that angles 
are equal to 
respectively and 
.
overlap with each other with 
on 
.
on rays 
and 
respectively. We notice that this covers all quadrilaterals 
is strictly increasing at a slower rate than 
as 
and we notice that if 
,
and 
.
so we have 
and similarly 
so 
and by symmetry, 
.
,
which would imply 
so 
and 
are collinear. Then 
implies that 
under the condition 
.
and 
.
cyclic. Similarly 
cyclic. Hence, 
,
.
on 
concur, and 
are isogonal in 
.
;
concur. 
and 
.
.
.
and 
.
.
,
and 
The problem is bashy angle chasing from here: 
so adding them together immediately implies the result.
would even let us define it as the inverse, which made the chasing cyclicity easier.
,
,
to 
,
,
,
is cyclic. We can now rewrite the problem with respect to this new quadrilateral, and the length condition rewrites with LoS as 
.
and 
.
cyclic and 
cyclic. Define 
.
are collinear by radical axis. Angle chasing also yields 
is a line. Define 
symmetrically to 
and 
are cyclic, so 
,
.
cyclic. Now we can rewrite the length condition from earlier with a lot more LoS to get that it suffices to show 
.
and 
.
.
and symmetric variants. This allows us to rewrite our length condition as 
.
to intersect 
at 
,
intersect 
at 
.
.
and we can get 
by straightforward angle chasing. This completes the proof.
![\[\measuredangle A^* = \measuredangle D^*A^*X + \measuredangle X^*A^*B^* = \measuredangle XDA + \measuredangle ABX = \measuredangle XBC + \measuredangle ABX = \measuredangle B,\]](http://latex.artofproblemsolving.com/5/2/8/528e1e4594cff8719dc8cc2a3e920da6d430a64c.png)
![\[A^*B^* \cdot C^*D^* = \frac{AB \cdot CD}{XA \cdot XB \cdot XC \cdot XD} = B^*C^* \cdot D^*A^*,\]](http://latex.artofproblemsolving.com/4/a/9/4a992aeaf1b154f8c55a06269dca9e53473def7c.png)
.
corresponds to the isogonal conjugate of of 
we have 
since 
cancels on each side.
and 
as well since we are fixing the angles of the quadrilaterals. WLOG 
decreases as 
and 
.
,
and 
(
.
.
so 
.
maps 
(
and 
.
.
.
,
,
,
.
.
so swapping 
and 
and multiplying gives
Similarly,
The same equations hold for 
and 
.
and 
.
,
and 
,
,
.
and 
must be close to 
.
intersect at the solution corresponding to 
with multiplicity 
,
and 
which in fact finishes, as then 
criterion. In the new quadrilateral, 
and let 
.
is equivalent to 
is equivalent to 
,
is equivalent to 
.
and 
and 
.
and 
with group operation 
which has constant point 
intersect 
and 
intersect 
and 
intersect 
and 
and 
intersections of the cubics 
,
and 
if and only if 
if and only if 
,
and 
and 
.
,
,
.
and that 
,
.
and 
are isogonal in 
,
,
has an isogonal conjugate in 
,
,
in the 
which also goes through 
.
in 
and 
are isogonal in 
if and only if 
if and only if 
and 
bisect 
if and only if 
,
,
so essentially ![[asy]
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[/asy]](http://latex.artofproblemsolving.com/5/a/3/5a31e9a9acef1721dce7c2f170cef478017ae6bd.png)
be the inverted image of 
,
.
so 
cyclic. Similarly, 
cyclic. This in fact immediately pins 
,
.
and 
are cyclic. Hence we have:
In fact 
is also equal to 
by similar logic plus the fact that 
,
and 
to complete the quadrilateral.
.
suffice. ![\[ XAB \sim YCB, \quad XBC \sim YDC, \quad XCD \sim YAD, \quad XDA \sim YBA \]](http://latex.artofproblemsolving.com/0/3/0/030d2be84246f2d9fdc297b378bb5177fa2647aa.png)
are each opposite similarities, if and only if 
.![\[ |{(a-b)(c-d)}| = |{(b-c)(d-a)}|. \]](http://latex.artofproblemsolving.com/3/e/4/3e432f695805f84e43a51f9f9c0c7d8d750cee46.png)
This expands as ![\[ (a-b)(c-d)(\overline{a}-\overline{b})(\overline{c}-\overline{d}) = (b-c)(d-a)(\overline{b}-\overline{c})(\overline{d}-\overline{a}) \]](http://latex.artofproblemsolving.com/1/4/6/146d76e9c0c6c1d86b0c310487b0bbe16e5cc1f9.png)
Grouping terms prioritizing the conjugated ones, we have ![\begin{align*} (\overline{a}\overline{b}+\overline{c}\overline{d})(a-d)(b-c) + (\overline{a}\overline{d}+\overline{b}\overline{c})(a-b)(c-d) &= (\overline{a}\overline{c}+\overline{b}\overline{d})\bigl[(a-b)(c-d)+(a-d)(b-c)\bigr] \\ &= (\overline{a}\overline{c}+\overline{b}\overline{d})(a-c)(b-d). \end{align*}](http://latex.artofproblemsolving.com/f/e/c/fecde11592d604744ec28edc835f8cf3c9a3ee89.png)
and two more symmetric equations. These imply ![\[ x = \frac{(a\overline{b}-b\overline{c})(b-c) - (b\overline{c}-c\overline{d})(a-b)}{(b-c)(\overline{b}-\overline{c}) - (a-b)(\overline{c}-\overline{d})} = \frac{a\overline{b}(b-c) + b\overline{c}(c-a) + c\overline{d}(a-b)} {\overline{b}(b-c) + \overline{c}(c-a) + \overline{d}(a-b)}. \]](http://latex.artofproblemsolving.com/6/c/f/6cf3e422674aa6b6a9d845aa4e10e7ae7a0a4a38.png)
We want to show the above expression is invariant under cycling 
.![\[ \frac{a\overline{b}(b-c) + b\overline{c}(c-a) + c\overline{d}(a-b)} {\overline{b}(b-c) + \overline{c}(c-a) + \overline{d}(a-b)} \overset{?}= \frac{b\overline{c}(c-d) + c\overline{d}(d-b) + d\overline{a}(b-c)} {\overline{c}(c-d) + \overline{d}(d-b) + \overline{a}(b-c)}. \]](http://latex.artofproblemsolving.com/1/b/a/1ba516e9f9c8f1945c610046494df49d9e58a8cc.png)
Expanding, 
Some terms cancel, leading to 
After factoring out 
,![\[ (a-d)(b-c)(\overline{a}\overline{b}+\overline{c}\overline{d}) + (a-b)(c-d)(\overline{b}\overline{c}+\overline{a}\overline{d}) \overset{?}= (a-c)(b-d)(\overline{a}\overline{c}+\overline{b}\overline{d}), \]](http://latex.artofproblemsolving.com/f/e/3/fe3cd21ed4e86562e43cd7fb09844f7b1b07c703.png)
.![\[ \measuredangle XAB = \measuredangle BCY = \measuredangle XCD \quad \text{and} \quad \measuredangle XBC = \measuredangle CDY = \measuredangle XDA, \]](http://latex.artofproblemsolving.com/b/f/b/bfb15e7fd2d5783c3acdb769e84998d04b4ee9d3.png)
so by previous reasoning we have 
.
