AoPS Academy
such that 
is unbounded and![\[2f(m)f(n)-f(n-m)-1\]](http://latex.artofproblemsolving.com/0/f/7/0f77b5addbf8f5e603eb1b6923cdac795cd132e9.png)
is a perfect square for all integer 
be positive integers such that 
. Prove that 
.
is similar with 
, thus there exists 2 points 
on the perpendicular bisector of 
such that 
is similar with , and 
is similar with .
and 
, but 
beeing similar with 
, we have 
, and similar 
, thus we have 
, therefore 
lies on both perpendicular bisectors of 
and 
, therefore 
.
.
be a convex quadrilateral. The perpendicular bisectors of its sides 
and 
meet at 
. Denote by 
a point inside the quadrilateral such that 
and 
. Show that 
.
to be on the perpendicular bisector of and such that 
. Define point 
to be such that 
. Note that since and are on the perpendicular bisector of , 
and 
.
and 
. This yields by spiral similarity that 
and 
. Therefore,![\[\frac{ZX}{DY'}=\frac{AZ}{AY'}=\frac{BZ}{BY'}=\frac{ZX}{CY'}\]](http://latex.artofproblemsolving.com/a/5/a/a5a261494b576a3278e22682a85258e626db6d76.png)
and therefore that 
is the intersection of the perpendicular bisectors of and . Therefore 
.
are midpoints of the sides and and 
, 
. Then since 
we get that 
, 
and ![\[ \frac{AX}{BX}=\frac{DX}{CX}. \]](http://latex.artofproblemsolving.com/9/d/6/9d6d311ed7177e2722398670c67bedfac4c613d1.png)
So ![\[ \frac{AP}{BP}=\frac{DQ}{CQ} \]](http://latex.artofproblemsolving.com/f/1/f/f1facae39b577879003a3708acd9f2601249f495.png)
and the parallel lines pass through and 
to 
and 
respectively, intersect on 
. Analoguosly, the parallel lines pass through and to 
and 
respectively, intersect on 
. Hence 
.
and 
be 
. Let 
be point such that 
and 
.
we have 
~
so 
and since 
we have that 
~
. Similary we have 
~.
and 
and 
and 
.
and 
we have 
. Now by Law of Cosine we have
which implies
so is the intersection of perpendicular bisectors of 
.
be a convex quadrilateral. The perpendicular bisectors of its sides and meet at . Denote by a point inside the quadrilateral such that and . Show that .
intersect. Let 
be midpoints of 
respectively. Let 
be the spiral center 
. Construct 
outside 
with 
and 
. Observe that 
with spiral center again. Further, we see 
. Let 
be the point on 
with 
. Suppose 
is the point inside with 
.
and 
.
under spiral similarity with pivot . Apply 
hence 
. Combined with 
we obtain 
. Now reflect in to get 
. Again linearity gives 
and we're done. 
hence 
and we're done here.
aren't parallel; otherwise the problem doesn't make sense.
be the reflections of over 
and 
be the projections of onto . Let be the midpoints of . Clearly 
are directly similar kites, so they're also similar to their direct average 
. Since 
and aren't parallel, this implies 
are the only such points on 
which divide segments in the same ratio with 
.
with 
. Once again 
are directly similar kites, so for any 
their weighted average 
is also a kite. Since 
we can choose so that this resulting kite contains ; then the other two vertices of the kite must lie on , divide those two segments in the same ratio, and create a segment perpendicular to 
. This implies these other two vertices are precisely by our earlier work, hence 
and 
; the other equality follows similarly.
.
and 
meet at , and 
meets 
at 
. We claim that coincides with .
, and 
. Also line 
is the internal angle bisector of 
, so it suffices to prove that 
.. Denote by 
by the image of any point 
.
, and 
, and 
bisects both 
and 
. Assume that line meets 
at 
, and a line perpendicular to passing meets at 
, respectively. Then,
so and 
coincides with 
. Thus 
, which implies 
. 
first and let 
denote its circumcenter.
be as in the original problem. such that 
and 
.
such that 
and 
.
(oppositely oriented). This will solve the problem.![[asy] pair A = dir(125); pair D = dir(275); pair X = dir(-5); pair O = origin; draw(unitcircle);
pair Y = 1.2*dir(-20); pair B = Y+(A-Y)*X/A;
pair C = Y+(D-Y)*X/D; filldraw(A--O--X--cycle, invisible, deepgreen); filldraw(A--B--Y--cycle, invisible, deepgreen); filldraw(D--O--X--cycle, invisible, blue); filldraw(D--C--Y--cycle, invisible, blue); filldraw(B--X--C--cycle, invisible, grey); draw(circumcircle(B, X, C), grey);
dot("$A$", A, dir(A)); dot("$D$", D, dir(D)); dot("$X$", X, dir(X)); dot("$O$", O, dir(-X)); dot("$Y$", Y, dir(Y)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C));
/* TSQ Source:
A = dir 125 D = dir 275 X = dir -5 O = origin R-X unitcircle
Y = 1.2*dir(-20) B = Y+(A-Y)*X/A
C = Y+(D-Y)*X/D A--O--X--cycle 0.1 lightgreen / deepgreen A--B--Y--cycle 0.1 lightgreen / deepgreen D--O--X--cycle 0.1 lightblue / blue D--C--Y--cycle 0.1 lightblue / blue B--X--C--cycle 0.1 yellow / grey circumcircle B X C grey
*/ [/asy]](http://latex.artofproblemsolving.com/7/2/f/72fbace6dce5f874f3398c87c2f954a2742e70aa.png)
and 
. We break symmetry now and set up as follows. Since we have a similarity 
, there should be a single complex number 
such that 
Next since 
we have ![\[ c = y + \frac{(d-y)x}{d} = \frac{d(x+y)-xy}{d}. \]](http://latex.artofproblemsolving.com/e/5/d/e5dfe61f4def3b0f41236623715a42aac634be55.png)
Now, we simply calculate ![\begin{align*} \frac{c-b}{x-b} &= \frac{\frac{d(x+y)-xy}{d}-[t(x-a)+a]}{x-[t(x-a)+a]} \\ &= \frac{dx + (d-x)[(1-t)a] - dt(x-a)-da}{d(1-t)(x-a)} \\ &= \frac{(1-t)[(d-x)a+dx-da]}{d(1-t)(x-a)} \\ &= \frac{(1-t)x(d-a)}{(1-t)d(x-a)} = \frac{x(d-a)}{d(x-a)} \\ &= \frac{\frac1d-\frac1a}{\frac1x-\frac1a} = \overline{\left( \frac{d-a}{x-a} \right)} \end{align*}](http://latex.artofproblemsolving.com/3/e/f/3ef71aa8ab50bf4c8d7914790e68611d15b0ffe8.png)
as desired.
be the circumcircle of 
then use rotation and similarity properties. 
, both inside , be points on the perp. bisector of such that 
(note there is only one choice for and ). For our setup, set 
be the unit circle, let 
, and let 
be the free variables; note 
, and 
is real. We desire to prove 
, or 
.
is the spiral center of 
, by the spiral center formula we have 
, which we can rearrange to get 
. Thus, we have 
, so changing 
for 
in this expression, we get 
. Thus, it suffices to prove that 
, or 
. But multiplying both sides by , we find the equation indeed holds, so we are done.
.
and 
be two similar triangles. Construct points , , such that triangles 
, 
and 
are similar. Then 
.
there exist linear function of complex plane mapping 
. Also consider identity function 
as a linear function of complex plane. Then to get to points , , you just take weighted average of these functions, but average of linear functions is linear, hence 
. 
reflection of by and 
reflection of by . Note that 
, construct such that 
. Note that 
, hence if we can complex glide them to 
, hence 
. Because triangles and 
are isosceles, we get and from similarities we got the angle condition, that problem wanted. 
be the cirumcenter of . Redefine such that 
, hence 
. We want to show lies on the perpendicular bisector of 
. Let 
be the unit circle, so 
. We fix 
and as arbitrary points, and will calculate and based on these points.
, so 
is the real axis. From the spiral similarity sending 
to , we have![\[ \frac{y-a}{0-a} = \frac{b-a}{x-a} \implies y=\frac{-a(b-a)}{x-a}+a=a\cdot \frac{1-b}{1-a}\]](http://latex.artofproblemsolving.com/8/d/5/8d58a96541fcf4949eab6a5b23781859aa46f7ce.png)
Let 
denote the reflection over real axis of 
. The angle conditions imply that there exists a spiral similarity sending 
and 
. Hence
Since we wanted to show is on the perpendicular bisector of , it suffices to prove that 
satisfies
So if 
, we need to prove 
. Magically, 
Hence 
. The end.
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[/asy]](http://latex.artofproblemsolving.com/6/0/a/60a19cf1b83b94de37887b75930b9261a275e1f1.png)
be the reflection of in . Redefine as the point such that is the center of spiral similarity which sends 
to 
, then 
and 
. Therefore, it suffices to show 
.
and 
, then 
.
Changing the subject we have
Now notice that is the center of spiral similarlity sending 
to 
. Therefore,
Hence 
Now 
Meanwhile
It is easy to see that they have the same modulus, so we are done.
![[asy]
unitsize(2cm);
pair A, B, C, D, X, Y;
X=(0,0);
A=(-1,2);
B=(-2,-2);
C=(3,-1);
D=A*(2*foot(C/B,(0,0),(1,0))-C/B);
Y=extension((A+B)/2,circumcenter(A,B,C),(C+D)/2,circumcenter(B,C,D));
draw(A--B--C--D--A--X--D--C--X--B);
draw(A--Y--B--C--Y--D);
label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, NE);
label("$X$", X, W);
label("$Y$", Y, S);
[/asy]](http://latex.artofproblemsolving.com/9/2/1/92120d6521582ab438f8ec5f41a2d969ba74cfa5.png)
and , this means that 
, so 
. since is inside . Assume without loss of generality . Then, we have 
.
, we must have
Similarly, we have
Therefore, we have
We need to show that 
is real. We have 
Similarly, we have
Therefore, we have 
. Since , we also have 
. Therefore, this means that
so 
Therefore, it suffices to show that 
This is equivalent to
We will show that 
. Adding one to each side gives 
which is equivalent to 
Therefore, it suffices to show 
. Since 
, this means that we have
.
be points on perpendicular bisector of 
such that 
so 
Finally 
as desired.
. There exists a unique spiral similarity that takes onto line 
. Let 
and 
. We have that 
so 
. We also have 
and 
. Thus,![\[\frac{DP}{CP}=\frac{AP}{BP}\cdot \frac{QX}{AQ}\cdot \frac{BQ}{QX}=1\]](http://latex.artofproblemsolving.com/6/3/9/639813fea1efd46695e72afc172bc330f222be46.png)
so 
, implying 
and the result follows.
to the clawson schmidt conjugate of and let 
Angle chasing gives 
bisects both 
and 
Now invert at 
After inversion if we define 
to be the intersections of the line through perpendicular to with then prism lemma and the angle bisector harmonic bundles give 
Thus 
so 
is a bisector of and 
as well. Thus 
so is the same point as given in the problem. Now 
!!! be the Clawson-Schmidt Conjugate of . We will prove that 
. All angles are directed in this solution. and share an angle bisector.
be the antipode of in 
and let 
be the antipode of in 
. By the definition of perpendicular bisectors, 
are respective midpoints of arcs 
of their respective circles, not containing 
bisects and 
bisects and 
are diameters by symmetry, 
, and 
. Thus, 
are collinear, hence and share an angle bisector.
with radius . Let 
be the inverse of 
. Note 
are cyclic. Hence,![\[A^*,Y,B^* \quad C^*,Y,D^* \quad A^*,X'^* ,D^* \quad B^*,X'^* ,C^*\]](http://latex.artofproblemsolving.com/b/a/b/babeb115cb8d770748845a0c31bf6186260a04c7.png)
are collinear (within each triple).
.
is the intersection of the angle bisector of with 
, and similarly 
is the intersection with 
. Recall that by Thales, 
, and since 
is the angle bisector, it is well-known that the two conditions imply :![\[(A^*,B^*,E^*,Y)=-1=(C^*,D^*,Y,F^* )\]](http://latex.artofproblemsolving.com/8/a/b/8ab83d93d8424a89fbe137a32947ebcf5e434403.png)
By the Prism Lemma,
are collinear looking at our complete quadrilateral. This means that 
are collinear.
Hence works. exists satisfying the properties in the problem, which is . such 
, without the acute angle restraint. Construct the point 
such that 
is a parallelogram. Let 
and 
are collinear,![\[\angle PAX_1'=180^\circ-\angle X_1 AD-\angle X_1' AX_1=180^\circ-\angle X_1 BX_1'-\angle CBX_1=\angle X_1' BP\]](http://latex.artofproblemsolving.com/1/d/1/1d132193d9355ca10c9df8b0443e020e6e4f6f7c.png)
Let 
be the function that takes the isogonal conjugate of an object with respect to 
Then, we have![\[\angle \Phi(X_1' )AB=\angle PAX_1'=\angle X_1' BP=\angle AB\Phi(X_1' )\]](http://latex.artofproblemsolving.com/6/2/c/62c1b418945d533bc3842eaa0537c033560f532e.png)
This implies that 
lies on 
, which we define as the perpendicular bisector of . Since isogonal conjugation is an involution, and obviously all of the before steps hold for the converse, for any 
on 
which satisfies the properties in the question.
be the desired locus, we have![\[\Phi(\mathcal{L}_1 )=\ell_1\]](http://latex.artofproblemsolving.com/2/9/f/29f2600ba71c727d05bc06ec195dc732461aea99.png)
Theorem: The isogonal conjugate of a line with respect to a triangle is a circum-conic. intersect at 
and 
such that is the midpoint of minor arc , and the midpoint of major arc . is a rectangular hyperbola.. We have:![\[\angle APM_1=\angle M_1 PB=\angle M_1 AB=\angle PA\Phi(M_1 )\]](http://latex.artofproblemsolving.com/7/5/1/751ea83f447aa26f30a1d221793672e0388f6b37.png)
By the Incenter Excenter Lemma, thus 
. Similarly,![\[\angle M_1 PB=\angle APM_1=\angle ABM_1=\angle \Phi(M_1 )BP\]](http://latex.artofproblemsolving.com/2/c/d/2cd7f8660a7326b025c3568b4ba632d2cffa63ee.png)
Hence 
. Note that 
are collinear as 
is the angle bisector of 
. This means 
is the point at infinity along , and this lies on 
are clearly collinear by Thales as 
![\[\angle M_2 AB+\angle M_1 PA=(180^\circ-\angle BPA)/2+(\angle BPA)/2=90^\circ\]](http://latex.artofproblemsolving.com/a/f/f/affcf087dc7452fd97375679ef7fbd17f63fcc31.png)
Hence 
which implies 
. Analogously, we get 
. This implies that 
is the point at infinity along 
. and belong to as 
lie on 
. This means that has two distinct points at infinity (Note the parallel lines were orthogonal). As we also know that is a conic, must be a hyperbola. be the center of the hyperbola. By definition, 
and 
are asymptotes of . But is parallel to 
which is orthogonal to by Thales as 
is a diameter of 
. Yet is parallel to , hence
is a rectangular hyperbola.![\[\angle PAX_1'=\angle X_1' BP\quad \Leftrightarrow\quad \angle X_1' AD=\angle CBX_1'\]](http://latex.artofproblemsolving.com/8/c/f/8cfc630137a690c66d89a6fabf21b86dd13a1fb2.png)
Hence also lies on . Note that constructing is equivalent to reflection over the midpoint of by the properties of parallelograms.. From our previous result, this must be a component of \mathcal{L}_1. Assume for the sake of contradiction that this reflection intersects the initial branch at some point , other than the midpoint of . Then , hence it is precisely the other branch of the hyperbola. This implies that the center of , , is indeed the midpoint of . Note that even when we have as the midpoint of , for to lie on , it must have multiplicity 
, one count from each branch of . Hence if we consider multiplicity, the midpoint of is still the center of reflection hence the center of . is always the midpoint of . are more specifically the lines respectively parallel to, and perpendicular to the angle bisector of through . such that 
. From all our previous reasoning, say by relabeling:![\[A\Leftrightarrow D,B\Leftrightarrow C,X_1\Leftrightarrow X_2,X_1'\Leftrightarrow X_2'\]](http://latex.artofproblemsolving.com/c/6/3/c6360a25abb753ec6b4afbc0eaaee773f21defbf.png)
And so on. We have an analogous result for the locus of , which in our original diagram is a rectangular hyperbola 
with center , which is the midpoint of . Its asymptotes are respectively the lines parallel to, and perpendicular to, the angle bisector of through . satisfying the constraints in the problem is simply the intersection of and .
and 
have no component in common, they have 
intersection points, counted with their multiplicity, and including points at infinity and points with complex coordinates.. Hence by Bézout’s Theorem, there are 
intersection points. First of all, the points at infinity along the angle bisector of and the orthogonal line, both of which lie on both and contribute to two intersection points, and are distinct. Further, as and are collinear, also lies on and . This gives another intersection point, which lies on a different plane to the points at infinity hence is distinct.. We already know our point from before satisfies the problem’s conditions, and since we constructed this point it must exist. If this is in the exterior of our quadrilateral, then no other point satisfying the conditions of the problem, can exist in the interior of as we already have counted intersection points, hence the problem is vacuously true. Else, it is in the interior of the quadrilateral. Then, there are exactly 
intersection points already, which is the exact number allowed by Bézout, hence is the unique point inside that works.
by uniqueness.![\[\angle APE=\angle ADX\quad \quad \quad \quad \quad \angle EPB=\angle XCB\]](http://latex.artofproblemsolving.com/a/8/1/a815f41fa0b874cb127928a10b717a709bcc357a.png)
Hence,![\[\angle AYB=\angle APB=\angle ADX+\angle XCB=2\angle ADX\]](http://latex.artofproblemsolving.com/1/d/1/1d1c994fb264fbb0c94583a9fa3dccfed371b360.png)
Because 
from the problem, and since 
is cyclic.
and 
.
and 
Prove that
and 
Prove that
Prove that 
and 
.
.
and 
be the points so that 
.
be the circumcenter of 
.
,
is cyclic, which means that 
,
,
,
.
.
,
and 
.
.
,
,
,
,
.
,
.
by reflecting 
was to involve the circumcenter of 
,
,
!
,
simplified to just 
,![\[ \frac{y-d}{y-c} = \frac{d-o}{x-o} = d\]](http://latex.artofproblemsolving.com/a/1/1/a11e0176c692bae0fe3b3a339e7b2910fac909d0.png)
since 
,
from our redefinition. In retrospect, we ``had to have'' 
the images of 
under the spiral similarity taking 
so 
.![\[ \angle ZKD = \angle ZKA + \angle AKD = \angle ZKA + \angle ABX = \angle ZKA + \angle ABZ + \angle ZBX = \angle KAB + \angle ZBX = \angle XBC + \angle ZBX = \angle ZBC \]](http://latex.artofproblemsolving.com/a/8/7/a87e8d911f7007edf49b87f44fa4d6b1e7540dd8.png)
, so \triangle ZBC \cong \triangle ZK
Z
.
be the Clawson-Schmidt conjugate of 
we will be done since 
.
,
,
.
,
.
and 
be the intersections of the line through 
and 
.
So 
and 
.