Preliminaries
Lemma 1 : Given a
with isogonal conjugate
Then
and
are isogonal conjugate WRT
Proof : Let
be the intersection of
with
respectively. Since
so
are isogonal conjugate WRT
Similarly,
are isogonal conjugate WRT
respectively, so we conclude that
are isogonal conjugate WRT
Lemma 2 : Given a
and a point
lying on the perpendicular bisector of
Let
be the point such that
Then
are collinear.
Proof : Let
be the reflection of
in the perpendicular bisector of
then
are isogonal conjugate WRT
so we conclude that
are collinear.
Lemma 3 : Given a
with circumcircle
and 9-point center
Let
be the isogonal conjugate WRT
and
be the midpoint of
Then
Proof : Let
be the circumcevian triangle of
WRT
and
be the reflection of
in
respectively, then from
Properties of Hagge circle (
Property 1,
Corollary 1.1,
Property 3) we get
the reflection
of
in
is the circumcenter of
and
where
is the orthocenter of
so
Main result
Notation : Given a with centroid 
circumcenter 
orthocenter 
9-point center 
symmedian point 
and ![$ \theta \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ]. $](//latex.artofproblemsolving.com/c/0/3/c03e5480ddebf50a7c9980f185fb6b35d4c5afc5.png)
Let 
be the Kiepert triangle of with angle 
be the perspector of 
and let 
be the isogonal conjugate of 
WRT respectively.
Property 1 : 
Proof : Let
be the tangential triangle of
Note that
so we conclude that
i.e.
are collinear.
Corollary 1.1 : 
Proof : From
Property 1 we get
is the intersection of
so by
Lemma 1 we conclude that
lies on
Property 2 : 
Proof : Simple angle chasing yields
so by
Lemma 2 we get
lies on
Similarly, we can prove
By Desargues's theorem for
and
are concurrent, so by Desargues's theorem for
and
we conclude that
lies on
Corollary 2.1 : 
Proof : From
Property 2 we get
is the intersection of
so by
Lemma 1 
lies on
Analogously, we can prove
Property 3 : Given ![$ \color{blue} \phi, \sigma \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ]. $](//latex.artofproblemsolving.com/9/f/6/9f6112dcca2b4cfe5b020534848449790307f283.png)
Then 
Proof : Note that for a fixed
![$ \tau \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ] $](//latex.artofproblemsolving.com/5/2/0/52005eb5fdc74f40420c61dbb4cc764f9273d748.png)
the mapping
is an involution on the Kiepert hyperbola of
so
passes through the pole
of
for all
Consider the case when
and
we conclude that
by
Corollary 1.1 and
Property 2 
Property 4 : Let 
be the intersection of 
with 
Then 
Proof : Let
be the intersection of
with the perpendicular bisector of
and let
then from
we get
is the image of
under the inversion WRT
so
On the other hand, by
Lemma 2 we get
so
hence by Menelaus' theorem for
and
we conclude that
Corollary 4.1 : 
are concurrent on 
Proof : From
Property 4,
are concurrent at
By Desargues's theorem for
and
we get
are concurrent, so by Desargues's theorem for
and
we conclude that
lies on
Analogously, we can prove
Property 5 : 
Proof : Let
be the midpoint of
respectively. First, note that
are parallelogram, so
is the medial triangle of
On the other hand, simple angle chasing yields the second intersection of
with
lies on
where
is the circle with center
and containing
so
Analogously, we can prove
so by Sondat's theorem for
and
we conclude that
Application
Notation : Let 
be the centroid, circumcenter, 9-point center, symmedian point, 1st Fermat point, 2nd Fermat point, 1st Isodynamic point, 2nd Isodynamic point of respectively.
Theorem (
Basic properties of Fermat points and Isodynamic points)
:
(1) 
(2) 
is the G-symmedian of
(3) (
Lester circle)
are concyclic.
Proof : (1) is a consequence of the results in the previous section. Let
be the intersection of
and
be the 1st Napoleon triangle, 2nd Napoleon triangle of
respectively. First, note that
lies on
so
is the reflection of
in
Since
is the center of
lies on the circumcircle of
and
are isogonal conjugate of
for
so by
Lemma 3 we get
hence
is the G-symmedian of
and
(2) is proved. Finally, from the previous conclusion and
we get
is tangent to
at
so
and hence we conclude that
are concyclic. i.e.
(3) is proved.
Remark : It is possible to prove
(2) or
(3) directly.
1. Proof of
(2) without using
(1) :
Since
are antigonal conjugate WRT
so the center of the Kiepert hyperbola
of
is the midpoint of
the isogonal conjugate of
WRT
is the perpendicular bisector of
hence the tangent of
at
is the G-symmedian of
On the other hand,
is the isotomic conjugate of
WRT
so
is tangent to
at
and hence
is the G-symmedian of
2. Proof of
(3) without using
(1) and
(2) :
Lemma : Given a
and a circle
passing through
and
Let
be the intersection of
with
respectively and let
be the points such that
Suppose that
is the intersection of the tangent of
at
with
respectively, then
Proof : Let
be the points such that
and let
be the intersection of
Clearly,
are homothetic and
so
hence by Menelaus' theorem we conclude that
Back to the main problem :
Let
be the 1st Napoleon triangle, 2nd Napoleon triangle of
respectively and let
be the intersection of
with the tangent of
at
Since
is the reflection of
in
respectively, so by
Lemma for
we get the reflection
of
in
lies on the perpendicular bisector of
Similarly, we can prove
lies on the perpendicular bisector of
so
and hence we conclude that
are concyclic.
by EmersonSoriano, Apr 5, 2025, 6:32 PM 
such that 
is a divisor of 
.
be a triangle with circumcenter 
. Let 
be a point on the perpendicular bisector of 
(see figure) and 
, 
be the intersections of the perpendicular bisectors of 
and 
, respectively, with 
and 
. Prove that the midpoints of 
and 
and the point 
![[asy]
import graph;
size(100);
// Define constants
real radius = 0.3;
real v_dist = sqrt(3)/2;
pen coinPen = black+0.8;
pen linePen = gray+dotted+0.6;
pen labelPen = black;
pen fillPen = lightgray+opacity(0.5); // Semi-transparent fill
real labelScale = 0.8;
// Function to get coin position
pair P(int i, int j) {
return ( (j - (i+1.0)/2.0), -(i-1)*v_dist );
}
int n = 4;
pair[][] coinPos = new pair[n+1][];
for(int i=1; i<=n; ++i) {
coinPos[i] = new pair[i+1];
for(int j=1; j<=i; ++j) {
coinPos[i][j] = P(i,j);
}
}
// Highlight the two flipped triangles T(2,2) and T(3,5)
// T(2,2) involves (2,1), (2,2), (3,2) - Downward
pair pA1 = P(2,1); pair pA2 = P(2,2); pair pA3 = P(3,2);
fill(pA1--pA2--pA3--cycle, fillPen);
// T(3,5) involves (3,3), (4,3), (4,4) - Upward (using problem's naming convention)
// Or more standard: (i=3, j=3) up -> (3,3), (4,3), (4,4)
pair pB1 = P(3,3); pair pB2 = P(4,3); pair pB3 = P(4,4);
fill(pB1--pB2--pB3--cycle, fillPen);
// Draw connecting lines
for(int i=1; i<n; ++i) {
for(int j=1; j<=i; ++j) {
draw(coinPos[i][j] -- coinPos[i+1][j], linePen);
draw(coinPos[i][j] -- coinPos[i+1][j+1], linePen);
}
}
for(int i=1; i<=n; ++i) {
for(int j=1; j<i; ++j) {
draw(coinPos[i][j] -- coinPos[i][j+1], linePen);
}
}
// Draw coins (after fills)
for(int i=1; i<=n; ++i) {
for(int j=1; j<=i; ++j) {
draw(Circle(coinPos[i][j], radius), coinPen);
}
}
// Add H/T labels based on the example image
label(scale(labelScale)*"H", coinPos[1][1], labelPen);
label(scale(labelScale)*"T", coinPos[2][1], labelPen);
label(scale(labelScale)*"T", coinPos[2][2], labelPen);
label(scale(labelScale)*"H", coinPos[3][1], labelPen);
label(scale(labelScale)*"T", coinPos[3][2], labelPen);
label(scale(labelScale)*"T", coinPos[3][3], labelPen);
label(scale(labelScale)*"H", coinPos[4][1], labelPen);
label(scale(labelScale)*"H", coinPos[4][2], labelPen);
label(scale(labelScale)*"T", coinPos[4][3], labelPen);
label(scale(labelScale)*"T", coinPos[4][4], labelPen);
[/asy]](http://latex.artofproblemsolving.com/7/8/4/7841e4b3d462255902a214d8a3fc2658045281b2.png)
![[asy]
import graph;
size(150); // Increased size for clarity
// Define constants
real radius = 0.3;
real v_dist = sqrt(3)/2;
pen coinPen = black+0.8;
pen linePen = gray+dotted+0.6;
pen labelPen = black;
pen fillPen = lightgray;
real labelScale = 0.6; // Smaller label scale
// Function to get coin position
pair P(int i, int j) {
return ( (j - (i+1.0)/2.0), -(i-1)*v_dist );
}
// Function to draw the grid and empty coins
void drawGrid(int n) {
pair[][] coinPos = new pair[n+1][];
for(int i=1; i<=n; ++i) {
coinPos[i] = new pair[i+1];
for(int j=1; j<=i; ++j) {
coinPos[i][j] = P(i,j);
}
}
// Draw connecting lines
for(int i=1; i<n; ++i) {
for(int j=1; j<=i; ++j) {
draw(coinPos[i][j] -- coinPos[i+1][j], linePen);
draw(coinPos[i][j] -- coinPos[i+1][j+1], linePen);
}
}
for(int i=1; i<=n; ++i) {
for(int j=1; j<i; ++j) {
draw(coinPos[i][j] -- coinPos[i][j+1], linePen);
}
}
// Draw coins (empty)
for(int i=1; i<=n; ++i) {
for(int j=1; j<=i; ++j) {
draw(Circle(coinPos[i][j], radius), coinPen);
}
}
}
// Function to draw a shaded UP triangle move and its label
void drawMoveUp(int i, int j, string lbl) {
pair p1 = P(i,j);
pair p2 = P(i+1,j);
pair p3 = P(i+1,j+1);
fill(p1--p2--p3--cycle, fillPen);
// Redraw circles on top
draw(Circle(p1, radius), coinPen);
draw(Circle(p2, radius), coinPen);
draw(Circle(p3, radius), coinPen);
// Label
label(scale(labelScale)*lbl, (p1+p2+p3)/3, labelPen);
}
int n=4;
drawGrid(n);
// Draw the 6 moves shown in the first diagram
drawMoveUp(1, 1, "$T(1,1)$");
drawMoveUp(2, 1, "$T(2,1)$");
drawMoveUp(2, 2, "$T(2,3)$"); // Note index from problem
drawMoveUp(3, 1, "$T(3,1)$");
drawMoveUp(3, 2, "$T(3,3)$"); // Note index from problem
drawMoveUp(3, 3, "$T(3,5)$"); // Note index from problem
[/asy]](http://latex.artofproblemsolving.com/6/e/8/6e8a9d60edc162308f478e0f6cc241a20e462665.png)
![[asy]
import graph;
size(150); // Increased size for clarity
// Define constants
real radius = 0.3;
real v_dist = sqrt(3)/2;
pen coinPen = black+0.8;
pen linePen = gray+dotted+0.6;
pen labelPen = black;
pen fillPen = lightgray;
real labelScale = 0.6; // Smaller label scale
// Function to get coin position
pair P(int i, int j) {
return ( (j - (i+1.0)/2.0), -(i-1)*v_dist );
}
// Function to draw the grid and empty coins (same as before)
void drawGrid(int n) {
pair[][] coinPos = new pair[n+1][];
for(int i=1; i<=n; ++i) {
coinPos[i] = new pair[i+1];
for(int j=1; j<=i; ++j) {
coinPos[i][j] = P(i,j);
}
}
// Draw connecting lines
for(int i=1; i<n; ++i) {
for(int j=1; j<=i; ++j) {
draw(coinPos[i][j] -- coinPos[i+1][j], linePen);
draw(coinPos[i][j] -- coinPos[i+1][j+1], linePen);
}
}
for(int i=1; i<=n; ++i) {
for(int j=1; j<i; ++j) {
draw(coinPos[i][j] -- coinPos[i][j+1], linePen);
}
}
// Draw coins (empty)
for(int i=1; i<=n; ++i) {
for(int j=1; j<=i; ++j) {
draw(Circle(coinPos[i][j], radius), coinPen);
}
}
}
// Function to draw a shaded DOWN triangle move and its label
void drawMoveDown(int i, int j, string lbl) {
// Coins involved are (i,j), (i,j+1), (i+1, j+1)
pair p1 = P(i,j);
pair p2 = P(i,j+1);
pair p3 = P(i+1,j+1);
fill(p1--p2--p3--cycle, fillPen);
// Redraw circles on top
draw(Circle(p1, radius), coinPen);
draw(Circle(p2, radius), coinPen);
draw(Circle(p3, radius), coinPen);
// Label
label(scale(labelScale)*lbl, (p1+p2+p3)/3, labelPen);
}
int n=4;
drawGrid(n);
// Draw the 3 moves shown in the second diagram
drawMoveDown(2, 1, "$T(2,2)$"); // Corresponds to coins (2,1), (2,2), (3,2)
drawMoveDown(3, 1, "$T(3,2)$"); // Corresponds to coins (3,1), (3,2), (4,2)
drawMoveDown(3, 2, "$T(3,4)$"); // Corresponds to coins (3,2), (3,3), (4,3)
[/asy]](http://latex.artofproblemsolving.com/3/1/a/31a1668332654ec625ce6a7ca414ef266fe18dbe.png)
is given. Let 
be set of all 
segments of real line of type ![$[i, j]$](http://latex.artofproblemsolving.com/a/6/4/a644edd67caa59eaded765c417b6d9822f782ec2.png)
, where 
and 
are integers, 
. A subset 
is said to be valuable if the intersection of any two segments from 
is either empty, or is a segment of nonzero length belonging to 
, diagonals 
are equal, and they intersect at 
. Perpendicular bisectors of 
intersect at point 
, and perpendicular bisectors of 
intersect at point 
. Prove that 
.
denote the set of positive rational numbers. Find, with proof, all functions 
such that, for all positive rational numbers 
and 
we have ![\[f(x)=f(x+y)+f(x+x^2f(y)).\]](http://latex.artofproblemsolving.com/9/e/a/9ea71d2ee277dc3c689ba2d791b451e29353edb3.png)
with isogonal conjugate 
Then 
and 
are isogonal conjugate WRT 
be the intersection of 
with 
respectively. Since 
so 
are isogonal conjugate WRT 
Similarly, 
are isogonal conjugate WRT 
respectively, so we conclude that 
lying on the perpendicular bisector of 
Let 
be the point such that 
Then 
are collinear.
be the reflection of 
in the perpendicular bisector of 
then 
are isogonal conjugate WRT 
so we conclude that 
and 9-point center 
Let 
be the isogonal conjugate WRT 
be the midpoint of 
Then 
be the circumcevian triangle of 
WRT 
be the reflection of 
in 
respectively, then from Properties of Hagge circle (Property 1, Corollary 1.1, Property 3) we get 
the reflection 
in 
is the circumcenter of 
and 
where 
is the orthocenter of 
so 
circumcenter 
orthocenter 
9-point center 
symmedian point 
and ![$ \theta \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ]. $](http://latex.artofproblemsolving.com/c/0/3/c03e5480ddebf50a7c9980f185fb6b35d4c5afc5.png)
Let 
be the Kiepert triangle of 
be the perspector of 
and let 
be the isogonal conjugate of 
WRT 
be the tangential triangle of 
so we conclude that 
i.e. 
are collinear. 
so by Lemma 1 we conclude that 
lies on 
so by Lemma 2 we get 
Similarly, we can prove 
By Desargues's theorem for 
and 
are concurrent, so by Desargues's theorem for 
and 
we conclude that 
so by Lemma 1 
lies on 
Analogously, we can prove 
![$ \color{blue} \phi, \sigma \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ]. $](http://latex.artofproblemsolving.com/9/f/6/9f6112dcca2b4cfe5b020534848449790307f283.png)
Then 
![$ \tau \in \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ] $](http://latex.artofproblemsolving.com/5/2/0/52005eb5fdc74f40420c61dbb4cc764f9273d748.png)
the mapping 
is an involution on the Kiepert hyperbola of 
passes through the pole 
of 
for all 
and 
we conclude that 
by Corollary 1.1 and Property 2 
be the intersection of 
with 
Then 
be the intersection of 
with the perpendicular bisector of 
and let 
then from 
we get 
under the inversion WRT 
so 
On the other hand, by Lemma 2 we get 
so 
hence by Menelaus' theorem for 
and 
we conclude that 
are concurrent on 
are concurrent at 
By Desargues's theorem for 
and 
we get 
are concurrent, so by Desargues's theorem for 
and 
we conclude that 
Analogously, we can prove 
be the midpoint of 
respectively. First, note that 
are parallelogram, so 
is the medial triangle of 
On the other hand, simple angle chasing yields the second intersection of 
with 
lies on 
where 
and containing 
so 
Analogously, we can prove 
so by Sondat's theorem for 
be the centroid, circumcenter, 9-point center, symmedian point, 1st Fermat point, 2nd Fermat point, 1st Isodynamic point, 2nd Isodynamic point of 
is the G-symmedian of 
are concyclic.
and 
be the 1st Napoleon triangle, 2nd Napoleon triangle of 
so 
Since 
lies on the circumcircle of 
and 
are isogonal conjugate of 
so by Lemma 3 we get 
hence 
is the G-symmedian of 
and (2) is proved. Finally, from the previous conclusion and 
we get 
is tangent to 
at 
and hence we conclude that 
are antigonal conjugate WRT 
of 
hence the tangent of 
passing through 
and 
Let 
be the intersection of 
respectively and let 
be the points such that 
Suppose that 
is the intersection of the tangent of 
at 
respectively, then 
be the points such that 
and let 
be the intersection of 
Clearly, 
are homothetic and 
so 
hence by Menelaus' theorem we conclude that 
be the 1st Napoleon triangle, 2nd Napoleon triangle of 
be the intersection of 
in 
respectively, so by Lemma for 
we get the reflection 
of 
and hence we conclude that 
satisfying 
, prove that![\[ \frac{a^4+5b^4}{a(a+2b)} + \frac{b^4+5c^4}{b(b+2c)} + \frac{c^4+5a^4}{c(c+2a)} \geq 1- ab-bc-ca \]](http://latex.artofproblemsolving.com/f/e/8/fe85cf52b883c214806ab837a2eb25cebb511d7b.png)
such that there exists a unique integer 
such that 
with the following property:![\[ n!\mid a^n + 1
\]](http://latex.artofproblemsolving.com/e/2/3/e23d39aab685888993d355d8c93d4e6475f53487.png)
May 2023
May 2021
