AoPS Academy
, the points 
and 
are taken such that the point lies between the points 
and , and 
. On segments 
and 
, points 
and 
are taken so that 
and 
. Prove that 
.
and points and on sides 
and 
, respectively, such that 
. Let 
and 
be intersection points of lines 
and 
with circumcircle 
of triangle 
, and 
and 
intersection points of lines 
and 
with side 
. If 
, prove that circle contains intersection point of angle bisector of 
with
such that 
divides 
for all positive integers 
and 
with 
.
such that 
Follows for all reals 
.
denote the set of positive integers. Find all functions 
such that for positive integers and 
![\[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]](http://latex.artofproblemsolving.com/c/1/f/c1f8ffe04cfcc45b498f3931a4796d5c56dc04d0.png)
be an integer. Rowan and Colin play a game on an 
grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of 
, how many orderly colorings are there? is selected inside acute 
so that 
and 
. Point is chosen on ray 
so that 
. Let be the midpoint of . is tangent to the circumcircle of triangle 
.
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
be a triangle for which there exists an interior point 
such that 
. Let the lines 
and 
meet the sides and at and respectively. Prove that ![\[ AB+AC\geq4DE. \]](http://latex.artofproblemsolving.com/3/a/4/3a49c5f8bd2930adca7bde2601dd045d4bf05ff6.png)
be the images of 
through the homothety of center and ratio 
. We have to show that 
, so it would be enough to show 
. Again, we notice that it's enough to show 
. Let be the vertex of the equilateral triangle 
, lying on the opposite side of 
as . Clearly, 
, so 
is equivalent to 
(the last equality is well-known, and it follows from Ptolemy's equality applied to the cyclic quadrilateral 
) or, in other words, 
. In terms of areas, this means 
, and this is clear since for fixed 
, the area 
reaches its maximum when 
, and in this case we have equality in the above inequality.
, which is, as we have shown, equivalent to the initial problem.
and 
on the exterior of the triangle 
.
, therefore the points 
are concyclic.
, so 
.
.![$ \frac {FP}{FD} = 1 + \frac {DP}{FD} = 1 + \frac {[APC]}{[AFC]}\geq 4$](http://latex.artofproblemsolving.com/8/d/8/8d817466ab09ffb2a3158b4396a83c3f244a4a73.png)
, and the equality occurs when 
is the midpoint of 
.
, and 
.
, and thus the problem is solved.
![$\frac{[APC]}{[AFC]} \ge 3$](http://latex.artofproblemsolving.com/0/6/1/0612eb9cf1e42a8b00d84e29081fbd6c271e8d09.png)
.![$(AF-CF)^2 \ge 0 \Rightarrow AF^2+CF^2+AF*CF \ge 3AF*CF \Rightarrow AC^2 \ge 3AF*CF \Rightarrow AP*CP\sin60^{\circ} \ge 3AF*CF\sin120^{\circ} \Rightarrow \frac{[APC]}{[AFC]} \ge 3$](http://latex.artofproblemsolving.com/e/9/f/e9faeb8cf1b77e532eab02c37fe115972864dcfe.png)
.
are all equal and sum up to 
they must each be equal to 
Construct a point 
outside such that 
is equilateral. Define point 
analogously. Since 
points 
are concyclic. Furthermore, since 
points 
are collinear.
This is equivalent to![\begin{align*}
[\triangle AB'C] & \geq 3[\triangle APC] \\
\iff AB' \cdot B'C \cdot \sin (60^{\circ}) & \geq 3 \cdot AF \cdot CF \cdot \sin (120^{\circ}) \\
\iff AB' \cdot B'C & \geq 3 \cdot AF \cdot CF .\end{align*}](http://latex.artofproblemsolving.com/4/d/7/4d73410941cf9b8502e04d93cf755162eab3fbe9.png)
Hence, it suffices to show that
which map to to and 
to have ratio at least 4, so 
By triangle inequality, we have that![\[AB'+A'C \geq B'C' \implies AB+AC \geq 4DE. \quad \blacksquare\]](http://latex.artofproblemsolving.com/1/9/d/19debbb6248a41142ca1725f1206dc38f559c9ec.png)
that is, the Fermat point must be the circumcenter of 
This is possible iff is equilateral, because 
and similarly 
and 
outwardly on the sides of . It is well known that 
and 
.
and 
.
and 
and using 
,
and other part is analogously proved.
and 
and 
outside of , so it's well known that 
and 
are lines. 
is cyclic, so consider the tangent at the point 
, the antipode of , labeled as line 
. Note that 
, so then 
with equality only when 
, and similarly 
. Let and be the points on segments 
and 
, respectively, such that 
and 
. Then since 
and 
, 
, as desired.
be a triangle for which there exists an interior point such that . Let the lines and meet the sides and at and respectively. Prove that 
Now, erect equilateral triangles 
on the sides, externally. Then 
are cyclic. Hence, 
and so 
are collinear. We get two more symmetric results and so is teh Fermat point given by 
Hence, it suffices to show
Let 
Then it suffices to show
which is true. 
and so we get 
as desired. 
be a triangle for which there exists an interior point such that . Let the lines and meet the sides and at and respectively. Prove that 
forming equilateral triangles 
, 
, 
sticking out from the triangle. Evidently, is the intersection of the three circumcircles of these equilateral triangles.
, and in particular is the concurrence point of 
, 
, and 
. Note now that 
.
As varies along 
with length fixed, note that the maximum length of 
occurs when 
, and this is also the case for the minimum length of 
. Thus 
.
as desired.
and 
, so that 
, 
, 
, 
are collinear and 
, , 
, 
are collinear.![[asy] size(7cm); defaultpen(fontsize(10pt)); pair A,B,C,Y,Z,F,D,EE; A=dir(110); B=dir(220); C=dir(320); Y=A+(C-A)*dir(60); Z=B+(A-B)*dir(60); F=extension(B,Y,C,Z); D=extension(B,Y,A,C); EE=extension(C,Z,A,B);
draw(D--EE); draw(B--Y,gray); draw(C--Z,gray); draw(circumcircle(A,F,C),linewidth(.3)); draw(circumcircle(A,F,B),linewidth(.3)); draw(C--Y--A--Z--B); draw(A--B--C--cycle,linewidth(.7));
dot("\(A\)",A,dir(105)); dot("\(B\)",B,S); dot("\(C\)",C,S); dot("\(F\)",F,dir(265)); dot("\(Y\)",Y,dir(30)); dot("\(Z\)",Z,dir(150)); dot("\(D\)",D,dir(-5)); dot("\(E\)",EE,dir(210)); [/asy]](http://latex.artofproblemsolving.com/9/5/8/95800899e16e3c862af157470fc4afdeb0cd2ac6.png)
and 
. It follows that 
as needed.
be a triangle for which there exists an interior point such that . Let the lines and meet the sides and at and respectively. Prove that 
is the Toricelli/1st Fermat point of . It is a well-known property of that if 
and 
are equilateral triangles erected outward from and , respectively, 
are collinear and 
is cyclic (similarly 
are collinear and 
is cyclic). is on minor arc , the minimum possible value of 
occurs when is on the midpoint of the arc, as this maximizes 
and minimizes 
. In that case, it is easy to show that 
and thus 
, hence it must be true that 
and similarly 
.
and 
, where 
. Since 
, by LoC on 
we get 
Using LoC on 
, we get
, finishing the problem.
be such that 
are equilateral. We have that 
and 
are collinear.
. Let 
, 
. Note that 
, the circumradius of 
, is equal to 
.
.
.
.
, as claimed. .
, as desired. 
be on 
extended such that 
Let 
be on extended such that 
Note that 
is cyclic. Also, 
and 
so 
Thus, 
is equilateral. Similarly, 
is equilateral. Now, 
Since 
is obtuse, it suffices to show 
and 
to prove that 
by AA so 
which implies the result that . Similarly, 
Now, WLOG suppose the parallel line through parallel to 
lies outside of 
Then this line intersects 
at 
as desired.
be a triangle for which there exists an interior point such that . Let the lines and meet the sides and at and respectively. Prove that 
such that 
and 
are equilateral triangles. Clearly, 
are cyclic quads, and 
implies 
are collinear. Similarly, 
and 
are collinear. Now 
so it suffices to show that 
.
and 
. It suffices to prove the following two lemmas to finish:
lies on arc 
of the circumcircle of equilateral triangle 
not containing and line 
meets 
at point . Then 
.
. Now 
is larger than the -median of 
and 
is smaller than the length it achieves when is antipodal to . For rigour, this follows as 
is fixed by shooting lemma. When is antipodal, equality is achieved, proving the lemma. with obtuse angle at , points 
lie on rays 
such that 
and 
. Then 
. to assume 
and 
. Now 
as 
and 
as 
, so 
unless 
and 
, proving the claim.
for points and , the conclusion follows.
and 
so that both are not in the of . Then it is well known that 
, , and , , are collinear, and moreover 
, 
are cyclic. Now we can obtain 
, 
and hence by LOC 
. To finish,![\[AB+AC=AX+AY\ge XY\ge 4DE\]](http://latex.artofproblemsolving.com/0/4/1/0418da0f0f823fd8604aa22376964955fd28bb47.png)
as desired. 
is 
and construction can easily be found from the theorem thonk:/
is the Fermat Point. Thus, let and be points such that 
and 
are equilateral triangles lying outside .
and 
.
, by Law of Cosines.![\begin{align*}
\dfrac{[AXB]}{[AFB]} = \dfrac{\tfrac{\sqrt{3}}{4} \cdot AB^2}{\tfrac{1}{2} \cdot AF \cdot FB \cdot \sin\left(120^{\circ}\right)} = \dfrac{AF^2 + FB^2 + AF \cdot FB}{AF \cdot FB} = \dfrac{AF}{FB} + \dfrac{FB}{AF} + 1 \geq 3.
\end{align*}](http://latex.artofproblemsolving.com/e/b/8/eb8e72d19667ff4cab4357a90cb0650f38383920.png)
Thus, ![$[AXB] \geq 3 \cdot [AFB]$](http://latex.artofproblemsolving.com/f/2/f/f2fc41d3ab479aeac141536d89025dc718ea6d1e.png)
, thus 
and . Then, follows, as desired. 
. But, 
, by triangle inequality. Therefore 
, as desired.
Erect equilateral triangles 
and 
outside of and 
Observe that 
is cyclic as 
Thus by Ptolemy on 
we get 
Since 
(by simple angle-chasing), we get 
so 
Therefore, 
by AM-GM on the denominator. Similarly, 
Therefore, 
as desired.
![\[EF = AE \cdot \frac{BF}{AB} = AB \cdot \frac{AF}{AF+BF} \cdot \frac{BF}{AB} = \frac{AF \cdot BF}{AF+BF}.\]](http://latex.artofproblemsolving.com/0/e/9/0e95f843678860f9b5a0f821c2a427bb846d01a2.png)
Now let 
, 
, and 
, and observe that 
while 
. From here, it is very much feasible to directly expand 
using Law of Cosines, but here is a comparatively nicer finish.
, , and 
outside triangle such that 
, and note that 
, et cetera. So ![\[DE^2 = EF^2+DF^2 + DE \cdot EF \leq \frac{FZ^2+FY^2 + FZ \cdot FY}{16} = \frac{ZY^2}{16} \leq \frac{(AB+AC)^2}{16}\]](http://latex.artofproblemsolving.com/9/e/b/9eb26563c7f139bb1ee53d9c43e11723741d9a0e.png)
as needed.
respectively. Then 
.
but decreases 
and thus 
as well.
on an obtuse triangle, 
is maximized when it is the longest side.
outside 
.
;
is a segment inside obtuse triangle 
,
.
,
.
.
.
,
and solving for 
.
.![\[
DE^2 = DF^2 + EF^2 + DF \cdot EF = \left ( \frac{ab}{a + b} \right)^2 + \left (\frac{ac}{a + c} \right)^2 + \frac{ab}{a + b} \cdot \frac{ac}{a + c}.
\]](http://latex.artofproblemsolving.com/2/4/0/2401372738f4da1ff9e019191ac1dc753758c180.png)
Thus it remains to show the inequality, for reals 
,
As this inequality is homogenous, we set 
.
and by GM-HM, this inequality is weaker than
By AM-GM, this inequality is weaker than![$$\sqrt[4]{(a^2 + b^2 + ab)(a^2 + c^2 + ac)} \ge \sqrt[4]{(ab + ac + a\sqrt{bc})^2}.$$](http://latex.artofproblemsolving.com/3/8/d/38dc91c403f97475aca3ef8e44e4b219594418c7.png)
Take both sides to the fourth power, and divide both sides by 
to get
recalling that 
which after rearranging is equivalent to
which is obviously true. 
so that![\[AB+AC=AX+AY\ge XY\ge 4DE.\]](http://latex.artofproblemsolving.com/5/8/4/5840c5169c74ac52b383ed43c5926a52d4282b75.png)
We are done.