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.
be the set of all positive integers. Find all functions 
such that for any positive integers 
and 
, the following two conditions hold:
, and
, 
, and 
are equal.
be a triangle with 
. The 
-excircle touches side 
and line 
at 
and 
, respectively. The 
-excircle touches side 
at 
. Let lines 
and 
meet at 
. Prove that 
.
be three diameters of the circumcircle of an acute triangle . Let be an arbitrary point in the interior of 
, and let 
be the orthogonal projection of on 
, respectively. Let 
be the point such that is the midpoint of 
, let 
be the point such that is the midpoint of 
, and similarly let 
be the point such that is the midpoint of 
. Prove that triangle 
is similar to triangle .
of positive integers such that 
.
players. Every player played every other player once, with no draws. In addition, each player had a numerical rating before the tournament began, with no two players having equal ratings. It turns out there were exactly 
games in which the lower-rated player beat the higher-rated player. Prove that there is some player who won no less than 
and no more than 
games.
, 
, and 
be positive integers. Prove that ![\[ \sum_{i = 1}^n \min\left(\left\lfloor \frac{x}{i} \right\rfloor, m \right) = \sum_{i = 1}^m \min\left(\left\lfloor \frac{x}{i} \right\rfloor, n \right). \]](http://latex.artofproblemsolving.com/b/7/a/b7aebfe11f646d604d2e8cb9d5bd70ec796949e8.png)
of points in the plane is balanced if, for any two different points 
and in , there is a point in such that 
. We say that is centre-free if for any three different points , and in , there is no points in such that 
.
, there exists a balanced set consisting of points. for which there exists a balanced centre-free set consisting of points.
such that
are all powers of 
. be an integer and let 
be the complete graph on vertices. Each edge of is colored either red, green, or blue. Let denote the number of triangles in with all edges of the same color, and let denote the number of triangles in with all edges of different colors. Prove![\[ B\le 2A+\frac{n(n-1)}{3}.\]](http://latex.artofproblemsolving.com/6/5/9/659d0199c3d685d43f1ced3704dfa59b02b586b3.png)
(The complete graph on vertices is the graph on vertices with 
edges, with exactly one edge joining every pair of vertices. A triangle consists of the set of 
edges between 
of these vertices.)
such that the equation ![\[x^{2}-(a^{2}+b^{2}+c^{2}+d^{2}+1)x+ab+bc+cd+da=0 \]](http://latex.artofproblemsolving.com/9/0/4/904d1b23d5c744118a7d68df1e2e94bddac8c1cd.png)
has an integer solution. Prove that the other solution is integer too and both solutions are perfect squares.
be a triangle such that![\[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 = \left( \dfrac{6s}{7r} \right)^2, \]](http://latex.artofproblemsolving.com/6/9/b/69b1964be5f92eb44a36b0b8604bf473fe27e210.png)
and 
denote its semiperimeter and its inradius, respectively. Prove that triangle is similar to a triangle 
whose side lengths are all positive integers with no common divisors and determine these integers.
be an acute-angled triangle with circumference 
. Let the angle bisectors of 
and 
intersect again at 
and 
. Let 
be the intersection point of these angle bisectors. Let 
and 
be the respective reflections of and in and . Prove that the center of the circle passing through , , lies on the altitude of triangle from ., there is exactly one card in the deck that has written on it. Now two players draw disjoint sets and of 
cards each from this deck. We would like to define a rule that declares one of them a winner. This rule should satisfy the following conditions:
cards: if the cards are laid down in increasing order face down and we are told which card belongs to which player, but not what numbers are written on them, we can still decide the winner.
and 
, and 
for all 
, then beats .
from the deck, beats and beats then also beats . and such that beats according to one rule, but beats according to the other.
and 
for 
. Prove that the sequence 
is increasing.
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.
be the point outside of , and 
is equilateral, by simple angle chase, we have that colinear, and lies on 
. Define similarly, notice by triangle inequality, 
Claim:, the line passes through is 
, let 
, 
, then 
, equality case holds iff 
.
be the perpendicular to from , and 
. So 
is a diameter, and it is trivial that 
, so 
and 
. Now we can see that 
is the hypotenuse of rt
, so 
, also we have 
, thus 
, so 
, so 
as desired. The equality only holds when 
, so when .
, WLOG let 
, let the parallel line to passing through intersecting 
at 
, then it is trivial that 
, equality holds when 
. Now we can finish the problem:
as desired.
be outside the triangle such that 
are equilateral. Then 
. It suffices to prove 
, however both of these are trivial upon taking altitudes from to , to and the symmetric version, the altitude from has length 
, the altitude from to lies on 
on the other side and thus has length at most one third of the altitude from , which gives the desired inequality. The symmetric variant finishes the other side, so we are done.
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.