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 an acute triangle with orthocenter 
and circumcenter 
. Let 
and 
be points on the line segments 
and 
respectively such that 
is a parallelogram. Prove that 
.
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
. If is an integer, then 
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 
, and that, for every such pair 
, the "next" such pair can be calculated as
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?
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
, 
and 
. Find 
and 
interesting if is composite and for every divisor 
of at least one of 
and 
is also a divisor of with 
be a triangle, and let 
and 
be points on segment symmetric with respect to the midpoint of . Let 
denote the circle passing through and tangent to line at 
. Similarly, let 
denote the circle passing through and tangent to line 
at 
. Let the circles and intersect again at point 
(
). Prove that 
.
be an acute triangle with 
the feet of the altitudes lying on 
respectively. One of the intersection points of the line 
and the circumcircle is 
The lines 
and 
meet at point 
Prove that 
cities, 
airline companies in a country. Between any two cities, there is exactly one -way flight connecting them which is operated by one of the two companies. A female mathematician plans a travel route, so that it starts and ends at the same city, passes through at least two other cities, and each city in the route is visited once. She finds out that wherever she starts and whatever route she chooses, she must take flights of both companies. Find the maximum value of .
be inscribed in the circumcircle 
and circumscribed about the incircle 
, with 
. The incircle touches the sides , 
, and at 
, , and , respectively. A line through 
, perpendicular to 
, intersects , , and at 
, 
, and 
, respectively. The line meets at 
(distinct from ). The circumcircle of triangle 
intersects at 
(distinct from ). Let 
be the midpoint of the arc 
of . The line cuts segments and 
at 
and 
, respectively, and the tangents to the circle 
at and intersect at 
. Prove that 
.
known by both players, Arnaldo picks a number 
but doesn't tell it to anyone. Bernaldo thens pick an arbitrary positive integer 
(not necessarily in ). Then Arnaldo tells the number of divisors of 
. Show that Bernaldo can choose in a way that he can find out the number 
chosen by Arnaldo.
be a set of 2025 positive real numbers. For a subset 
, define 
as the median of 
when all elements of are arranged in increasing order, with the convention that 
. Define![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set of 2025 positive real numbers, the following inequality holds:![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in .
in 
colors, there exists an arithmetic progression of size 
. How can we derive Van der Waerden's theorem for 
from this?
be circumcenter of a non-isosceles triangle and be a point in the interior of 
. Let 
be foots of perpendicular lines from to 
. Suppose that 
is cyclic and is the circumcenter of , 
. Prove that 
bisects
) is equilateral. For example: Let D, E, F be the feet of normals from Q to BC, CA, AB. The A-Apollonius circle is a locus of points with the ratio of distances 
from the triangle vertices B, C and similarly for the B- and C-Apollonius circles. Hence 
The quadrilateral AEQF is cyclic with the right angles at the vertices E, F, hence its circumradius is 
so that 
and similarly 
Consequently,
Let e = DE = EF = FD be the side length of the equilateral pedal triangle 
. The area S of the triangle with circumradius R is![$S = \frac 1 2 [AQ \cdot BQ\ \sin(C + 60^\circ) + BQ \cdot CQ\ \sin(A + 60^\circ) + CQ \cdot AQ\ \sin(B + 60^\circ)] =$](http://latex.artofproblemsolving.com/f/3/6/f366f9eb8156b02d601f6cd494d616f35574f9d4.png)
![$= \frac{e^2}{2} [\frac{\sin(C + 60^\circ)}{\sin A \sin B} + \frac{\sin(A + 60^\circ)}{\sin B \sin C} + \frac{\sin(B + 60^\circ)}{\sin C \sin A} ] =$](http://latex.artofproblemsolving.com/b/4/9/b497432f2723c116cdbe466bcf92a8e260616e39.png)
![$= \frac{4R^3e^2}{abc} [\sin A \sin(A + 60^\circ) + \sin B \sin(B + 60^\circ) + \sin C \sin(C + 60^\circ)] =$](http://latex.artofproblemsolving.com/b/7/4/b74ad02beacd85573e0c414d547197b7b7ccb6b7.png)
![$= \frac{R^2e^2}{S} [\frac 1 2 (\sin^2 A + \sin^2 B + \sin^2 C) + \frac{\sqrt 3}{2}(\sin A \cos A + \sin B \cos B + \sin C \cos C)] =$](http://latex.artofproblemsolving.com/3/2/7/3270af3188a9b94a30a0a9d4aae5657f802e48d6.png)
![$= \frac{e^2}{8S} [a^2 + b^2 + c^2 + \frac{4 \sqrt 3 R^2}{2}\ (\sin 2A + \sin 2B + \sin 2C)] =$](http://latex.artofproblemsolving.com/6/b/1/6b19d60832fd954fbe948204b71d304f084f1826.png)
of the 1st isodynamic point Q. Any other equilateral triangle 
inscribed in the triangle 
so that 
is obviously obtained from the equilateral pedal triangle by a spiral similarity with the center Q and similarity coefficient greater than 1, hence its side e' = D'E' is greater than the side e = DE.
is the fermat point of then the inequality reduces to :
twice its area...
say, if 
Then the inequality is reduced to
(inside its circumcircle) and the angle 
the convex one, is equal to
as before. The area S of the triangle is still![$S = \frac{1}{2}\left[AQ \cdot BQ \sin(C+60^\circ)+BQ \cdot CQ \sin(A+60^\circ)+CQ \cdot AQ \sin(B+60^\circ)\right] =$](http://latex.artofproblemsolving.com/8/4/9/8494330566806bb1f9fe5b4703afc3e6344d9faf.png)
the triangle 
homothetic to the outer Napoleon triangle of has the maximum area (side length). Hence, its side is twice the measure of the side of the outer Napoleon triangle.![$ L = \sqrt {\frac {_1}{^6}(a^2 + b^2 + c^2) + \frac {2\sqrt {3}}{3}[\triangle ABC]}.$](http://latex.artofproblemsolving.com/3/8/f/38f00e14b3555b0d9d202c45123d50330889958e.png)
we can associate an equilateral triangle 
homothetic to 
and inscribed in 
Thus, by Gergonne-Ann theorem, it follows that ![$ [\triangle ABC]^2 = [\Delta][\Delta'].$](http://latex.artofproblemsolving.com/f/e/9/fe98a550d91a0534c7c1d16ff53fd4dbabdd1fde.png)
The area (side L') of will be minimum if the area (side) of 
is maximum. In other words, ![$ [\triangle ABC]^2 = \frac {_3}{^{16}}(L')^2(4L^2).$](http://latex.artofproblemsolving.com/7/a/8/7a823b1149b01cff0f09b7b7c6f181b4bc92a8dd.png)
of the Napoleon triangle yields:![$ [\triangle ABC]^2 = \frac {_1}{^8}(L')^2(a^2 + b^2 + c^2 + 4\sqrt {3}[\triangle ABC])$](http://latex.artofproblemsolving.com/f/b/f/fbfc29c5b328442761ed375b59845d837b6bce4f.png)
![$\Longrightarrow L' = \frac {2\sqrt {2}[\triangle ABC]}{\sqrt {a^2 + b^2 + c^2 + 4\sqrt {3}[\triangle ABC]}}.$](http://latex.artofproblemsolving.com/e/7/9/e7903113a4d5f3e3a9ebc041537ec98ff653a17c.png)
![$ S = [AEFT] + [CDTE] + [BDTF] \ldots$](http://latex.artofproblemsolving.com/d/6/7/d675fd8bd845dd39b8d1383d9a8ac8a9fbc70548.png)
are less than 
. Let the point be the Torricelli point of the triangle. So, 
. So,
,
,
.
.. This combined with the first inequality, gives the result.
. Take T to be the point lying on the same side of as such that 
and 
.
such that 
of an equilateral triangle lie on the sides 
respectively of a triangle 
If 
are the respective lengths of these sides, and 
the area of 
prove that![\[ DE \geq \frac{2 \cdot \sqrt{2} \cdot S}{\sqrt{a^2 + b^2 + c^2 + 4 \cdot \sqrt{3} \cdot S}}. \]](http://latex.artofproblemsolving.com/6/5/3/653eea326c5850b8019b2115a11c97c2fc9d3ba3.png)
the Toricelly point of triangle 
.
![$ S=[AEFT]+[CDTE]+[BDTF]$](http://latex.artofproblemsolving.com/5/b/3/5b3579f430d2acbe16b98245353c7681c753ffb3.png)
≤
≥
.
Is this inequality true?