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 a convex pentagon such that![\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.
\]](http://latex.artofproblemsolving.com/d/3/d/d3d9a82f4318190298a8f008d417a364e03f1fca.png)
The diagonals 
and 
meet at 
. Prove that the line 
bisects the side 
.
be an acute-angled triangle with 
. Let 
be the orthocenter of triangle 
, and let 
be the midpoint of the side 
. Let 
be a point on the side 
and 
a point on the side 
such that 
and the points , , are on the same line. Prove that the line 
is perpendicular to the common chord of the circumscribed circles of triangle and triangle 
.
, 
, and 
are fixed on a line in this order. Let 
be a circle passing through and whose center does not lie on the line . Denote by the intersection of the tangents to at and . Suppose meets the segment 
at 
. Prove that the intersection of the bisector of 
and the line does not depend on the choice of .
and 
be two points inside triangle 
such that![\[ \angle MAB = \angle NAC\quad \mbox{and}\quad \angle MBA = \angle NBC.
\]](http://latex.artofproblemsolving.com/2/0/4/20410178c70693881f348c67454f01057a3fc40c.png)
![\[ \frac {AM \cdot AN}{AB \cdot AC} + \frac {BM \cdot BN}{BA \cdot BC} + \frac {CM \cdot CN}{CA \cdot CB} = 1.
\]](http://latex.artofproblemsolving.com/b/b/6/bb64f80eccb4634b9c9e15cade2f7c0b87b60923.png)
the set of all positive real numbers.
which have the property:![\[f(x)f(y)=2f(x+yf(x))\]](http://latex.artofproblemsolving.com/4/f/f/4ffbde35ff082585a188c1a4368d39338d1b07a4.png)
and 
.
+1 w
questions, each with 
answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?
, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
, 
the following inequality holds:
When does equality hold?
with 
, the foot of altitudes from 
to the sides 
are 
, respectively. is the orthocenter. is the midpoint of segment . Lines 
and 
intersect at 
. Let the tangents drawn to circumcircle 
from and intersect at 
. Prove that 
are colinear
be an integer and let 
be positive integers. In a group of 
people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following two conditions to hold at the same time:
.
with 
, there is someone who has played exactly 
games of chess.
be an acute triangle with orthocenter and circumcircle 
. Let be the midpoint of side . Point is chosen from the minor arc on such that 
. Let be a point on such that 
is perpendicular to 
, and 
be a point on line such that 
is perpendicular to . Lines 
and intersect at point 
, and point 
is the reflection point of with respect to .
.
prime number, there exists only one positive number 
that makes the equation 
a perfect square of a positive integer
satisfying the following equations: 
be triangle, inscribed in parabola. Tangents in points forms triangle . Prove that 
. (
is area of triangle ).
be triangle, inscribed in parabola. Tangents in points forms triangle . Prove that . ( is area of triangle ).
be any point on the Steiner circumellipse of 
and let 
be the cevian triangle of WRT 
Then ![$[DEF]=2 \cdot [ABC].$](http://latex.artofproblemsolving.com/3/3/e/33ec9a123e2b3b63c128a4fe7fe4aa12551261a1.png)
be the antimedial triangle of (
against ). The Steiner circumellipse 
of is obviously the Steiner inellipse of tangent to its sides at 
are collinear on the polar of WRT and similarly 
and 
are respectively collinear. Thus by Pappus theorem for the hexagon 
it follows that 
is at infinity, i.e. 
and similarly we get 
![$[DEF]=[DYZ]=[CYZ]=2\cdot [ABC]. \ \blacksquare$](http://latex.artofproblemsolving.com/b/4/0/b4028b012dc90ae29c0bc707cbf447409f6ab058.png)
be the perspector of the parabola. Since the center of any inconic is the isotomcomplement of its perspector (in the parabola case at infinity), then it follows then that the isotomic conjugate of WRT must be on the line at infinity is on the Steiner circumellipse of and the result follows from the lemma.
