1.证明（mavropnevma ）Let and
. The region and
is the triangle . But is for (equality
, thus the triangles of largest possible area are
(equality for disallowed). is for ).
EDIT. Referring to the next post - it is interesting to figure out the envelope of the lines is the parabola functions for a fixed , with triangular area , but in some way it is irrelevant, since the problem fixed on the parabola for some . , and the
being delimited by the tangent to that parabola at
1.证明（kunny） parabola Which touches The domain points at is envelope of the lines .
, thus the
of the family of the lines swept by .
is shown by the shaded region, excluded two
Edit:I was misreading the context problem, thank you for pointing put it, mavropnevma.
I have just attached another figure.Needless to say, mavropnevma's solution is perfect. P.S.I remember that the similar problem original problem has been posed in 1970's in Tokyo University entrance exam/Science. According to my memory, let region be the maximum area of any triangle which is involved in the to find the extrema of .
in original problem, then draw the graph of
Here is the similar problem posed in Tokyo University entrance exam/Science, second round, 1978
be the part which is correspond to .
of the parabola
that is to say,
Let the tangent Line of the line Let . at .
intersects with the line
and intersects with
. We are to consider the questions as below in the range of
(1) Let denote the area of triangles of such that .
, respectively. Find the range
(2) Let Note :
be the domain enclosed by line segments contains line segments and .
be the maximum area of the triangle with a vertex and draw the graph, then find the exterme value.
which is contained in
Find the function
, which means for all points
which is closed to
holds. We call local maximum, local minimum as extreme value.
2.证明(Luis Gonzá lez)Let of touching at
and Clearly but since and
Let be the incircle are homothethic with incircles ( is
are symmetric about the midpoint of and
the M-excircle of MAB), it follows that with corresponding cevians 2.证明(Andrew64)As shown in the figure below. is the intersection of and .
meet at the same point
link17.jpg [ 37.23 KiB | Viewed 52 times ] 3.证明(mavropnevma)Since it is irrelevant which persons of the same gender know each other, we may assume there ore none such, and consider the bipartite graph having the left shore made of the boys and the right shore made of the girls, with an edge connecting a boy and a girl if they know each other. The condition means does not contain any induced cycle of length , and the requirement is to show the number of edges satisfies .
Thus it is an extremal graph theory question, for bipartite
graphs with forbidden
symmetry we should also have Denote by
. the set of girls that each and
the set of girls that each knows exactly one boy, and by and
knows more than one boy; take
. We obviously have
. Let us count the number boys, and one girl knows both
of objects . For each of the
is a girl,
are distinct there is at most
knowing them both (by the condition), so
. Moreover, by pigeonhole
, by Jensen's inequality. We thus need , so
. Finally, we get .
For equality to be reached we obviously need , namely for each pair of boys having exactly one girl knowing both of them; and then we need .
3.证明(crazyfehmy)Another solution: Consider the bipartite graph where there are girls boys and denote the girls by 's and boys by 's as vertices. Let denote the number of edges from the vertex the girl to set . If and is connected to some
then for any
must not connected to both
. Now let us count such
pairs. For every girl
many pair of edges. Since all such edge pairs must be
distinct for all girls, and since there are at most
such pairs, we have
or equivalently Now assume that or . Then we have are greater than or equal to and are
and we need to show that Since for we have
. and hence .
Now by Cauchy-Schwarz inequality we have
, we have
and we need to show that
. Then we have
which means that
which is a contradiction. So, we are done.
4.解 （crazyfehmy） Let
be called a nice pair if
satisfies the conditions
stated in the problem. Firstly, we shall prove a lemma:
is a nice pair then
for all integers
for some integers such that
. Then we can find another pair
. (The proof is easy) Then consider the system
and Remainder Theorem this system has a solution an element of
. By the Chinese such that both and is is not a nice pair because is divisible by .
which means that
Now, we shall show that if solution for all integers
has a such that
. For the proof assume there exists an integer
has no solution in integers. Then it is easy to see that has also no solution for all integers which are not divisible
. Now take
and consider the numbers
. For all of these numbers,
we have no solution and there are (take ) and there are
such numbers. Since all square residues have solutions square residues modulo including zero, this means that for the number we have a
all nonsquare residues
, the equation must have no solution. However, for so for the number
is not a square residue but solution which is a contradiction. For
is not a square residue but
has a solution since obtain a contradiction so the second lemma is also proved.
is a square residue. Hence we again
can take every value modulo and if
we must have
and also we must have
is a nice pair. So,
can take only three
values. We will consider each case separately:
then we must have
If means If means
then we must have
then we must have
4.解（dinoboy）First, remark that it suffices for For modulo For simply note that we require or we
to be injective modulo . require for some
Now, what values can values modulo
? It is a simple exercise to show all and then as can be
can be obtained (just express it as
we can transform the problem to what values modulo expressed as works out is if
, which is known to be all of them). Therefore the only way this problems and .
Therefore we simply require that of modulo , 10 modulo .
and and modulo
. For each value of
so the answer should simply be
is not hard, but I'm lazy and felt like reducing it to an already solved problem.
，so satisfies the problem.
So the answer should be
6.解（crazyfehmy）If since has always a solution
then is equivalent to
satisfies which and since is odd. be the and let
in the set
Now we will show that if elements of the set
then the condition does not satisfy. Let
. Consider the sums
. Since are also different modulo
's are different modulo
, the numbers
. On the other hand, none of
's can be equivalent to
modulo is a
because otherwise we would have two equivalent terms. Hence permutation of and by adding up these which equations we
. Now do the same procedure for all . . Then we have Let
's to get and
. So, we have
many numbers equivalent to each other modulo . However, we know that there are many numbers modulo which are all equivalent to each other modulo . Hence in order for 's to be different modulo . So Hence all possible , we must have and we are done. values are which means that . and hence
7.证明 （Luis Gonzá lez）Let then
is the exsimilicenter of of
is midpoint of the arc
bisects externally is midpoint of the arc of is external bisector of and Note that is a Thebault circle of the cevian of externally tangent to its circumcircle By Sawayama's lemma passes through its C-excenter is C-excenter of is M-isosceles, i.e. circumcenter of 7.证明（Andrew64）As shown in the figure. Let be the intersection of and . It's fairly obvious So we have , and So Thus Consequently . , and , are concyclic. Hence is
is the bisector of
link18.jpg [ 31.39 KiB | Viewed 93 times ] 8.证明（duanby） hint:(a-b)(c-b)(a-d)(c-d) in detail: product (a-b)(c-b)(a-d)(c-d) for every a,c be the number on , b,d be the number on for point x,y if they are not ajjectent then in the product, it will occur twice, if it's ajjectent it's appears only once, and also chick the point that are on and then we get it. iampengcheng1130 2013 中国女子数学奥林匹克第 7 题的解答