2013中国女子数学奥林匹克试题及其解答

2013中国女子数学奥林匹克试题及其解答

1.证明（张云华）

1.证明（mavropnevma ）Let
and
, . The region
and
is the triangle
. But
is for
(equality
).
,
for
,
, 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 ss 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 examScience.
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 examScience, second round,
1978

In the -plane, let be the part which is correspond to
.
of the parabola ,
that is to say,

Let the tangent Line of
the line
Let
.

at .
at intersects with the line at 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

(2) Let

Note :

Let
.
by , respectively. Find the range
be the domain enclosed by line segments
contains line segments and .
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

Note : A function has local minimum (or local maximum) at a point
, which means for all points which is closed to ,
holds. We call local maximum, local minimum as extreme
value.



2.证明(Luis González)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 .
are homothetic

Therefore ,

Attachments:
, meet at the same point .

[ 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 's; by
symmetry we should also have .

Denote by the set of girls that each knows exactly one boy, and by
knows more than one boy; take and
the set of girls that each
and . We obviously have
.
Let us count the number
boys, and knows both
of objects
. For each of the
, where is a girl, are distinct
there is at most doubletons
one girl knowing them both (by the condition), so . Moreover, by pigeonhole

we have .
On the other hand, we have
, 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 and
boys and denote the girls by 's and boys by 's as vertices.

Let denote the number of edges from the vertex to set . If is connected to some and
then for any the girl must not connected to both and . Now let us count such
pairs. For every girl there are 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

and if

, we have
and we need to show that . Assume . Then we have
which means that


4.解（crazyfehmy） Let

which is a contradiction. So, we are done.
be called a nice pair if satisfies the conditions
stated in the problem.
Firstly, we shall prove a lemma:

If

Proof: Let or and divides for some integers
such that
and
and
is a nice pair then for all integers and .
. 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
and
. By the Chinese
such that both and is
is not a nice pair because which means that
is divisible by

Now, we shall show that if
solution for all integers
or then
.
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

by . Now take
we have no solution and there are
(take ) and there are
and consider the numbers . For all of these numbers,
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
is not a square residue but
solution which is a contradiction. For the number
so for
is not a square residue but
has a solution since
obtain a contradiction so the second lemma is also proved.

Now, since
is a square residue. Hence we again
can take every value modulo
and if
and we must have and
and also we must have is a nice pair. So, can take only three
values. We will consider each case separately:

If

If
means
If
means

If we

count these possibilities,
.
4.解（dinoboy）First, remark that it suffices for to be injective modulo
For modulo simply note that we require
For modulo or we require
.
for some that
and .
if i am not wrong, we get
then we must have

then we must have which
which
then we must have which means

Now, what values can
values modulo
take 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

Therefore we simply require that
of

Note: To show takes all values modulo
modulo , 10 modulo
.
and
and modulo
, which is known to be all of them). Therefore the only way this problems
and .
. For each value of there are values
so the answer should simply be
without relying on
is not hard, but I'm lazy and felt like reducing it to an already solved
problem.

4.解（yunxiu）
hence

So the answer should be

6.解（crazyfehmy）If
since
has always a solution
，so
satisfies the problem.
.
，
let then
is equivalent to
satisfies
which
and since is odd.
be the
and let
in the set for all
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 's
. 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 equations we get
which means

. 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 González） Let
then
cut again at Since 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
.



Namely is the bisector of

Attachments:
.
, and
,
Hence is



are concyclic.

[ 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