. 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.

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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