Các bài toán tự sáng tác

If are non-negative numbers, no two of which are zero, then Loi giai :Nguyen Dinh Thanh Cong.. Ta con the su dung SOS de giai bai toan nay.. là điểm đẳng giác của T qua Loi giai , goi T’

If are non-negative numbers, no two of which are zero, then

Loi giai :Nguyen Dinh Thanh Cong

Nhan 2 ve voi 2(a+b+c)

Chu y bdt sau :

Recalling the known inequality:

Ta chi can CM :

Ta dung BCS va xet 2 TH de CM BDT nay

Ket thuc CM

Ta con the su dung SOS de giai bai toan nay

Bai Toan : Cho tam giác ABC , T là điểm Torixeli ( T thuộc miền trong tam giác ) AT,BT,CT cắt BC,CA,AB tại là điểm đẳng giác của T qua

Loi giai , goi T’ la diem dang giac cua T Ta doi xung voi T qua BC Ta Cm Ta thuoc

AT de suy ra T’ thuoc AT1

CM su dung luong giac , dinh ly Sin

Bai toan :Cho x,y thuoc [2009’-2009] thoa man (x^2-Y62-2xy)^2=4

Tim gia tri nho nhat cua x^2+y^2

Goi y :Su dung pt pell

Lá thôi bay trong cơn mưa mùa hạ

Chút tinh khôi để lại còn vẹn nguyên

Huyền nhung thoáng ngây thơ 1 cơn gió

Hồ mơ lặng bóng dáng buồn xa xăm.

Bai toan: Given triangle Let be an arbitray point on A-altitude

, Prove that the midpoints of and are collinear

Loi giai : is collinear

from

True

Done

Bai toan : ABC is an acute-angled triangle The incircle (I) touches BC at K The altitude

AD has midpoint M The line KM meets the incircle again at N

Show that Loi giai : NK is the bisector of angle BNC, hence

Then

Cach khac: Let be the points of tangent of and and It

is well-known that Let be the midpoint of Since is the polar

are similar But are respectively the median of and so and are also similar By this, we have However, is the tangent of so either is By Newton's identity in a harmonic division we have

This means is also the tangent of the circumcircle of

at Therefore, the triangles and are similar, form the ratios,

we deduce that Finally, so by Mc Laurin's identity, take

, which means that , id est is the angle bisector

Bai toan: Let be the set of all positive integers

solve the following equation in :

Loi giai:Tran Nam Dung

WLOG we can assume that We are to prove that Suppose the contradiction that Consider cases:

must not be a perfect square , contradiction

are consecutive integers Rewrite the equation in the following form:

Thus, we can deduce that:

Contradiction

In conclusion we have and the solution is followed

Bai toan : ,and a point is Cevian triangle of are

Generalization:

concur at which lies on

Solution:

respectively then are collinear

Applying Menelaus's theorem we obtain:

We will show that

(it's right from Menelaus's theorem) Therefore Similarly we are done

Bai toan : Problem (zaizai-hoang):

Let a,b,c are positive number such that: Prove that:

Loi giai : Nguyen Cong

using AM-GM, we get

so ETS

Here, from Schur ineq case and AM-GM, we get

so the lemma is proved

Let's start to prove the problem !

so the condition can't be satisfied Therefore,

Using lemma, ETS

Nhan xet: Bai toan tren con co cach giai khac la dung BDT Schur ket hop voi pqr nhung loi giai co nhung bien doi kha dai Khong dung voi ve dep cua bDT

such that

congruent

In , we have:

But so Thus

, and we are done Let be trhee non-negative real numbers Prove that