Các bài toán tự sáng tác

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. Let for which and Consider th[r]

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

Loi giai :Nguyen Dinh Thanh Cong Nhan ve voi 2(a+b+c)

Chu y bdt sau :

Recalling the known inequality:

Ta chi can CM :

Ta dung BCS va xet TH de CM BDT Ket thuc CM

Ta the su dung SOS de giai bai toan

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

BC,CA,AB.CMR : đồng quy điểm

Loi giai , goi T’ la diem dang giac cua T Ta doi xung voi T qua BC Ta Cm Ta thuoc AT de suy 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 bay mưa mùa hạ Chút tinh khơi để lại cịn vẹn ngun Huyền nhung thống ngây thơ gió Hồ mơ lặng bóng dáng buồn xa xăm.

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

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

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

of with respect to , , thus , which yields and

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

the geometrical length, we have and ,

divides those two equations we deduce that Hence

, which means that , id est is the angle bisector

of So By this, Q.E.D

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:

1 If then Hence

must not be a perfect square , contradiction

2 If , notice that must be odd, since which and

are consecutive integers Rewrite the equation in the following form:

We have so either or is divisible by This

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

midpoints of cut at

Prove that are concurrent

Generalization:

lie on such that concur at Then

concur at which lies on Solution:

Let be the intersections of and and and ,

respectively then are collinear

We have then concur at

Denote

Applying Menelaus's theorem we obtain:

Denote

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

Lemma for ,

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 !

If then

so the condition can't be satisfied Therefore,

Using lemma, ETS

Nhan xet: Bai toan tren 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

Let for which and Consider the point

such that

and the point so that Prove that

Let be the circumcircle of and We have:

, i.e., and are

congruent

Since, and then

But so Thus