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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]
(1)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 ,
(2)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
(3)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
(4)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 ,
(5)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
(6)But so Thus