Các bài toán tự sáng tác

<span class='text_page_counter'>(1)</span><div class='page_container' data-page=1>

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

BC,CA,AB.CMR : đồng quy tại 1 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 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 khơi để lại cịn vẹn ngun

Huyền nhung thống 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.
,

</div>
<span class='text_page_counter'>(2)</span><div class='page_container' data-page=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.

</div>
<span class='text_page_counter'>(3)</span><div class='page_container' data-page=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
<i>solve the following equation in : </i>

<i>Loi giai:Tran Nam Dung</i>

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

</div>
<span class='text_page_counter'>(4)</span><div class='page_container' data-page=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.

<b>Generalization: </b>

lie on such that concur at . Then

concur at which lies on .
<b>Solution: </b>

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

<b>Lemma. for </b> ,

</div>
<span class='text_page_counter'>(5)</span><div class='page_container' data-page=5>

using AM-GM, we get

so ETS

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

so the lemma is proved.

<b>Let's start to prove the problem ! </b>

If then

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 .

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 .

</div>
<span class='text_page_counter'>(6)</span><div class='page_container' data-page=6>

But so . Thus

</div>

<!--links-->