## Archive for Tháng Mười Một, 2010

## The generalization of Newton line

Posted in Download on Tháng Mười Một 27, 2010| Leave a Comment »

## Fontene’s theorem and some corollaries

Posted in Download on Tháng Mười Một 27, 2010| Leave a Comment »

Fontene theorem and some corollaries

Problem by Le Duc Tam, A2k43 Highshool for gifted student, Hanoi University of Science:

## An approach to prove the concurrency

Posted in Download on Tháng Mười Một 27, 2010| Leave a Comment »

This article was published in the magazine **Mathematics and Youth **05/2010:

## 8 incenters are concyclic

Posted in Home on Tháng Mười Một 19, 2010| Leave a Comment »

**Problem:** Given a quadrilateral which is inscribed in such that . The tangents of through intersect each other and make a circumscribed quadrilateral . intersects at . Prove that the incenters of triangles are concyclic.

**Proof:**

Denote the incenters of triangles , respectively. It’s well-known that concur at .

Since we get or is a cyclic quadrilateral.

Then . But so , which follows that . This means is the bisector of or .

Similarly, .

We have . Therefore .

Thus

It is equivalent to are concyclic. Similarly, are concyclic. So . Similar for . We are done.

## A problem about Miquel point

Posted in Home on Tháng Mười Một 19, 2010| Leave a Comment »

**Problem** (by **Lamoen (van) F**. and **Grinberg D.**): A line intersects sides (or their extensions) ,and of triangle at points , and , respectively; let,,and be the centers of the circumscribed circles of triangles ,, and ,respectively; let ,,and be the respective orthocenters of these triangles. Prove that:

a)

b)The midperpendiculars to segments ,, and meet at one point.

**My proof:**

a, Let be the Miquel point of the completed quadrilateral . We have then . Similarly we are done.

b, First we will express two lemmas:

**Lemma 1:** Given two circles and intersect each other at and . A line through and perpendicular to intersects and at and . Let be the arbitrary points on and , respectively. The lines through and parallel to cuts the lines through and parallel to at . Let be the projection of on . Then are concyclic.

**Proof:**

Let be the second intersections of and and . cuts at cuts at . Note that and so .Similarly, . This means is the orthocenter of triangle .

Therefore or .

Construct two parallelograms and . Let be the projection of on . We have:

. Thus is a cyclic quadrilateral or . Similarly, .

We get are the altitudes of triangle . Therefore they concur at .

Hence lie on the circle with diameter . We are done.

**Lemma 2:** Given a completed quadrilateral . Let be the orthocenters, and be the circumcenters of triangles and . are the midpoints of and . Then the lines through and perpendicular to , through and perpendicular to and concur at . Let be the projection of on then are concyclic.

**Proof:**

We only solve the lemma in 1 case as in the figure.

For the concurrency see the topic **IMO Shortlist 2009 – Problem G6**.

Let be the midpoints of . We have lie on the 9-point circle of triangles , respectively. So .

On the other side,

But Therefore . We obtain . Hence . This follows that.

Thus are concyclic.

**Back to our problem: **

The Steiner’s theorem is stated that are collinear. Let be the intersection of the line through and perpendicular to and the line through and perpendicular to . Applying lemma 2 we have

Let be the intersection of the perpendicular bisector of and the line through and perpendicular to . intersects the line through and parallel to at . From lemma 1 we can easy show that lies on the circle with diameter . Applying lemma 2 we have are concyclic so .

Thus the perpendicular bisectors of and the line through and perpendicular to concur at . Similarly we are done.

## My problems about concurrent circles

Posted in Home on Tháng Mười Một 19, 2010| Leave a Comment »

**Problem 1:** Given a triangle . Let be an arbitrary point in the plane, be the pedal triangle of wrt . Let be the center of ,$latex L$ be an arbitrary point on . intersects at , respectively. Prove that are concurrent.

Note that this is the generalization of the **Steinbart**‘s point.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=366219

**Problem 2:** Given a triangle . Let be a point in the plane such that the pedal triangle of wrt is the cevian triangle of wrt . Let be the isogonal conjugate of wrt . intersects at ,respectively. Prove that are concurrent.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=366218

## The 9-point center lies on the altitude

Posted in Home on Tháng Mười Một 19, 2010| Leave a Comment »

**Problem** (by** Lym**): Prove that the 9-point centers of the line (k) perpendicular to Euler line of triangle ABC forming 3 triangles with are on the altitudes of .

**My proof:**

We only need to prove in case passes through . Other cases we can use the homothetic to finish proof.

In the figure below, let intersect at at . The line through and perpendicular to intersects at . The line through and perpendicular to intersects at . We will show that is the 9-point center of triangle .

Since lie on the circle then

Therefore is the bisector of or

On the other side, let be the intersection of and .

We have

Therefore

But then

This means is the center of 9-point circle of triangle . We are done.