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Archive for Tháng Một, 2011

Problem: Let ABC be a triangle, O is the circumcenter, \left(N\right) is the nine point circle and F_{a}F_{b}F_{c} is the Feuerbach triangle. The perpendicular to the line BC through the point N intersects \left(N\right) at the point Q_{a} (the points A and Q_{a} lie on the same side of the line BC) and similarly construct the points Q_{b}, Q_{c}. Prove that the lines F_{a}Q_{a}, F_{b}Q_{b}, F_{c}Q_{c} are concurrent and the point of concurrency lies on
the line ON.

(See also: http://forumgeom.fau.edu/FG2005volume5/FG200521.pdf )

My proof:
Construct the circumcircle of triangle ABC. Let U be the midpoint of the arc BC which does not contain A. Let M_a be the midpoint of BC. I_aM_a intersects the A-altitude of triangle ABC at X_a, similar with X_b, X_c. Let D, E, F be the tangencies of (I_a) and AC, BC, AB respectively.
Note that U is the circumcenter of triangle BCI_a then by Thales’s theorem: \dfrac{UM_a}{AX_a}=\dfrac{I_aU}{I_aA}=\dfrac{UB}{I_aA}.
But \Delta UBM_a\sim \Delta I_aAD \Rightarrow \dfrac{UB}{I_aA}=\dfrac{UM_a}{I_aD}. Therefore \dfrac{UM_a}{AX_a}=\dfrac{UM_a}{I_aD} or AX_a=r_a.
Let K be the reflection of E wrt I_a then AX_aI_aK is a parallelogram. We get AI_a intersects KX_a at T such that T is the midpoint of AI_a. From my proof at the last corollary of Fontene’s theorem we can apply for the case P\equiv I_a and get F_a lies on the circle with diameter ET. If F'_a is the projection of E on KX_a then F'_a\in (ET) and F'_a\in (I_a) so F'_a\equiv F_a, or F_a\in X_aK. But it is easy to see that K, F_a, Q_a are collinear then X_a, F_a, Q_a are collinear.
Let Q'_a be the reflection of Q_a wrt N. Construct the tangents of (N) through Q'_a, Q_b, Q_c, they intersect each other and make triangle A'B'C'. Note that A'B'//AB, B'C'//BC, C'A'//AC and D,E,F; Q_b,F_a,Q_c are the tangencies of A and A’-excircle with their sides so Q_bQ_c//DF or Q_bQ_c//X_bX_c. Similarly, we claim that \Delta Q_aQ_bQ_c and X_aX_bX_c have Q_cQ_a//X_cX_a, Q_aQ_b//X_aX_b. This means X_aQ_a, X_bQ_b, X_cQ_c concur at their homothetic center. (1)
Let L, P be the tangencies of (I_b) and BC, AB; J, Q be the tangencies of (I_c) and BC, AC.
From my alticle at here we obtain LP, QJ and A-altitude of triangle ABC are concurrent and denote X'_a as their concurrency. Because BJ=CL=p-a then M_a is the midpoint of JL.
Note that \Delta BI_aC\sim \Delta LX'_aJ so by Thales’s theorem, X'_aI_a must passes through the midpoint of BC, or X'_a\equiv X_a.
Let G be the projection of X_a on I_bI_c, H_a be the projection of A on BC. We have:
\angle JX_aH_a=90^o-\angle X_aJC=\angle C/2=90^o-\angle ACI_b
=90^o-\angle X_aB_2C_2=\angle B_2X_aG.
Note that X_bX_c//I_bI_c then X_aG is the altitude of triangle X_aX_bX_c, which follows that X_aA passes through the circumcenter of triangle X_aX_bX_c. This means the orthocenter H of triangle ABC is the circumcenter of triangle X_aX_bX_c. (2)
From (1) and (2) we get the homothetic center of two triangles X_aX_bX_c and Q_aQ_bQ_c lies on HN. Our problem is solved.

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Ten concyclic points

Problem (nsato): Let A_1, A_2,A_3, A_4 and A_5 be five concyclic points.  For 1 \le i < j \le 5, let X_{i,j} be the intersection of the Simson lines of A_i and A_j with respect to the triangle formed by the other three points.  Show that all ten such points X_{i,j} are concyclic.

My proof:

Lemma 1: Given 4 points A_1, A_2, A_3, A_4 on the circle (O). Then the Simson lines d_1, d_2, d_3, d_4 of A_1, A_2, A_3, A_4 wrt \Delta A_2A_3A_4, A_1A_3A_4, A_1A_2A_4, A_1A_2A_4, respectively, concur at the orthocenter X of A_1A_2A_3A_4. Let G be the centroid of A_1A_2A_3A_4 then \vec{OG}=\frac{1}{2}\vec{OX}.

Proof: Denote H_2, H_3 the orthocenters of the triangles A_1A_3A_4 and A_1A_2A_4. Then A_2H_3//=A_3H_2. We get A_2H_3H_2A_3 is a parallelogram. Therefore A_2H_2, A_3H_3 concur at the midpoint X of A_2H_2 and A_3H_3. It’s easy to see that X also lies on d_2 and d_3 Similarly we get d_1,d_2,d_3,d_4 concur at X.
Let M, N be the midpoints of A_1A_3 and A_2A_4 then G is the midpoint of MN.
We have XN//=\frac{1}{2}A_4H_2//=OM so XNOM is a parallelogram. We get G is the midpoint of OX.

Lemma 2: Given 5 points A_1, A_2, A_3, A_4, A_5 on the circle (O). Denote H_1, H_2, H_3, H_4 , H_5 the orthocenters of A_2A_3A_4A_5, A_1A_3A_4A_5, A_1A_2A_4A_5, A_1A_2A_3A_5, A_1A_2A_3A_4, respectively. Then 5 points H_1, H_2, H_3, H_4, H_5 are concyclic.
Proof: Let G_1, G_3,G_4 be the centroids of A_2A_3A_4A_5, A_1A_2A_4A_5, A_1A_2A_3A_5, respectively. According to lemma 1 we obtain G_1G_4//H_1H_4. Let M, N, P be the midpoints of A_3A_5, A_2A_4, A_1A_2 then G_1G_4 is the midline of the triangle MNP. This means G_1G_4//NP//A_1A_4. Similarly, G_3G_4//A_3A_4. Therefore \angle H_1H_4H_3=\angle A_1A_4A_3.
Similarly \angle H_1H_5H_3=\angle A_1A_5A_3=\angle A_1A_4A_3=\angle H_1H_4H_3. So H_1,H_3,H_4,H_5 are concyclic. Similarly we are done.

Lemma 3: Given a triangle ABC with its circumcircle (O). Let E, F be arbitrary points on (O) then the angle between the Simson lines of two points E,F is half the measure of the arc EF.

This lemma is trivival, so I will leave it to you.

Back to our problem: Let H_1,H_2, H_3, H_4, H_5 be the orthocenters of 5 triangles A_iA_jA_k (1\leq i<j<k\leq 5). Applying the lemma 2 we obtain H_1H_2H_3H_4H_5 are cyclic. Applying the lemma 1 we get the Simson lines d, l of A_3, A_1 wrt \Delta A_2A_4A_5, A_2A_4A_5 passes through H_1, H_3. Let F be the intersection of d and l then applying the lemma 3 : \angle H_1FH_3=\angle A_1A_4A_3=\angle H_1H_4H_3, which follows that F lies on (H_1H_2H_3H_4H_5). Similarly we are done.

Another corollary of lemma 2:

Problem 2 (Nguyen Van Linh): Given a cyclic pentagon ABCDE. Let A',B',C',D',E' be the second intersections of the 9-point circles of triangles ABC,BCD,CDE,DEA,EAB. Prove that A',B',C',D',E' are concyclic.

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Problem 1(Tran Quang Hung): Let ABC be a triangle with its circumcircle (O). Let P and Q be arbitrary points such that P, O, Q are collinear. A_1B_1C_1 is the pedal triangle of P wrt \Delta ABC, A_2B_2C_2 is the circumcevian triangle of Q wrt \Delta ABC. Show that (PA_1A_2), (PB_1B_2), (PC_1C_2) are coaxal.

Denote O_a, O_b, O_c the centers of (PA_1A_2), (PB_1B_2), (PC_1C_2); X the intersection of the line through A_2 and perpendicular to PA_2. Similarly we define Y, Z.
Note that O_a is the midpoint of PX therefore (O_a), (O_b), (O_c) are coaxal iff O_a, O_b, O_c are collinear iff X, Y, Z are collinear. (1)
We have \dfrac{XB}{XC}=\dfrac{S_{XBA_2}}{S_{XCA_2}}=\dfrac{\sin \angle XA_2B. A_2B}{\sin \angle XA_2C. A_2C}=\dfrac{\sin \angle XA_2B}{\sin \angle XA_2C}.\dfrac{\sin \angle BAQ}{\sin \angle CAQ}
But \dfrac{\sin \angle XA_2B}{\sin \angle XA_2C}=\dfrac{\cos \angle BA_2P}{\cos \angle CA_2P}.
Let A_3 be the intersection of A_2P and (O), A_4 be the intersection of A_3O and (O). Similarly we define B_3, B_4, C_3, C_4.
We have \dfrac{\cos \angle BA_2P}{\cos \angle CA_2P}=\dfrac{\cos \angle BA_4A_3}{\cos \angle CA_4A_3}=\dfrac{\sin \angle BA_3A_4}{\sin \angle CA_3A_4}=\dfrac{\sin \angle BAA_4}{\sin \angle CAA_4}
Hence \dfrac{XB}{XC}=\dfrac{\sin \angle BAQ}{\sin \angle CAQ}.\dfrac{\sin \angle BAA_4}{\sin \angle CAA_4}.
Do the same with \dfrac{YC}{YA}, \dfrac{ZA}{ZB} then applying Ceva-sine theorem we obtain (1) \Leftrightarrow AA_4, BB_4, CC_4 are concurrent.
On the other side, according to Pascal’s theorem for 6 points A, B, A_3, B_3, A_2, B_2 we claim M, P, Q are collinear (M is the intersection of AB_3 and BA_3)
Applying Pascal’s theorem again for 6 points A, B, A_3, B_3, A_4, B_4 we get the intersection of AA_4 and BB_4 lies on PQ. Similarly the intersection of CC_4 and BB_4 lies on PQ. This means AA_4, BB_4, CC_4 are concurrent. We are done.


(1) (Nguyen Van Linh): Given triangle ABC and its circumcircle (O). Denote M_a, M_b, M_c the midpoints of BC, CA, AB, respectively, P an arbitrary point inside \Delta ABC, A',B',C' the second intersections of AP, BP, CP and (O). Prove that (A'OM_a), (B'OM_b), (C'OM_c) concur at two points.

(2) ( 77ant) : For \triangle ABC with its orthocenter H, \triangle DEF is its orthic triangle.
AK, BL, CM are perpendicular to EF, FD, DE respectively.
Prove that three circumcircles of \triangle KHD, \triangle LHE, \triangle FHM are coaxal.

But it is not the end of the line, I found that the generalization of Steinbart’s theorem is the application of problem 1:
Problem 2 (Nguyen Van Linh): Given a triangle ABC. Let P be an arbitrary point in the plane, A_1B_1C_1 be the pedal triangle of P wrt \Delta ABC. Let O be the center of (A_1B_1C_1),L be an arbitrary point on PO. A_1L, B_1L, C_1L intersects (A_1B_1C_1) at A_2,B_2,C_2, respectively. Prove that AA_2, BB_2, CC_2 are concurrent.


Let Q be the isogonal conjugate of P wrt \Delta ABC. X_3, Y_3, Z_3 be the projections of P onto B_2C_2, C_2A_2, A_2B_2.
Consider the homothetic with center P and ratio 2: O\mapsto Q, A_2\mapsto X, B_2\mapsto Y, C_2\mapsto Z, A_1\mapsto X_2, B_1\mapsto Y_2, C_1\mapsto Z_2, X_3\mapsto X_1, Y_3\mapsto Y_1, Z_3\mapsto Z_1; L\mapsto T.
Then Q is the center of (XYZ) and X_1Y_1Z_1 is the pedal triangle of P wrt \Delta XYZ.
We have T, L, P are collinear therefore T, P, Q are collinear. Moreover, T is the intersection of XX_2, YY_2, ZZ_2.
By problem 1, we get (PX_1X_2), (PY_1Y_2), (PZ_1Z_2) are coaxal, which follows that the intersections of B_2C_2 and BC, A_2C_2 and AC, A_2B_2 and AB are collinear. According to Desargues’s theorem we claim AA_2, BB_2, CC_2 are concurrent.

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