View topic - Beautiful • Art of Problem Solving

Page 1 of 1

[ 7 posts ]

Author

Message

delegat

Posts: 533
Location: Sarajevo-Bosnia and Herzegovina

Posted: Jun 29, 2007, 8:42 am • # 1

IN triangle $ABC$ , let $I$ be the incenter and let $I_{a}$ be the excenter opposite $A$ . Suppose that $II_{a}$ meets $BC$ and the circumcircle of triangle $ABC$ at $A'$ and $M$ , respectively. Let $N$ be the midpoint of arc $MBA$ of the circumcircle of triangle $ABC$ . Let lines $NI$ and $NI_{a}$ intersect the circumcircle of triangle $ABC$ again at $S$ and $T$ , respectively. Prove that $S$ , $T$ , and $A'$ are collinear.

_________________
SALEM MALIKIĆ Sarajevo college secondary school - University of Sarajevo

Virgil Nicula

Posts: 4860
Romania

Blog: View Blog

Posted: Jun 30, 2007, 11:22 am • # 2

delegat wrote:

In a triangle $ABC$ , let $I$ be the incenter and let $I_{a}$ be the exincenter opposite $A$ . Suppose that $II_{a}$ meets $BC$ and the circumcircle of triangle $ABC$ at $A'$ and $M$ , respectively. Let $N$ be the midpoint of arc $MBA$ of the circumcircle of triangle $ABC$ . Let lines $NI$ and $NI_{a}$ intersect the circumcircle of triangle $ABC$ again at $S$ and $T$ , respectively. Prove that $S$ , $T$ , and $A'$ are collinear.

Lemma (a remarkable property of a harmonical division). Given are four collinear points $A,C,B,D$ in this order for which $\frac{CA}{CB}=\frac{DA}{DB}$ , i.e. the division $(A,C,B,D)$ is harmonically. Denote the middlepoint $M$ of the segment $[CD]$ . For a circle $w$ which passes through the points $A,M$ denote the its diameter $[NP]$ which is perpendicularly on the its chord $[AM]$ and the second intersections of the circle $w$ with the mentioned lines $\left\{\begin{array}{c}T_{1}\in ND\cap w\ ,\ T_{2}\in PD\cap w\\\ S_{1}\in NC\cap w\ ,\ S_{2}\in PC\cap w\end{array}$ . Then $B\in T_{1}S_{1}\cap T_{2}S_{2}$ .

Proof of the proposed problem. The division $(A,A',I,I_{a})$ is harmonically because $\frac{AI}{AI_{a}}=\frac{p-a}{p}=\frac{r}{r_{a}}=\frac{A'I}{A'I_{a}}$ . It is well-known that $MI=MI_{a}=MB=MC$ . Remain only to apply the previous lemma.

pohoatza

Posts: 1100
Location: Bucharest
Romania

Posted: Jun 30, 2007, 12:06 pm • # 3

I was also thinking to this problem and I found a synthetic solution, also based on some basic harmonical division facts.

So the quadrupoint $(AIA'I_{a})$ is harmonical, i.e $(AIA'I_{a})=-1$ , thus $N(AIA'I_{a})=-1$ and by intersecting it with the circumcircle, we have that the quadrilateral $ASDT$ is harmonical, where $D \in NA' \cap w$ . $(1)$ .

On other hand, $NN \| AM$ , where $NN$ is the tangent in $N$ to the circumcircle, thus $N(IMI_{a}N)=-1$ , i.e the quadrilateral $NTMS$ is harmonical. $(2)$ .

By $(1)$ , the lines $SS$ , $TT$ and $AD$ are concurrent, and by $(2)$ , the lines $SS$ , $TT$ , $MN$ are concurrent, thus the all four lines $SS$ , $TT$ , $AD$ , $MN$ are concurrent in $X$ . $(3)$

Now because $(ASDT)=-1$ , we have also $A'(ASDT)=-1$ , which implies that the quadrilateral $MS'NT'$ is harmonical, i.e the lines $S'S'$ , $T'T'$ and $MN$ are concurrent.

Similar from $(NTMS)=-1$ , we have also $A'(NTMS)=-1$ , which gives that the quadrilateral $DS'AT'$ is harmonical, i.e the lines $S'S'$ , $T'T'$ and $AD$ are concurrent.

Therefore again all fourlines $S'S'$ , $T'T'$ , $AD$ and $MN$ are concurrent also in $X$ . $(4)$ (because in $(3)$ we defined $X \in AD \cap MN$ .)

Thus by $(3)$ we have that $ST$ is the polar of $X$ and again by $(4)$ , $S'T'$ is also the polar of $X$ , thus $S' \equiv S$ and $T' \equiv T$ .

Attachments:

incenter collinearity.png [ 74.65 KiB | Viewed 91 times ]

_________________
Cosmin Pohoata, Bucharest, Romania

delegat Posts: 533 Location: Sarajevo-Bosnia and Herzegovina	Posted: Jun 30, 2007, 12:10 pm • # 4 Ok Thank you Where can I learn more about this harmonical division? Is there any solution by using some elementary theorems like Menelaus, Ceva etc. _________________ SALEM MALIKIĆ Sarajevo college secondary school - University of Sarajevo

pohoatza

Posts: 1100
Location: Bucharest
Romania

Posted: Jun 30, 2007, 2:00 pm • # 5

Virgil Nicula wrote:

As the title of this nice problem says, "Beautiful" generalization, Virgil Nicula.

My idea as in the proof above works again, so denote $N' \in NB \cap w$ .

Because of $N(ACBD)=-1$ , we have that the quadrilateral $AS_{1}N'T_{1}$ is harmonical, thus the lines $S_{1}S_{1}$ , $T_{1}T_{1}$ and $AN'$ are concurrent in a point $Z$ .

Now because $NN \| AM$ and $MC=MD$ , we have that $N(C,M,D, \infty)=-1$ , i.e $(NN, ND, NM, NC)=-1$ , and by intersecting the pencil with $w$ , we have that the quadrilateral $NT_{1}N'S_{1}$ is harmonical.

Thus the lines $S_{1}S_{1}$ , $T_{1}T_{1}$ and $NN'$ are concurrent also in $Z$ .

On other hand, let's denote $S_{3}\in S_{1}B \cap w$ and $T_{3}\in T_{1}B \cap w$ , therefore the pencils $B(AS_{1}N'T_{1})$ and $B(NT_{1}N'S_{1})$ are harmonical,

And by intersecting them again with $w$ , we have that the quadrilaterals $NT_{3}MS_{3}$ and $AS_{3}N'T_{3}$ are harmonical, i.e the pairs of lines ( $MN$ , $S_{3}S_{3}$ , $T_{3}T_{3}$ ) and ( $AN'$ , $S_{3}S_{3}$ , $T_{3}T_{3}$ ) are concurrent, i.e the all four lines $S_{3}S_{3}$ , $T_{3}T_{3}$ , $AN'$ and $MN$ are concurrent in $Z$ .

Thus $S_{3}T_{3}$ is the polar of $Z$ , but also $S_{1}T_{1}$ is the polar of $Z$ , i.e $S_{1}\equiv S_{3}$ and $T_{1}\equiv T_{3}$ .

Therefore the points $S_{1}$ , $B$ , $T_{1}$ are collinear and similar prove that $S_{2}$ , $B$ and $T_{2}$ are collinear, thus $B \in S_{1}T_{1}\cap S_{2}T_{2}$ .

Attachments:

virgil nicula's generalization.png [ 51.67 KiB | Viewed 67 times ]

_________________
Cosmin Pohoata, Bucharest, Romania

Virgil Nicula

Posts: 4860
Romania

Blog: View Blog

Posted: Jun 30, 2007, 4:40 pm • # 6

Quote:

An equivalent enunciation. Let $ABCD$ be a cyclic convex quadrilateral inscribed in the circle $w$ so that $AB=AD$ . Denote : the intersection $M\in AC\cap BD$ ; the reflection $N$ of the point $M$ w.r.t. the point $D$ ; the second intersection $E$ of the circle $w$ with the line $AN$ ; the point $P\in (MN)$ for which $\frac{PM}{PN}=\frac{BM}{BN}$ . Prove that $P\in CE$ .

Proof.

$\left\{\begin{array}{c}m(\widehat{ABD})=x\\\ m(\widehat{CAD})=y\\\ m(\widehat{DBE})=z\end{array}\right\|$ $\implies$

$\left\{\begin{array}{c}\frac{CM}{CA}=\frac{DM}{DA}\cdot\frac{\sin \widehat{CDM}}{\sin\widehat{CDA}}=\frac{DM}{DA}\cdot\frac{\...$ $\implies$

$\frac{CM}{CA}\cdot\frac{EA}{EN}\cdot\frac{PN}{PM}=1$ $\implies$ $\boxed{\ P\in CE\ }\ \blacksquare$

Virgil Nicula

Posts: 4860
Romania

Blog: View Blog

Posted: Jun 30, 2007, 11:36 pm • # 7

Lemma (prove easily !) Let $ABC$ be a triangle inscribed in the circle $w$ . For a point $M$ which belongs to the sideline $BC$ denote the second intersection $N$ of the circle $w$ with the line $AM$ . Then exists the relation $\boxed{\frac{MB}{MC}=\frac{AB}{AC}\cdot\frac{NB}{NC}}$ . If $N: =A$ , i.e. the line $AM$ is tangent to the circle $w$ then $\frac{MB}{MC}=\left(\frac{AB}{AC}\right)^{2}$ .

Quote:

A very nice equivalent problem.
Let $ABCDE$ be a cyclic convex pentagon. Denote the points $\left\{\begin{array}{c}M\in AC\cap BD\\\ N\in AE\cap BD\\\ P\in BD\cap CE\end{array}$ .
Consider the following sentencies : $\left\{\begin{array}{c}S1.\ \blacktriangleright AB=AD\\\\ S2.\ \blacktriangleright DM=DN\\\\ S3.\ \blacktriangleright \frac{B...$ .

(The last sentence means that the division $(B,M,P,N)$ is harmonically.)

Prove that if two from these sentencies are truly then and the other sentence is truly.

Proof.

Observe that $\boxed{\frac{BM}{BN}=\frac{AM}{AN}\cdot\frac{BC}{BE}}\ \ (1)$ and $\left\{\begin{array}{c}CM=\frac{DM}{DA}\cdot BC\\\\ EN=\frac{DN}{DA}\cdot BE\end{array}$ . Apply the Menelaus' theorem to the transversal $\overline{CPE}$ in $\triangle AMN$ : $\frac{CM}{CA}\cdot\frac{EA}{EN}\cdot\frac{PN}{PM}=1$ , i.e. $\boxed{\frac{PM}{PN}=\frac{DM}{DN}\cdot\frac{BC}{BE}\cdot\frac{AE}{AC}}\ \ (2)$ . Observe that $S1\Longleftrightarrow\widehat{CMN}\equiv\widehat{ABC}\equiv\widehat{CEN}\Longleftrightarrow \widehat{CMN}\equiv\widehat{CEN}$ , i.e. the quadrilateral $CMEN$ is cyclically, i.e. $AM\cdot AC=AE\cdot AN$ . Thus, $\left\{\begin{array}{ccc}S1 & \Longleftrightarrow & \frac{AM}{AN}=\frac{AE}{AC}\\\\ S2 & \Longleftrightarrow &...$ . Therefore,

$S1\wedge S2\implies$ $\frac{BM}{BN}\ \begin{array}{c}(1)\\ =\end{array}\ \frac{AM}{AN}\cdot\frac{BC}{BE}\ \begin{array}{c}(S1)\\ =\end{array}\ \fra...$ $S3$ .

$S1\wedge S3\implies\frac{DM}{DN}\ \begin{array}{c}(S3)\\ =\end{array}\ \frac{AM}{AN}\cdot\frac{AC}{AE}\ \begin{array}{c}(S1)\...$ $S2$ .

$S2\wedge S3\implies\frac{AM}{AN}\ \begin{array}{c}(S3)\\ =\end{array}\ \frac{DM}{DN}\cdot\frac{AE}{AC}\ \begin{array}{c}(S2)\...$ $S1$ .

Last edited by Virgil Nicula on Jul 01, 2007, 11:21 am, edited 2 times in total.

Display posts from previous: Sort by

Share on Facebook

Previous topic | Next topic

Community » Olympiad Section » Geometry » Geometry Proposed & Own Problems

All times are UTC - 8 hours [ DST ]

Moderators: yetti, MithsApprentice, N.T.TUAN, Peter, darij grinberg, orl, pohoatza, pbornsztein, High School Olympiad Moderators

Page 1 of 1

[ 7 posts ]

Beautiful

Login