View topic - a probably new characterization of AC/AB • Art of Problem Solving

Page 1 of 1

[ 11 posts ]

Author

Message

pohoatza

Posts: 1100
Location: Bucharest
Romania

Posted: Jun 08, 2007, 10:17 am • # 1

Let $ABC$ be a triangle, let $E, F$ be the tangency points of the incircle $\Gamma(I)$ to the sides $AC$ , respectively $AB$ , and let $M$ be the midpoint of the side $BC$ . Let $N = AM \cap EF$ , let $\gamma(M)$ be the circle of diameter $BC$ , and let $X, Y$ be the other (than $B, C$ ) intersection points of $BI$ , respectively $CI$ , with $\gamma$ . Prove that
$\frac {NX} {NY} = \frac {AC} {AB}.$
Author: Cosmin Pohoata

_________________
Cosmin Pohoata, Bucharest, Romania

Last edited by pohoatza on Aug 16, 2008, 5:42 pm, edited 5 times in total.

Virgil Nicula

Posts: 4860
Romania

Blog: View Blog

Posted: Jun 08, 2007, 1:30 pm • # 2

pohoatza wrote:

Let $ABC$ be a triangle such that $b>c$ and denote by $D,E,F$ the intersections of the its incircle with the sides $BC$ , $AC$ , $AB$ respectively.
Consider the circle $w=C(M)$ with the diameter $[BC]$ . Denote the intersections $\left\{\begin{array}{c}X\in w\cap (NF\\\ Y\in w\cap (NE\\\ N \in AM \cap EF\end{array}$ . Prove that $\frac{NX}{NY}=\frac{AB}{AC}$ .

Proof I. The well-known properties $\left\|\boxed{\begin{array}{c}IN\perp BC\\\\ X\in CI\ ,\ Y\in BI\end{array}}\right\|$ $\implies$ $\left\{\begin{array}{c}m(\widehat{YXI})=\frac{B}{2}\ ,\ m(\widehat{XYI})=\frac{C}{2}\\\\ m(\widehat{NIX})=90^{\circ}-\frac{C}...$ $\implies$

$\frac{NX}{NY}=\frac{IX}{IY}\cdot\frac{\sin \widehat{NIX}}{\sin \widehat{NIY}}=$ $\frac{\sin \widehat{XYI}}{\sin\widehat{YXI}}\cdot \frac{\sin \widehat{NIX}}{\sin \widehat{NIY}}=$ $\frac{\sin\frac{C}{2}}{\sin\frac{B}{2}}\cdot\frac{\cos\frac{C}{2}}{\cos\frac{B}{2}}$ $\implies$ $\frac{NX}{NY}=\frac{\sin C}{\sin B}$ $\implies$ $\boxed{\ \frac{NX}{NY}=\frac{AB}{AC}\ }$ . Remark. $\frac{NF}{NE}=\frac{AC}{AB}$ .

Proof II. Denote the intersection $Z\in BX\cap CY$ . Observe that $Z\in ID$ and the point $I$ is the orthocenter of the triangle $BZC$ , i.e. the triangle $XDY$ is the orthic triangle of $\triangle BZC$ . From the well-known property "the ray $[DI$ is the bisector of the angle $\widehat{XDY}$ " obtain $\frac{NX}{NY}=\frac{DX}{DY}$ . Prove easily that $\left\{\begin{array}{c}m(\widehat{DXY})=B\\\ m(\widehat{DYX})=C\end{array}\right\|$ $\implies$ $\triangle XDY\sim\triangle BAC$ $\implies$ $\frac{DX}{DY}=\frac{AB}{AC}$ . In conclusion, $\frac{NX}{NY}=\frac{AB}{AC}$ .

A general remark. If you know the above mentioned remarkable properties, then this very nice Pohoatza's problem is easily !

Virgil Nicula wrote:

A similar problem. Let $C(I)$ be the incircle of $\triangle ABC$ . Denote the second intersections $M$ , $N$ , $P$ of the circumcircle $C(O)$ of $\triangle ABC$ with the lines $AI$ , $BI$ , $CI$ respectively. Prove that $AM\perp NP$ and $R\in OM\cap NP\Longrightarrow$ $\frac{RN}{RP}=\frac{AB}{AC}$ .

cancer Posts: 329	Posted: Jul 19, 2007, 5:58 am • # 3 Virgil Nicula wrote: $X\in CI\ ,\ Y\in BI$ Please, Virgil Nicula, I have no idea how this comes!

Virgil Nicula Posts: 4860 Blog: View Blog	Posted: Jul 19, 2007, 11:22 am • # 4 cancer wrote: Virgil Nicula wrote: $X\in CI\ ,\ Y\in BI$ Please, Virgil Nicula, I have no idea how this comes! See www.artofproblemsolving.com/Forum/viewtopic.php?t=155800

cancer Posts: 329	Posted: Jul 19, 2007, 7:19 pm • # 5 What? I can't see how it's related... are there any tangent circles?

Virgil Nicula Posts: 4860 Blog: View Blog	Posted: Jul 19, 2007, 9:53 pm • # 6 See the definitions of the points $X$ , $Y$ from the above enunciation !!.

cancer Posts: 329	Posted: Jul 20, 2007, 12:27 am • # 7 Virgil Nicula wrote: See the definitions of the points $X$ , $Y$ from the above enunciation !!. I already see! Unless the incircle and $w$ are tangent??

Virgil Nicula Posts: 4860 Blog: View Blog	Posted: Jul 20, 2007, 12:39 am • # 8 NO ! 1. $N\in AM\cap EF$ . 2. $(NE$ and $(NF$ are rays (semiline) ! 3. The circle $w$ has the center in the middle $M$ of the side $[BC]$ and the length of the its radius is equal to $MB$ ! Sorry, this is the all ! Now please, ask you Pohoatza to explain away ...

April

Posts: 1271
Location: Hanoi, Vietnam
Viet Nam

Posted: Oct 04, 2008, 3:39 pm • # 9

pohoatza wrote:

We have some well-known results in this configuration:
1. $X$ , $Y$ lie on line $EF$ .
2. $N$ lies on $ID$ , where $D$ is the tangency point of $\Gamma(I)$ to the side $BC$ .

Now, denote by $P$ the intersection of $AI$ with $BC$ . It's easily to see that $\angle NIX=\angle PIC$ , therefore triangles $NIX$ and $PIC$ are similar. It implies $\frac{NX}{NI}=\frac{PC}{PI}$ . Similarly, we conclude that $\frac{NI}{NY}=\frac{PI}{PB}$ .

Hence, we have $\frac{NX}{NY}=\frac{NX}{NI}\cdot\frac{NI}{NY}=\frac{PC}{PI}\cdot\frac{PI}{PB}=\frac{PC}{PB}=\frac{AC}{AB}$ . This completes our solutions. :lol:

_________________
"Mathematics, mathematics, mathematics. This much mathematics? No, more!" - Grigore Moisil

thaithuan_GC Posts: 216	Posted: Oct 05, 2008, 3:00 am • # 10 Dear April, Your proof is Cosmin's. It was introduced in an aritcile of his.

mathVNpro

Posts: 467
Location: Le Hong Phong Highschool for the Gifted, Ho Chi Minh city, Viet Nam

Posted: Nov 18, 2009, 6:36 am • # 11

Let $D$ be the tangency point of $BC$ and $(I)$ and $\mathcal {K}\equiv EF\cap BC$ . Also, let $X',Y'$ respectively be the intersections of $BI,CI$ with $EF$ . It is well known that $(\mathcal {K}DBC) = - 1$ $\Longrightarrow$ $(X'\mathcal {K},X'D,X'B,X'C) = - 1$ . In the other hand, since $X'$ lies on the perpendicular bisector of $DF$ , therefore, $X'B$ is the bisector of $\angle \mathcal {K}X'D$ , thus, $BX'\perp CX'$ , which implies $X'\equiv X$ . With the same argument, we can also conclude that $Y'\equiv Y$ . As the consequence, $E,F,X,Y$ are collinear.

Let $A'$ be the intersection of $ID$ and the line $\triangle$ passes through $A$ parallel to $BC$ . Let $N'\equiv ID\cap EF$ and $\mathcal {K'}$ be the intersection of $EF$ with $\triangle$ . Since $\angle AA'I = 90^{\circ}$ $\Longrightarrow$ $A'\in (AEF)$ . Therefore, $A'N'$ is the bisector of $\angle EA'F$ , but $A'N'\perp A'\mathcal {K'}$ $\Longrightarrow$ $(A'E,A'F,A'N',A'\mathcal {K'}) = - 1$ , which implies $(EFN'\mathcal {K'}) = - 1$ $\Longrightarrow$ $A(CBN'\mathcal {K'}) = - 1$ . But $A\mathcal {K'}\|BC$ , therefore, $AN'$ passes through $M$ , which implies $N'\equiv N$ $\Longrightarrow$ $I,D,N$ are collinear.

Now, it is easy to prove that $\triangle YDX$ $\cong$ $\triangle BAC$ (a.a) and $DI$ is the bisector of $\angle YDX$ , then $\frac {NX}{NY} = \frac {DX}{DY} = \frac {AC}{AB}$ . $\square$

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

[ 11 posts ]

a probably new characterization of AC/AB

Login