exinscribed circle and its tangency pts

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exinscribed circle and its tangency pts

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Source: Russian olympiad 2003/problem 10.7; IMAR test, october 2003

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#1 h Jun 12, 2004, 11:11 AM • 2 Y

Y by Adventure10, Mango247

The exinscribed circle of a triangle $ABC$ corresponding to its vertex $A$ touches the sidelines $AB$ and $AC$ in the points $M$ and $P$ , respectively, and touches its side $BC$ in the point $N$ . Show that if the midpoint of the segment $MP$ lies on the circumcircle of triangle $ABC$ , then the points $O$ , $N$ , $I$ are collinear, where $I$ is the incenter and $O$ is the circumcenter of triangle $ABC$ .

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#2 h Jun 12, 2004, 11:58 AM • 2 Y

Y by Adventure10, Mango247

this should be easy with trilinears

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#3 h Jun 12, 2004, 12:00 PM • 2 Y

Y by Adventure10, Mango247

Maybe, but while trying to solve it I also tried not to spoil it with trilinears or trigonometry

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#4 h Jun 19, 2004, 9:13 AM • 3 Y

Y by Mr_pi, Adventure10, Mango247

Cuzzez! I have been thinking upon this problem for two hours, and the only solution I found is a kind of trig bash. However, since nobody has posted any synthetic solution, here is mine:

We know that if $I_{a}$ , $I_{b}$ , $I_{c}$ are the excenters of triangle ABC, then

- the lines $AI_{a}$ , $BI_{b}$ , $CI_{c}$ are the three altitudes of the triangle $I_{a}I_{b}I_{c}$ ;
- the triangle ABC is the orthic triangle of the triangle $I_{a}I_{b}I_{c}$ ;
- the incenter I of the triangle ABC is the orthocenter of the triangle $I_{a}I_{b}I_{c}$ ;
- the circumcircle of the triangle ABC is the nine-point circle of triangle $I_{a}I_{b}I_{c}$ .

We also know that

- the circumcenter V of the triangle $I_{a}I_{b}I_{c}$ lies on the line $I_{a}N$ (and the two similar lines through $I_{b}$ and $I_{c}$ );
- this circumcenter V is the reflection of the incenter I of triangle ABC in the circumcenter O of triangle ABC;
- the circumradius of triangle $I_{a}I_{b}I_{c}$ equals $I_{a}V = 2R$ , where R is the circumradius of triangle ABC.

(Note that the circumcenter V of triangle $I_{a}I_{b}I_{c}$ is called the Bevan point of triangle ABC.)

Since the points M and P are the points of tangency of the A-excircle of triangle ABC with the sidelines AB and AC, these points M and P must be symmetric to each other with respect to the angle bisector of the angle CAB. Hence, the midpoint of the segment MP must lie on the angle bisector of the angle CAB. But this angle bisector is the line $AI_{a}$ , and this line is the $I_{a}$ -altitude of triangle $I_{a}I_{b}I_{c}$ . So we see that the midpoint of the segment MP lies on the $I_{a}$ -altitude of triangle $I_{a}I_{b}I_{c}$ . On the other hand, by our hypothesis, the midpoint of the segment MP lies on the circumcircle of triangle ABC, i. e. on the nine-point circle of triangle $I_{a}I_{b}I_{c}$ . Hence, the midpoint of the segment MP is the point of intersection of the $I_{a}$ -altitude of triangle $I_{a}I_{b}I_{c}$ with the nine-point circle of this triangle. But, by the definition of a nine-point circle, the $I_{a}$ -altitude of triangle $I_{a}I_{b}I_{c}$ intersects the nine-point circle of this triangle at two points: at the foot A of the $I_{a}$ -altitude, and at the midpoint of the segment joining the vertex $I_{a}$ of the triangle $I_{a}I_{b}I_{c}$ with the orthocenter I of this triangle. Since the midpoint of the segment MP is distinct from A, we can therefore conclude that the midpoint of the segment MP is the midpoint of the segment joining the vertex $I_{a}$ of the triangle $I_{a}I_{b}I_{c}$ with the orthocenter I of this triangle. In other words, the midpoint of the segment MP is the midpoint of the segment $I_{a}I$ .

Therefore, the quadrilateral $M I_{a}P I$ is a parallelogram, and $MI\parallel I_{a}P$ . But $I_{a}P \perp CA$ ; hence, $MI \perp CA$ , and thus the triangle AYM is right-angled, where Y is the point where the incircle of triangle ABC touches the side CA. Hence, $AY=AM\cdot \cos \measuredangle YAM=AM\cdot \cos A$ . Now, since the points M and P are symmetric to each other with respect to the angle bisector of the angle CAB, we have AM = AP, so that $AY=AP\cdot \cos A$ , and

$\cos A=\frac{AY}{AP}$ .

But Thales shows

$\frac{AY}{AP}=\frac{r}{r_{a}}$ ,

where r, $r_{a}$ , $r_{b}$ , $r_{c}$ are the inradius and the exradii of triangle ABC. Thus,

$\cos A=\frac{r}{r_{a}}$ .

Now, what do we want to prove? We want to show that the points O, N, I are collinear. This will obviously hold if we show that N = V (since we know that the points O, V, I are collinear, for V is the reflection of I in O). The point V lies on the line $I_{a}N$ ; hence, in order to obtain N = V, it is enough to show that the segments $I_{a}N$ and $I_{a}V$ are equal. But $I_{a}N = r_{a}$ and $I_{a}V = 2R$ . Hence, we must show that $r_{a}= 2R$ .

Here the battle begins. Denoting x = cos A, y = cos B, z = cos C, we know the formulas

$r=R\left( x+y+z-1\right)$ ;
$r_{a}=R\left(-x+y+z+1\right)$ .

Also, we have given

$x=\cos A=\frac{r}{r_{a}}$ ,

so that $r=xr_{a}$ . Thus, we can substitute the above formulas to get

R (x + y + z - 1) = x R (- x + y + z + 1).

After some algebra, this simplifies to

(x - 1) (y + z - x - 1) = 0.

Now, since $x-1\neq 0$ (the cosine of a triangle's angle is never 1), this becomes y + z - x - 1 = 0, so that y + z = x + 1, and

$r_{a}=R\left(-x+y+z+1\right) =R\left(-x+x+1+1\right) =2R$ ,

qed..

The converse of the problem also holds and can be proved just by inverting the observations above.

EDIT: See also http://www.mathlinks.ro/Forum/viewtopic.php?t=105101 for a discussion of this problem.

Darij

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#5 h Aug 23, 2006, 9:02 PM • 2 Y

Y by Adventure10, Mango247

Only for the fans and not for the professionals.

This post has been edited 34 times. Last edited by Virgil Nicula, Aug 24, 2006, 3:17 AM

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#6 h Aug 23, 2006, 10:02 PM • 2 Y

Y by Adventure10, Mango247

Let $X$ be midpoint of $MP$ ,WE know $I_{a}X.I_{a}A=OI_{a}^{2}-R^{2}=2Rr_{a}=r_{a}^{2}\rightarrow r_{a}=2R$
let $D,D'$ be reflection of $I,O$ on the $BC$ we have $OX\perp BC$ and $D'D=D'N$ so cause $OX=R\ ,\ I_{a}N=2R$ and $X$ is midpoint of $II_{a}$ hence $O,I,N$ are collinear.

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yetti

#7 h Aug 24, 2006, 4:36 AM • 2 Y

Y by Adventure10, Mango247

Inversion in the excircle $(I_{a})$ carries the vertices A, B, C into the midpoints X, Y, Z of the excircle chords MP, MN, PN and the circumcircle (O) of $\triangle ABC$ into the 9-point circle of $\triangle MNP.$ If $X \in (O),$ the circumcircle is carried into itself, hence it is the 9-point circle of $\triangle MNP$ (and $r_{a}= 2R)$ . But (O) is also the 9-point circle of the triangle $\triangle I_{b}II_{c},$ always centrally similar to the triangle $\triangle MNP$ (with parallel sides). Thus in case $X \in (O),$ these 2 triangles are congruent. The external bisector $I_{b}I_{c}\perp AI_{a}\equiv AX$ of $\angle A$ meets (O) again at the midpoint Q of $I_{b}I_{c}$ and XQ is a diameter of (O), hence the midpoint O of XQ is the similarity center of $\triangle MNP \cong \triangle I_{b}II_{c},$ so that I, N, O are collinear and O is the midpoint of IN.

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#8 h Mar 24, 2007, 4:32 PM • 2 Y

Y by Adventure10, Mango247

Virgil Nicula wrote:

$2.\blacktriangleright$ If $l_{a}$ is the length of the bisector of the angle $\widehat{BAC}$ , then $\boxed{\ L\in w\ }\Longleftrightarrow AL\cdot l_{a}=bc\Longleftrightarrow$
$p\cos\frac{A}{2}\cdot \frac{2bc\cos\frac{A}{2}}{b+c}=bc$

Why is this?

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#9 h Mar 24, 2007, 4:59 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

$2.\blacktriangleright$ If $l_{a}$ is the length of the bisector of the angle $\widehat{BAC}$ , then is well-known that $l_{a}=\frac{2bc}{b+c}\cdot\cos \frac{A}{2}$ . Denote $S\in AI\cap BC$ . Then $\boxed{\ L\in w\ }$ $\Longleftrightarrow$ $\left\{\begin{array}{c}\triangle ABL\sim \triangle ASC\\\ LA=RA\cdot \cos \frac{A}{2}\\\ RA=p\end{array}\right\|$ $\Longleftrightarrow$ $\left\{\begin{array}{c}AL\cdot l_{a}=bc\\\ LA=p\cdot \cos \frac{A}{2}\end{array}\right\|$ $\Longleftrightarrow$
$p\cos\frac{A}{2}\cdot \frac{2bc\cos\frac{A}{2}}{b+c}=bc$ .

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