three circles tangent both to the incircle and the circumcir

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grobber

7849 posts

grobber

#1 h May 11, 2005, 3:18 PM • 2 Y

Y by Adventure10, Mango247

Let $(I),(O)$ be the incircle, and, respectiely, circumcircle of $ABC$ . $(I)$ touches $BC,CA,AB$ in $D,E,F$ respectively. We are also given three circles $\omega_a,\omega_b,\omega_c$ , tangent to $(I),(O)$ in $D,K$ (for $\omega_a$ ), $E,M$ (for $\omega_b$ ), and $F,N$ (for $\omega_c$ ).

a) Show that $DK,EM,FN$ are concurrent in a point $P$ ;

b) Show that the orthocenter of $DEF$ lies on $OP$ .

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mecrazywong

606 posts

mecrazywong

#2 h May 11, 2005, 4:36 PM • 2 Y

Y by Adventure10, Mango247

I don't have time to finish it right now, but I guess the following approach works.
For part a, just note that DK bisects the arc BC than containing A.
For part b, use the well known fact that O lies on the Euler line of triangle DEF.

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prowler

312 posts

prowler

#3 h May 12, 2005, 5:45 PM • 2 Y

Y by Adventure10, Mango247

$D$ - point which is center of omothety of one $\omega_a$ and incircle,
$K$ - point which is center of omothety of one $\omega_a$ and circumcircle.
It rezults that point which is center of omothety of circumcircle and incircle lies on $DK$ .
Hence $P$ is exactly point of similitude of incircle and circumcircle (I think so called).
So it lies on $OI$ and from Iran 1995 it contains ortocenter of $DEF$ .

This post has been edited 1 time. Last edited by prowler, May 13, 2005, 8:01 PM

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erdos

312 posts

erdos

#4 h May 12, 2005, 7:17 PM • 2 Y

Y by Adventure10, Mango247

Can anyone give a clear solution please ??

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darij grinberg

6555 posts

darij grinberg

#5 h May 12, 2005, 10:35 PM • 2 Y

Y by Adventure10, Mango247

For Erdos, I am rewriting Prowler's solution:

grobber wrote:

Let $(I),(O)$ be the incircle, and, respectiely, circumcircle of $ABC$ . $(I)$ touches $BC,CA,AB$ in $D,E,F$ respectively. We are also given three circles $\omega_a,\omega_b,\omega_c$ , tangent to $(I),(O)$ in $D,K$ (for $\omega_a$ ), $E,M$ (for $\omega_b$ ), and $F,N$ (for $\omega_c$ ).

a) Show that $DK,EM,FN$ are concurrent in a point $P$ ;

b) Show that the orthocenter of $DEF$ lies on $OP$ .

Solution. Let Q be the internal center of similitude of the circumcircle (O) and the incircle (I) of triangle ABC. Then, of course, this point Q lies on the line OI which joins the centers O and I of these two circles.

By the Monge theorem, the internal center of similitude of the circles (O) and (I), the internal center of similitude of the circles (I) and $\omega_a$ , and the external center of similitude of the circles $\omega_a$ and (O) are collinear. But the internal center of similitude of the circles (O) and (I) is the point Q; the internal center of similitude of the circles (I) and $\omega_a$ is the point D (since these two circles externally touch at D), and the external center of similitude of the circles $\omega_a$ and (O) is the point K (since these two circles internally touch at K). Hence, the points Q, D and K are collinear; in other words, the point Q lies on the line DK. Similarly, the point Q lies on the lines EM and FN. Thus, the lines DK, EM and FN are concurrent at the point Q. This solves part (a) of the problem, and shows that the point of concurrence P of the lines DK, EM and FN is the point Q.

Now, in order to prove part (b) of the problem, it is enough to show that the orthocenter of triangle DEF lies on the line OQ. Since the point Q lies on the line OI, the line OQ is the line OI; hence, it remains to show that the orthocenter of triangle DEF lies on the line OI. In other words, it remains to show that the orthocenter of triangle DEF, the incenter I of triangle ABC and the circumcenter O of triangle ABC are collinear. But this was shown in http://www.mathlinks.ro/Forum/viewtopic.php?t=6228 .

Problem solved.

Darij

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erdos

312 posts

erdos

#6 h May 12, 2005, 11:07 PM • 2 Y

Y by Adventure10, Mango247

Thanks Darij

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juancarlos

161 posts

juancarlos

#7 h May 13, 2005, 8:17 PM • 2 Y

Y by Adventure10, Mango247

Also $AK,BM,CN$ are concurrent.

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jayme

9775 posts

jayme

#8 h Nov 24, 2011, 10:58 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,
an article concerning the “Ayme’s theorem’’ and this question have been put on my website.
http://perso.orange.fr/jl.ayme vol. 20 p. 17, 18
You can use Google translator
Sincerely
Jean-Louis

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thecmd999

2860 posts

thecmd999

#9 h Jan 25, 2014, 2:26 AM • 1 Y

Y by Adventure10

a

b

Notice the similarity to 2003 Vietnam TST #2, which concerns the incenter of the orthic triangle rather than the orthocenter of the intouch triangle.

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khanhnx

1618 posts

khanhnx

#10 h Apr 21, 2019, 9:49 AM • 1 Y

Y by Adventure10

Here is my solution for this problem
Solution
a) We define again point $K$ is second intersection of $\bigodot(AI)$ with ( $O$ )
Let $X$ , $Y$ , $Z$ be midpoint of $\stackrel\frown{BC}$ which not containing $A$ , $\stackrel\frown{CA}$ which not containing $B$ , $\stackrel\frown{AB}$ which not containing $C$
Then: $K$ is Miquel point of quadrilateral $BCEF$
So: $\triangle$ $KBF$ $\stackrel{+}{\sim}$ $\triangle$ $KCE$
Hence: $\dfrac{KB}{KC}$ = $\dfrac{BF}{CE}$ = $\dfrac{BD}{CD}$ or $KD$ is internal bisector of $\widehat{BKC}$
Let $J$ $\equiv$ $OK$ $\cap$ $ID$
We have: ( $KD$ ; $KJ$ ) $\equiv$ $\dfrac{\pi}{2}$ $-$ ( $CK$ ; $CX$ ) $\equiv$ $\dfrac{\pi}{2}$ $-$ ( $DK$ ; $DB$ ) $\equiv$ ( $DJ$ ; $DK$ ) (mod $\pi$ )
So: ( $J$ ; $JK$ ) $\equiv$ $\omega_a$
Then: $DK$ passes through $X$
Similarly: $Y$ $\in$ $EM$ , $Z$ $\in$ $FN$
But: $EF$ $\parallel$ $YZ$ , $FD$ $\parallel$ $ZX$ , $DE$ $\parallel$ $XY$ then: $DX$ , $EY$ , $FZ$ concurrent at point $P$ is external homothetic center of ( $I$ ) and ( $O$ ) or $DK$ , $EM$ , $FN$ concurrent at $P$
b) Let $H$ be orthocenter of $\triangle$ $DEF$
Since: $I$ is orthocenter of $\triangle$ $XYZ$ , we have: $I$ , $H$ , $P$ are collinear
But: $P$ , $I$ , $O$ are collinear then $I$ , $H$ , $P$ , $O$ are collinear of $H$ $\in$ $PO$

This post has been edited 2 times. Last edited by khanhnx, Apr 21, 2019, 9:51 AM

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