Parallel lines..

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Parallel lines..

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Source: Kazakhstan National Olympiad 2014 P3 D1 10 grade

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134 posts

ts0_9

#1 h Mar 26, 2014, 10:58 AM • 4 Y

Y by kazakhboy, HWenslawski, Adventure10, and 1 other user

The triangle $ABC$ is inscribed in a circle $w_1$ . Inscribed in a triangle circle touchs the sides $BC$ in a point $N$ . $w_2$ — the circle inscribed in a segment $BAC$ circle of $w_1$ , and passing through a point $N$ . Let points $O$ and $J$ — the centers of circles $w_2$ and an extra inscribed circle (touching side $BC$ ) respectively. Prove, that lines $AO$ and $JN$ are parallel.

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jayme

#2 h Mar 26, 2014, 11:27 AM • 2 Y

Y by HWenslawski, Adventure10

Dear Mathlinkers,
can you please proposed a figure for this problem...
Sincerely
Jean-Louis

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136 posts

#3 h Mar 27, 2014, 7:53 AM • 4 Y

Y by mathuz, Adventure10, Mango247, and 1 other user

Let $O_1$ , $I$ and $O$ be centers of $w_2$ , $w_1$ and circumcircle of $\triangle{ABC}$ . Then we must prove that $AO_1||JN$ . For this it is enough to prove $\frac{AI}{JI}=\frac{IO_1}{IN}$ Let $w_2$ touch to circumcircle of $\triangle{ABC}$ at $M$ and $w_1$ touch to $AB$ , $AC$ at $C_1$ , $B_1$ . Since $\frac{BM}{MC}=\frac{BN}{NC}=\frac{BC_1}{CB_1}$ and $\angle{MBA}=\angle{MCA}$ we know that $\triangle{MBC_1}$ and $\triangle{MCB_1}$ are similar. $\To$, $\angle{MC_1A}=\angle{MB_1A}$ which means $IC_1MAB_1$ is cyclic. $\TO$ $\angle{IMA}=\angle{IC_1A}=90^{\circ}$ . if $AA_1$ is diametr of circumcircle of $\triangle{ABC}$ then $M,I,A_1$ are on a line.Let $L$ be midpoint of arc $BC$ then it is also midpoint of $IJ$ and $M,N,L$ are on a line. $K$ is intersection of $MI$ and $OL$ . Then by menelau's theorem on $\triangle{ALO}$ : $\frac{AI}{JI}=\frac{AI}{2LI}=\frac{OK}{KL}=\frac{IO_1}{IN}$ SO we are done!

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#4 h Mar 27, 2014, 6:08 PM • 2 Y

Y by Adventure10 and 1 other user

jayme wrote:

Dear Mathlinkers,
can you please proposed a figure for this problem...
Sincerely
Jean-Louis

http://vk.com/doc85099109_283730556?hash=04aad86df41a5182fe&dl=8f48255e53b58ddc25

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mathuz

#5 h Apr 9, 2014, 6:47 AM • 3 Y

Y by kazakhboy, Adventure10, Mango247

here another nice proof:
Let $\Omega$ -circumcircle of $ABC$ , $S$ -tangency point of $\omega_2$ and $\Omega$ . $P$ is midpoint of arc $BC$ , $S,N,P$ are collinear. Let $M$ is midpoint the altitude $AH$ . If the tangent line to $\Omega$ at the point $S$ intersect $BC$ at $X$ then $XN^2=XS^2=XB\cdot XC$ $\Rightarrow$ $M,N,J$ are collinear and $XI\perp MN$ . So we need to prove $XI\perp AO$ . Let $XI\cap AO=K$ . We have that the radius of $\omega_2$ equal to $\frac{AH}{2}$ . So $AM=ON$ and $AM\parallel ON$ $\Rightarrow$ $AMNO$ is parallelogram. $\angle HAK=\angle INM=\angle HXK$ and $A,X,H,K$ are concyclic, $\angle AKX=90^\circ$ .

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5 posts

#6 h Jan 24, 2016, 3:30 PM • 2 Y

Y by Adventure10, Mango247

why is the radius equal to AH/2

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43 posts

#7 h Feb 23, 2021, 5:41 PM

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ts0_9 wrote:

The triangle $ABC$ is inscribed in a circle $w_1$ . Inscribed in a triangle circle touchs the sides $BC$ in a point $N$ . $w_2$ — the circle inscribed in a segment $BAC$ circle of $w_1$ , and passing through a point $N$ . Let points $O$ and $J$ — the centers of circles $w_2$ and an extra inscribed circle (touching side $BC$ ) respectively. Prove, that lines $AO$ and $JN$ are parallel.

I doubt the lines $AO$ and $JN$ are parallel. Can someone confirm? Perhaps I am just misunderstanding the wording of the problem. If possible, can someone attach the diagram?

This post has been edited 1 time. Last edited by BatyrKHAN, Feb 23, 2021, 5:44 PM

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113 posts

#8 h Jul 14, 2021, 10:09 AM

Y by

Amazing problem. Let $S$ be the center of $\omega_1$ , $E$ the point where $\omega_2$ touches $\omega_1$ . Obviously $E$ is the center of homothety between $\omega_2$ and $\omega_1$ so $E,D,N$ are collinear where $D$ is midpoint of an arc $BC$ of $\omega_1$ which doesn't contain $A$ .

We will prove that points $O,I,N$ are collinear. Let tangent to $\omega_1$ and $\omega_2$ at $E$ intersect line $BC$ at $T$ and let $AI\cap BC={G}$ . From $DG \cdot DA=DB^2=DE \cdot DN$ we have that quadrilateral $EAGN$ is cyclic. Now we have that $\angle ENT=\angle GND=\angle EAG=\angle TEN$ so we have $TE=TN$ which implies that $TN$ is tangent to $\omega_2$ . Now line $BC$ is the common tangent of $\omega_2$ and the incircle so $O,I,N$ are collinear.

For proving $AO \parallel JN$ it's enough to prove $\frac{OI}{IN}=\frac{AI}{IJ} \Leftrightarrow \frac{ON-IN}{IN}=\frac{AJ-IJ}{IJ} \Leftrightarrow \frac{ON}{IN}=\frac{AJ}{IJ}$ . From the homothety of $\omega_1$ and $\omega_2$ we have that $\frac{ON}{R}=\frac{EN}{ED}$ where $R$ is radius of $\omega_1$ so it's enough to prove that $R\cdot IJ\cdot EN=AJ\cdot IN\cdot DE$ or using that $IJ=2DI$ we have to prove that $2R\cdot DI\cdot EN=AJ\cdot IN\cdot DE$ .

Now this is probably bashable but we want to avoid dirty work. We use some synthetic observations. It's well-known that $\angle AEI=90^{\circ}$ because $E$ is $A$ -Sharky devil point so if $EI$ intersects again $\omega_1$ at $A'$ we know that $A'$ is the antipode of $A$ in $\omega_1$ . We have that $\frac{DE}{2R}=sin \angle EAD=\frac{IE}{AI}$ so we have to prove that $DI\cdot AI\cdot EN=AJ\cdot IN\cdot IE$ . From the power of the point $I$ with the respect to $\omega_1$ we have that $DI\cdot AI=IE\cdot IA'$ so what we have to prove reduces to $IA'\cdot EN=AJ\cdot IN$ .

It's well-known that $D$ is the midpoint of the line segment $IJ$ so from $A'D \perp IJ$ we have that $IA'=JA'$ . Now we have that $\angle A'AJ=\angle A'AD=\angle A'ED=\angle IEN$ and $\angle AJA'=\angle IJA'=\angle JIA'=\angle AIE=90^{\circ}-\angle EAD=\angle OED=\angle OEN=\angle ONE=\angle INE$ so we have that triangles $AJA'$ and $EIN$ are similar which gives us $\frac{EN}{IN}=\frac{AJ}{JA'}=\frac{AJ}{IA'}$ as we wanted.

Attachments:

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620 posts

#9 h Jul 28, 2021, 2:39 PM

Y by

First of all; Let $T$ be the intersection of $NI$ with the parallel to $BC$ through $A$ . It's well-known that the circle with diameter $TI$ is exactly our circle and is tangent to the circumcircle at the sharky point. Let $H$ be the foot of the altitude from $A$ . It's well known that $NJ$ bisects segment $AH$ (say at $M$ ). Now just notice the parallelogram $AMNO$ . $\blacksquare$
Proof the well known things

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This post has been edited 1 time. Last edited by Steve12345, Jul 28, 2021, 3:20 PM
Reason: proved well known things

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13 posts

#10 h May 31, 2025, 11:09 PM

Y by

Let:

$AB = c, \quad AC = b,$
Let $F$ be the point of tangency of the incircle $\omega$ with $BC$ , $E$ the point of tangency of the $A$ -excircle with $BC$ , $AH$ the altitude, $M$ the midpoint of $AH$ , and $K$ the midpoint of arc $BC$ . Let $p$ be the semi-perimeter.

By the Midpoint Altitude Lemma, we have:

$M - N - J \quad \text{are collinear},$
and by Archimedes’ Lemma:

$F - N - K.$
We want to prove:

$2ON = \frac{FN}{\sin \angle FON / 2} = \frac{FN}{\sin \angle KNC} = AH \cdot \sin \angle KNC = FN.$
⸻

From power of a point:

$FN = \frac{BN \cdot CN}{NK}.$
Also:

$NK \cdot \sin \angle KCN = KC \cdot \sin \frac{A}{2},$
so:

$AH \cdot KC \cdot \sin \frac{A}{2} = BN \cdot CN = (p - b)(p - c).$
⸻

From the Law of Sines:

$AH \cdot KC \cdot \sin \frac{A}{2} = (\sin C \cdot b) \cdot \left( \frac{\sin^2 \frac{A}{2} \cdot c}{\sin C} \right),$
$= (p - b)(p - c).$
Hence, the identity:

$\sin \frac{A}{2} = \sqrt{ \frac{(p - b)(p - c)}{bc} }.$
$\boxed{\text{Q.E.D.}}$

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