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Source: 2018 China North Mathematical Olympiad Grade 10 Test 1 P1

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348 posts

#1 h Jul 26, 2018, 12:55 AM • 1 Y

Y by Adventure10

In triangle $ABC$ , let the circumcenter, incenter, and orthocenter be $O$ , $I$ , and $H$ respectively. Segments $AO$ , $AI$ , and $AH$ intersect the circumcircle of triangle $ABC$ at $D$ , $E$ , and $F$ . $CD$ intersects $AE$ at $M$ and $CE$ intersects $AF$ at $N$ . Prove that $MN$ is parallel to $BC$ .

This post has been edited 2 times. Last edited by Devastator, Aug 14, 2018, 12:46 AM
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348 posts

#2 h Jul 26, 2018, 1:04 AM • 3 Y

Y by Durjoy1729, Adventure10, Mango247

hint

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jayme

#3 h Jul 26, 2018, 11:17 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,

Pascal's theorem in a degenerated case work very well..

Sincerely
Jean-Louis

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4 posts

#4 h Aug 10, 2018, 1:46 PM • 2 Y

Y by Adventure10, Mango247

Obviously,
$\angle BAO = \angle CAH$
Since $\angle BAI = \angle CAI$ ,
then $\angle DAI = \angle HAI$
$\implies \angle DAE = \angle DCE$
then $\angle EAF = \angle DCE$
$\implies$ AMNC cyclic
$\implies \angle MNE = \angle MAC = \angle BAE = \angle BLE$
thus $MN || BC$

This post has been edited 2 times. Last edited by Pikachu1729, Aug 10, 2018, 1:47 PM

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1618 posts

#5 h Jul 7, 2019, 1:38 AM • 1 Y

Y by Adventure10

We have: $\angle{MCN} = \angle{EAD} = \angle{MAN}$
Then: $A$ , $C$ , $M$ , $N$ lie on a circle
So: $\angle{ANM} = 180^o - \angle{ACM} = 180^o - 90^o = 90^o$
But: $AN$ $\perp$ $BC$ then: $MN$ $\parallel$ $BC$

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#6 h Aug 4, 2019, 4:00 AM • 2 Y

Y by Adventure10, Mango247

This is relatively easy for a G10 problem, isn't it (considering no special theorems were used)? Btw I also got the same solution as Pikachu1729

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zuss77

#7 h Aug 4, 2019, 11:42 AM • 2 Y

Y by Adventure10, Mango247

It also works on any isogonal lines $AD, AF$ .
$\angle MCN = \angle MAN$ $\implies ACMN$ - cyclic.
By Reim $MN \parallel DF \parallel BC$ .

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#8 h Mar 13, 2020, 9:59 PM

Y by

Trivial

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1900 posts

#9 h Jul 4, 2020, 5:39 AM

Y by

Complex numbers work very well too if you're too lasy to even draw a diagram.
Let $a,b,c\in\mathbb{C}$ be such numbers lying on unit circle centered at $0$ such that $A=a^2, B=b^2, C=c^2, E=-bc$ We have $D=-a^2, F=-\frac{b^2c^2}{a^2}$ Hence $M=\frac{CD(A+E)-AE(C+D)}{CD-AE}=\frac{b(c^2-a^2)+c(bc-a^2)}{b-c}$ and using analogous formula $N=\frac{a^2bc(c-b)+c(b^2c^2-a^4)}{a^2(b-c)}$ Thus $\frac{M-N}{B-C}=\frac{b (c^2 - a^2)(a^2 - b c)}{a^2(b-c)^2(b+c)}$ Clearly $\frac{M-N}{B-C}=\overline{\left(\frac{M-N}{B-C}\right)}$ QED

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#10 h Apr 24, 2022, 7:32 PM • 1 Y

Y by centslordm

Since $O$ and $H$ are isogonal conjugates with respect to $\triangle ABC,$ $\measuredangle NCM=\measuredangle ECD=\measuredangle EAD=\measuredangle FAE=\measuredangle NAM$ and $ACMN$ is cyclic. Hence, $\measuredangle BCE=\measuredangle BAE=\measuredangle MAC=\measuredangle MNC.$ $\square$

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#11 h Feb 19, 2023, 12:40 PM

Y by

Applying pascal on $DCEEAF$ , we have $DF\cap EE$ lies on $MN$ . However , $\angle AFD=90^{\circ}$ and $AF\perp BC$ so $BC\parallel FD$ . Thus, we need to prove that $EE \parallel FD$ which is equivalent to $EF=ED$ . And it’s true because $E$ is the midpoint of arc $BC$ not containing $A$ and arc $BF$ = arc $DC$ due to $AH,AO$ being isogonal conjugate.

This post has been edited 3 times. Last edited by Quidditch, Feb 19, 2023, 12:41 PM

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