Vmo 2006 a2

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Vmo 2006 a2

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modular arithmetic

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#1 h Feb 28, 2006, 5:10 AM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a convex quadrilateral. Take an arbitrary point $M$ on the line $AB$ , and let $N$ be the point of intersection of the circumcircles of triangles $MAC$ and $MBC$ (different from $M$ ). Prove that:

a) The point $N$ lies on a fixed circle;

b) The line $MN$ passes though a fixed point.

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110 posts

#2 h Feb 28, 2006, 6:38 PM • 1 Y

Y by Adventure10

the circles MAC and MBC intersect at points M and C.
So maybe its MAD circle ?

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#3 h Mar 3, 2006, 9:16 AM • 2 Y

Y by Adventure10, Mango247

a) Let $T={AC}\cap{BD}$ then $C, T, D, N$ are concyclic (angle chasing).

b) Suppose $MN$ intesects $(CTD)$ in point $X$ then $NC, NT, NX, ND$ form $r$ -corelative
harmonic fascicle on a fixed circle (again angles), hence $X$ is a fixed point.

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Sailor

#4 h Mar 4, 2006, 3:45 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

For b) you can just observe that $\angle{DNX}=\angle{DBA}$ and $\angle{CNX}=\angle{CAB}$ and thus $X$ is a fixed point.

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#5 h Apr 8, 2006, 7:35 PM • 2 Y

Y by Adventure10, Mango247

For a)
$\angle DNC=\angle DNM+\angle MNC=180^o-\angle A+180^o-\angle B=\angle C+\angle D$
which is constant.
But $DTC\neq 180^o-\angle D-\angle C$ such that the quadrilateral $CNTD$ to be cyclic.. Why i'm wrong?

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607 posts

xirti

#6 h Apr 8, 2006, 7:40 PM • 2 Y

Y by Adventure10, Mango247

For b) On the specified fixed circle, denote by $T_M$ the second intersection of $MN$ with the circle. The arc $TD$ always have measure $180^o-\angle A$ . Hence $T$ is a fixed point on $MN$ .

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#7 h Apr 22, 2009, 5:31 AM • 3 Y

Y by Shinichi-123, Adventure10, Mango247

This problem is nice but easy

. I think to prevent from pointing out the case of the figure of this problem (At least there must be 2 cases), I would prefer using the directed angle to kill it. Here is my solution:
a) Let $O$ be the intersections of the $2$ diagonals of $ABCD$ . $(OA,OB)\equiv (OC,OD)\equiv - (DO,DC) - (CD,CA)\equiv (DM,DB) + (CA,CM)\equiv (NM,NB) + (NA,NM)\equiv (NA,NB)\pmod\pi$ . Therefore, $N$ blongs to $(OAB)$ , which is obviously fixed. (Done!)
b) Let $P$ be the intersection of $MN$ with $(OAB)$ . Then, we get, $(PN,PO)\equiv (AN,AO)\equiv (AN, AC)\equiv (MN,MC) \pmod \pi$ , which leads to the fact that $PO//CD$ , but $CD$ is fixed, $O$ is fixed, $(OAB)$ is fixed. Hence, $P$ is fixed too, which completes our proof.

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#8 h Jan 31, 2011, 4:20 AM • 2 Y

Y by Adventure10, Mango247

@Mod please post the prob in it's correct form

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1618 posts

#9 h Apr 21, 2019, 3:04 PM • 2 Y

Y by AlastorMoody, Adventure10

Here is my solution for this problem
Solution
a) We have: ( $NC$ ; $ND$ ) $\equiv$ ( $NC$ ; $NM$ ) + ( $NM$ ; $ND$ ) $\equiv$ ( $AC$ ; $AB$ ) + ( $BA$ ; $BD$ ) (mod $\pi$ )
Then: $N$ lies on fixed circle is the circle passes through $C$ , $D$ which satisfies: ( $NC$ ; $ND$ ) $\equiv$ ( $AC$ ; $AB$ ) + ( $BA$ ; $BD$ ) (mod $\pi$ )
b) Let $P$ $\equiv$ $AC$ $\cap$ $BD$
Draw line through $P$ and parallel with $AB$ intersects ( $PCD$ ) at $S$ ( $S$ $\ne$ $P$ )
We have: ( $NC$ ; $ND$ ) $\equiv$ ( $NC$ ; $NM$ ) + ( $NM$ ; $ND$ ) $\equiv$ ( $AC$ ; $AB$ ) + ( $BA$ ; $BD$ ) $\equiv$ ( $AP$ ; $AB$ ) + ( $BA$ ; $BP$ ) $\equiv$ ( $PA$ ; $PB$ ) $\equiv$ ( $PC$ ; $PD$ ) (mod $\pi$ )
So: $P$ , $C$ , $N$ , $D$ lie on a circle
But: ( $NM$ ; $NS$ ) $\equiv$ ( $NM$ ; $ND$ ) + ( $ND$ ; $NS$ ) $\equiv$ ( $BM$ ; $BD$ ) + ( $PD$ ; $PS$ ) $\equiv$ ( $PS$ ; $PD$ ) + ( $PD$ ; $PS$ ) $\equiv$ 0 (mod $\pi$ )
Then: $M$ , $N$ , $S$ are collinear
Hence: $MN$ passes through fixed point $S$ when $M$ moves on $AB$

This post has been edited 1 time. Last edited by khanhnx, Apr 21, 2019, 3:28 PM

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