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Source: Chinese TST 2007 5th quiz P1

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237 posts

#1 h Jan 4, 2009, 3:08 PM • 2 Y

Y by Adventure10, Mango247

Let convex quadrilateral $ABCD$ be inscribed in a circle centers at $O.$ The opposite sides $BA,CD$ meet at $H$ , the diagonals $AC,BD$ meet at $G.$ Let $O_{1},O_{2}$ be the circumcenters of triangles $AGD,BGC.$ $O_{1}O_{2}$ intersects $OG$ at $N.$ The line $HG$ cuts the circumcircles of triangles $AGD,BGC$ at $P,Q$ , respectively. Denote by $M$ the midpoint of $PQ.$ Prove that $NO = NM.$

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467 posts

#2 h Jan 5, 2009, 3:17 PM • 2 Y

Y by Adventure10, Mango247

It is easy to prove that O1OO2G is a parallelogram
So GN=NO
If we can prove OM⊥PQ then the problem is solved => all we have to do is to prove M is the midpoint of XY => we need XP = YQ

∠YCB+∠XAD=∠YGB=∠XGD=∠PAD=∠XAD+∠XAP => ∠YCB=∠XAP => ∠XAD=∠QCY
PX/sin∠PAX=AP/sin∠AXY
PX=AP/sin∠AXY* sin∠PAX=AP/sin∠GBY*sin∠YCB=AP/GY*BY
YQ/sin∠QCY=YC/sin∠YQC
YQ= YC/sin∠YQC* sin∠QCY=YC/sin∠CAD*sin∠XAD=YC/CD*XD
AP/GY*BY=AP/GD*XD YC/CD*XD
PX/YQ=(AP/GD)/ (YC/CD)
But we have AP/GD= sin∠PGA/ sin∠CAD YC/CD= sin∠YDC/ sin∠CYD
∠CAD=∠CYD ∠PGA=∠YDC=∠YBC
=> PX=YQ
=> M is the midpoint of XY => OM⊥PQ
DONE!

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46 posts

#3 h Jan 18, 2009, 3:05 AM • 3 Y

Y by Adventure10, Mango247, Mango247

I think the idea for this problem isn’t new. We can solve this problem by using a well know lemma (It’s similar a problem in IMO 1977): Let (O) and (O’) have common AB, a line throughs A, intersect two circles resepective at M, N. We have MN always pass a fixed point T, T is midpoint of CD ( C in (O), D in (O’) and CD is perpendicular to AB at B)

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#4 h Jan 22, 2009, 8:52 AM • 2 Y

Y by Adventure10, Mango247

I think the idea for this problem isn’t new. We can solve this problem by using a well know lemma (It’s similar a problem in IMO 1977): Let $(O)$ and $ (O^')$ have common $AB$ , a line throughs $A$ , intersect two circles resepective at $M, N$ . We have $MN$ always pass a fixed point $T$ , $T$ is midpoint of $CD$ ( $C$ in $(O)$ , $D$ in $ (O^')$ and $CD$ is perpendicular to $AB$ at $B$ )

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#5 h Sep 10, 2009, 2:19 PM • 2 Y

Y by Adventure10, Mango247

Can't we draw ${XN }$ ⊥ ${QP}$ then try to prove ${GX=MX}$ ${\rightarrow}$ ${MO\parallel XN}$ to get it ?Dear Mathlinkers ?
By this way I find that :when ${H}$ is not on ${AB \cap CD}$ we also have : ${MN=ON}$ [See the figure ,I think the conclusion is right

]

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#6 h Sep 10, 2009, 5:00 PM • 2 Y

Y by Adventure10, Mango247

By angle chasing we have ∠P’AX’=∠Y’CB
P’X’=AP’*sin∠P’AX’/ sin∠AX’Y’=AP’* sin∠Y’CB/ sin∠AX’Y’=AP’*BY’/AY’
Y’Q’=CY’* sin∠Y’CQ’/ sin∠Y’Q’C=CY’* sin∠X’AD/ sin∠DBC=CY’*X’D/CD
P’X’/Y’Q’= (AP’*BY’/AY’) / (CY’*X’D/CD)
Notice that ⊿X’CD∽⊿Q’CY’ => CD/CY’=CX’/CQ’ =>
P’X’/Y’Q’=( AP’ *BY’* CX’)/( CQ’*X’D*AY’)
Again notice ⊿BCQ’∽⊿DAP’ => AP’/CQ’=AD/BC => P’X’/Y’Q’=(AD*BY’* CX’)/( BC *X’D*AY’)
It is well-known if BD,AC and X’Y’ are concurrent in a circle , then we have
AD*BY’* CX’= BC *X’D*AY’ (It can be proved by sin law)
Thus we acquire P’X’=Y’Q’=> M’is the midpoint of X’Y’ => OM’⊥ X’Y’
From the above , we have O1OO2G is a parallelogram So GN=NO
Which means M’N=NO

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411 posts

campos

#7 h Sep 12, 2009, 5:13 AM • 2 Y

Y by Adventure10, Mango247

we can apply this result (these points are different from those of the problem):
let $XYZW$ a parallelogram and let $l$ be a line through $X$ . consider the intersections $P,Q$ of the circles of center $Y$ and $W$ of radius $XY$ and $XW$ , resp, with $l$ . then $Z$ lies on the perpendicular bisector of $PQ$ .
proof:
let $l$ be the x-axis, $X = (0,0)$ , $Y = (a,b)$ , $W = (c,d)$ , then, $P = (2a,0)$ , $Q = (2c,0)$ and $Z = (a + c,b + d)$ , and we're done...

applying the lemma to parallelogram $OO_1GO_2$ and any line $l$ , it follows that $OP = OQ$ ... this implies that $OM$ is perpendicular to $l$ ... this implies that triangle $OMG$ is right-angled, and we conclude that $ON = MN$ ...

(there's no need for $H$ to be on $l$ )

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859 posts

#8 h Sep 13, 2009, 11:24 AM • 2 Y

Y by Adventure10, Mango247

This well-known result came from Russia 1995, Romanian TST 1996 and others :
Let $T$ be the second intersection of $(O_1)$ and $(O_2)$ then $HO\perp GT$ and $H,T,O$ are collinear.
Since $OO_2\perp BC, O_1G\perp BC$ then $OO_2//O_1G$ , similarly we get $GO_1OO_2$ is a parallelogram thus they intersect at the midpoint $N$ of $OG$ .
We only need to show that the perpendicular bisector of $PQ$ passes through $O. (*)$
Let $HO\cap (O_1)=\{J\}. \cap (O_2)=\{F\}. OO_2//GJ$ then $OO_2$ is midline parallel to $GJ$ of triangle $GJF.$
It follows that $O$ is the midpoint of $JF$ and $(*)$ is right.

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178 posts

#9 h Dec 16, 2016, 8:41 AM • 1 Y

Y by Adventure10

As mentioned above $N$ is the midpoint of $OG$ .

Let $(GAB),(GCD)$ intersect at $M'$ , then since $AB,M'G,CD$ are concurrent, $M'$ lies on $HG$ . It is well known that $\angle OM'G=90^{\circ}$ , and from here, we see that $M'$ is the midpoint of $PQ\implies M\equiv M'$ . Hence $N$ is circumcenter in right triangle $\triangle OMG\implies NO=NM$ .

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