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 O1,O2 be the circumcenters of triangles AGD,BGC. O1O2 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.
Problem
Source: Chinese TST 2007 5th quiz P1
Tags: geometry, circumcircle, parallelogram, perpendicular bisector, geometry proposed
05.01.2009 17:17
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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18.01.2009 05:05
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)
22.01.2009 10:52
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)
10.09.2009 17:19
Can't we draw XN ⊥QP then try to prove GX=MX → MO∥XN to get it ?Dear Mathlinkers ? By this way I find that :when H is not on AB∩CD we also have :MN=ON[See the figure ,I think the conclusion is right ]
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10.09.2009 20:00
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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12.09.2009 08:13
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 OO1GO2 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)
13.09.2009 14:24
This well-known result came from Russia 1995, Romanian TST 1996 and others : Let T be the second intersection of (O1) and (O2) then HO⊥GT and H,T,O are collinear. Since OO2⊥BC,O1G⊥BC then OO2//O1G, similarly we get GO1OO2 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∩(O1)={J}.∩(O2)={F}.OO2//GJ then OO2 is midline parallel to GJ of triangle GJF. It follows that O is the midpoint of JF and (∗) is right.
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16.12.2016 11:41
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 ∠OM′G=90∘, and from here, we see that M′ is the midpoint of PQ⟹M≡M′. Hence N is circumcenter in right triangle △OMG⟹NO=NM.