Generalization of APMO-2012-1
- Tahmid Hasan
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Let $P$ be a point in the interior of a triangle $ABC$.Let $AD,BE,CF$ be cevians through $P$.Prove that $(PAF)+(PBD)+(PCE)=(PBF)+(PCD)+(PAE)\;\;$,where $(ABC)$ denotes the area of triangle $ABC$.[Adib showed a solution using mass point geometry but I didn't understand ]
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Re: Generalization of APMO-2012-1
Which is actually wrong.Tahmid Hasan wrote:Let $P$ be a point in the interior of a triangle $ABC$.Let $AD,BE,CF$ be cevians through $P$.Prove that $(PAF)+(PBD)+(PCE)=(PBF)+(PCD)+(PAE)\;\;$,where $(ABC)$ denotes the area of triangle $ABC$.[Adib showed a solution using mass point geometry but I didn't understand ]
[and also, this is my 800th post ]
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Re: Generalization of APMO-2012-1
Also, you can WLOG take $(APF) > \frac 12(ABC)$, which can disprove the claim.
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- Tahmid Hasan
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Re: Generalization of APMO-2012-1
Oops,big mistake.Now an edited version and most probably this'll be right-Let $P$ be a point in the interior of a triangle $ABC$.Let $AD,BE,CF$ be cevians through $P$.Prove that $(PAF)+(PBD)+(PCE)=(PBF)+(PCD)+(PAE)$ iff $P$ lies on one of the medians.[$(ABC)$ denotes the area of triangle $ABC$.]Tahmid Hasan wrote:Let $P$ be a point in the interior of a triangle $ABC$.Let $AD,BE,CF$ be cevians through $P$.Prove that $(PAF)+(PBD)+(PCE)=(PBF)+(PCD)+(PAE)$,where $(ABC)$ denotes the area of triangle $ABC$.[Adib showed a solution using mass point geometry but I didn't understand ]
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Re: Generalization of APMO-2012-1
Yep, the if part is true since the followings are true:
$\triangle PBD=\triangle PCD$
By Ceva,$\triangle PFB=\triangle PEC,\triangle PAF=\triangle PAE$.
$\triangle PBD=\triangle PCD$
By Ceva,$\triangle PFB=\triangle PEC,\triangle PAF=\triangle PAE$.
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- Phlembac Adib Hasan
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Re: Generalization of APMO-2012-1
When did I do so?Tahmid Hasan wrote:Let $P$ be a point in the interior of a triangle $ABC$.Let $AD,BE,CF$ be cevians through $P$.Prove that $(PAF)+(PBD)+(PCE)=(PBF)+(PCD)+(PAE)\;\;$,where $(ABC)$ denotes the area of triangle $ABC$.[Adib showed a solution using mass point geometry but I didn't understand ]
I think I was not able to give you the full idea of mass-point in a short time.Look, mass point only works with ratios.There are barely a few problems having a lot of intersecting straight lines have deal with mass point.For example, Ceva can be handled easily by mass point.In this problem there is a sum, not ratio.So here mass point asserts $(PAF)+(PBD)+(PCE)=k[(PBF)+(PCD)+(PAE)]\;\;$ where $k$ is an arbitrary positive real, which is, beyond doubt, true.There is no extra bounding to fix $k=1$, which you are saying.And areal co-ordinate method can disprove it much more easily.
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- Tahmid Hasan
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Re: Generalization of APMO-2012-1
Well I remember you telling me the outlines but as I didn't understand,I thought you gave me the solution.Sorry for the misunderstanding.And yes,'BARY'ing this problem really works magic.Phlembac Adib Hasan wrote: When did I do so?
I think I was not able to give you the full idea of mass-point in a short time.Look, mass point only works with ratios.There are barely a few problems having a lot of intersecting straight lines have deal with mass point.For example, Ceva can be handled easily by mass point.In this problem there is a sum, not ratio.So here mass point asserts $(PAF)+(PBD)+(PCE)=k[(PBF)+(PCD)+(PAE)]\;\;$ where $k$ is an arbitrary positive real, which is, beyond doubt, true.There is no extra bounding to fix $k=1$, which you are saying.And areal co-ordinate method can disprove it much more easily.
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- Tahmid Hasan
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Re: Generalization of APMO-2012-1
Sanzeed already posted the if solution and my is quite similar to that so there is no need to post it.
I've come up with a computational solution of the only if part.here it is
Let $(ABC)=S,(APF)=S_1,(BPF)=S_2,(BPD)=S_3$
,$(CPD)=S_4,(CPE)=S_5,(APE)=S_6$
Let $\frac {AF}{BF}=x,\frac {BD}{CD}=y,\frac {CE}{AE}=z$
So by Ceva's theorem,$xyz=1$
now $\frac {AF}{BF}=x$or,$\frac {AB}{BF}=x+1$[componendo]
hence,$\frac {BF}{AB}=\frac {1}{x+1}=\frac {S_2+S_3+S_4}{S}$
similarly,$\frac {DC}{BC}=\frac {1}{y+1}=\frac {S_4+S_5+S_6}{S}$
and $\frac {AE}{AC}=\frac {1}{z+1}=\frac {S_1+S_2+S_6}{S}$
adding these we get $\frac {1}{x+1}+\frac {1}{y+1}+\frac {1}{z+1}=\frac {S_1+S_2+S_6+S_4+S_5+S_6+S_2+S_3+S_4}{S}=\frac {3}{2}$
multiplying by $2(x+1)(y+1)(z+1)$ we get $2(y+1)(z+1)+2(z+1)(x+1)+2(x+1)(y+1)=3(x+1)(y+1)(z+1)$
simplifying we get $xy+yz+zx+3xyz=x+y+z+3$ or,$xy+yz+zx=x+y+z$[since $xyz=1$]
which can be written into $xyz-xy-yz-zx+x+y+z-1=0$
factorizing we get $(x-1)(y-1)(z-1)=0$
which means one of $x,y,z$ must be $1$ meaning $P$ is on one of the medians.[Proved]
I've come up with a computational solution of the only if part.here it is
Let $(ABC)=S,(APF)=S_1,(BPF)=S_2,(BPD)=S_3$
,$(CPD)=S_4,(CPE)=S_5,(APE)=S_6$
Let $\frac {AF}{BF}=x,\frac {BD}{CD}=y,\frac {CE}{AE}=z$
So by Ceva's theorem,$xyz=1$
now $\frac {AF}{BF}=x$or,$\frac {AB}{BF}=x+1$[componendo]
hence,$\frac {BF}{AB}=\frac {1}{x+1}=\frac {S_2+S_3+S_4}{S}$
similarly,$\frac {DC}{BC}=\frac {1}{y+1}=\frac {S_4+S_5+S_6}{S}$
and $\frac {AE}{AC}=\frac {1}{z+1}=\frac {S_1+S_2+S_6}{S}$
adding these we get $\frac {1}{x+1}+\frac {1}{y+1}+\frac {1}{z+1}=\frac {S_1+S_2+S_6+S_4+S_5+S_6+S_2+S_3+S_4}{S}=\frac {3}{2}$
multiplying by $2(x+1)(y+1)(z+1)$ we get $2(y+1)(z+1)+2(z+1)(x+1)+2(x+1)(y+1)=3(x+1)(y+1)(z+1)$
simplifying we get $xy+yz+zx+3xyz=x+y+z+3$ or,$xy+yz+zx=x+y+z$[since $xyz=1$]
which can be written into $xyz-xy-yz-zx+x+y+z-1=0$
factorizing we get $(x-1)(y-1)(z-1)=0$
which means one of $x,y,z$ must be $1$ meaning $P$ is on one of the medians.[Proved]
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- Tahmid Hasan
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Re: Generalization of APMO-2012-1
What a shock!!This problem appeared in USA TST 2003(Problem 2)
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