Let $X,\ Y$ and $Z$ be the midpoints of sides $BC,\ CA$, and $AB$ of the triangle $ABC$, respectively. Let $P$ be a point inside the triangle. Prove that the quadrilaterals $AZPY,\ BXPZ$, and $CYPX$ have equal areas if, and only if, $P$ is the centroid of $ABC$.
Problem
Source: Gulf MO 2012, Problem 1
Tags: geometry, analytic geometry, geometry proposed
WakeUp
14.06.2012 15:30
If the areas are equal clearly $P$ lies in the interior of $\triangle XYZ$. If it isn't, it is on the segments $(XY),(YZ),(ZX)$ or in the interior of one of the triangles $AYZ,BZX,CXY$. If it is on one of the segments $(XY),(YZ),(ZX)$ (say on the segment $(XY)$) then $[CYPX]=[CYX]=\frac{1}{3}[ABC]$. But $P$ lies on the opposite side of the segment $(YZ)$ to $A$ hence $[AZPY]=[AZY]+[ZPY]=\frac{1}{3}[ABC]+[ZPY]>\frac{1}{3}[ABC]>[CYPX]$, a contradiction. If $P$ lies in the interior of one of the triangles $AYZ,BZX,CXY$ (again suppose it is $CXY$) then use the same argument, except $[AZPY]>\frac{1}{3}[ABC]>[CYPX]$ to reach another contradiction. Hence $P$ lies in the interior of triangle $XYZ$. Hence $[AZPY]=[BXPZ]=[CYPX]$ is equivalent to $[PXY]=[PYZ]=[PZX]$. Areal coordinates now turn this into algebra. If $P$ has areal coordinates $(k,l,m)$ then we find that $b+c-a=c+a-b=a+b-c=\frac{1}{3}$ which gives $a=b=c=\frac{1}{3}$ so $P$ is the centroid of triangle $XYZ$ and hence the centroid of triangle $ABC$. Conversely if $P=G$ then the areas of the quadrilaterals are clearly equal.
Mahi
15.06.2012 05:55
A shorter solution: The if part is trivial. For the only if part, we know that $[AZY]=[BZX]=[CXY]= \frac 14 [ABC]$ so the equality in the statement implies $[PZY]=[PZX]=[PXY]$ and thus $P$ is the centroid of $XYZ$. But centroid of $\triangle ABC$ and $\triangle XYZ$ is the same, which implies $P$ is the centroid of $\triangle ABC$. (If the part "$P$ is inside $XYZ$" is necessary-) $ [AZPY]=\ [BXPZ] =\ [CYPX] = \frac 13[ABC] > [AZY],[BZX],[CXY]$ so all the areas $[PXY],[PYZ],[PZX]$ are positive.
Particle
16.06.2012 14:35
Let $G$ is the centroid of $ABC$.For $P=G$ it's very simple to show $(AZPY)=(BXPZ)=(CYPX)$.(Here $(AZPY)$ denotes area of $AZPY$.)So this part is skipped.
Lemma:Let $P$ and $P'$ be two points inside triangle $ABC$.Let the distance between $P$ and $BC$ is $d_1$ and between $P$ and $CA$ is $d_2$.Also let the distance between $P'$ and $BC$ is $d_1'$ and between $P'$ and $CA$ is $d_2'$.If $d_1=d_1'$ and $d_2=d_2'$ then $P=P'$.
Proof:Draw parallel lines passing through $P$ to $BC$ and $CA$.The lemma is straightforward now.
Drop perpendiculars form $P$ to $BC,CA,AB$ at points $X',Y',Z'$,respectively.Furthermore drop perpendiculars form $G$ to $BC,CA,AB$ at points $X'',Y'',Z''$,respectively.As $X$ is the mid-point of $BC$, so $(BPX)=(CPX)=x$ and similarly $(CPY)=(APY)=y$, $(APZ)=(BPZ)=z$.The given condition implies now $x+y=x+z=z+x$.It follows directly $x=y=z$.So $x=y=z=\frac {1}{6}(ABC)=(BXG)=(CYG)$.So $PX'=GX''$ and $PY'=GY''$.Now from our lemma $P=G$ as desired.
Mahi
16.06.2012 17:43
Particle wrote: Lemma:Let $P$ and $P'$ be two points inside triangle $ABC$.Let the distance between $P$ and $BC$ is $d_1$ and between $P$ and $CA$ is $d_2$.Also let the distance between $P'$ and $BC$ is $d_1'$ and between $P'$ and $CA$ is $d_2'$.If $d_1=d_1'$ and $d_2=d_2'$ then $P=P'$. It can also be obtained (easily in my opinion) by SAS congruence.
mavropnevma
16.06.2012 17:54
Another, slightly different, approach. Once it is argued $P$ must be interior to $\triangle XYZ$ (because the areas of the quadrilaterals are $\frac {1} {3}$ of the area of $\triangle ABC$, while the areas of the triangles $\triangle AYZ$, $\triangle BZX$, $\triangle CXY$ are $\frac {1} {4}$ of the area of $\triangle ABC$), hence the areas of the triangles $\triangle XPY$, $\triangle YPZ$, $\triangle ZPX$ must be equal, notice that such a point is uniquely determined, since it has to lie on parallels at determined distances from the sides of $\triangle XYZ$. Since the centroid $G$ satisfies, it must be that $P$ coincides with $G$.