Centroid and area

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Source: Gulf MO 2012, Problem 1

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Malik

#1 h Jun 14, 2012, 8:16 AM • 1 Y

Y by Adventure10

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 $ABC$ .

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WakeUp

#2 h Jun 14, 2012, 12:30 PM • 1 Y

Y by Adventure10

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.

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52 posts

Mahi

#3 h Jun 15, 2012, 2:55 AM • 2 Y

Y by Adventure10, Mango247

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.

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#4 h Jun 16, 2012, 11:35 AM • 2 Y

Y by Adventure10, Mango247

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.
my proof

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Mahi

#5 h Jun 16, 2012, 2:43 PM • 1 Y

Y by Adventure10

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.

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15142 posts

#6 h Jun 16, 2012, 2:54 PM • 2 Y

Y by Adventure10, Mango247

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 $G$ .

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