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The question is as follows:

If a line cuts a triangle into two pieces of equal area, must that line go through the centroid of the triangle? Explain your answer.

I stated yes because we know that the centroid (formed by the intersection of the triangle's medians) create six smaller triangles that will each have $\frac{1}{6}$ of the whole area of the triangle. If you group three of those smaller triangles that have areas of $\frac{1}{6}$ each (which can be seen on either side of a median line), then it will be equal to $\frac{1}{2}$ and the other side will also be $\frac{1}{2}$ of the whole area of the triangle.

Yet I still can't help but second-guess my answer. Is there any other point, other than the centroid, that a triangle can be divided into two polygons of equal area? I have seen posts on StackExchange regarding this problem, but I was not able to understand its complexities for I am just a high-schooler. Any help will be greatly appreciated.

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  • $\begingroup$ Think about a line parallel to an edge of the triangle and cutting it into two parts of equal area. $\endgroup$ – Lord Shark the Unknown Oct 7 '17 at 17:24
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This might not always be true

Consider the following case. Line $EF||BC$ and $AGD$ is perpendicular to $BC$. Assume that area $AEF$ is equal to area $EFCB$.

enter image description here

Let $AE/EB=AF/FC=AG/GD=x:1$. Then, $EF:BC=x:(x+1)$ (why?).

Now, area $AEF$ is half of area $ABC$ (why?) So, $$1/2\cdot AG\cdot EF=1/4\cdot AD \cdot BC\rightarrow 2x^2=(x+1)^2$$ Notice $x\neq 2$

Now, construct a median $AIH$ on $BC$:

enter image description here

If line $EGF$ passes through a centroid, then the centroid must be the point $I$ (why?).

But, since $AI:IH=AG:GD=x\neq2$, hence, line $EGF$ does not pass through a centroid, while still dividing $\triangle ABC$ into two equal areas.

QED.

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  • $\begingroup$ So there are at least three lines through the centroid cutting the triangle in half.... $\endgroup$ – Lord Shark the Unknown Oct 7 '17 at 17:25
  • $\begingroup$ You solved a particular case of the stated problem: just the case that the line must pass through a vertex. $\endgroup$ – Andrei Oct 7 '17 at 17:27
  • $\begingroup$ I agree, i am looking into other solutions. $\endgroup$ – Gaurang Tandon Oct 7 '17 at 17:29
  • $\begingroup$ Solution has been updated $\endgroup$ – Gaurang Tandon Oct 7 '17 at 17:46
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Let $O$ be the centroid of triangle $ABC$, and let $L$ be the line through $O$ parallel to $BC$. Let $L$ meet $AB$ at $P$ and $AC$ at $Q$. Then $|AP|=\frac23|AB|$ (why?) and $|AQ|=\frac23|AC|$. So the area of triangle $APQ$ is $\frac49$ of that of $ABC$.

So, if you have a line $L'$ parallel to $BC$ bisecting the triangle's area, it won't pass through $O$.

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  • $\begingroup$ I still don't understand how you came to the conclusion that "line $L'$ is parallel to $BC$" and it bisects the triangle's area but it won't go through O. $\endgroup$ – geo_freak Oct 7 '17 at 17:36
  • $\begingroup$ @geo_freak Because if it did pass through $O$ it would be $L$. $\endgroup$ – Lord Shark the Unknown Oct 7 '17 at 17:37
  • $\begingroup$ So are you saying that the line must go through the centroid to bisect the triangle's area? I am very confused... $\endgroup$ – geo_freak Oct 7 '17 at 17:38
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Let $h$ be a height of a triangle $\triangle ABC$ from vertex $A$ to a base $BC$. Draw a line paralel to $BC$ passing a centroid of the triangle. The line is $\frac 13h$ apart from $BC$; however, the line which halves the triangle's area is $(1/\sqrt 2)h$ apart from $A$, so it's $(1-1/\sqrt 2)h$ apart from $BC$.

Hence the two lines are different.

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  • $\begingroup$ Thanks so much! $\endgroup$ – geo_freak Oct 7 '17 at 18:30

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