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Consider triangle $ABC$ with base $AB$ fixed in length and position. As the vertex $C$ moves on a straight line (which hasn't to be parallel to line $AB$),what is the locus of the centroid of the triangle ?

My effort:

After a little bit of playing i see that the locus of the centroid is a line parallel to the base $AB$,but how would i prove it with means of synthetic geometry .

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    $\begingroup$ In the question stated, you seemed to have the assumption that the line $C$ moves along is parallel to $AB$. It would be better to state it clearly in the question. $\endgroup$
    – Element118
    Oct 23 '15 at 11:13
  • $\begingroup$ I've taken it as it is written on my book, btw i'm editing now . $\endgroup$
    – Nameless
    Oct 23 '15 at 11:19
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    $\begingroup$ If "the vertex $C$ moves on a straight line (which hasn't to be parallel to line $AB$", then "triangle $ABC$ has always the same area" is not true. $\endgroup$
    – mathlove
    Oct 23 '15 at 11:24
  • $\begingroup$ I see ...I've misinterpreted the problem,how would i prove it then with basic geometry ? $\endgroup$
    – Nameless
    Oct 23 '15 at 11:34
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$$ G = \frac{A+B+C}{3} = \frac{A+B}{3}+\frac{C}{3}, $$ hence if $A,B$ stay fixed and $C$ moves on some line $l$, $G$ moves on a line parallel to $l$.

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Geometrical Approach: Let the fixed vertices of $\triangle ABC$ be $A(0, 0)$ (at the origin) & $B(a, 0)$ be lying on the x-axis & the vertex $C(x, b)$ is movable parallel to the base $AB$ say on the line $y=b$

where, $x$ is variable & $a, b$ are arbitrary constants.

then the centroid $G(h, k)$ of $\triangle ABC$ is given as $$(h, k)\equiv\left(\frac{0+a+x}{3}, \frac{0+0+b}{3}\right)\equiv \left(\frac{x+a}{3},\ \frac{b}{3}\right)$$ Comparing the corresponding coordinates, we get $h=\frac{x+a}{3}$ & $k=\frac{b}{3}$ which shows that the locus of the centroid $G(h, k)$ is a straight line $\color{red}{y=\frac{b}{3}}$ which is parallel to the fixed base $AB$ i.e. the centriod $G$ moves on a straight line parallel to the fixed base $AB$

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There is another way to think about this question.

Let $E$ be the midpoint of side $BC$.

Using Ceva's theorem on $\triangle CDB$ and line $AXE$,

$\frac{CX}{XD}\frac{DA}{AB}\frac{BE}{EC}=1$

$\frac{CX}{XD}=\frac{2}{1}$ (which is a well-known result concerning medians)

Another way this can be stated is: $\frac{CD}{XD}=\frac{3}{1}$

It follows that a scaling of factor $\frac{1}{3}$ with centre $D$ will bring the locus of $C$ to the locus of $X$.

Hence the locus of $X$ is a straight line parallel to the locus of $C$ where the distance from the locus of $X$ to $D$ is $\frac{1}{3}$ the distance from the locus of $C$ to $D$.

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  • $\begingroup$ How would i prove it without Ceva's Theorem ? $\endgroup$
    – Nameless
    Oct 23 '15 at 11:35
  • $\begingroup$ Well, we need to get some information about ratio about lengths, so without Ceva's theorem, I don't think this proof can be done. Menelaus' theorem is another related theorem which can fix this gap. $\endgroup$
    – Element118
    Oct 23 '15 at 11:43

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