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Let $ABCD$ be a parallelogram and $G$ the center of gravity(the intersection point of the medians) for the $\triangle ABC$. $M \in AD$ and $D \in NC$.

Prove that $G,M,N$ are collinear if and only if $\displaystyle \frac{CN}{ND}-\frac{AM}{MD}=\frac{1}{2}.$

I don't know anything about this problem, it seems to hard for me. How can I use the notion of collinearity knowing that raport ?

enter image description here Thanks!

Iuli
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  • What do you call "medians" to in a parallelogram? Perhaps you meant "diagonals"? – DonAntonio Jan 11 '13 at 13:22
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    Do you mean "$N \in DC$" (instead of "$D\in NC$")? If not, then where is $N$? – Blue Jan 11 '13 at 13:43
  • Reminds me of this too much, although it can't be used directly http://en.wikipedia.org/wiki/Menelaus%27_theorem – Bojan Serafimov Jan 11 '13 at 12:17
  • @Bojan: I made your comment a, well, comment. – Willie Wong Jan 11 '13 at 13:44
  • @DonAntonio A median in a quadrilateral connects the midpoints of two opposite sides. – Maesumi Jan 11 '13 at 13:53
  • @Maesumi, thanks. I call that "midline or midsegment", as "median" in a triangle is the segment joining a vertex with the middle point of the opposite side. – DonAntonio Jan 11 '13 at 14:09
  • This is odd: I, just as blue, think it should have been $,N\in CD,$ , yet then $,G,M,N,$ cannot not be collinear in any case, as $,G,$ isn the half parallelogram determined by $,AC,$ whereas $,M,N,$ are on the other half and in adjacent sides. – DonAntonio Jan 11 '13 at 14:14
  • Perhaps $,M\in AC,$...? – DonAntonio Jan 11 '13 at 14:15
  • @DonAntonio You are right. The usage apparently is not standard. So sometimes it is called median, sometimes bimedian. At any rate the problem says $G$ is center of triangle $ABC$, not of the parallelogram. – Maesumi Jan 11 '13 at 14:40
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    Point $N$ is used before being defined. Statement $D \in NC$ needs correction. – Maesumi Jan 11 '13 at 14:42
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    If $N$ is on $CD$ extended beyond the parallelogram then $G, M, N$ can be collinear and $D \in NC$. – Peter Phipps Jan 11 '13 at 14:45
  • I think we shall wait until Iuli addresses all these doubts...perhaps, as has happened many times, there's some mistake(s) in the question. – DonAntonio Jan 11 '13 at 15:10
  • @DonAntonio No, no . there are not mistake. I verified in the book and it's ok how I wrote it. – Iuli Jan 11 '13 at 18:54
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    Well @Freddy, "everything's fine" now that you gave your personal interpretation and explanation, *not before" when we had to guess what/where is N...The OP is supposed to pose clear questions, it's not other members' task to guess. – DonAntonio Jan 11 '13 at 20:53
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    No @Iule, it's not OK as you wrote it at the beginning. NOW, after you added a diagram, it is ok and clear. – DonAntonio Jan 11 '13 at 20:57

2 Answers2

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The setup of the problem is "affine invariant", which implies that we may assume $ABCD$ to be a square in the $(x,y)$-plane. In other words, we may introduce coordinates $(x,y)$ such that $$A=(0,0), \ B=(1,0), \ C=(1,1), \ D=(0,1), \ N=(u,1), \ M=(0,v)\ ,$$ where the numbers $u$, $v\in{\mathbb R}$ have certain values. It follows that $G=\bigl({2\over3},{1\over3}\bigr)$ and therefore $$\vec{GM}=\left(-{2\over3}, v-{1\over3}\right),\quad \vec{GN}=\left(u-{2\over3}, {2\over3}\right)\ .$$ The three points $G$, $M$, $N$ are in line iff the "vector product" $\vec{GM}\wedge\vec{GN}=0$, which means that $u$ and $v$ have to satisfy $$-uv +{2\over3} v+{1\over3}u-{2\over3}=0\ .\qquad(*)$$ On the other hand $${CN\over ND}={u-1\over -u}\ ,\quad {AM\over MD}={v\over 1-v}\ ,$$ and it is easily verified that ${CN\over ND}-{AM\over MD}={1\over2}$ is equivalent with $(*)$.

Christian Blatter
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Let the line through $G$ parallel to $AB$ intersect $BC$ at $X$, and the line through $G$ parallel to $BC$ intersect $AB$ at $Y$. Mark the length of $GX$ as $a$, and the length of $GY$ as $b$. Its easy to prove that $a = AB/3$ and $b = BC/3$ by solving the system of equations of lines AG and CG.

Let $T$ be the intersection of $GY$ and $CD$. By using the similar triangles $NDM$ and $NTG$ we get:

$\frac{ND}{MD} =\frac{NT}{GT} =\frac{ND + 2a}{2b} = \frac{ND}{2b} + \frac{a}{b}$

multiply by: $\frac{3b}{ND}$

$\frac{3b}{MD} = \frac{3}{2} + \frac{3a}{ND}$ (1)

Rearranging the original equation and using (1) we get:

$\frac{CN}{ND} - \frac{AM}{MD} = \frac{3a + ND}{ND} - \frac{3b - MD}{MD} = \frac{3a}{ND} - \frac{3b}{MD} + 2 = \frac{1}{2}$

This was the 'only if' part of the 'if and only if'. For the 'if' part, because we only used equalities, and every equation was equivalent to the previous, we can say that from

$\frac{CN}{ND} - \frac{AM}{MD} = \frac{1}{2}$

follows that

$\frac{ND}{MD} =\frac{NT}{GT}$

This, along with the fact that the corresponding angles are equal due to parallel lines, shows that triangles $NDM$ and $NTG$ are similar.

From the similarity, we get that the angle $\angle MND = \angle GNT$.

$M$, $D$, and $T$ are collinear, so $\angle GND = \angle GNT$.

From the last two sentences: $\angle MND = \angle GND$, which is a proof that $N$, $M$, and $G$ are collinear.

Bojan Serafimov
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