7

$G$ is the centroid of $\triangle\mathit{ABC}$. $X$, $Y$, and $Z$ are points on the sides (or lines through the sides) $\overline{\mathit{BC}}$, $\overline{\mathit{AC}}$, and $\overline{\mathit{AB}}$, respectively, such that $G$, $X$, $Y$, and $Z$ are collinear. If $Y$ is between $G$ and $Z$, \begin{equation*} \frac{1}{\mathit{GX}} = \frac{1}{\mathit{GY}} + \frac{1}{\mathit{GZ}} . \end{equation*}

How can I show this without using vectors (directed line segments) and Menelaus' Theorem?

A solution that I have seen assumes that X is a point on $\overline{\mathit{BC}}$ and constructs points U and V that trisect the side; U is half as far from B as it is from C. (We assume that X is distinct from both U or V so that the line through G and X intersects the other two sides.) Since G is half as far from the endpoint of the median on $\overline{\mathit{BC}}$ as it is from A, and since U is half as far from B as is $\overline{\mathit{GU}}$ is parallel to side $\overline{\mathit{AB}}$. Likewise, $\overline{\mathit{GV}}$ is parallel to side $\overline{\mathit{AC}}$. \begin{equation*} \frac{GX}{GZ} = \frac{UX}{BU} \qquad \text{and} \qquad \frac{GX}{GY} = \frac{VX}{CV} . \end{equation*} That is fine.

Now the solution says that $\mathit{UX} = \mathit{UZ} - \mathit{XZ}$ so that \begin{equation*} \frac{\mathit{GX}}{\mathit{GZ}} = \frac{\mathit{UX}}{\mathit{BU}} = \frac{\mathit{UZ} - \mathit{XZ}}{\mathit{BU}} \end{equation*} and \begin{equation*} \frac{\mathit{UZ}}{\mathit{BU}} = \frac{\mathit{GX}}{\mathit{GZ}} + \frac{\mathit{XZ}}{\mathit{BU}} . \end{equation*} Likewise, since $\mathit{VX} = \mathit{VZ} - \mathit{XZ}$, \begin{equation*} \frac{\mathit{VZ}}{\mathit{CV}} = \frac{\mathit{GX}}{\mathit{GY}} + \frac{\mathit{XZ}}{\mathit{CV}} . \end{equation*}

I would appreciate an elementary explanation of these last two equalities - involving only similar triangles and proportions - without mentioning $\mathit{UX} = \mathit{UZ} - \mathit{XZ}$ and $\mathit{VX} = \mathit{VZ} - \mathit{XZ}$.

Michael Rozenberg
  • 194,933
A gal named Desire
  • 251

1 Answers1

1

Let $Z$ be placed on the line $AB$ such that $A$ is placed between $Z$ and $B$, $X$ be placed on the side $BC$,

$Y$ be placed on the side $AC$, $BD$ be a median of $\Delta ABC$, $GX=x$, $GY=y$ and $GZ=z$.

Thus, $$[Z,Y,G,X]=[A,Y,D,C],$$ which says $$\frac{ZG}{ZX}:\frac{YG}{YX}=\frac{AD}{AC}:\frac{YD}{YC}$$ or $$\frac{z}{x+z}\cdot\frac{x+y}{y}=\frac{\frac{AC}{2}}{AC}\cdot\frac{YC}{YC-\frac{1}{2}AC}$$ or $$\frac{(x+y)z}{(x+z)y}=\frac{1}{2-\frac{AC}{YC}}.$$ Now, let $GE||BC$ and $E\in DC$.

Thus, $$\frac{EC}{CY}=\frac{GX}{XY}$$ or $$\frac{\frac{1}{3}AC}{CY}=\frac{x}{x+y},$$ which says $$\frac{AC}{CY}=\frac{3x}{x+y}.$$ Id est, $$\frac{(x+y)z}{(x+z)y}=\frac{1}{2-\frac{3x}{x+y}}$$ or $$\frac{z}{(x+z)y}=\frac{1}{2y-x}$$ or $$2yz-xz=xy+yz$$ or $$\frac{1}{x}=\frac{1}{y}+\frac{1}{z}$$ and we are done!

Actually. By definition, for four points $A$, $B$, $C$ and $D$ which they are placed on the same line we have: $$[A,B,C,D]=\frac{AC}{AD}:\frac{BC}{BD}.$$ You can read about this here: https://en.wikipedia.org/wiki/Cross-ratio

Michael Rozenberg
  • 194,933
  • $Y$ is a point on side $\overline{\mathit{AC}}$. So, $Y$ is between $A$ and $C$. – A gal named Desire Jul 22 '19 at 16:14
  • You said "$C$ (is) placed between $A$ and $Y$." – A gal named Desire Jul 22 '19 at 16:17
  • Also, you said "$[Z,G,Z,Y]=[A,D,C,Y]$." You must have a typographical error here. First, you have the point $Z$ twice in the expression on the left side of the equals sign. Second, you said $\overline{\mathit{BD}}$ is a median of the given triangle. That means $D$ is a point on side $\overline{\mathit{AC}}$. $Y$ is also a point on side $\overline{\mathit{AC}}$. So, all four points on the right side of the equals sign are collinear. – A gal named Desire Jul 22 '19 at 16:23
  • 1
    I am driving now. Wait please, I'll fix and explain all – Michael Rozenberg Jul 22 '19 at 16:32
  • @A gal named Desire I added something. See now. – Michael Rozenberg Jul 22 '19 at 17:33
  • The points $X$,$Y$, and $Z$ in your explanation are different than in my post. With your labeling, $Y$ is the point on the line through the centroid of $\triangle\mathit{ABC}$ and off the triangle. – A gal named Desire Jul 22 '19 at 18:53
  • That is fine - I can adapt to your labeling. – A gal named Desire Jul 22 '19 at 18:54
  • May you offer me a reference to a textbook in Geometry to "the cross-ratio of four collinear points" on the Euclidean plane? – A gal named Desire Jul 22 '19 at 18:57
  • @A gal named Desire Now I see that in my picture $X$ is placed between $G$ and $Y$ and $G$ is placed between $Z$ and $X$. If you want I am ready to write a proof, when $Y$ is placed between $G$ and $Z$. The proof that the cross ratio is the same it's just using of the law of sines. – Michael Rozenberg Jul 22 '19 at 22:43
  • Please clarify your last comment. "The proof that the cross ratio is the same" … same as what? I think you are saying that the equality between two cross ratios is derived from the Law of Sines. May you verify that assertion? There is no circle. There are only lines! – user143462 Jul 23 '19 at 19:00
  • @user143462 I meant that we can prove that $[A,B,C,D]=[A',B',C',D']$ (see the picture here: https://en.wikipedia.org/wiki/Cross-ratio ) by law of sines.I am ready to show a proof if you want. – Michael Rozenberg Jul 24 '19 at 01:55
  • @Michael Rozenberg Yes, I understand your argument. It is quite lucid, actually. I especially like your computation of $\mathit{GX}/\mathit{GY}$ using $\triangle\mathit{YGE} \sim \triangle{YXC}$! – A gal named Desire Jul 26 '19 at 16:45
  • @Michael Rozenberg This is the first time that I have seen the concept of cross ratio. Yes, I would like to see an explanation for $[A,B;C,D] = [A^{\prime}, B^{\prime}; C^{\prime},D^{\prime}]$ in the picture on Wikipedia. I asked it in the following post. – A gal named Desire Jul 26 '19 at 16:45
  • https://math.stackexchange.com/questions/3304826/cross-ratio-concept-from-euclidean-geometry – A gal named Desire Jul 26 '19 at 16:45