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The medians of a triangle intersect in a trisection point of each.

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  • $\begingroup$ You should see the pure geometry proof. The proof analytically is easier than the pure geometry proof. $\endgroup$ – Inceptio Jun 29 '13 at 8:33
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    $\begingroup$ @Inceptio the proofs using vectors depend on the fact that vectors encode information about the Euclidean Plane. It is quite easy to prove trisection using similar triangles. The fact that certain lines are parallel and certain angles are equal is implicit in the fact that we can describe the situation using vectors. $\endgroup$ – Mark Bennet Jun 29 '13 at 9:49
  • $\begingroup$ @MarkBennet: I agree with you. But the idea of Analytical geometry was invented only to ease the difficulties of pure geometry(Not this question in particular). I might be wrong, but whole of conic section would be very interesting without the Cartesian-plane. :) $\endgroup$ – Inceptio Jun 29 '13 at 10:01
  • $\begingroup$ @Inceptio Conics were around long before Descartes. $\endgroup$ – Mark Bennet Jun 29 '13 at 10:07
  • $\begingroup$ @MarkBennet: Yes, Apollonius and his students had even given many geometrical proof for his theorems. But later, application of the Cartesian plane over conics made the topic uninteresting. $\endgroup$ – Inceptio Jun 29 '13 at 14:24
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Let the vertices of the triangle be at the points (vectors) $u$, $v$, and $w$.

The midpoint $m$ of the side joining $u$ and $v$ is $\frac{u+v}{2}$.

The point on the line segment $mw$, one-third of the way up from $m$ towards $w$, is $$\frac{2m}{3}+\frac{w}{3}.\tag{1}$$

Substitute $\frac{u+v}{2}$ for $m$ in (1). We get $$\frac{1}{3}(u+v+w).$$ This expression is symmetric in $u$, $v$, and $w$, so lies on all three medians.

Remark: To make the proof look more "analytic" we could uglify it. Let the coordinates of the vertices be $(u_x,u_y)$, $(v_x,v_y)$, $(w_x,w_y)$. Then as above we compute the coordinates of $m$, and then of trisection point.

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    $\begingroup$ nice approach $\large{+1}$ $\endgroup$ – M.H Jun 29 '13 at 6:45
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Hint: Consider the analytic formulas relating the $\color{#00A000}{\text{vertices}}$, $\color{#C00000}{\text{midpoints}}$, and $\color{#0000FF}{\text{median}}$:

$\hspace{3.5cm}$enter image description here

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I can propose the following evidence. Let $ABC$ be an arbitrary triangle of a Cartesian plane. Let its vertices $A$, $B$, and $C$ have the coordinates $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3, y_3)$ respectively. Then the midpoints $A_1$, $B_1$, and $C_1$ of the sides $BC$, $CA$, and $AB$ respectively have the coordinates $(\frac{x_2+x_3}2, \frac{y_1+y_3}2)$, $(\frac{x_1+x_3}2, \frac{y_1+y_3}2)$ and $(\frac{x_1+x_2}2, \frac{y_1+y_2}2)$ respectively. Shifting, if necessary, the zero $O$ of the Cartesian plane, without loss of generality, we may assume that $x_1+x_2+x_3=y_1+y_2+y_3=0$. Then all the determinants

$\left|\begin{matrix} x_1 & y_1 \\ \frac{x_2+x_3}2 & \frac{y_2+y_3}2\\ \end{matrix} \right|,$ $\left|\begin{matrix} x_2 & y_2 \\ \frac{x_1+x_3}2 & \frac{y_1+y_3}2\\ \end{matrix} \right|$, and $\left|\begin{matrix} x_3 & y_3 \\ \frac{x_1+x_2}2 & \frac{y_1+y_2}2\\ \end{matrix} \right|$

are equal to zero. Therefore all the triangles $AOA_1$, $BOB_1$, and $COC_1$ have zero area. Therefore $O$ is the intersection point of the medians of the triangle $ABC$. Moreover, ${\bf AO}=(-x_1,-y_1) =2(\frac{x_2+x_3}2, \frac{y_1+y_3}2)=2{\bf OA_1}$. Similarly, ${\bf BO}=2{\bf OB_1}$ and ${\bf CO}=2{\bf OC_1}$. Thus $O$ is the trisection point of each of the segments $AA_1$, $BB_1$, and $CC_1$.

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Here is the ugly solution:

First, note that $a,b,c$ are affinely independent, that is, $\binom{1}{a},\binom{1}{b},\binom{1}{c}$ are linearly independent. (Otherwise the three points line on a line.)

Pick the vertices $a,b$ first, and the corresponding midpoints on the opposite sides. We want to find the intersection of the medians, that is, find $\lambda_1, \lambda_2$ such that $$(1-\lambda_1)a + \frac{\lambda_1}{2}b + \frac{\lambda_1}{2}c = \frac{\lambda_1}{2}a + (1-\lambda_1)b + \frac{\lambda_1}{2}c$$ Rearranging gives $$ a(1-\lambda_1-\frac{\lambda_2}{2}) + b(\frac{\lambda_1}{2}-(1-\lambda_2))+c(\frac{\lambda_1}{2}-\frac{\lambda_2}{2}) = 0$$ A quick computation shows that $$1(1-\lambda_1-\frac{\lambda_2}{2}) + 1(\frac{\lambda_1}{2}-(1-\lambda_2))+1(\frac{\lambda_1}{2}-\frac{\lambda_2}{2}) = 0$$ from which we get the coefficients are all zero (because of affine independence). Solving gives $\lambda_1 = \lambda_2 = \frac{2}{3}$, and hence the intersection is $\frac{1}{3}(a+b+c)$. Repeating with the other two pairs results in the same answer.

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  • $\begingroup$ The formatting may not intentional. I guess. $\endgroup$ – silentboy Aug 3 '15 at 4:51
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The medians of an equilateral triangle must trisect by symmetry. Any other triangular shape can be found by performing a linear transformation on the vertices of the equilateral triangle, and such transformations preserve the medians along with their trisection point.

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