Let $ABC$ be an arbitrary triangle with vertices $A,B,C$. Denote by $M_{AB}$ the midpoint of the line $AB$, and analogously define $M_{AC}, M_{BC}$. The center/ barycenter /center of mass $Z$ is defined as common intersection of lines $AM_{BC}, BM_{AC}$ and $CM_{AB}$.

Question: Is there a conceptual method (= closed expression in dependence of angles and side lengths of $ABC$) to calculate e.g. the angles of "type" $\angle ZAB$ and of type $ \angle M_{AC}ZA $? Is it even possible to derive it using synthetic geometry only, i.e. coordinate free & without analytic expressions à la sin, cos, etc.? Even though it looks rather elementary I haven't found a way to approach it.

#ADDED LATER: The latter issue on the possibility to derive it by means of synthetic geometry (esp. without making use of analytic expressions) appears to me rather subtle. For example, it is rather easy to determine corresponding angles synthetically if we would e.g. instead consider construction of subdivision of $ABC$ into say orthic triangle, or say subdivision triangles coming in analogous manner from the intersection point of bisectors.

In both latter cases the angles of appearing triangles in these subdivisions can be easily determined by playing with synthetic methods. But even on heuristic level, why does it appear to be "unplausible" to have such nonanalytic expressions for angles of subdivision triangles constructed with centroid?

So the metaquestion becomes why subdividing a triangle via centroid point is the only " canonical" subdivision where the angles of obtained subtriangles cannot be expressed without analytic expressions?

Rmk.: By " canonical" i mean arising from "choiceless constructions", eg we not subdivide the triangle randomly into subtriangles, but by construction based on "canonical" choices, eg form bisection lines and intersect them.
For instance, choosing bisection line of an angle is "canonical" choice, choosing a line dividing an given angle at say - randomly picked $35$ degrees - is not canonical, it's kind of "taste depending", another person would maybe pick another degree at which it would like to split the angle.

  • $\begingroup$ Can you be more specific with what you mean by 'conceptional'? $\endgroup$
    – AnCar
    Commented May 21 at 22:18
  • 1
    $\begingroup$ @AnCar: as closed expressions depending of the angles and side lengths of $ABC$. Thanks for bringing this to attention, I should have specified it $\endgroup$
    – user267839
    Commented May 21 at 22:25
  • $\begingroup$ So the line $AM_{BC}$ is perpendicular to $A$, and similarly for the others? $\endgroup$ Commented May 21 at 22:31
  • $\begingroup$ @H.sapiensrex: No, why? How can a line be perpendicular to a point? See en.m.wikipedia.org/wiki/Centroid for this construction $\endgroup$
    – user267839
    Commented May 21 at 22:35
  • $\begingroup$ Whoops, my mistake. I thought $A,B,C$ referred to the sides $\endgroup$ Commented May 21 at 22:48

2 Answers 2


Let assume that we have a system of coordinates with $A=(0,0)$, $B=(x_b,0)$, $C=(x_c,y_c)$, then the coordinates of the centroid $Z$ are given by

$$ Z=\frac{1}{3}(x_b+x_c,y_c) $$

So it is possible to calculate all the angles by calculating the angolar coefficient of the various lines, e.g. the angle $\angle ZAB$ is given by $$ \angle ZAB=\arctan\left(\frac{y_c}{x_b+x_c}\right) $$

Synthetic method

Let call $a=AB$, $b=BC$, $c=AC$,

I will focus on $\angle ZAB$, the other angles can be obtained in similar way. By using the formula for the medians we have $$ m_b=\sqrt{2a^2+2c^2-b^2}/2 $$ $$ m_c=\sqrt{2a^2+2b^2-c^2}/2 $$ By the properties of the centroid we have $$ ZA=\frac{2}{3} m_b $$ $$ ZB=\frac{2}{3} m_c $$

By using the law of cosine to the triangle $\triangle AZB$ we obtain

$$ \angle ZAB=\arccos\left(\dfrac{a^2+(ZA)^2-(ZB)^2}{2a(ZA)}\right) $$ or $$ \angle ZAB=\arccos\left(\dfrac{3a^2+c^2-b^2}{2a\sqrt{2a^2+2c^2-b^2}}\right) $$

  • $\begingroup$ Do you maybe know if it is posible to derive it only with synthetic methods? (especially coordinate free & without analytic expressions (arctan, sin, ...))? $\endgroup$
    – user267839
    Commented May 21 at 22:45
  • 1
    $\begingroup$ @user267839 I added the synthetic method proof, but still, as there is no right angle (orthocenter) and the angles are not split in half (incenter) an $\arccos$ appears as a consequence of the cosine law $\endgroup$
    – Marco
    Commented May 22 at 11:42
  • 1
    $\begingroup$ aren't both $ZA$ and $ZB$ always $2/3$ of their medians? see eg en.m.wikipedia.org/wiki/Centroid#Of_a_triangle (...so I not understand what you mean by switch between $1/3$ and $2/3$) ...Or did I misunderstood your notation? $\endgroup$
    – user267839
    Commented May 22 at 12:55
  • 1
    $\begingroup$ ...Also a philosophical note: I think that the analytic expressions appear every time in triangle geometry when we pass between involved angles and proportions of side lengths. In your proof the $\angle ZAB$ angle is expressed in terms of lengths of sides. Of course we could also try to express it in terms of angles of $ABC$ but this procedure would reduce to: (1): express side lengths in terms a scalar factor times sine, cosine fractions of angles (2): insert these in your formula. The problem is that the arccos is not "neutralized" by sine and cosine expressions encoding the $\endgroup$
    – user267839
    Commented May 22 at 13:32
  • 1
    $\begingroup$ Do you know Conway triangle notation ? $\endgroup$
    – Jean Marie
    Commented May 23 at 21:00


enter image description here

After sketching the median $m_a$ solve the half triangle using Sine Rule of sides

$$(c,a/2,m_a=\sqrt{2b^2+2 c^2-a^2}/2)$$

for the required angles.

  • $\begingroup$ I don't understand the meaning of your answer. $\endgroup$
    – Jean Marie
    Commented May 23 at 21:01
  • 1
    $\begingroup$ I lefts out some explanation in the sketch containing the median $m_a$. $\endgroup$
    – Narasimham
    Commented May 24 at 23:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .