Problem: Show that, in a convex hexagon, there exists a diagonal which cuts off a triangle of area not more than one-sixth of the hexagon.

My attempt: Suppose we have a hexagon $ABCD$. There are two possible cases: either the main diagonals are concurrent, or they are not. enter image description here

If the main diagonals $AD, BE, CF$ concur at a point $G$, then the main diagonals cut the hexagon into $6$ triangles, atleast one of which has area $\leq \frac 16 [ABCDEF]$ Suppose one such triangle is $DEG$. Thus one of the triangles $DEF$ or $DEC$ has area $\leq[DEG]$, and we are done.

But suppose the main diagonals are not concurrent, i.e., they form a triangle $PQR$. How can I prove the statement in this case?

  • $\begingroup$ Choose any of the 3 points p,q,r. Everything about the rest of your argument holds. $\endgroup$
    – Doug M
    May 19, 2016 at 6:48
  • $\begingroup$ @DougM: No it doesn't. Suppose he chooses R. Then FER certainly has area leas than a sixth of the entire hexagon, but the argument that one of AFE or FED will have smaller area doesn't work because ARD are now not collinear. $\endgroup$ May 19, 2016 at 7:06
  • $\begingroup$ How to rigorously prove that one of the triangles has an area less than or equal to the triangle which is less than one sixth the area ? Is it because the point $G$ lies on a straight line with the other two and hence the area of a triangle must be maximum and minimum at one of the end points ? $\endgroup$
    – Saikat
    May 17, 2017 at 16:33
  • $\begingroup$ @user230452 because the hexagon is partitioned into 6 disjoint triangles. If all triangles have $\geq \frac{1}{6}[ABCDEF]$ area then the cumulative area will exceed $[ABCDEF]$. Hence atleast one of them will have have an area less than one-sixth $\endgroup$ May 19, 2017 at 15:30
  • $\begingroup$ I understood that. I was asking about the other part where you proved one of the diagonals cuts of a smaller or equal area ... I got it cleared in the comment section of the answer. $\endgroup$
    – Saikat
    May 20, 2017 at 3:07

1 Answer 1


Consider the six triangles ABQ, BCQ, CDR, DER, EFP, FAP. They are disjoint and cover an area smaller than the entire hexagon. And each of them reaches up to a diagonal that doesn't touch their base.

hexagon diagram with colors

  • $\begingroup$ Can you please explain a bit more ? I didn't understand how to find such a diagonal because all the triangles you described require two diagonals to make ... $\endgroup$
    – Saikat
    May 17, 2017 at 16:31
  • $\begingroup$ @user230452: Suppose we find that DER has area less than $\frac 16$ of the hexagon. Then, depending on the relative angles between ED and FC, either FED or CED will have even smaller area than DER. And each of FED and CED is cut off from the hexagon by a single diagonal. (This reasoning was not in the answer, because it closely mimics what the OP was already doing in his Case I). $\endgroup$ May 17, 2017 at 17:32
  • $\begingroup$ Actually, I didn't understand Case 1. Can you explain how to rigorously prove that one of those triangles has an area less than or equal to the triangle chosen ? Is it because when a side of a triangle is kept constant, and the third point is varied along a line, the area must be maximal and minimal at it's end points or be equal throughout the line ? $\endgroup$
    – Saikat
    May 18, 2017 at 2:24
  • $\begingroup$ @user230452: Let's consider DER. The area of a triangle whose one side is DE is the length of DE times half the perpendicular distance from the third corner to DE (produced indefinitely in both directions). Now as we move along the line FRC, the perpendicular distance to any fixed line (such as DE) will vary monotonically, so F and C cannot both be farther from DE than R is. So either DEF or DEC has area $\le$ that of DER. $\endgroup$ May 18, 2017 at 10:39
  • $\begingroup$ [Note that this property of perpendicular distances between a line and points on another line is not true in hyperbolic geometry -- and indeed, in the hyperbolic plane there are hexagons that don't have the property being proved here. So a rigorous proof needs to depend on the parallel postulate somehow]. $\endgroup$ May 18, 2017 at 10:52

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