Starting with a convex pentagon P, we define the "middle polygon" Q, whose vertices are the middle points of the sides of the initial pentagon. The ratio between the areas of this polygons seem to always satisfy: 1/2 < area(Q)/area(P) < 3/4

The lower bound is easy to obtain. I don't see how to get the upper bound.

This problem is equivalent to the following one. Just forget about the middle polygon for a moment. Start with a convex pentagon and consider also all his 5 diagonals. You will obtain a central pentagon. Prove that the area of the new central pentagon is less than the sum of the areas of the five small triangles which have a side adjacent to the sides of this central polygon.

  • 5
    $\begingroup$ Your "solution" deals with the regular pentagon case. The question asked is for any convex pentagon. $\endgroup$ – Manuel Silva Nov 8 '09 at 3:50
  • 13
    $\begingroup$ Can someone explain to me why the geometry of plane pentagons is not a fit subject for mathematical research? And how does your argument apply to papers such as jstor.org/pss/3647745 ? $\endgroup$ – David Eppstein Nov 8 '09 at 4:17
  • 4
    $\begingroup$ I think there is a danger of overdoing the "not appropriate here" smackdowns. The present problem is probably easily answered by some arguments from plane geometry; one past problem could have easily be answered by "Schanuel's lemma does the trick". Not convinced that we should be too dogmatic/hasty about depth $\endgroup$ – Yemon Choi Nov 8 '09 at 8:12
  • 5
    $\begingroup$ I would like to add that there is very interesting differential geometry/dynamics surrounding the study of precisely this sort of problem as a dynamical system. For instance, if you do this projectively, you can think that you are not drawing a new smaller pentagon, (so the limiting pentagon as you iterate becomes smaller and smaller) but replacing the original. The limit of such systems is studied by people like Serge Tabachnikov at Penn State, and is very interesting. $\endgroup$ – David Jordan Nov 9 '09 at 14:56
  • 4
    $\begingroup$ The inequality is stated on page 412 of the the book "Recent Advances in Geometric Inequalities" by Mitrinovic, Pecaric, and Volenec (which is available online). They state it without a proof as a theorem of Skljarskiy, Cencov, and Jaglom. The reference they give is a russian(?) book or article which I unfortunately cannot find. $\endgroup$ – Philipp Lampe Nov 16 '09 at 18:18

I used qepcad to compute that the intersection of the set of possible area ratios with the interval [1/2, 3/4] is (1/2, 3/4). Since the set of possible area ratios is the image of a connected space under a continuous function, and we know the set contains (1/2, 3/4), but not 1/2 or 3/4, it must equal (1/2, 3/4). Here is a log of the qepcad session.

                Quantifier Elimination                 
            Elementary Algebra and Geometry            
      Partial Cylindrical Algebraic Decomposition      

               Version B 1.53, 16 Jul 2009

                       Hoon Hong                       

With contributions by: Christopher W. Brown, George E. 
Collins, Mark J. Encarnacion, Jeremy R. Johnson        
Werner Krandick, Richard Liska, Scott McCallum,        
Nicolas Robidoux, and Stanly Steinberg                 
Enter an informal description  between '[' and ']':
[ area of middle pentagon ]
Enter a variable list:
Enter the number of free variables:
Enter a prenex formula:
(E x1)(E y1)(E x2)(E y2)[ a >= 1/2 /\ a <= 3/4 /\
                          x1 > 0 /\ y1 > 0 /\ 1 - x1 - y1 < 0 /\
                          x2 > 0 /\ x2 y1 + y2 - x1 y2 - y1 < 0 /\ x1 + x2 y1 - x2 - x1 y2 < 0 /\
                          a (1/2)(y1 + x1 y2 - x2 y1 + x2) = (1/8)(0 - 1 + x1 + 2 x2 + 2 y1 + y2 + 2 x1 y2 - 2 x2 y1) ].


Before Normalization >

An equivalent quantifier-free formula:

2 a - 1 > 0 /\ 4 a - 3 < 0

=====================  The End  =======================

12 Garbage collections, 473385670 Cells and 0 Arrays reclaimed, in 8158 milliseconds.
1345504 Cells in AVAIL, 40000000 Cells in SPACE.

System time: 79624 milliseconds.
System time after the initialization: 79028 milliseconds.
  • $\begingroup$ I mentioned this problem to a friend of mine and he also solved it by computer. He mentioned that he could also prove that for convex polygons with more than 5 sides, the range was (1/2,1) [independent of the number of sides]. He also made the cryptic remark that the 3/4 bound for n=5 (the bound we are struggling with here), could be obtained by hand using coordinates (and this would remove all computers from the proof of the question at hand). His only hint was "WLOG A=(0, 0), B = (1, 0) and D = (0, 1)" (which I guess one achieves by a linear transformation). Ouch! $\endgroup$ – Kevin Buzzard Nov 15 '09 at 21:25
  • $\begingroup$ Reid: if you change 1/2 to 1/2+epsilon does "FALSE" become "TRUE"? That would be a check to see if you'd done the translation to coordinates correctly. $\endgroup$ – Kevin Buzzard Nov 15 '09 at 21:28
  • $\begingroup$ I improved my query so that it checks that ratios strictly between 1/2 and 3/4 are attainable. It would be nice to remove the conditions on a altogether, but unfortunately that causes qepcad to run for longer than I was willing to wait (> 1 hour). $\endgroup$ – Reid Barton Nov 15 '09 at 22:03
  • $\begingroup$ Reid, it should be possible to prove by explicit constructions like the ones I mentioned that ratios strictly between 1/2 and 3/4 are attainable. $\endgroup$ – Michael Lugo Nov 16 '09 at 0:35

This was originally a comment on Kristal Cantwell's answer; it is now a complete solution of its own.

As Kristal and others have pointed out, we can assume that the vertices of the pentagon are, in cyclic order, $(1,0)$, $(0,0)$, $(0,1)$, $(x\_1, y\_1)$, $(x\_2, y\_2)$. As she describes, we compute

$$\mathrm{Area}(P) = (1/2) \left( x\_1 + [x\_2 y\_1 - x\_1 y\_2] + y_2 \right)$$


$$\mathrm{Area}(Q) = (1/8) \left( x\_1 + [(x\_1+x\_2)(1+y\_1) - x\_1(y\_1 + y\_2)] \right.$$ $$\phantom{\mathrm{Area}(Q)=} \left.+[(1+x\_2)(y\_1+y\_2) - y\_2 (x\_1 + x\_2)] + y\_2 -1 \right).$$

This simplifies to $$\mathrm{Area}(Q) = (1/8) \left( 2 x\_1 + x\_2 + y\_1 + 2 y\_2 + 2 [ x\_2 y\_1 - x\_1 y\_2] -1 \right).$$

In a previous version of her comment; Kristal and I had different formulas at this point, and I had some discussion of that here. We now seem to both agree on the formulas above.

We want to show that $$(3/4) \mathrm{Area}(P) - \mathrm{Area}(Q) \geq 0$$.

Plugging in the above formulas, clearing out the $8$ in the denominator, and simplifiying, we want to show that $$x\_1 - x\_2 - y\_1 + y\_2 + [x\_2 y\_1 - x\_1 y\_2] +1 \geq 0.$$ Rearranging, the left hand side is $$[(x\_2 -1)(y\_1 -1) - (x\_1-1)(y\_2-1)] + 1$$

The quantity in square brackets is twice the signed area of the triangle $(x\_1, y\_1)$, $(1,1)$, $(x\_2, y\_2)$. (Signed area means positive if the triangle is oriented counter-clockwise, negative if clockwise.) If this is positive, we are done. By the convexity hypothesis, the triangle $(x\_1, y\_1)$, $(0,0)$, $(x\_2, y\_2)$ is positively oriented. So the only way for the term in square brackets to be negative is if $(0,0)$ and $(1,1)$ are on opposite sides of the line $(x\_1, y\_1)$, $(x\_2, y\_2)$. Assume from now on that this is the case.

If we slide $(x\_1, y\_1)$ and $(x\_2, y\_2)$ apart, keeping them on the same line, then the oriented area of $(x\_1, y\_1)$, $(1,1)$, $(x\_2, y\_2)$ will only grow more negative. So we may assume that we have slid them as far apart as possible. That is to say, that $x\_1 = 0$ and $y \_2 =0$. Since $x\_1+y\_1$ and $x\_2 + y\_2$ are $\geq 1$, this means that $x\_2$, $y\_1 \geq 1$.

Then $$[(x\_2 -1)(y\_1 -1) - (x\_1-1)(y\_2-1)] + 1 \geq$$ $$ 0 \cdot 0 - (-1)(-1) + 1 =0.$$

  • $\begingroup$ Where you diverged from me I was wrong. I tried to correct that in a rewrite. $\endgroup$ – Kristal Cantwell Nov 16 '09 at 19:03
  • $\begingroup$ Ah, good. I'll edit. $\endgroup$ – David E Speyer Nov 16 '09 at 19:10
  • 1
    $\begingroup$ Those are also the formulas in my input to qepcad, except with (x1, y)1) and (x2, y2) swapped. $\endgroup$ – Reid Barton Nov 16 '09 at 20:21

I think I can prove this. I start by getting three consecutive points of the pentagon ABC equal to (0, 1),(0, 0) (1, 0) and the remainder with coordinates (x_1,y_1) and (x_2,y_2) both with sum of coordinates greater than one and both with both coordinates positive using area preserving transformations as mentioned in a comment to another post.

I expand one dimension and contract an orthogonal direction by the same amount. Doing this I first get the angle to be right then after this I use the same type of transformation contracting along one ray of the now right angle and expanding along the other ray. Doing this I can any ratio I want. Then I translate the middle point to the origin and rotate till I have the three desired points in the desired position and the other conditions on the remaining points come from the convexity of the polygon which is preserved by the transformations. I think I also have to note that the following proof will not be changed by an expansion of both coordinates by the same amount because while I can get an equilateral right triangle by area preserving transformations The area of the original triangle may not be 1/2.

Then I use the fact that the area in the plane of the triangle formed by any two points with coordinates (x_1,y_1) and (x_2,y_2) and the origin is the one half of the determinant of the matrix containing the two points. I think you have to be careful the points are going in clockwise order or else you will get a negative number. The points being in clockwise order means that y_1/x_1 > y_2/x_2. I apply this to the pentagon and get the area the 1/2 of determinant of (x_1,y_1) and (x_2,y_2) plus 1/2(x_1) + 1/2(y_2). I sweep clockwise from the origin and count the areas of the five coordinates formed by the midpoints this overestimates by 1/8 which I will have to subtract. I look at the determinants. For the second and fourth of these triangles it becomes clearer if I subtract one row from another. In any case the result of all this is the area of the midpoint polygon is 1/4 of the determinant of (x_1,y_1) and (x_2,y_2) plus 1/4(x_1) + 1/4(y_2) +1/8(x_1)+1/8(y_2)-1/8. To get this is less than 3/4 of the area of the polygon I need
1/8 the determinant of (x_1,y_1) and (x_2,y_2) + 1/8(x_1) + 1/8(y_2) + 1/8 is greater than 1/8(x_2)+1/8(y_1) or or 1/8 the determinant of (x_1,y_1) and (x_2,y_2) + 1/8(x_1) + 1/8(y_2) + 1/8-1/8(x_2)-1/8(y_1) +1/8

The above expression is 1/4 of the area of the triangle with vertices (x_1,y_1),(1,1),(x_2.y_2) if the point (1,1) is on the same side of line connecting (x_1,y_1) and (x_2.y_2) plus 1/8. If this holds then the expression is positive and we are done. So we can assume that the point (1,1) is on the same side of the line connecting (x_1,y_1) and (x_2.y_2) is on the other side of the line and the above expression is the negative of the area of the triangle mentioned above. Now the farther apart the points are the greater the area of the triangle so if we assume maximum separation we get x_1 and y_2 = zero and substituting this into the above we have 1/8(-x_2-y_1 + (x_1)(y_1) -1 is less than zero or 1/8(y_1-1)(x_2-1) is less than zero but since x_1+y_1 is greater than one and x_1 is zero we have y_1 is greater than one similarly we have x_2 is greater than one and the product is positive. This follows Speyer's so far proof, I thought I could get the 1/8 as the difference between the midpoint polygon and 3/4 the original polygon but in some cases it is less.

So we have the following the area of the midpoint polygon is less than 3/4 of the original polygon furthermore the difference between 3/4 of the original polygon and the area of the midpoint polygon is 1/8-plus or minus the area of the triangle formed by (x_1,y_1),(1,1),(x_2.y_2) depending on which side of the line connecting (x_1,y_1) and (x_2.y_2) is (1,1).I want to find the difference between 3/4 of the area and the area of the midpoint polygon in terms of the areas of geometric figures in the original polygon.

Tracing this back to the original polygon the area of the midpoint polygon is 3/4 of the original polygon -1/4 of the area of triangle ABC plus another triangle which has a point N I need to construct. Here is the construction take the midpoint of CA, M connect it to B then extend the line BM to twice its length to get point N then if N is on the same side of DE as A subtract 1/4 of the area of END otherwise add it from the above the result will always be less than 3/4 of the area of the original polygon or less from the above. This should hold for any three consecutive vertices of any polygon.

  • 2
    $\begingroup$ Holding three corners of the pentagon fixed and multiplying the other two by a large constant, the area of the middle pentagon should tend towards 1/2 -- the pentagon is almost a triangle, two quarters of which are filled by the middle pentagon. I'm also worried about your calculation of the area of the original pentagon -- it seems to me that it need not be more than 1/2 the determinant of (x_1,y_1) and (x_2,y_2): for example, if (x_1,y_1) is close to the x-axis and (x_2,y_2) is close to the y axis. $\endgroup$ – Hugh Thomas Nov 16 '09 at 3:51
  • $\begingroup$ There were mistakes I think I have fixed them. $\endgroup$ – Kristal Cantwell Nov 16 '09 at 19:01
  • $\begingroup$ I think I see the problem with what I did. The triangle NDE appears with another term in Speyer's proof. I think that term corresponds to triangle MBN. I think I could eventually get a geometric equivalent to the difference between the 3/4 area and the area of the half point pentagon but there is a complication that it involves the area of two triangles at least one of which may have to be subtracted in some cases. $\endgroup$ – Kristal Cantwell Nov 17 '09 at 1:17
  • $\begingroup$ I finally got this to work and got the difference between 3/4 of the area of the original polygon and the area of the midpoint polygon as an expression of geometrical figures in the original polygon everything before this didn't work because I was trying to get too big a difference between areas. $\endgroup$ – Kristal Cantwell Nov 19 '09 at 21:35

For pentagons, area(Q)/area(P) can take any value in the open interval (1/2, 3/4). Let the vertices of P be A, B, C, D, E in cyclic order. We can get arbitrarily close to 3/4 from below if we let P degenerate to a triangle by letting the pair (A, B) approach one vertex of a triangle, (C, D) approach a second vertex of the triangle, and E approach a third. (It doesn't seem to matter which triangle we take.) The interval is open because I am not allowing the case where the vertices actually coincide.

Similar constructions show that if pentagons are replaced with an n-gon for n ≥ 6, we can get area(Q)/area(P) arbitrarily close to 1.

If instead we let (A, B, C) degenerate to a single vertex of the triangle, then area(Q)/area(P) can be made arbitrarily close to 1/2; this construction actually works for n ≥ 4.

It's obvious that area(Q)/area(P) = 1/4 if P is a triangle. I think area(Q)/area(P) = 1/2 if P is a quadrilateral.

  • 1
    $\begingroup$ It seems like this shows that you can get any value in (1/2,3/4), but why can't you get values outside of that interval? $\endgroup$ – Anton Geraschenko Nov 10 '09 at 15:34
  • $\begingroup$ Anton, I don't know. I was hoping someone else would. In particular, maybe there's an argument that extrema of this ratio are at the degenerate polygons. $\endgroup$ – Michael Lugo Nov 10 '09 at 15:59

It's called a "midpoint polygon". The problem seems to be addressed (with proofs for triangles and quadrilaterals but only a conjecture for the pentagon) here:


(The other construction mentioned is a {5/2} star polygon inscribed in a pentagon.)

  • 1
    $\begingroup$ I doubt that the problem is solved over there. They prove that the area of the midpoint pentagon is equal to P/2 plus one fourth of the area of the star pentagon. But for the area of the star polygon they count the area of the middle pentagon twice (for some strange reason). In my opinion the arguments presented in the link show the equivalence between the two formulations of the problem given by Manuel. $\endgroup$ – Philipp Lampe Nov 10 '09 at 16:21
  • 1
    $\begingroup$ Ah yes, they switch from proof to conjecture for the upper bound once they get to pentagons. $\endgroup$ – Jason Dyer Nov 10 '09 at 16:28

I asked this problem to a friend of mine, and he gave a proof of both the lower and the upper bounds that, to me, look cleaner than any offered so far. People's opinions may differ on this. I am now going to cut and paste from his email. Note that both arguments are rather short but it's easy to check the details and they look fine to me. The argument establishes the correct bounds for an n-gon for any n>=5.

The area lost in the process of joining midpoints is 1/4 of the sum of the areas of ABC, BCD, etc., and it's easy to see that the only pairs of those triangles that intersect are consecutive pairs, from which you get that the total of those areas is at most twice the area of the original pentagon for 4 or more side (and it's easy to see it's strict inequality for 5 or more sides). So the lower bound is easy, geometrically (and for 6 or more sides, cluster two or more vertices near each of the vertices of a triangle to see that 1 is the right upper bound).

As for 3/4 ... you want to prove the total of the areas of ABC, BCD etc. is greater than the area of the pentagon, or equivalently, cutting two triangles out of the pentagon, that area(EAB) + area(ABC) + area(CDE) > area(ABD). Putting E (x_1, y_1) and C (x_2, y_2), convexity means y_1, y_2 > 0, x_1 < 0, x_2 + y_2 > 1, EAB alone is big enough if y_1 > 1 and ABC is big enough if y_2 > 1 (while if either = 1 you have one of the areas equal to area(ABD) while the others are positive). So you can assume 0 < y_1, y_2 < 1, want to prove y_1 + y_2 + x_1(y_2 - 1) - x_2(y_1 - 1) > 1 and the LHS is minimised for x_1 as large as possible and x_2 as small as possible (we know the signs of y_2 - 1 and y_1 - 1), which minimum turns out to be 1 + y_1y_2 > 1.


I think we no loss the generality if we take the vertices (0, 1), (0, 0), (1, 0), (1, y), (x, 1), in couterclockwise, with $0 < x < 1, 0 < y<1$. In this case the pentagon is still convex, and

$Area(P)=\frac{1}{2}+\frac{x+y-xy}{2}$ and $Area(Q)=\frac{P}{2}+\frac{1}{8}$.

Than the ratio $R=\frac{Area(Q)}{Area(P)}=\frac{1}{2}+\frac{1}{4}\frac{1}{1+x+y-xy}$.

From $ 0< x < 1, 0 < y < 1$ \implies $1/x + 1/y$ > $2$ \implies $1+x+y-xy$ > $1+xy$ > $1$

\implies $R$ < \frac{3}{4}.

Also, we can take $x=1-\alpha$, $y=1-\beta$, $0 < \alpha < 1, 0<\beta<1$

\implies $1+x+y-xy=2-\alpha\beta<2$.

Then $R > \frac{1}{2}$ + $\frac{1}{8}$ = $\frac{5}{8}$ a new inferior limit.

  • $\begingroup$ I think there is a problem with this. For one thing there are examples where the ratio is 3/4. Your maximum might hold for all pentagons with parallel sides though. $\endgroup$ – Kristal Cantwell Nov 19 '09 at 19:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.