Question : Letting $S{^\prime}$ be the area of the inner pentagon made by the five diagonals of a convex pentagon whose area is $S$, then find the max of $\frac{S^\prime}{S}$.

                   three pentagons, each with the their five diagonals drawn, and the inner pentagonal region enclosed by those diagonals marked

I've been interested in this simple question. It seems that a regular pentagon and its affine images would give the max, but I'm facing difficulty.

Remark : This question has been asked previously on math.SE without receiving any answers.

  • 1
    $\begingroup$ You might want to look up "pentagram map." $\endgroup$
    – user25199
    Oct 1, 2013 at 9:00
  • 2
    $\begingroup$ @Carl: Thanks, but I'm not sure this is a clue. $\endgroup$
    – mathlove
    Oct 2, 2013 at 10:56
  • $\begingroup$ @mathlove: just out of curiosity, have you experimented with a dynamic geometry program to test the conjecture? $\endgroup$ Dec 4, 2013 at 12:56
  • $\begingroup$ @alvarezpaiva: Well, sort of. To be honest, I would like anyone who knows well about this kind of experimentation to test this conjecture. $\endgroup$
    – mathlove
    Dec 5, 2013 at 6:26
  • $\begingroup$ Thank you everyone. I'm going to give my $+50$ reputation to Vít Tuček. By the way, if we consider the same question for a convex $n$-gon, only the $n=5$ (pentagon) case might be meaningful since in the $n\ge 6$ cases $S^\prime/S$ do not seem to have the max. $\endgroup$
    – mathlove
    Dec 15, 2013 at 7:07

4 Answers 4


Not an answer, just a long comment. This comment slowly evolved into a partial numerical solution: the regular pentagon is the local maximum.

I've used Maple to produce the formula for $\frac{S'}{S}$ and the result is a rational function on $\mathbb{R}^{10}$ with both numerator and denominator of degree 12. The numerator has 11495 monomial terms while the denominator is factored into product of six degree two polynomials. This means that the symmetry of the problem is somehow lost during the computation and so one must be a bit careful when using such programs.

Nevertheless, restricting the pentagon to have the first side on the $x$-axes I get numerator that factorizes as product of $x_2$ and a polynomial with 182 monomials. While analytical solution is still most probably out of the question, it may be possible to use this for numerical experiments.

For example, working on $\mathbb{R}^{10}$, I was able to verify that the regular pentagon is a stationary point. To my surprise, Maple computed the eigenvalues of the Hessian at this point analytically. Four of the eigenvalues are negative, but unfortunately there are six eigenvalues that are numerically zero ($\approx 10^{-9}$).

I think it could be possible to "prove" this way that the regular pentagon is a local maximum. I'll try to look at it again tomorrow.

Edit: Let me summarize the comments so far.

As David Speyer mentioned, all pentagons are affinely congruent to those whose vertices lies on a conic. If one uses the rational parametrization of the circle $t \to \left(\frac{t}{1+t^2}, \frac{1-t^2}{1+t^2}\right)$, then the degree of the resulting polynomials in the Jacobian rise to more than 30. Nevertheless, one can factor out terms like $t$ or $t^2 + 1$. Of course, we can still fix one vertex and surprisingly, it matters (at least in Maple) which one. Fixing $(0,1)$ as a vertex leads to better results with degrees $(29,29,29,29)$ and number of monomials $(2348,2754,2754,2348)$. This is still unsolvable (at least in Maple on my computer). But weep not for we can at least compute the Hessian at the regular pentagon and get it's eigenvalues (in this case only numerically, analytic computation timed out): $-0.174, -0.075, -0.001, -0.003$.

We can conclude (if we believe Maple) that the regular pentagon is the local maximum among "circle pentagons".

Dag Oskar Madsen reminded me that I didn't use all the freedom we have in the problem. By fixing $(0,0), (1,0)$ and $(0,1)$ I obtained rational function in four variables with numerator of degree 8 with 53 terms and denominator of degree 10 factored into a product of 6 terms. The numerators of the Jacobian have the following structure: "Degrees:", [16, 16, 15, 15], "Terms:", [637, 633, 424, 510] from which we see that this resulted in smallest system so far. On the other hand, any symmetry in the equations was most probably destroyed and thus it may be harder to solve it.

Fixing vertices $1,2,3$ of the regular pentagon with edge $12$ on the $x$-axis I verified that the eigenvalues of the Hessian are indeed negative, namely $-0.027, -0.012, -0.148, -0.136$. This establishes numerically that the regular pentagon is a local maximum in the set of all pentagons.

I've tried to play with reformulation by Noam D. Elkies and treat it like a constrained maximization problem. That is $$ S = \frac{1}{2}\left(a + \sqrt{a^2 - 4b}\right) \\ S' = S - a + \sum_{i \text{ mod } 5} \frac{A_{i-2}A_{i+2}}{S-A_i},\\ \text{where}\\ a = \sum_{i \text{ mod } 5} A_i, \quad b = \sum_{i \text{ mod } 5} A_iA_{i+1} $$ and $A_i$ is the area of triangle formed by three consecutive vertices $P_{i-1}P_iP_{i+1}$. The problem is to maximize $S'$ relative to $S=1$.

The system for Lagrange multipliers reduces to 5 rational equations with numerators of degree 12 with 1544 terms $$ \frac{\partial L}{\partial a_i} = f_i + \left(\lambda + \sum_{j \text{ mod } 5} \frac{A_{j-2}A_{j+2}}{(1-A_j)^2} \right)g_i =0, $$ where $$ g_i = \frac{1}{2}\left( 1 + \frac{A_i - 4A_{i-1} - 4A_{i+1}}{2-a}\right)\\ f_i = g_i -1 + \sum_{j \text{ mod } 5} \frac{ (A_{j-2}\delta_{i,j+2} + A_{j+2}\delta_{i,j-2})(1-A_j) - A_{j-2}A_{j+2}(g_i - \delta_{i,j}) }{(1-A_j)^2}; $$ plus the constraint $$ a - b = 1, $$ which is quadratic with eleven terms.

I encourage everybody to try this out in their CAS of choice. I got curious about the current state of the art in Groebner basis and asked another question.

  • $\begingroup$ One idea I had for making the computation simpler was to WLOG that the conic through the vertices of the pentagon is the unit circle, and the first vertex is at $(1,0)$. We can then parametrize the other $4$ points as $(\frac{2 t_i}{1+t_i^2}, \frac{1-t_i^2}{1+t_i^2})$, getting a rational function of $4$ variables. This still wasn't simple enough to do by hand, but does it turn out nicely when plugged into your function? $\endgroup$ Dec 9, 2013 at 5:30
  • $\begingroup$ There are two things I don't understand. First, why is it sufficient to constraint on the circle? And second, after plugin it it the numerator has degree 2 less than the denominator. The numerator has degree 20 and the denominator has degree 22. Both are factorized into 6 terms. $\endgroup$ Dec 9, 2013 at 5:56
  • 1
    $\begingroup$ Technically the circle isn't enough: We need to try a circle, a parabola and a hyperbola. Any 5 points are on a conic, and a conic up to affine transformation is either $x^+y^2=1$, $y=x^2$ or $y^2-x^1=1$. I figured I'd do the circle first because it includes the case we believe to be optimal. Your result, however, suggests that this is not a useful suggestion. $\endgroup$ Dec 9, 2013 at 14:06
  • 2
    $\begingroup$ An alternative is to fix three neighbouring vertices at $(0,1)$, $(0,0)$ and $(1,0)$. (We can do this without loss of generality since we only need to consider pentagons up to affine transformation.) Does this make the formula simpler? $\endgroup$ Dec 9, 2013 at 15:37
  • 2
    $\begingroup$ Fixing the three vertices at $(0,0)$, $(1,0)$ and $(0,1)$ to be in positions $1$, $3$ and $4$ seems a little better: denominator is degree $7$, a product of $6$ terms; numerator is degree $8$ with $49$ terms. $\endgroup$ Dec 9, 2013 at 22:01

[Corrected (two typos noted by V.T.) and expanded (alternative coordinates)]

Also not an answer, but this time a simpler algebraic formulation.

All indices are cyclic mod $5$. Let the vertices be $P_i$ ($i \bmod 5$), and let $A_i$ be the area of triangle $P_{i-1} P_i P_{i+1}$. Then $S$ is the larger root of the quadratic equation $$ S^2 - S \sum_{i\,\bmod\,5} A_i + \sum_{i\,\bmod\,5} A_i A_{i+1}, $$ and $$ S' = S - \sum_{i\,\bmod\,5} A_i + \sum_{i\,\bmod\,5} \frac{A_{i-2} A_{i+2}}{S-A_i}. $$ So we are to prove that as $A_i$ vary over all positive numbers the ratio $S'/S$ is maximized when the $A_i$ are all equal.

The formula for $S$ was obtained by choosing coordinates so that the $P_i$ are at $(0,0)$, $(0,1)$, $(1,0)$, $(x_2,y_2)$, $(x_3,y_3)$ (as suggested by Dag Oskar Madsen) and seeking a simple relation between $S$ and the $A_i$ that has the appropriate symmetry in the $A_i$. In retrospect this amounts to the quadratic relation on the six Plücker coordinates for a $2$-dimensional subspace of ${\bf R}^4$.

To prove the formula for $S'$ we can then argue as follows. Let $Q_i$ be the vertex of the inner pentagon opposite $P_i$. Then $S - \sum_{i\,\bmod\,5} A_i$ almost gives the area $S'$ of the inner pentagon except we must add the areas of the five triangles $P_{i-2} Q_i P_{i+2}$ which were subtracted twice. We evaluate the area of this triangle by using the following observation. Let $WXYZ$ be a convex quadrilateral whose diagonals intersect at $O$. Then $$ {\rm Area}(XOY) = \frac{{\rm Area}(WXY) {\rm Area}(XYZ)} {{\rm Area}(WXYZ)}. $$ (Proof: use the area formula $\frac12 ab\sin C$ for the areas of $WOX$, $XOY$, $YOZ$, $ZOW$.) In our setting $WXYZ$ is $P_{i+1}P_{i+2}P_{i-2}P_{i-1}$ and its area is $S-A_i$.

The formula above for $S'$ is less pleasant to work with than it might look because each denominator $S-A_i$ contains a square root. It might be better to use as coordinates the quadrilateral areas $$ B_i := S-A_i = \frac12(P_{i+1} - P_{i-2}) \times (P_{i-1} - P_{i+2}) $$ instead of $A_i$. Since the relation between $S$ and the $A_i$ is homogeneous quadratic, the relation between $S$ and the $B_i$ is also homogeneous quadratic, and indeed it turns out to be given by the same polynomial $$ S^2 - S \sum_{i\,\bmod\,5} B_i + \sum_{i\,\bmod\,5} B_i B_{i+1} = 0, $$ though this time $S$ is the smaller root. So now we are to prove that the ratio between $$ S' = S - \sum_{i\,\bmod\,5} (S-B_i) + \sum_{i\,\bmod\,5} \frac{(S-B_{i-2}) (S-B_{i+2})}{B_i}. $$ and $S$ is maximal when all the $B_i$ are equal.

  • $\begingroup$ By $C$ you mean $S$ or is it something different? Is it clear that if the areas $A_i$ are equal, then the pentagon must be regular? $\endgroup$ Dec 10, 2013 at 17:14
  • $\begingroup$ Good questions. 1) Sorry, you're right: "formula for $C$" should read "formula for $S$". I'll fix it in the next edit. 2) Yes, if all the $A_i$ are equal then the pentagon must be an affine image of a regular pentagon. More generally, the $A_i$ determine the pentagon up to area-preserving affine transformation. That's because once we've solved for $S$ you know the areas of the other five triangles spanned by the vertices (e.g. $P_0 P_2 P_3$ has area $S - A_1 - A_4$), [cont'd] $\endgroup$ Dec 10, 2013 at 17:30
  • $\begingroup$ ... and then once you've placed three of the points, say $P_0$ and $P_{\pm 1}$, you can locate each of the others by intersecting lines parallel to $P_0 P_1$ and $P_0 P_{-1}$. $\endgroup$ Dec 10, 2013 at 17:31
  • $\begingroup$ Huh? Lines parallel to $P_0P_1$ and $P_0P_{-1}$? Anyway, I am inclined to believe the result since we need to place two more points and we have at least three conditions. By the way, there is another $C$ in your post in $\frac{1}{2}ab\sin C$. $\endgroup$ Dec 10, 2013 at 18:44
  • $\begingroup$ Yes, parallel. Once you've placed $P_0$ and $P_1$, the area of triangle $P_0 P_1 P_i$ is proportional to the distance from $P_i$ to the line $P_0 P_1$, so knowing the area puts $P_i$ on a line parallel to $P_0 P_1$. If you also know the area of $P_0 P_{-1} P_i$ then $P_i$ is the intersection of that line with one parallel to $P_0 P_{-1}$. And yes, I know about the other $C$, but that's just the usual notation for the angle of triangle $ABC$ at vertex $C$. $\endgroup$ Dec 10, 2013 at 19:53

Here's a possible strategy:

Consider deformations of a vertex of the convex (outer) pentagon which preserve area. The vertex is constrained to move along a line parallel to the segment connecting the adjacent vertices, and clearly must remain in a compact interval if the pentagon is to remain convex. The area of the inner pentagon may be written as a sum and difference of areas of triangles, each of which has at most one vertex move along a line segment during the deformation, so has area proportional to the length of an edge. The vertices of the inner pentagon are determined by projective transformations of the moving vertex, so have coordinates along the fixed segments given by linear fractional transformations of the displacement. Thus, the area of the pentagon is a sum of rational functions, and it turns out each is concave down on the interval, so the area is a concave rational function with a unique maximum. One should therefore get 5 constraints on the pentagon vertices.

Up to affine transformations, there are a 4-parameter family of pentagons (e.g. by fixing 3 vertices, as David Speyer suggested in a comment). So one obtains an overdetermined system, which hopefully has a unique (or finitely many) solutions, one of which is the (affine) regular pentagon.

  • $\begingroup$ If there is a unique maximum, then I think we are more or less done. I've already computed in Maple that the regular pentagon is a local maximum in the set of "circular pentagons" and right now I am trying to extend this result to all pentagons. What bothers me though is your claim that "The area of the inner pentagon may be written as a sum and difference of areas of triangles, each of which has at most one vertex move along a line segment during the deformation." I can't imagine not using e.g. area of $P_1P_2Q_4$ (with notation as in the Noam Elkies's answer). $\endgroup$ Dec 11, 2013 at 16:07
  • $\begingroup$ @VítTuček: the area formula I'm using is $P_0P_2P_3-P_0P_1P_4+P_2P_3Q_0+P_0P_4Q_2+P_0P_1Q_3-P_2P_3Q_1-P_3P_2Q_4$. The first 3 areas are constant, and the others have a single vertex moving along a diagonal. If I computed the linear fractional transformations correctly, then the signed areas of these last 4 triangles are concave down as a function of the displacement of $P_0$ along the line parallel to $P_1P_4$. $\endgroup$
    – Ian Agol
    Dec 11, 2013 at 20:12
  • $\begingroup$ Also, I'm only claiming concavity in certain directions in the space of pentagons - this does not prove concavity globally, but it might give enough constraints to characterize the maximum. $\endgroup$
    – Ian Agol
    Dec 11, 2013 at 20:21
  • $\begingroup$ Nice! Your formula actually explains why it's better to fix $P_0, P_2, P_3$ instead of three consecutive vertices. $\endgroup$ Dec 12, 2013 at 3:15

A complete solution is available at https://arxiv.org/abs/1812.07682

On the polygon determined by the short diagonals of a convex polygon, Jacqueline Cho, Dan Ismailescu, Yiwon Kim, Andrew Woojong Lee


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