# Max area: Quadrilateral with fixed perimeter and interior angle

Edit (May 1, 2018): There appears to be an implicit assumption related to convexity in several of the answers posted thus far; see the answer and image posted here.

So: Let us restrict to the case in which the quadrilateral is convex.

My post is based on a tweet linking to a blog post where the following problem is posed:

A [convex] quadrilateral has perimeter $60$ and a $30^{\circ}$ angle. What is the maximum possible area?

In the comments, there is a link to a GeoGebra applet for exploring this problem. Tinkering with the applet led me to believe that the maximization occurs when the other three angles are each $110^{\circ}$, but I neither possess a proof that this is optimal, nor have computed the exact area (in closed form) for this case.

I would appreciate an answer that broaches the above special case, although I think it is reasonable to ask about a natural generalization for quadrilaterals; in either scenario, a proof or a link to an already existing one would be great.

Generalization:

A [convex] quadrilateral has perimeter $N$ and a $d^{\circ}$ angle. What is the maximum possible area?

(So, as a well-known example, for the case in which $d = 90$, the maximum possible area occurs for a square with side-length $N/4$; therefore, the max area is $N^2/16$.)

• You can restrict your parameters such that the quadrilateral is convex. If you have one angle $\theta > \pi$ radians then replacing that angle with $2\pi - \theta$ will give you a quadrilateral with the same perimeter but a larger area.
– John
Apr 30, 2018 at 19:13
• Maybe try Lagrange multipliers? $x$ coordinate of $B$ (length of $AB$ is one unknown. Length of $AD$ is another unknown. This determines where $A,B,D$ are based on your definition of $AB$ on the $x$ axis and $\angle DAB$ as the $30$ deg angle. Then the $x,y$ coordinate of $C$ are what's left. From there you can constrain the perimeter and then take the area of the two triangles $ADC$ and $ACB$ as the function to maximize.
– John
Apr 30, 2018 at 19:29
• I feel that the quadrilateral should be convex, or else the maximum area is non-existent, as it can approach a equilateral triangle while always increasing the area.
– user211599
May 1, 2018 at 20:24
• @JefferyOpoku-Mensah: I agree, and have edited in this assumption! Thanks. May 1, 2018 at 20:46

Here is an overview of what I am about to do.

• Show that the quadrilateral must be convex.
• Prove $\triangle BCD$ must be isosceles.
• Prove $\triangle BAD$ must also be isosceles.
• Note that single variable calculus may be used to find the maximum area.
• Use calculus to show that all three other angles must be equal.

Clearly, $ABCD$ must be convex, as replacing changing such an angle from $\theta \mapsto 360^\circ - \theta$ preserves the perimeter, while increasing the area.

EDIT: This only works with angle $\angle BCD$. I consider the problem to be ill-defined if the quadrilateral is not convex, as in, I do not believe there is a maximum (I do not have the appropriate room or time to prove this currently).

Consider the ellipse $\mathcal{E}$ with foci $B, D$ containing $C$. Moving $C$ along $E$, we see that the perimeter of $ABCD$ is unchanged. The area of $ABCD$ is $[BAD] + [BCD]$, and $[BAD]$ also remains unchanged. The area of $\triangle BCD$ varies, this is $$[BCD] = \frac{1}{2} \cdot BD \cdot CC'$$ where $C'$ is the foot of the altitude from $C$. Clearly, $BD$ is invariant, and $CC'$ is the only varying quantity. It obtains a maximum when it becomes a semi-axis of $\mathcal{E}$. When this happens, $\triangle BCD$ isosceles.

Assume $AD > AB$. Let $E$ be on segment $\overline{AD}$ such that $AE = AB$. Furthermore, let $D'$ be the midpoint of $\overline{ED}$, and $B'$ be a point on ray $\overrightarrow{AB}$ such that $AB' = AD'$. In this way, the perimeters of $\triangle ABD$ and $\triangle AB'D'$ are equal. Also, $EBB'D'$ is an isosceles trapezoid, let the height be $h$. Then $[BB'D'] = h\cdot D'B'$ and $[BED'] = h \cdot EB$, so $[BB'D'] > [BED']$. Furthermore, $ED' = D'D$, and using the height from $B$ to line $\overline{AD}$, $[BED'] = [BD'D]$. As a result, $$[BB'D'] > [BD'D] \implies [ABD'] + [BB'D'] > [ABD'] + [BD'D] \implies [B'AD'] > [BAD]$$ Note that $B'AD'$ was isosceles. In this way the perimeter of $ABCD$ remains constant, yet the area of $\triangle BAD$ is maximized. This happens when $\triangle BAD$ is isosceles.

Equipped with the knowledge that $\triangle BCD$ an $\triangle BAD$ are isosceles, Phil H's argument may suffice.

I am personally intrigued by the fact that the angles were all $110^\circ$. It appears using calculus on the length of the sides may not be the best tool, as there are untapped symmetries still hidden in the problem.

Consider $\triangle ABC$, and use the normal naming conventions. By using the sine formula twice, we obtain $$[ABCD] = \frac{a^2\sin(2\beta) + b^2\sin(2\alpha)}{2}$$ But consider the fact that $a+b = 30$ and $$\frac{a}{\sin \alpha} = \frac{b}{\sin \beta}$$ Then $$b = \frac{30\sin(\beta)}{\sin(\alpha) + \sin(\beta)}$$ and $$a^2 = b^2 \frac{\sin(\alpha)^2}{\sin(\beta)^2}$$ Our area becomes $$[ABCD] = b^2 \left(\frac{\sin(\alpha)^2\sin(2\beta)+\sin(\beta)^2\sin(2\alpha)}{2\sin(\beta)^2} \right)$$ Which is $$\left(\frac{30\sin(\beta)}{\sin(\alpha) + \sin(\beta)}\right)^2 \left(\frac{\sin(\alpha)^2\sin(2\beta)+\sin(\beta)^2\sin(2\alpha)}{2\sin(\beta)^2} \right)$$ "reducing" to $$\frac{450(\sin(\alpha)^2\sin(2\beta)+\sin(\beta)^2\sin(2\alpha))}{(\sin(\alpha) + \sin(\beta))^2}$$ Taking the derivative yields $$-\frac{900\sin(\alpha)(\sin(\alpha) (\cos(\beta)\sin(2\beta)-(\sin(\beta)+\sin(\alpha))\cos(2\beta))-\sin(2\alpha)\cos(\beta)\sin(\beta))}{(\sin(\alpha)+\sin(\beta))^3}$$ Substituting $\beta = 60^\circ - \alpha/3$ gives that the derivative is zero. (I really can't bother to check further, so let's believe that it is the only extremum, and that it is a maximum).

Therefore, we have arrived at the fact that all three other angles are equal. From here, computing the area is trivial. For $\alpha = 15^\circ$ the area is

$$\frac{450\left(\frac{(\sqrt{3}-1)^2\sin(110^\circ)}{8} + \frac{\sin(110^\circ)^2}{2}\right)}{\left(\sin(110^\circ)+\frac{\sqrt{3}-1}{2^\frac{3}{2}}\right)^2} \approx 154.3$$

• It seems important to double check that the quadrilateral must be convex; see the updated question. May 1, 2018 at 19:02
• Indeed, I saw elsewhere that if the quadrilateral is concave, we may approach an equilateral triangle, but then does the maximum actually exist?
– user211599
May 1, 2018 at 20:26
• I agree that the problem is not well-formulated without imposing a convexity condition. May 1, 2018 at 20:43

There's no requirement given that the quadrilateral must be convex. People are assuming that the result must be convex, but that's not correct. If the quadrilateral is concave at $D$, that cannot be switched to a convex quadrilateral without violating the constraint that the angle $A$ is $30$ degrees.

A quadrilateral with an arbitrarily short side $AD$, and with the other three sides equal in length, will have an area approaching the area of an equilateral triangle: $\frac{\sqrt{3}}{4} \cdot 20^2$ or about $173.21$. This is significantly larger than any other solution proposed so far.

Below is an image from the linked GeoGebra app in which $D$ is dragged towards $A$; by zooming in, one can see that the $30^{\circ}$ angle is maintained, but the result is much higher than the previously claimed maximum for convex figures.

Write an expression for the area of the quadrilateral as the sum of the areas of two isosceles triangles, $ADB$ and $DCB$, as the special case such that $AD + DC = AB + BC = 30$.

Hence $AD = AB = x$ and $DC = BC = 30 - x$, which gives a formula for the area:

$$A = \sin(15)x\Big(\cos(15)x+\sqrt{(30-x)^2-(\sin(15)x)^2}\Big)$$

Setting the derivative $dA/dx = 0$ and solving for $x$ yields:

$$x = 22.797\ldots, \text{ hence } A = 154.303\ldots$$

For the general case, having non-symmetry about $AC$ is not optimal. What works for maximizing triangle $ADC$ will apply to triangle $ABC$.

An analysis of the previous statement follows:

For the general case, establishing a maximum area for triangle $ABD$ is key. For two sides and an included angle, the maximum area of the resulting triangle, given the sum of the two sides, say $(40)$, and the included angle $(30)$, is......... $$A = 0.5\sqrt{x^2+(40-x)^2-(2x(40-x)cos(30))}(40-x)sin(30+sin^{-1}(\frac{sin(30)(40-x)}{\sqrt{x^2+(40-x)^2-(2x(40-x)cos(30))}}))$$ Where $0 \le x \le 20$

Note: the limited domain considers all combinations of the two adjacent side lengths.

Setting the derivative $dA/dx = 0$ and solving for x yields:

$$x = 20\ldots, \text{ hence } A_{max} = 100$$

In summary, an isosceles triangle is optimal, and it follows therefore that back to back isosceles triangles is optimal for the general case.

• A proof of symmetry being optimal would involve showing how making the triangles dissimilar (example: making the angles 14 and 16 and changing the ratio of AD to AB) reduces the area in both directions. May 1, 2018 at 0:08
• Good fix :) I added an analysis to show isosceles triangles are optimal for the convex solution. I'm still coming up to speed using MathJax. May 1, 2018 at 20:00

This approach is inspired by Jeffery Opoku-Mensah's observation that in the optimal configuration, the other three angles of the quadrilateral all equal to $110^\circ$.

To attack this problem, we will generalize it a little bit and look for conditions to achieve optimality.

• Let $n$ be any integer $\ge 3$ and $[n] = \{ 0, 1, \ldots, n - 1 \}$.
• Let $v_0, v_1, \ldots, v_{n-1}$ be the vertices of any simple $n$-gon with perimeter $p$ and area $\mathcal{A}$.
• Extend definition of $v_k$ by periodicity. For any $k = qn+r$ where $r \in [n]$, $v_k$ is an alias of $v_r$.
• Let $\ell_k$ be the length of the edge joining $v_{k-1}$ and $v_k$.
• Let $\alpha_k$ be the internal angle $\angle v_{k-1} v_k v_{k+1}$.

The problem at hand can be rephrased as.

Given $n = 4$, $\alpha_0 = \frac{\pi}{6}$ and $p = 60$, what is the maximum value of $\mathcal{A}$?

Instead of fixing $p$ and maximize $\mathcal{A}$, it is more convenient to consider following generalization:

Given $n$ and $m$ of the angles $\alpha_{i_1}, \alpha_{i_2}, \ldots, \alpha_{i_m}$, what are the relations among $\ell_k$, $\alpha_k$ when the ratio $\frac{\mathcal{A}}{p^2}$ is maximized?

We can divide the edges $v_{k-1}v_k$ of the $n$-gon into two groups, those that are free (i.e. both $\alpha_k$ and $\alpha_{k-1}$ are not fixed) and those don't.

For any free edge $v_{k-1}v_k$, consider following perturbation of the $n$-gon which shift $v_k$ along the line $v_{k+1}v_k$ for a small distance $\epsilon$.

$$v_j \rightarrow v_j(\epsilon) = \begin{cases} v_{k} + \frac{\epsilon}{\ell_{k+1}}(v_k - v_{k+1}), & j = k\\v_j, & j \ne k\end{cases}$$

Treating $p$ and $\mathcal{A}$ as functions of $\epsilon$. For small $\epsilon$, it is easy to verify

\begin{align} p(\epsilon) &= p(0) + \epsilon( 1 + \cos\alpha_k) + O(\epsilon^2)\\ \mathcal{A}(\epsilon) &= \mathcal{A}(0) + \frac12\ell_k (\epsilon\sin\alpha_k) + O(\epsilon^2) \end{align}

If the $n$-gon is in a configuration which maximizes the ratio $\frac{\mathcal{A}}{p^2}$, we need

$$\left.\frac{d}{d\epsilon}\frac{A(\epsilon)}{p(\epsilon)^2}\right|_{\epsilon=0} = 0 \quad\implies\quad \frac{\ell_k\sin\alpha_k}{2\mathcal{A}} = \frac{2(1+\cos\alpha_k)}{p} \quad\iff\quad \frac{p\ell_k}{4\mathcal{A}} = \cot\frac{\alpha_k}{2}\tag{*1a}$$ Similarly, if we perturb the polygon by shifting $v_{k-1}$ along the line $v_{k-2}v_{k-1}$ for small distance $\epsilon$:

$$v_j \rightarrow v_j(\epsilon) = \begin{cases} v_{k-1} + \frac{\epsilon}{\ell_{k-1}}(v_{k-1} - v_{k-2}), & j = k-1\\v_j, & j \ne k-1\end{cases}$$

We obtain following relation $$\frac{p\ell_k}{4\mathcal{A}} = \cot\frac{\alpha_{k-1}}{2}\tag{*1b}$$

Comparing $(*1a)$ and $(*1b)$, we conclude $\alpha_{k-1} = \alpha_{k}$ whenever the edge $v_{k-1}v_k$ is free.

In particular, if only one of the angle is given, then in order to maximize the ratio $\frac{\mathcal{A}}{p^2}$, we need all remaining angles equal to each other.

When the edge $v_{k-1}v_k$ isn't free, above perturbations become illegal as they are changing the angles $\alpha_{k-1}$ and $\alpha_k$. Instead, we can parallel shift the edge $v_{k-1}v_k$ outward for a distance $\epsilon$ while keeping other edges the same. If we do that, we obtain following relation which is valid for all edges.

$$\frac{p\ell_k}{4\mathcal{A}} = \frac12\left(\cot\frac{a_k}{2} + \cot\frac{a_{k-1}}{2}\right)\tag{*1c}$$ Summing over $k$, we find the maximum value of $\mathcal{A}$ satisfies:

$$\mathcal{A} = \frac{p^2}{4}\left(\sum_{k=0}^{n-1}\cot\frac{\alpha_k}{2}\right)^{-1}\tag{*2}$$

Apply this to the problem at hand with $v_0 = A$, we have $$p = 60,\quad\alpha_0 = \frac{\pi}{6}\quad\text{ and }\quad\alpha_1 = \alpha_2 = \alpha_3 = \frac13(2\pi - \alpha_0) = \frac{11\pi}{18}$$ The maximum value of area equals to

\begin{align}\mathcal{A} &= \frac{p^2}{4}\left(\cot\frac{\alpha_0}{2} + 3\cot\frac{2\pi - \alpha_0}{6}\right)^{-1} = \frac{60^2}{4}\left(\cot\left(\frac{\pi}{12}\right) + 3\cot\left(\frac{11\pi}{36}\right)\right)^{-1}\\ &\approx 154.3031702366152 \end{align}

• This is great! What you formalize is (I think) the strategy that I used, informally, in GeoGebra to find the maximum area: i.e., I perturbed individual vertices such that the area kept increasing, which resulted in the image in the OP. Question (not sure if this is obvious, or even makes sense! but I ask nevertheless): My understanding of this answer is that it involves a local approach of perturbing individual vertices to maximize area; if this is correct, then is it clear that the global max is achieved by addressing vertices one by one? Why does maximizing for each maximize for all? May 1, 2018 at 18:38
• @BenjaminDickman A global max is a local max. This answer essentially derives some necessary condition (but not a sufficient one) that a configuration is local max. The goal is to explain why the three angles need to equal equal to each other. May 2, 2018 at 0:35

Consider the quadrilateral $Q$ in the image below :

The area of $Q$ is given by $$f(a,b,c,d,A,C) = \frac{a d \sin A+bc \sin C}{2}$$

So here we want to maximize $f(a,b,c,d,A,C)$ subject to : \begin{align*} A &= \alpha \\ a+b+c+d &= N \\ a^2 + d^2 - 2ad \cos A &= b^2 + c^2 - 2bc \cos C \end{align*}

Note : As stated here, the third constraint imposes that two triangles share the side BD.

That is, we want to maximize $$f(a,b,c) = \frac{a (N-a-b-c) \sin \alpha +bc \sin \left( C(a,b,c)\right)}{2}$$ where $$C(a,b,c) = \pi - \cos^{-1}\left(\frac{a^2 + (N-a-b-c)^2 - 2a(N-a-b-c) \cos \alpha - b^2 + c^2}{ - 2bc } \right)$$

Solving $\nabla f = 0$ manually is quite challenging. With a software (e.g., Maple) and by setting $\alpha:=30^°$, $N:=60$:

f := (1/2)*a*d*sin(alpha)+(1/2)*b*c*sin(C)
C := arccos((a^2+d^2-2*a*d*cos(alpha)-b^2-c^2)/(-2*b*c))
d := N-a-b-c
eqns := [diff(f, a) = 0, diff(f, b) = 0, diff(f, c) = 0]
vars := {a, b, c}
fsolve(eval(eqns, [alpha = (1/6)*Pi, N = 60]), vars, complex)

$$a = 22.79705048 \\ b = 7.202949525 \\ c = 7.202949525$$

which yields: $$f(22.79705048,7.202949525,7.202949525) \approx 154.3031703$$ We can also compute the other parameters \begin{align*} C(a,b,c) &= \pi-\cos^{-1}(9.016974745-5.008487375\sqrt{3}) \approx 109.999999^{°}\\ d &= N-a-b-c \approx 22.79705048 \end{align*}

• Barring some mis-calculation, it seems that your formula yields $$A(\pi/6, 60) = \frac{1}{8} \frac{60^2 \sin \frac{\pi}{6}}{\sin \frac{\pi}{6} + 1} = 150$$ which is less than the approximate $154.3$ in the GeoGebra image above. Apr 30, 2018 at 21:45
• Indeed, I have edited to correct this. May 1, 2018 at 10:23
• Thanks for the correction; although, as you may imagine from your approximation (and can see in the other responses) the exact value of this parameter should be $110^{\circ}$. Still, +1! May 1, 2018 at 14:52