Max area: Quadrilateral with fixed perimeter and interior angle

Question

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.

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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 $N^2/16$.)

Answer 1

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

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

Answer 2

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*}

Answer 3

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.

Answer 4

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.

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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).

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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.

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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.

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

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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.

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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).

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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

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

Answer 5

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.