Can I determine the angles of a quadrilateral if I know the lengths of the sides and the difference between the diagonals?

Question

I know the lengths of the four sides of a quadrilateral and the difference between the diagonals (but I do not know the actual lengths of the diagonals). My instinct is that this information ought to be sufficient to determine the angles of the quadrilateral, because a specific difference between the diagonals constrains it to a single, fixed shape.

example measurements: L side length = 326mm; R side length = 325mm; bottom length = 677mm; top length = 675mm; diagonal from bottom L to top R is 7mm longer than from bottom R to top L

Answer 1

Let's call the vertices of the quadrilateral $A,B,C,D$ in counter clockwise direction. Let $a = AB , b = BC, c = CD , d = DA $. We can can let $A = (0,0)$ on the cartesian plane, and $B = (a, 0) $. Further, let $C = (x_1, y_1) $ and $D = (x_2, y_2)$. Now we have

$ b^2 = (x_1 - a)^2 + y_1^2 $

$ c^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2$

$ d^2 = x_2^2 + y_2^2$

And if $e = AC - BD$ , then

$ e = \sqrt{ x_1^2 + y_1^2 } - \sqrt{ (x_2 - a)^2 + y_2^2 } $

we now have

$ x_1^2 + y_1^2 = b^2 - a^2 + 2 a x_1$

$ (x_2 - a)^2 + y_2^2 = d^2 + a^2 - 2 a x_2 $

Hence,

$ e = \sqrt{ b^2 - a^2 + 2 a x_1 } - \sqrt{ d^2 + a^2 - 2 a x_2 } $

Squaring,

$ e^2 = b^2 + d^2 + 2 a (x_1 - x_2) - 2 \sqrt{ (b^2 - a^2 + 2 a x_1)(d^2 + a^2 - 2 a x_2 ) } $

So that

$ \bigg( e^2 - b^2 - d^2 - 2 a (x_1 - x_2) \bigg)^2 = 4 \bigg(b^2 - a^2 + 2 a x_1\bigg) \bigg(d^2 + a^2 - 2 a x_2 \bigg)$

So now we have four quadratic equations in $4$ unknowns $x_1,y_1, x_2, y_2$.

The actual dimension of the problem can be reduced to $2$ utilizing equations (1) and (3), which are equations of circles centered at $(a, 0)$ and $(0,0)$ respectively.

Specifically, we have two parameters, $\theta$ and $\phi$, and we have $ x_1, y_1, x_2, y_2$ defined in terms of them as follows:

$ x_1 = a + b \cos \theta $

$ y_1 = b \sin \theta $

$ x_2 = d \cos \phi $

$ y_2 = d \sin \phi $

With this we have eliminated equations (1) and (3). Now the second equation becomes

$ c^2 = ( a + b \cos \theta - d \cos \phi)^2 + ( b \sin \theta - d \sin \phi)^2 $

And the last equation becomes

$ \bigg( e^2 - b^2 - d^2 - 2 a (a+b \cos\theta - d \cos \phi) \bigg)^2 \\= 4 \bigg(b^2 - a^2 + 2 a (a + b \cos \theta)\bigg) \bigg(d^2 + a^2 - 2 a d \cos \phi \bigg)$

These two equations in $\theta$ and $\phi$ can be solved numerically via the Newton-Raphson (multivariate) method.

And that's what I have done. I implemented the Newton-Raphson method for the specific two equations above. With $a= 7, b = 4, c = 3, d = 5, e = 1 $, I got the following two solutions. Note that in the second solution, the order of subtraction of the diagonals is reversed, in other words, the program finds solutions such that the difference of the diagonals in absolute value is $e$.

The program listing follows. It is written in Microsoft Excel VBA (Visual Basic for Applications).

Public Sub quadrilateral_()
Dim x(2), y(2), jac(2, 3) As Double
Dim a, b, c, d, e As Double
Dim sol_mat(160, 2) As Double
Dim vers(5, 2) As Double
a = 7
b = 4
c = 3
d = 5
e = 1
For it1 = 0 To 59
t1 = it1 / 60 * (2 * p)
x(1) = t1
For it2 = 0 To 59
t2 = it2 / 60 * (2 * p)
x(2) = t2
success = 0
For ic = 1 To 30
Call find_y_quad(a, b, c, d, e, x, y)
If norm(y, 2) < 0.000001 Then
success = 1
For i = 1 To 2
x(i) = x(i) - 2 * p * Int(x(i) / (2 * p))
Next i
t = x(1)
s = x(2)
vers(1, 1) = 0
vers(1, 2) = 0
vers(2, 1) = a
vers(2, 2) = 0
vers(3, 1) = a + b * Cos(t)
vers(3, 2) = b * Sin(t)
vers(4, 1) = d * Cos(s)
vers(4, 2) = d * Sin(s)
vers(5, 1) = vers(1, 1)
vers(5, 2) = vers(1, 2)
' make sure that quadrilateral is convex
For i = 2 To 4
j = i + 1
For k = 1 To 2
u1(k) = vers(i, k) - vers(i - 1, k)
u2(k) = vers(i + 1, k) - vers(i, k)
Next k
Call cross(u1, u2, u3)
If u3(3) < 0 Then GoTo lnext_it2
Next i
Call append_cmat(sol_mat, x)
Exit For
End If
Call find_jac_quad(a, b, c, d, e, x, jac)
For i = 1 To 2
jac(i, 3) = -y(i)
Next i
Call reduce_matrix(jac, 2, 3, 0.00000001, det)
If Abs(det) < 0.000001 Then
MsgBox ("jac not invertible")
Exit Sub
End If
For i = 1 To 2
x(i) = x(i) + jac(i, 3)
Next i
Next ic
If success = 0 Then
'MsgBox ("did not converge")
Exit Sub
End If
' MsgBox ("converged in " + Str(ic - 1) + " iterations")
lnext_it2:
Next it2
Next it1
MsgBox ("total no. of solutions found = " + Str(sol_mat(0, 0)))
k1 = 10
For i = 1 To sol_mat(0, 0)
For j = 1 To 2
ActiveSheet.Cells(i, j) = sol_mat(i, j)
Next j
t = sol_mat(i, 1)
s = sol_mat(i, 2)
vers(1, 1) = 0
vers(1, 2) = 0
vers(2, 1) = a
vers(2, 2) = 0
vers(3, 1) = a + b * Cos(t)
vers(3, 2) = b * Sin(t)
vers(4, 1) = d * Cos(s)
vers(4, 2) = d * Sin(s)
vers(5, 1) = vers(1, 1)
vers(5, 2) = vers(1, 2)
k = k1
narcs = 0
ipass = 2
Call draw_points(2, 1, 5, vers, 1)
Call draw_points(0, 1, 4, vers, 3)
k2 = k
If i = 1 Then
Call plot_curve(0, k1, 0, 0, 0, &quot;&quot;, &quot;&quot;)


Else
Call plot_curve(1, k1, 0, 0, 0, &quot;&quot;, &quot;&quot;)


End If
k1 = k2
Next i
End Sub

---

Public Sub find_y_quad(a, b, c, d, e, x, y)
Dim lhs, rhs As Double
t = x(1)
s = x(2)
y(1) = (a + b * Cos(t) - d * Cos(s)) ^ 2 + (b * Sin(t) - d * Sin(s)) ^ 2 - c ^ 2
lhs = (e ^ 2 - b ^ 2 - d ^ 2 - 2 * a * (a + b * Cos(t) - d * Cos(s))) ^ 2
rhs = 4 * (b ^ 2 - a ^ 2 + 2 * a * (a + b * Cos(t))) * (d ^ 2 + a ^ 2 - 2 * a * d * Cos(s))
y(2) = lhs - rhs
End Sub

---

Public Sub find_jac_quad(a, b, c, d, e, x, jac)
Dim y(2), y1(2), u(2) As Double
Call find_y_quad(a, b, c, d, e, x, y)
For j = 1 To 2
For i = 1 To 2
u(i) = x(i)
Next i
u(j) = u(j) + 0.01
Call find_y_quad(a, b, c, d, e, u, y1)
For i = 1 To 2
jac(i, j) = 100 * (y1(i) - y(i))
Next i
Next j
End Sub

---

Public Sub append_cmat(cmat, c)
Dim i, j, n As Integer, ns As Integer
n = cmat(0, 0)
n = n + 1
For i = 1 To 2
cmat(n, i) = c(i)
Next i
Call eliminate_duplicates_2d(n, 2, cmat, 0.000001, ns, cmat)
cmat(0, 0) = ns
End Sub

---

Answer 2

Yes, it is possible because there are five unknowns, the four angles, $\alpha, \beta,\gamma,\delta$ and one of the diagonals $D_1,D_2$ because $|D_1-D_2|=L$ where $L$ is data. But we have five equations.

Suppose $D_1\lt D_2$ so $D_2=D_1+L$, our five equations are $$D_1^2=a^2+d^2-2ad\cos(\alpha)\\D_1^2=b^2+c^2-2bc\cos( \gamma)\\(D_1+L)^2=a^2+b^2-2ab\cos( \beta)\\(D_1+L)^2=c^2+d^2-2cd\cos (\delta)\\\alpha+\beta+\gamma+\delta=2\pi$$

Answer 3

I'll add another answer just to show how this can be reduced to the solution of a single algebraic equation. Let's call $a = AB$, $b = BC$, $c = CD$, $d = DA$ the given sides of the quadrilateral and $y=BD$, $y+l=AC$ its unknown diagonals (but $l$ is given) .

From the cosine law we get: $$ \cos\angle ADB={y^2+d^2-a^2\over2dy}, \quad \cos\angle CDB={y^2+c^2-b^2\over2cy}. $$ Hence, using the cosine addition formula: $$ \begin{align} \cos\angle ADC= &\ \cos\angle ADB\cos\angle CDB-\sin\angle ADB\sin\angle CDB\\ =&\ {y^2+d^2-a^2\over2dy}\cdot{y^2+c^2-b^2\over2cy}- \sqrt{1-{(y^2+d^2-a^2)^2\over(2dy)^2}} \sqrt{1-{(y^2+c^2-b^2)^2\over(2cy)^2}} \end{align} $$ and finally, from the cosine rule: $$ (y+l)^2=c^2+d^2-2cd \left({y^2+d^2-a^2\over2dy}\cdot{y^2+c^2-b^2\over2cy}- \sqrt{1-{(y^2+d^2-a^2)^2\over(2dy)^2}} \sqrt{1-{(y^2+c^2-b^2)^2\over(2cy)^2}}\right), $$ which is an equation in the unkown $y$. We can rearrange terms and square both sides, to eliminate roots and denominators. The final result is a sixth degree equation: $$ 2 y^6+6 l y^5 -\left(a^2+b^2+c^2+d^2-7 l^2\right)y^4 -2\left(a^2 + b^2 + c^2 + d^2 -2 l^2\right)l y^3 +\left(a^2 b^2+b^2 c^2+c^2 d^2+d^2 a^2 -2 a^2 c^2-2 b^2 d^2 -(a^2 +b^2 +c^2 +d^2) l^2+l^4\right)y^2 \\ +2\left(a^2 b^2 - a^2 c^2 - b^2 d^2 + c^2 d^2 \right) l y \\ +a^4 c^2+a^2 c^4+b^4 d^2+b^2 d^4 -a^2 b^2 c^2-a^2 b^2 d^2-a^2 c^2 d^2-b^2 c^2 d^2\\ +(c^2 d^2 +a^2 b^2-a^2 c^2-b^2 d^2) l^2 =0. $$ Inserting here, for instance, the data given in the question: $$ a = 677,\quad b = 325,\quad c = 675,\quad d = 326,\quad l = 7, $$ we obtain only two positive solutions: $y\approx746.775$ and $y\approx588.986$, the second one corresponding to a self-intersecting quadrilateral.

Answer 4

If it is not required that the quadrilateral be convex, then in some cases the lengths of the four sides and the difference between the diagonals are insufficient to constrain the quadrilateral to a single, fixed shape.

To illustrate this, suppose we are given $AB = DA = 5$, $BC = CD = 3$ and $BD - AC = 1$. Consider the diagram below where $B,C,D$ are in line yielding a quadrilateral consisting of two $3-4-5$ triangles:

Obviously this does not meet the requirements since $BD-AC=2$. Now imagine $C$ moved gradually to the right while holding point $A$ and lengths $AB,BC,CD,DA$ fixed, so that $B$ and $D$ are pulled towards each other. Initially $BD=6$ and $AC=4$. Eventually, when $B$ and $D$ coincide, $BD=0$ and $AC=5+3=8$ so $BD-AC$ has reduced continuously from $2$ to $-8$. There must be an intermediate point with $C$ to the right of $BD$ at which $BD-AC=1$.

Note that the argument does not depend on an assumption that $BD$ and $AC$ change at the same rate or pro rata (such an assumption would be incorrect).

Now imagine, with the same conditions, $C$ moved gradually to the left until $B$ and $D$ coincide. In this case $BD$ falls continuously from $6$ to $0$ and $AC$ from $4$ to $5-3=2$, so $BD-AC$ falls from $2$ to $-2$. Again, there must be an intermediate point at which $BD-AC=1$ with in this case $C$ to the left of $BD$.

So there exist two distinct quadrilaterals, one convex and one non-convex, meeting the given requirements.