In about two months from now, Europe will inaugurate a new monument called the Arc Majeur (more info here http://www.arcmajeur.com/en/). This monument is a circle arc of $205.5^\circ$, has a total height of 60 meters and a total width of 75 meters.

I was wondering if it is possible to find the radius of this arc based on the information given on their website.

I drew the circle with unknown radius $r$ in some frame $O$ and represented the end points of the arc by the vector $\vec{p} = r\cdot\cos(\theta) \hat{i} + r\cdot\sin(\theta) \hat{j}$ and the vector $\vec{q} = R_{O}(\frac{103\pi}{120})\cdot\vec{p}$.

My first thought was that maybe the arc was oriented in a way such that the total width was equal to its diameter. This is the case for the orientation angle $\theta \in [0,\frac{17\pi}{120}]$. But for these values, the maximum total height yields about $53.6$ meters, so that does not quite work. The radius is therefore definitely longer than $37.5$ meters. This also conveniently eliminates the possibility of $r=30$ when considering the height to be equal to the diameter.

Next, I figured that the minimum height is achieved for $\theta = \frac{17\pi}{240}$ and similarly the minimum width is achieved by turning the arc a quarter turn in the anticlockwise sense, i.e. for $\theta = \frac{137\pi}{240}$ (both periodic by half a turn).

Finally, I tried to consider the orientation of the arc for $\theta \in [\frac{17\pi}{240}, \frac{\pi}{2}]$. In this range, it is possible to state the total height as $h=r\cdot{(1+\sin\theta)}$ and the total width as $w=r\cdot{[1+\cos{(\frac{17\pi}{120}-\theta)}]}$. Trying to solve this system of equations by hand seems very ambitious to say the least.

Is it simply the case that no solution can be calculated by hand or did I do something wrong? Also, I tried to plug that equation into Wolfram Alpha but I was not quite sure which angle would be the most realistic.

EDIT: The actual angle of the arc is 205.5°, rather than 202.5°, which means the numbers I use for the ranges of $\theta$ are slightly off!

EDIT 2: Rather than solving the equation for $\theta$, I simply asked Wolfram to solve the system of equations for the height and the width of my circle arc and it spit out the exact value $r=-15\cdot{(2\sqrt{10}-9)}$ which does seem reasonable. I used the wrong formulae so this answer is not correct!

EDIT 3: I plugged in the right numbers for the arc's angle, corrected my expressions for rotating vectors in two dimensions and corrected my formulae for the height and the width of the arc. Turns out an exact answer technically exists, but it is a horrible beast. Solving for $\theta$ first and then plugging the most plausible approximate answer into the equation to solve for $r$ yielded the value $r\approx37.8$ meters and it turns out the arc is tilted by an angle of $\theta \approx 35.93^\circ$. It is safe to say that @Aretino's method yields the same result and therefore his answer is correct.

Thanks everyone!

  • $\begingroup$ Te angle is 205.5°, not 202.5°. $\endgroup$ Aug 28, 2019 at 21:32
  • $\begingroup$ @Aretino you are absolutely right, I will edit that rightaway! $\endgroup$ Aug 28, 2019 at 22:20
  • 1
    $\begingroup$ But are you interested in the actual radius of the Arc Majeur, or are you just asking for the solution of an abstract and simplified case? Because you are ignoring that a part of the arc lies underground and mustn't be considered for the height. $\endgroup$ Aug 29, 2019 at 7:04

3 Answers 3


You can see below a sketch of the arc, taken from a screenshot of the video you linked. I then measured the angles importing the image into GeoGebra.

As you can see, we have four unknowns: angles $\angle BOF=\angle DOF=\alpha$, $\angle AOB=\beta$, $\angle COD=\gamma$ and the radius $r$ of the arc (note that in the figure there are two angles of width $\alpha$, corresponding to that part of the arc which is underground). Hence the three given data ($60\ $m, $75\ $m and $205.5°$) are not enough to determine all the unknowns. If we add, as I found here, that the height of the smaller arc is $28\ $m, then we can set up the system of equations: $$ \cases{ r[\cos\alpha-\cos(\alpha+\gamma)]=28 \\ \\ r[1+\sin(\alpha+\gamma)]=75 \\ \\ r[\cos\alpha-\cos(\alpha+\beta)]=60 \\ \\ 2\alpha+\beta+\gamma=205.5° } $$ This system can be solved numerically to give:

$$ r\approx37.975\ \text{m},\quad \alpha\approx16.3393°,\quad \beta\approx112.004°\quad \gamma\approx60.8174°. $$

This result, however, shouldn't be taken too seriously, because many uncertainties are involved: the arc is not an ideal line and has a finite width, affecting the given measurements. Moreover, the angles I found with GeoGebra are quite different from the above solutions.

enter image description here


If we consider $60\ $m as the complete height of the arc, including the underground part, then the solution can be found without using the extra information on the smaller arc. We need only solve the above system, but discarding the first equation and setting $\alpha=0$. The result in this case is: $$ r\approx37.8123\ \text{m},\quad \beta\approx125.929°\quad \gamma\approx79.5709°. $$

  • $\begingroup$ Very interesting, thanks. I tried to plug in some equations into Wolfram again and from its calculations the radius should be around $40.1$ meters. I guess the final set of equations for the height and width are simply transcendental and unsolvable in closed-form. Scratch that, Wolfram's exact answer is $-15\cdot{(2\sqrt{10}-9)}$ $\endgroup$ Aug 28, 2019 at 22:51
  • $\begingroup$ Which equations did you use? Did you take into account that a part of the arc lies underground? $\endgroup$ Aug 29, 2019 at 6:58
  • $\begingroup$ I had Wolfram solve for $h=r\cdot{(1+ \sin \theta)}$ and $w=r\cdot{(1+ \cos \theta)}$ for the values of $h=60$ and $w=75$. I did not take into account the part of the arc that lies underground because the height given on the website is the structure's absolute height, i.e. from the very top of the structure to the very bottom of the structure, including the part of the arc embedded in the ground. $\endgroup$ Aug 29, 2019 at 10:21
  • $\begingroup$ @JansthcirlU There is no part of the structure underground: the physical structure is composed of two arcs (one smaller than the other), mounted so that they belong to the same circle. See the figure I included in my answer. $\endgroup$ Aug 29, 2019 at 11:14
  • $\begingroup$ @JansthcirlU Moreover, the equations you wrote above cannot be right. $\endgroup$ Aug 29, 2019 at 11:22

Welcome. If the width of arc is a and its height i h, and the arc is a part of a circle due to wolfram we have:

$a=2\sqrt {h(2R-h)}$

Where R is the radius of circle. Putting a=75 and h=60 we get $R=41.72$

But R can not be less than h, this solution is not reasonable. However if $a=2\times 75=150$ then we get $R=80>h=60$ which is reasonable. in this case we have:

$Sin (\alpha)=\frac{75}{80}≈0.94$


And the angle opposite the arc is:


Note that this angle can not be greater than $180^o$.

  • $\begingroup$ Thank you for your answer. If I'm not mistaken, is that formula you used not the one to calculate the length of the chord? The width of the arc in my example exceeds the chord length for certain angles. If possible, could you link me to the definitions or the sources you used to get to this result? $\endgroup$ Aug 28, 2019 at 22:33
  • $\begingroup$ Just type 'circular segment' and see Wolfram Mathworld. I assumed that the two edges of arc touch the grond. if angle is $202^o$ then a part of arc must locate under the ground. To me ,total height is from ground to maximum point of arc. $\endgroup$
    – sirous
    Aug 29, 2019 at 7:04
  • $\begingroup$ Thank you for clarifying, however, the height given on the website is measured from the very top of the structure to the very bottom of the structure, i.e. it is not just the height of the arc sticking out above the ground. $\endgroup$ Aug 29, 2019 at 10:31
  • $\begingroup$ Please see my edit. may be a=150 is correct value. $\endgroup$
    – sirous
    Aug 29, 2019 at 10:38
  • $\begingroup$ Why must $h$ be greater than $r$? Given a circle segment can't you have $h \in [0, 2r]$ for arc angles $\theta \in [0, 2\pi]$? $\endgroup$ Aug 29, 2019 at 10:42

Yes it is, if and only if the values of height and width are in the correct range proportion, thus the radius doesn't matter, only the proportionality of the height/width.

It is possible to state that a circle arc $\beta$ starting at $\alpha$ from cero is framed by a rectangle proportionally to the radius, shuch as

$ \forall (\alpha, \beta) \in \mathbb{R} \rightarrow h/w = ( 1 + \sin\alpha ) / ( 1 + \cos(\alpha + \beta)) \bigg| \alpha \geq (180-\beta)/2 \land \alpha \leq (270 - \beta)/2 $

As $\beta$, as well as $w,h$ (width and height) are known, the only unknown value is $\alpha$, and it seems easy to put on the left. If $\alpha$ is not in the range of the condition $\alpha \geq (180-\beta)/2 \land \alpha \leq (270 - \beta)/2$ then the values of width and height are not correct.

Rectangle frame of an arc

  • $\begingroup$ I'm not quite sure if I understand that definition you posted, but I'll see if I can make sense of it and apply it to my situation, thanks! $\endgroup$ Aug 28, 2019 at 22:35
  • $\begingroup$ If you want I can try to explain better that. Note that $\beta$ in this equation/proposition is the difference with the complete circle and $\alpha$ is the rotation of the arc, See the draft in the link. For more details don't heasitate to ask me for. $\endgroup$
    – Izar Urdin
    Aug 29, 2019 at 4:06

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