I came across the flag of Trinidad and Tobago and it got me thinking about the geometry of that diagonal. Picture below.

Flag of Trinidad and Tobago

If you look at the diagonal, you'll see it doesn't just go from one corner to the opposite. It's placed in such a way that the lower border of the stripe touches one corner, and the upper border touches the opposite corner.

I found a few other flags with diagonals placed in that way, like the flags of Brunei, Republic of Congo, or the Democratic Republic of Congo.

I was thinking, how would you draw that diagonal if you only knew the flag's width and height, and the stripe width. I haven't found an easy way to solve this, and I was wondering if someone has thought about this before, or wanted to give it a try.

I've also read this post before posting this question, thanks to a recommendation from "similar questions". That one is actually the same setup, but the questions asked are different, so I'm still unsure how to solve this problem.

What I've done so far

I'll start with some definitions. The dimensions I've defined can be found on the figure below.

Flag diagram with the dimensions I've defined

The bottom-left corner detail with more dimensions below:

A detail of the diagram above, with more dimensions defined

  • Flag's width: $w$.
  • Flag's height: $h$.
  • Flag's diagonal: $d$. Note this is also the stripe's diagonal.
  • Stripe's width, or "thickness": $t$.
  • Stripe's length: $l$. Note this is the distance $\overline{AD}$, or $\overline{BC}$.
  • The angle between the stripe and the bottom border of the flag: $\alpha$.
  • Vertical distance $\overline{AB}$: $y$.
  • Horizontal distance $\overline{AB}$: $x$.

I ultimately want to know how to define $x$ and $y$ in terms of $w$, $h$, and $t$. i.e. $x(w, h, t)$ and $y(w, h, t)$.

One thought that I have in mind, but haven't been able to write mathematically is the relation between the thickness and the angle. It should be clear that: $$ \begin{align*} t &= 0 \to \alpha = \arctan(h / w), &B \equiv A\\ t &= w \to \alpha = \pi/2, &B \equiv G \end{align*} $$

The relations I've found: $$ \begin{align} t^2 + l^2 = d^2 \tag{1} \\ w^2 + h^2 = d^2 \tag{2} \\ x^2 + y^2 = t^2 \tag{3} \\ l \cos{\alpha} = w - x \tag{4} \\ l \sin{\alpha} = h + y \tag{5} \end{align} $$

Equations $(1)$, $(2)$ and $(3)$ come from applying the Pythagoras theorem to the triangles $\triangle{ABC}$, $\triangle{AGC}$, and $\triangle{AJB}$, respectively. Equations $(4)$ and $(5)$ come from defining the flag's width and height using the stripe dimensions.

With them, I've done the following steps:

  1. $(4)^2 + (5)^2$, to remove the $\sin$ and $\cos$, arriving to: $$ \begin{align} x^2 - 2wx + y^2 + 2hy + w^2 + h^2 - l^2 = 0 \tag{6} \\ \end{align} $$

  2. Rewrite $(1)$ as $l^2 = d^2 - t^2$, and substitute the value of $d$ from $(2)$, and the value of $t$ from $(3)$, arriving to: $$ \begin{align} l^2 = w^2 + h^2 - x^2 - y^2 \tag{7} \\ \end{align} $$

  3. Substitute $(7)$ in $(6)$, arriving to: $$ \begin{align} x^2 - wx + y^2 + hy = 0 \tag{8} \\ \end{align} $$

So far this looks good to me, but I'm not sure how to get to $x(w, h, t)$ and $y(w, h, t)$. I can do $x(y)$ as: $$ \begin{align} x = \frac{w\pm\sqrt{w^2-4(y^2 + hy)}}{2} \tag{9} \\ \end{align} $$

And from here I could substitute that value of $x$ in $(3)$, but I haven't found a way to isolate $y$ and reach $y(w, h, t)$.


  1. Am I heading in the right direction?
  2. Have I missed some relation that would make this simpler?


  • 1
    $\begingroup$ There is also this related question $\endgroup$ May 13, 2022 at 10:07
  • $\begingroup$ Thanks for the link @JaapScherphuis! Very useful. I think I'll be able to solve it with that info. I didn't take into account how the length $\overline{AE}$ relates with $w$, $h$, and $t$. That could be just the one I mentioned I didn't know how to write mathematically, between the angel and the thickness. $\endgroup$ May 13, 2022 at 12:03


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