Three rectangles all of different sizes, two aligned left and one right

Three rectangles all of different sizes, two aligned left and one right

Considering the image above, where:

  1. rectangles A, B and C must preserve their aspect ratio
  2. height(A) + X + height(B) = height(C) = frame height
  3. width(A) = width(B)
  4. width(A) + X + width(C) = frame width
  5. the frame aspect ratio must be preserved
  6. X is constant

How can I calculate the width and height of each rectangle (A, B, C) so that all the rules above are met, for any frame height or width?


I need this so that I can arrange photos in groups of three.

When I get a group of three photos, I need to always find the optimal arrangement for them, on a given space, maintaining their aspect ratios and so that they all align neatly.

The inner margin between the photos is constant. It's always the same amount regardless of the available space.

My fundamental problem is that the frame width and height and this 'x' (inner margin) are set in millimeters.

But from the photos I only know their width and height in pixels.

This initially seemed like a trivial problem but soon found out it's not. I'm struggling to find a solution to this.

  • $\begingroup$ What do you mean by preserving their aspect ratios? $\endgroup$ Commented Oct 7, 2021 at 17:45
  • $\begingroup$ Even if you know $X$ and the frame width and frame height, you cannot solve for the height of A, as you could just change it while changing the height of B by an opposite amount $\endgroup$
    – Henry
    Commented Oct 7, 2021 at 18:00
  • $\begingroup$ @RolandKillian, when the frame is scaled, all rectangles must preserve their respective ratio between width and height $\endgroup$
    – tsimmi
    Commented Oct 7, 2021 at 19:05
  • $\begingroup$ Did you try to assume aspect ratios and than solve the six resulting equations (assuming x is given)? $\endgroup$
    – Moti
    Commented Oct 8, 2021 at 1:51
  • $\begingroup$ @Moti, x is given yes. I did try that but I'm struggling a bit. How would you go about doing it? Many thanks $\endgroup$
    – tsimmi
    Commented Oct 8, 2021 at 11:27

1 Answer 1



The aim of our answer is to show analytically that what OP is striving for is viable only in a very special case of the described scenario. A set of equations to determine heights and widths of the photos for this case is given.

For brevity, we use simple letters $h$, $w$, $s$, and $r$ to denote heights, widths, scale factors, and aspect ratios of the geometrical entities we are dealing with. They are subscripted with capital letters $\bf{A}$, $\bf{B}$, $\bf{C}$, and $\bf{F}$ to show that they belong to photo-A, -B, -C, and frame respectively. The lateral gap between photos is denoted by $x$.

We wish to scale down (or up) the three photos encased in a frame, say Frame-0, (see $\mathrm{Fig.\space 1}$), so that they fit neatly as shown in $\mathrm{Fig.\space 2}$ inside a new frame, say Frame-1 while observing following constraints.

$\qquad 1.\space \text{The lateral gap between the photos is the same (i.e. $x$) for both scenarios.}$

$\qquad 2.\space \text{Scaling of photos must be distortionless (i.e. their aspect ratios should be preserved)}.$

Theoretically, the gap between photos and the scale factors can have any positive real values greater than zero. But in practice, we do not let scale factors to be equal to 1, since that means no scaling.

We hope to derive a relationship between the aspects ratios of the two frames as a function of scale factor of photo-A, (i.e. $s_A$), and gap between the photos, (i.e. $x$). To do this we use the three equations given in OP’s problem statement and several of ours.

$\qquad\qquad\text{Equations for the Frame-0:}$ $$h_\mathrm{F0} =h_\mathrm{C} = h_\mathrm{A} + x + h_\mathrm{B} \tag{1}$$ $$w_\mathrm{B} = w_\mathrm{A} \tag{2}$$ $$w_\mathrm{F0} = w_\mathrm{A} + x + w_\mathrm{C} \tag{3}$$

$\qquad\qquad\text{Equations for the Frame-1:}$ $$h_\mathrm{F1} =s_\mathrm{C} h_\mathrm{C} = s_\mathrm{A} h_\mathrm{A} + x + s_\mathrm{B} h_\mathrm{B} \tag{4}$$ $$ s_\mathrm{B} w_\mathrm{B} = s_\mathrm{A} w_\mathrm{A} \tag{5}$$ $$w_\mathrm{F1} = s_\mathrm{A} w_\mathrm{A} + x + s_\mathrm{C} w_\mathrm{C} \tag{6}$$

Equations (2) and (5) give us $$s_\mathrm{B} = s_\mathrm{A}. \tag{7}$$

Using (1), (2), and (5), we can show that $$h_\mathrm{F1} = s_\mathrm{A} h_\mathrm{F0}+ \left( 1 - s_\mathrm{A}\right) x \qquad\text{and} \tag{8}$$ $$s_\mathrm{C} = s_\mathrm{A} + \left( 1 - s_\mathrm{A}\right)\frac{x}{ h_\mathrm{F0}} .\qquad\qquad \tag{9}$$

Replacing the last term on the right hand side of (6) using (3) and (9) yields the following expression for $w_\rm{F1}$, $$w_\mathrm{F1} = s_\mathrm{A} w_\mathrm{A} + x + \bigg\{ s_\mathrm{A} + \left(1- s_\mathrm{A}\right) \frac{x}{ h_\mathrm{F0}} \bigg\} \left(w_\mathrm{F0}- w_\mathrm{A} -x\right), $$

which can be expanded and simplified to get, $$w_\mathrm{F1} = s_\mathrm{A} w_\mathrm{A} +\left(1- s_\mathrm{A}\right)\left(h_\mathrm{F0} + w_\mathrm{F0}- w_\mathrm{A} -x \right) \frac{x}{ h_\mathrm{F0}}. \tag{10}$$

Now, we can write down an expression for the aspect ratio, i.e. $\frac{h_{F1}}{w_{F1}}$ of the frame-1 in terms of the aspect ratio, i.e. $\frac{h_{F0}}{w_{F0}}$ of the frame-0 using (8) and (10). $$\frac{h_{F1}}{w_{F1}} = \frac{ s_\mathrm{A} h_\mathrm{F0}+ \left( 1 - s_\mathrm{A}\right) x}{ s_\mathrm{A} w_\mathrm{A} + x + \Big\{ s_\mathrm{A} + \left(1- s_\mathrm{A}\right) \frac{x}{ h_\mathrm{F0}} \Big\} \left(w_\mathrm{F0}- w_\mathrm{A} -x\right)} $$

When simplified, this reduces to $$r_\mathrm{F1}= \cfrac{h_\mathrm{F1}}{w_\mathrm{F1}} = \cfrac{r_\mathrm{F0}}{1+\cfrac{x\left(1 - s_\mathrm{A}\right)\left(h_\mathrm{F0}- w_\mathrm{A}-x\right)}{\Bigl\{ s_\mathrm{A} h_\mathrm{F0}+x\left(1 - s_\mathrm{A}\right)\Big\} w_\mathrm{F0}}}. \tag{11}$$

If we manage to maintain the aspect ratio of the frame while scaling, we have $r_\mathrm{F1}= r_\mathrm{F0}$. For this to happen, we must have, $$x\left(1 - s_\mathrm{A}\right)\left(h_\mathrm{F0}- w_\mathrm{A}-x\right)=0. \tag{12}$$

This has three diffrent solutions, each one is corresponding to a special cases of the scaling scenario, i.e. $$x = 0,\qquad\quad\tag{13}$$ $$s_\mathrm{A} = 1, \quad\text{and} \tag{14}$$ $$x= h_\mathrm{F0}- w_\mathrm{A}. \tag{15}$$

We have already stated at the beginning of this answer that we are not interested in the first two cases. Therefore, there remains only one case, namely $ x= h_\mathrm{F0}- w_\mathrm{A}$, which gives us exactly what OP want. However, whether this case can be realized in practice at will or not is another story. Please note that (15) implies $h_\mathrm{F0} \gt w_\mathrm{A}$.


We can give a test that will show whether a frame-0 with its three photos can be successfully scaled to obtain a frame-1 with all aspect ratios intact. As shown in $\mathrm{Fig.\space 3}$, draw the two diagonals of the rectangle-A and rectangle-B to intersect each other. If this point of intersection lands anywhere on the left vertical side of the rectangle-C, you can perform the scaling while preserving the aspect ratios of all entities.

The following three equations can be used to calculate the scale factor of each of the photos, using which the width and height of the individual photo can be determined. $$s_\mathrm{A} = s_\mathrm{B} = \frac{h_\mathrm{F1} - x}{h_\mathrm{F0} - x} \tag{16}$$ $$s_\mathrm{C}=\frac{ h_\mathrm{F1} }{ h_\mathrm{F0} }\tag{17}$$


Unless the photos and the frame of the original set satisfy the geometrical constrain (15), i.e. $$x = h_\mathrm{F0} - w_\mathrm{A},\quad\text{where}\quad h_\mathrm{F0} \gt w_\mathrm{A},$$ they cannot be scaled while preserving their respective aspect ratios to arrange them inside a new frame, the aspect ratio of which is equal to that of the original frame.

  • $\begingroup$ Fantastic answer. Thank you very for the thoughtful clarification. $\endgroup$
    – tsimmi
    Commented Oct 26, 2021 at 14:52

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