# What is the smallest dimension for an inscribed rectangle of known proportions?

For a circle with a diameter of d, what is the smallest dimension for an inscribed rectangle (maximum height for a horizontally-oriented rectangle or maximum width for a vertically-oriented rectangle) whose sides are proportional by 1920:1080?

In terms of application, what is the smallest dimension (maximum height if horizontally-oriented) of the largest possible camera sensor with a resolution of 1920*1080 pixels that will fit inside a telescope's image circle of diameter 1.25 inches (or any other arbitrary diameter, d)?

• Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos Jul 19 at 7:32
• I tried solving this myself in WXMaxima, but I get two different (real positive) results of either h = SQRT(81*d^2 - 256*w^2)/9 or h = 16*SQRT(d^2 - w^2)/9. Probably not analyzing the problem correctly: (%i2) solve(d^2=((1920/1080)*w)^2+h^2,h); (%o2) [h=-sqrt(81*d^2-256*w^2)/9,h=sqrt(81*d^2-256*w^2)/9] (%i4) solve(d^2=w^2+((1080/1920)*h)^2,h); (%o4) [h=-(16*sqrt(d^2-w^2))/9,h=(16*sqrt(d^2-w^2))/9] – Nicole Sharp Jul 19 at 7:32

There's only one rectangle with given proportions $$p:q$$ inscribed in a given circle (excluding rotation symmetries). The diameter $$d$$ of the circle is the size of an inner diagonal of such rectangle.

Let $$a$$ and $$b$$ be the sizes of the sides of such a rectangle; and let $$\theta$$ be the angle between a diagonal and a side of length $$b$$. Now a diagonal, a side of length $$b$$, and an opposing side of length $$a$$ form a right triangle.

Suppose $$a \leq b$$ and $$p \leq q$$. Given this configuration, we have

• $$\theta = \tan^{-1}(p/q)$$
• $$a = d*\sin(\theta)$$
• $$b = d*\cos(\theta)$$

• $$\theta = \tan^{-1}(1080/1920) \approx 0.512$$ radians
• $$a = d*\sin(\theta) \approx 0.490 d$$
• $$b = d*\cos(\theta) \approx 0.872 d$$

I figured it out. I got the following in WXMaxima to compute either the width or height of the inscribed rectangle:

(%i8) solve(d^2=((1920/1080)*h)^2+h^2,h);

(%o8) [h=-(9*d)/sqrt(337),h=(9*d)/sqrt(337)]

(%i9) solve(d^2=w^2+((1080/1920)*w)^2,w);

(%o9) [w=-(16*d)/sqrt(337),w=(16*d)/sqrt(337)]

or for a rectangle of proportion 1920:1080 inscribed in a circle of diameter d,

h = 9*d/SQRT(337) = ~0.490*d

w = 16*d/SQRT(337) = ~0.872*d