Rectangle inscribed in a circular sector of angle 60

Question

My apologies if this has been asked before.

Given a circular sector, say of radius $r$, with internal angle $60^{\circ}$, construct a rectangle inscribed in that sector so that the length of the rectangle is twice the width.

I am looking for a simple construction. This is from a chapter that talks about isometries and similarity transformations (primarily dilations). I have a method using trigonometry, but given this is not covered in the text (Libeskind) there must be a solution that is simpler and more elegant.

The ugly trigonometric method:

Let $O$ be the centre of the circle of which the sector is a portion. Let the desired rectangle be $ABCD$ with $A$ and $B$ on the straight segments of the sector, $C$ and $D$ on the arc.

We require $\angle DAB = 90$, $\angle ABD=30$.

Let $OA=d$. Then by symmetry $OB=d$ and $AB=d$ (as $\triangle OAB$ must be equilateral). Then $AD=2d$ and also $\angle OAD = 60 + 90 = 150$.

Then by the cosine rule in $\triangle OAD$ we get $r^2 = \sqrt{3} (\sqrt{3}+6)d^2 $.

So I could construct $d$ this way, but it would be very tedious.

Any help with an easier solution that does not use trigonometry would be great.

Answer 1

Start with your sector. $A$ is the center of the associated circle, $B$ and $C$ are the endpoints of the arc.

Construct line segment $BC$ and the bisector of angle $BAC$, which intersect at $D$. Bisect angle $ADB$; this bisector intersects $AB$ at $E$. From $E$ construct the line perpendicular to line $BC$ which intersects line segment $BC$ at $F$. Draw ray $AF$ which intersects arc $BC$ at $G$. From $G$ construct the line perpendicular to $AD$ which intersects arc $BC$ at a second point $H$. $GH$ measures a long side of the rectangle whose other two vertices lie on rays $AB$ and $AC$.

Answer 2

Analysis:-

We want to construct the rectangle ABCD inscribed in the $60^0$ - sector OPQ such that $AD = 2AB$.

Our 1st target is find $\theta (= \angle AOD)$.

Applying sine law to $\triangle OAD$, we have

$\dfrac {1}{\sin (30 - \theta)} = \dfrac {2}{\sin \theta}$

… Using compound angle formula, and special angle values and rationalization, we have …

$\tan \theta = \dfrac {\sqrt 3 - 1}{2}$ ($= 20.xxx$ degrees approx.)

Our next target is to construct such $\theta$.

1. Drop $PR \bot OQ$ cutting $OQ$ at $R$. (Note that $OR = 1, OP = 2, PR = \sqrt 3$.)

2. Draw circle $OSQ$ (centered at $R$, radius $= RO$) cutting $PR$ at $S$. Then $PS = \sqrt 3 – 1$.

3. Draw the blue circle using PS as radius.

4. Through P, draw the perpendicular to OP, cutting the blue circle at T (a point nearer to Q). Then, $\triangle POT$ meets our requirement with $\angle POT = \theta$.

5. The point where OT intersects the red arc is our point D.

6. Reflect D about the angle bisector of $\angle POQ$ to get C. It is not difficult to get B and A.

NB Construction of D is simple but the analysis is a bit involved.