34. It is required to take a rectangular frame in a horizontal position along a corridor bounded by vertical walls of which a horizontal cross-section is two concentric semicircles of radii $r$ and $r\ \sqrt{3}$; the frame is of length $2x$ and breadth $y$. One side of length $2x$ is tangential to the inner wall, and the two ends of the opposite side are in contact with the outer wall, as shown in Fig. Misc. Ex. 1-34. Prove that

$$x^{2} = 2r^{2} - 2ry - y^{2}.$$

Prove that, if $x$ and $y$ may vary, the greatest possible area enclosed by the frame is $\frac{1}{2}r^{2}\ \sqrt{3}.$

Fig. Misc. Ex. 1-34.

I noticed a few things that I believe are relevant. The first is that one of the sides of the rectangle is a chord of the larger circle and the second is that the angle between the radius of the smaller circle and the opposite side of rectangle is 90 degrees.

It's been a long time since I had fun with circles, but using Wolfram MathWorld I was able to piece together this much:

Annotated figure

$2x = 2r\sqrt{3}\sin\left(\dfrac{\theta}{2}\right) \Leftrightarrow x = r\sqrt{3}\sin\left(\dfrac{\theta}{2}\right)$

$h = r\sqrt{3} - (y + r) \Leftrightarrow y = r\sqrt{3} - r - h$

$y + r = r\sqrt{3}\cos\left(\dfrac{\theta}{2}\right) \Leftrightarrow y = r\sqrt{3}\cos\left(\dfrac{\theta}{2}\right) - r$

$x^{2} = 3r^{2}\sin^{2}\left(\dfrac{\theta}{2}\right)$

$y^{2} = 3r^{2}\cos^{2}\left(\dfrac{\theta}{2}\right) - 2r^{2}\sqrt{3}\cos\left(\dfrac{\theta}{2}\right) + r^{2}$

But that's about everything of use - does anybody have any hints?

Edit to incorporate the suggestion of André Nicolas:

May we form a triangle as shown in dashed pink and so derive $y = \sqrt{3r^{2} - x^{2}} - r$

enter image description here


A start: By the Pythagorean Theorem we have $y=\sqrt{3r^2-x^2}-r$, which can be rewritten as $y+r=\sqrt{3r^2-x^2}$. Square both sides and simplify.

For the area maximization, equivalently maximize $x^2y^2$, that is, $2r^2y^2-2ry^3-y^4$.

  • $\begingroup$ Thanks - is that equation for $y$ arrived at by forming a triangle consisting of one of my dashed blue lines (radius of the larger circle of length $r\sqrt{3}$), the dashed pink line and the dashed green line as far as the top side of the rectangle (length $y + r$) and the top side of the rectangle (length $x$?). In other words we say: $3r^{2} = (y + r)^{2} + x^{2}$ by the Pythagorean Theorem and rearrange? $\endgroup$ – Au101 Mar 18 '15 at 20:02
  • 1
    $\begingroup$ You are welcome. I have colour deficiencies. Let $AB$ be the chord of the big circle given by the outer "long" side of the rectangle. Let $M$ be the midpoint of that chord, and let $O$ be the centre of the circles. Then looking at $\triangle AOM$ we see that $(AO)^2=(AM)^2+(OM)^2$, so $(\sqrt{3} r)^2=x^2+(OM)^2$. Thus $OM=\sqrt{3r^2-x^2}$. The width $y$ of the triangle is $OM-r$. $\endgroup$ – André Nicolas Mar 18 '15 at 20:10
  • $\begingroup$ Yes, I think we are on the same lines, allow me to edit the question just to confirm. $\endgroup$ – Au101 Mar 18 '15 at 20:12

Place the circle centers at the origin of a Cartesian grid. Without loss of generality, we can assume optimal rectangle is oriented with its base parallel to the $X$ axis, touching the smaller circle at the base midpoint at $(r,0)$.

Then the coordinates of the upper right hand corner can be taken to be $(rp,rq)$ where, since it must be on the outer circle, $$ q = \sqrt{3-p^2}$$

Then the width of the rectagle is $2rp$ and the height is $r(q-1) = r(\sqrt{3-p^2}-1)$. The area is $$A=wh = 2r^2p(\sqrt{3-p^2}-1)$$ $$\frac{dA}{dp} = 2r^2\left[\sqrt{3-p^2}-1-\frac{p^2}{\sqrt{3-p^2}} \right]$$

We can set this to zero and solve for $p$ but since this is not easy I will show the steps. The expression is simpler if we solve for $q$, noting that $p^2 = 3-q^2$.

$$ q-1- \frac{3-q^2}{q}=0\\ q^2 - q - 3 + q^2 = 0 \\ 2q^2 -q+3 = 0 \\ q = \frac{1+\sqrt{1+4\cdot 3\cdot 2}}{4}= \frac{3}{2} \\p = \frac{\sqrt{3}}{2} \\A_{\max} = 2r^2p(q-1) = 2r^2\frac{\sqrt{3}}{2}\frac{1}{2}=\frac{\sqrt{3}r^2}{2} $$


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