1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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, $2r^2y^2-2ry^3-y^4$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $p^2 = 3-q^2$.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.