Circle over a rectangle - radius needed (opening is 60" vertical x 25" horizontal; strut is 2.5" wide)

[rich_a]

So here's my challenge - I have a rectangular opening that needs the longest possible square strut installed.

Contextually, I understand the geometry, but I am missing one important element - the radius of my circle.

Here goes - my opening is 60" vertical x 25" horizontal. My strut is 2.5" wide and cut square (no miters for fit) at both ends.

The way for me to solve this is to place a circle with its center-point at the "middle" of my opening where a chord that intersects the edges of my opening is 2.5".

All the geometry falls in place from there and, logically, a defined box (60"x25") can only have one circle with center in the middle and a 2.5" cord at the corners - but how to find that radius?? I'm stumped. Thanks for any insights!!

 

[lev888]

Not sure why the circle is necessary.
Here's my diagram (hope I understand the problem correctly). You can set up a system of 3 equations with 3 variables (a, b, x) and solve it in wolframalpha.com. I'm getting x=63.3018. I used triangle similarity and the Pythagorean theorem. You can test the solution using a cardboard cutout.

 

[Dr.Peterson]

Every way I try this, I get a fourth degree equation to solve. I let Wolfram Alpha solve it:

solve (bc-ax)^2+(bx-ac)^2=(x^2-c^2)^2, a=60, b=25, c=2.5 - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

If you want a general formula, you'll find it here (presumably the second of four):

solve (bc-ax)^2+(bx-ac)^2=(x^2-c^2)^2 - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

The length of the rectangle turns out to be 63.302; the resulting radius is 31.676, if that matters. (I didn't use the circle in my work, only in constructing the drawing.)

​

I am not bothering to show how I got the equation to solve; if anyone wants to see it, it is a little interesting.

 

[limiTS]

Easiest way, I think, is to use similar triangles. Going by lev888's diagram, the small triangle in the bottom right is similar to the big triangle to the right of x. Therefore,

[imath]\dfrac{a}{b}=\dfrac{25-b}{60-a}[/imath]

A second equation involving a and b comes from the Pythagorean theorem:

[imath]a^2+b^2=2.5^2[/imath]

Using a root-finding technique, I get [imath]a=0.895146185704521\ldots[/imath] and [imath]b=2.33424791018834\ldots[/imath]

To find x, use the Pythagorean theorem: [imath](60-a)^2+(25-b)^2=x^2[/imath]

[imath]x=63.3018172109304\ldots[/imath]

Then you can find the radius.

 

[Dr.Peterson]

Easiest way, I think, is to use similar triangles. Going by lev888's diagram, the small triangle in the bottom right is similar to the big triangle to the right of x. Therefore,

[imath]\dfrac{a}{b}=\dfrac{25-b}{60-a}[/imath]

A second equation involving a and b comes from the Pythagorean theorem:

[imath]a^2+b^2=2.5^2[/imath]

Using a root-finding technique, I get [imath]a=0.895146185704521\ldots[/imath] and [imath]b=2.33424791018834\ldots[/imath]

To find x, use the Pythagorean theorem: [imath](60-a)^2+(25-b)^2=x^2[/imath]

[imath]x=63.3018172109304\ldots[/imath]

Then you can find the radius.

My approach was more or less equivalent, but directly in terms of the dimensions of the rectangles.

Taking the outer rectangle as a by b, and the inner rectangle as c by x, and calling the smaller angle [imath]\theta[/imath], the same two triangles yield [math]x\sin(\theta)+c\cos(\theta)=a\\x\cos(\theta)+c\sin(\theta)=b[/math] We can eliminate \(\theta\ by isolating sin and cos, then using the Pythagorean identity, to obtain the equation I gave to WA.