What is the area of the shaded region in this geometry problem?

[guss]

Find the area of the shaded region if r = 1/4 and each side of the square is 1.

I know the solution can involve similar right triangles and systems of equations, but if I remember correctly this was a pretty messy way to do it.

Anyone have a simple (or more complicated) solution? You can assume everything in this image that looks relatively tangent is tangent, by the way.

 

[kramer733]

Thanks. I'll definitely try this one out.

 

[phinds]

Holy crap, Batman ... this is a VERY tricky problem. The answer is .3049

Hint ... if you do it the hard way, you'll end up gnawing your arm off to get out of the trap. Use some trig; that makes it less nasty.

 

[guss]

Holy crap, Batman ... this is a VERY tricky problem. The answer is .3049

Hint ... if you do it the hard way, you'll end up gnawing your arm off to get out of the trap. Use some trig; that makes it less nasty.

Nice work.

Can you please explain how you did it? Or at least give some good hints :)

 

[Bill Simpson]

Or at least give some good hints :)

Hint, without giving away too much: When does a quadratic have only one root?

Think about that for a while until you see how that might relate to this problem.

Hint: Once you see that the problem is simple.

 

[dodo]

In Fig. 1, the green triangle has an hypotenuse of h = sqrt(2) - sqrt(2 * 0.25^2) = sqrt(2) - sqrt(1/8), so the angle at A is arcsin(0.25 / h).

In Fig.2, considering the blue triangle, the angle B is PI/4 and the angle C is PI - A - B, while one side is given: the square's diagonal, sqrt(2).

The gray area, then, is half the area of the square minus the area of the blue triangle. I get 1/2 - (sqrt(17) - 1)/16, or 0.304805898.

 

[JJacquelin]

It is also possible thanks to analytic geometry.
This provides the general formula for any r (0 < r < 0.5)

 

[phinds]

Gee dodo, why don't you just GIVE him the answer. That way he's sure to learn something.

 

[kl0818]

http://img40.imageshack.us/img40/7513/wktxu.png

Note that \angle AOC = \angle DBC = \theta
Also, AB=BD=k

　　DC = DB tan\theta = k tan\theta
\Rightarrow OC = OD + DC = 0.25 + k tan\theta

　　\frac{OA}{OC} = cos\theta
\Rightarrow \frac{0.25}{0.25 + k tan\theta} = cos\theta
\Rightarrow k = \frac{1}{4}\frac{1-cos\theta}{sin\theta} = 0.25 tan\frac{\theta}{2}

Consider \Delta BEF ,
BE = 0.75 - k = 0.75 - 0.25 tan\frac{\theta}{2}

Now, \frac{FE}{BE} = tan\theta
\Rightarrow \frac{1}{0.75 - 0.25 tan\frac{\theta}{2}} = tan\theta
\Rightarrow \frac{4}{3-tan\frac{\theta}{2}} = tan\theta
\Rightarrow tan\frac{\theta}{2} = \frac{1}{2} (\sqrt{17} -3)

Area of Consider \Delta BEF = \frac{1}{2}(BE)(EF) = \frac{1}{2}(\frac{3}{4}-\frac{1}{4}\frac{1}{2} (\sqrt{17} -3))(1) = \frac{9-\sqrt{17}}{16}