Circle in pentagon

G

Gamest

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Jan 31, 2011
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Square SQUA with midpoints M and N of sides SQ and SA, respectively, has an area of 64 square units. What is the number of square units, rounded to the nearest integer in the area of the largest circle which can be drawn in pentagon MNAUQ?

I know that MN is 4*(2^1/2), but how does that help?

Thanks in advance
 
There certainly ar other solutions. This one is a bit analytical.

Place you structure on a set of coordinate axes with the origin at Point U.

The equation of the line defined by MN, then is x - y + 12 = 0.

Where is the center of the circle?

1) Wherever it is, it is equidistant from the x- axis and the y-axis.
2) It is also this same distance from the line defined by MN.
3) If 'a' is the distance to the three lines from the center of the circle...

\(\displaystyle \frac{|a - a + 12|}{\sqrt{1^{2}+1^{2}}} = a\)

You're almost done. Let's see what you get.
 
This is a wild guess.

If you inscribed the circle in the original square, it's radius would be 4. (Corrected)

Now, connect the midpoints of all four sides forming a square whose sides are \(\displaystyle 4\sqrt{2}\). A circle inscribed in that square has a radius of \(\displaystyle 2\sqrt{2}\). Would the circle you are after have a radius that is the average of the two radii just mentioned?
 
Well, I'm switching MN so that I get triangle NAM instead of NSM (I like top right quadrant!).

Sooo...following TK's shrewd idea: I have S(0,8), Q(0,0), U(8,0) and A(8,8).
Also M(8,4) and N(4,8). Equation of MN: y = -x + 12

Hark! MN touches circle at (6,6).

Ya'll impressed? :roll:
 
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