For those bored, a little math puzzle

[LennoxLewis]

What is the radius of the largest circle that you can fit in one quadrant of a bigger of radius 1?

 

[SticksandStones]

Real quick, I'm going to say sqrt(2)/4.

 

[Gokul43201]

1/(1+sqrt(2))

PS: aka sqrt(2)-1

 

[Topher925]

R_small = R_large x (2^0.5-1)

 

[LennoxLewis]

1/(1+sqrt(2))

PS: aka sqrt(2)-1

Good work, how did you get to it?

 

[Gokul43201]

Let the radial position of the center of the incircle be x. From symmetry, this center must lie on the radial line that bisects the angle of the quadrant, i.e., the line at angle pi/4. Draw two radii of the small circle: one ending on the arc of the quadrant (running along this line at pi/4) and the other ending on one of its two arms (running normal to the arm). Equating these radii gives you 1-x=x/sqrt(2), and the radius of the incircle, r=1-x.

 

[LennoxLewis]

Let the radial position of the center of the incircle be x. From symmetry, this center must lie on the radial line that bisects the angle of the quadrant, i.e., the line at angle pi/4. Draw two radii of the small circle: one ending on the arc of the quadrant (running along this line at pi/4) and the other ending on one of its two arms (running normal to the arm). Equating these radii gives you 1-x=x/sqrt(2), and the radius of the incircle, r=1-x.

Right, i used a similar but less effective way. By similar means, i set up the equation (1 - x)^2 = x^2 + x^2, asin Pythagoras's rule for that triangle.