This appeared in another thread.
Sector OAB is a quarter of a circle of radius 3. A circle is drawn inside this sector, tangent at three points. What is the radius of the inscribed circle? Express your answer in simplest radical form.
Let's see what geogebra can do.
1) Create point (0,3) and call it A. Create point (3,0) and call it B. Create point (0,0) and call it O.
2) Use the "Circular arc with center between two points," tool to create sector OAB.
3) Enter in the input bar f(x)=x.
4) Remove the x and y axes.
5) Draw line segments OA and OB.
6) Create a slider called c and min = 0 and max = 3. Create a point (f(c),f(c)) and call it C. This point will ride along f(x) when we adjust the slider.
7) Draw points (0,y(C)) and (x(C),0) they will be called D and E.
8) Draw line segments DC and CE. Check the algebra pane and the distances will be displayed.
9) Use the intersection tool to find the point on arc AB that intersects f(x). It will be called I.
10) Hide f(x) and draw line segment CI, your drawing should look like fig 1.
11) Enter in the input bar e + g - 2h. You should get in the number pane i = -1.171572875253812.
Remember to have set your options to 15 digits.
12) Select the slider and use the right arrow to click carefully until i becomes positive and then one click back with the left. The first stop will show the slider at 1.2
13) Adjust the slider to have an increment of .001 and then press the right arrow until i again just turns positive and the back it off by one. You should have i = -0.003093511065233.
14) Repeat this process until you get
i = -0.000000034374931
c = 1.242640680000005.
What we have here now is C being the center of the circle we want. c is the radius of that circle.
Your drawing should look like fig 2.
How can we get more from the number c = 1.242640680000005? That is easy, just take it over to Wolfram Alpha and enter this.
so that is the radius and we are done with geogebra.