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Topic review (newest first)

2012-05-03 17:28:57

Hi Vindaka;

Welcome to the forum.

2012-05-03 13:32:16

Cool stuff!  I hadn't seen that before.  From here, you can find the ratio of the area of the ccsbumririced circle to the area of the triangle.The triangle has area .5*AB*Sin(c)The circle has area Pi*D^2/4, with D = C/Sin(c) = B/Sin(b) = A/Sin(a).  D^2 = AB/[Sin(a)Sin(b)]. Taking all this and plugging it in together we get:A_circ/A_tri = Pi*AB*2/[4*Sin(a)*Sin(b)*AB*Sin(c)]= Pi/[2*Sin(a)*Sin(b)*Sin(c)]

2011-11-05 22:58:05


A unit square is inscribed in a circle which is inscribed in an equilateral triangle. What is the area of the triangle?

We can solve the problem using geogebra and some experimental math. Of course it can be solved analytically but let us just use these tools.

1) Create point A at (1,2)

2) Create point B at (8,2)

3) Use the regular polygon tool and click on A and then B and enter 3 in the upcoming input box. There is an equilateral triangle.

4) Use the midpoint tool and click A and B. Then click B and C. And then click C and A.

You will now have points D,E and F as midpoints of the sides.

5) Use the circle through 3 points tool and click F,E and D.

You will now have an inscribed circle in the triangle.

6) Draw line segment CD.

7) Use the intersect two objects tool and click on the circle and line segment CD.

Point G and H will be created. Make G invisible.

8) Get the midpoint or center of DH and point I will be created. This is the center of the circle.

9) Draw a line through I and parallel to AB.

10) Use the intersection tool and click the circle and this parallel line to create points J and K.

11) Use the regular polygon tool and click on points D and K and leave the 4 and press enter. You now have an inscribed square.

Make the points that are not necessary like L, M, E, F and I invisible. Do the same with the parallel line. Now we just have to get that square to have an area of 1 and we will have our answer.

Create a slider, call it P and give it a min of 2 and a max of 8.

12) Go into the algebra window and set B to look like this:
B = (P,2) and press enter.

13) Now using the slider move P until you get poly2 to be as close to 1 as you can without going under. You should get to 1.04.

14) Adjust the slider increment to .01 and repeat 13. Use the left and right arrow keys for finer motion. Make sure you have rounding set to 15 digits.

15)This time you will get to 1.00041666. Set increment to .001 and repeat.

Use smaller and smaller increments until you get poly2 to be as close to 1 as you can. I got 1.000 000 000 013 734. That is pretty close.

Read what poly1 is. I got 2.598076211389001

We have an empirical answer. We can go with that or we can try using modern techniques to get a little more. We take that number over to:

We enter it into the input box and run "Simple Lookup and Browser for any number." We find our number neatly sandwiched between two constants that are quite close. We take a guess and choose the one above it.

We say experimentally the area of the triangle is:

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