[bonanova]

Sketch a circle, and another circle whose center lies on the perimeter of the first.

At one intersection of the circles center a third circle - all three of the same radius.

The center of each circle lies on an intersection of the other two.

See that funny little curved triangle in the center?

Quickly compare its area to 1/4 of the area of one of the circles.

.

1. Less than?
2. Equal to?
3. Greater than?

.

Clue: Draw more circles.

[Guest]

equal

[Guest]

Less than.

The shape is clearly an equilateral triangle with 3 extra little slices of a circle added to the sides.

If the circles have radius r, then the equilateral triangle has an area = sqrt(3)(r^2)/4;

Looking at one of the circles and considering the sector of that circle between the centers of the other two circles, we see that the angle is 60 degrees, since

the triangle is equilateral. 60/360 = 1/6, therefore the sector's area is 1/6 the area of the entire circle or (pi * r^2)/6. If we subtract the area of the equilateral

triangle from the area of the sector, we get the area of one of the little slices of the circle.

r^2(pi/6 - sqrt(3)/4)

Since the figure is symmetrical, we know that the other slices have the same area.

We multiply this number by 3, and then add the area of the equilateral triangle.

3r^2(pi/6 - sqrt(3)/4) + sqrt(3)(r^2)/4

After simplification, we get (r^2/2)(pi-sqrt(3));

Multplying and dividing by pi/2 we get

(pi r^2)/4 * (2 - 2*sqrt(3)/pi).

The left part is 1/4 the area of the circle, which we were asked to compare to.

The right side is approximately equal to 0.897342209;

Since this number is less than one, we know that the area of the triangular thing is less than the area of 1/4 of the circle.

[Guest]

I'll go ahead and say it is greater than 1/4. To me it looks more like a third. The two smaller ones on either side look as if they are each about half the area of the center triangle, and the cresent shape across looks like it would be equal. That sums up to be 1/3. That's just me looking at it with no math. (not that I'd be able to do any sort of math to figure it out anyway

[Guest]

Less than.

The right side is approximately equal to 0.897342209;

Since this number is less than one, we know that the area of the triangular thing is less than the area of 1/4 of the circle.

I used the same way and found the same figure. So he must be correct. But maybe there exists a more simple way avoiding calculaters???

[Guest]

lesser

Edited August 20, 2009 by bonanova
spoiler added

[bonanova]

I used the same way and found the same figure. So he must be correct. But maybe there exists a more simple way avoiding calculaters???

Yes.

Check the OP - I added a clue

[plainglazed]

...less than by the amount of one segment left from the inscribed equalateral triangle formed at the three intersecting points that make up that "funny little curved triangle in the center". Or so it appears. If you connect all the points of intersection you get three equalateral triangles so the chord formed by the outermost points where circle two and circle three intersect circle one is actually a diameter (or just adding the three 60 degree angles of the three equilateral triangles). If we use this diameter as the line of symmetry and look at the bilaterally symmetrical result, we see the two halves overlap the same amount as four of the segments described above.

[Guest]

Yes.

Check the OP - I added a clue

I drew all 7 circles last night along with two perpendicular diameters of the center circle, one of which was tangent to another two other circles. Then, looking at one quadrant, it appeared that the area of the funny little convexish triangle not included in the funny little curved triangle in that quadrant is probably equal to the area of the chunk of the funny little curved triangle that bulges out of the other side of the quadrant. Thus, the area of the funny little curved triangle is equal to the are of a quadrant. Did I make myself clear?

Edited August 20, 2009 by Brighterthan1000rocks

[plainglazed]

...less than by the amount of one segment left from the inscribed equalateral triangle formed at the three intersecting points that make up that "funny little curved triangle in the center". Or so it appears. If you connect all the points of intersection you get three equalateral triangles so the chord formed by the outermost points where circle two and circle three intersect circle one is actually a diameter (or just adding the three 60 degree angles of the three equilateral triangles). If we use this diameter as the line of symmetry and look at the bilaterally symmetrical result, we see the two halves overlap the same amount as four of the segments described above.

looking at the drawing and the above proof(?) of the line connecting the outermost points where circle two and circle three intersect circle one is actually a diameter, we can see that half the area of the circle equals three equilateral triangles of length r (radius) and three segments formed by 60 degree angles. The area of the "funny little curved triangle in the center" is one equilateral triangle of length r and three of the segments. So which is greater, one such equilateral triangle or three such segments answers the question. It's easy to see that three such segments fit into the triangle with room to spare so the "funny little curved triangle in the center" is less than one fourth the area of the circle.

[Guest]

look at the sections and you will see that it is less than.

[Guest]

geometry I.bmp

If the two red zones are equal in area, and I believe them to be, though I am a long way from being able to do the math, then the area of "the triangle" equals a quadrant of the circle.

[bonanova]

The consensus has it.

Carve the circle into two types of shapes.

The triangle has 1/4 [3/12] of the circle's dark blue shapes.

But only 1/6 of its light blue shapes.