NeoPets Riddles (Lenny Conundrums) and Answers Here

Of course it's related to Math, gah I hate it. ;;

I think the fact that they are tangent to one another is a clue to finding out the area in between the circles.

I got 596729.8166 for the area of all of the circles combined. Is that right?

I think that is the easy part.

For the other part, you can use the three centers as vertices for a triangle and subtract three sectors that are formed.

Does anyone have the answer? I need the av i was too late last week please help?

this is tooooo much, I think I will just eyeball it and take a wild guess

This one is easy....i was waiting for something more ...difficult...good luck. It's not as difficult as it seems...I'll sumit the answer on my son's account. YOu need a bit of math...

Well, I got the formula for congruent tangent circles.. cause they make an equilateral triangle when you join the centres. But, these circles have different radii.

Is the answer 1240

I got 623671 for the whole thing, but you should verify that in case I made a mistake...

What formulas did you use to figure out the area in between the circles?

That's exactly what I got! I may actually be right...

They still make a triangle, it's just not equilateral...

Calculating the area of the circles is easy. Nine circles, apply A = pi * r^2 to calculate the area of each circle from the radius.

There are seven areas in between the circles. Look at the triangle formed by the lines between the centers of the three tangent circles. If the triangles have radii R1, R2 and R3, then the triangle has sides a = R1 + R2, b = R1 + R3 and c = R2 + R3. Heron's Formula tells us that the area of this triangle is A = sqrt(s*(s-a)*(s-b)*(s-c)) where s = (a + b + c) / 2.

This triangle includes the section between the triangles, but it also has a section of each triangle, so if we can figure out the areas of those sections, we can subtract them from the area of the triangle to get the area of the part between the triangles.

Let La, Lb and Lc be the angles opposite sides a, b and c, respectively. We can use the law of cosines to calculate the cosine of each of these angles. The law of cosines is a^2 = b^2 + c^2 - 2*b*c*cos(La). Solving for cos(La), we get cos(La) = (b^2 + c^2 - a^2) / (2*b*c). There are similar formulas for the other two angles. We can then take the arccosines to get La, Lb and Lc.

Now, assuming we express La, Lb and Lc in radians, and knowing that a circle contains 2*pi radians, we can find the area of a section of circle with radius R and angle L as A = (L/2)*R^2. La will be the angle in the circle with radius R3, Lb the angle in the circle with radius R2, and Lc the angle in the circle with radius R1. Calculate the areas of the three sections and subtract them from the area of the triangle that we calculated above to find the area of the part between the three circles.

Add the areas of the nine circles and the seven parts between the circles to get the total area.

Given three radii, r1, r2, and r3:

triangle area = SQRT((r1+r2+r3)*r1*r2*r3)

sector1 area = ACOS((r1^2+r1*r2+r1*r3-r2*r3)/((r1+r2)*(r1+r3)))*r1^2/2

sector2 area = ACOS((r2^2+r2*r1+r2*r3-r1*r3)/((r2+r1)*(r2+r3)))*r2^2/2

sector3 area = ACOS((r3^2+r3*r2+r3*r1-r2*r1)/((r3+r2)*(r3+r1)))*r3^2/2

Subtract the three sector areas from the triangle area.
Note: the angle returned by ACOS() is in radians.

That's exactly what I did, but with a ti-84 program I wrote.

I used Excel.

Wow, that's a lot of equations. Thank you for sharing your method -- it's gonna take me quite some time to get through so many calculations.

This is the same as the formulas I used, but expressed in terms of the radii of the three circles instead of defining the sides and angles of the triangle. Should be a bit easier to calculate than the process I used. I used Excel as well.