The chord length in the puzzler was 70’. I made a right triangle with the base being 1/2 of the chord and the height of the triangle running from where the chord touches the inner circle to the center of both the circles. The hypotenuse stretches from the center of the circle to where the one end of the chord touches the outer circle so the hypotenuse equals the radius of the large circle ® and the height of the triangle equals the radius of the small circle ®. The R*R=r*r+chord length /2. Therefore, R*R - r*r = 70/2 * 70/2 ft sq. The area needing painting = Pi * R*R - Pi * r*r, or A=Pi*(R*R-r*r). Therefore, A= Pi*70/2*70/2 sq.ft.=Pi*1,225 sq.ft. which is about 3,849 sq.ft. -David Duffy

You lost me at R. I won’t disclose which one.

Good job. It’s not intuitive at all that the area of the carousel floor is equal to the area of a circle with its diameter equal to a cord tangent to the inner circle.

It is definitely not intuitive. What I find interesting is that you can find the area of the “ring”, but not the sizes of the circles.

I just visualized pushing the chord sideways toward the center of the “ring” and intuitively knew that the area reduction of the ring had to match that of the hole. So, when the hole disappeared the chord was the diameter of what was once the ring. and pi r squared times half the diameter (35 ft) gives the same answer as you got without all that triangle stuff.

I don’t know how close of an increment of paint they were going to buy, measure A line across the carousel tangent to the center circle, at a perpindicular to the tape or whatever, measure the width of the center circle and add it to the tangent measurement of the ride, now you have the diameter of both the large and small circle, go from there.

Yes, that is very interesting you don’t need to know the diameter of either of the two circles to figure out the area of the annulus.

I was thinking the sol’n required to make another measurement, the distance from where the cord touched the inner circle to the outer circle, following the path (only in the opposite direction) described above. I think if you made that measurement, then you could figure it out too. But the OP’s solution is clearly better, less measuring to do.

OK. It’s been a long, long time since high-school plane geometry. Would someone please tell me how you know the radius of the inner circle if you can’t measure it?

You don’t know the radius of the inner circle, and don’t need to. That is the crux of the solution. There is an infinite number of circle diameters that work, the inner and outer just have to have the correct relationship.

Ah. After re-reading the solution it made sense. That’s great! I love stuff like this. Thx

The solution given above is incorrect because the painter cannot find the center of the circles.

He needs to draw a second chord that intersects the first and is also tangent to the inner circle. He then needs to measure the angle between the two chords. Call the distance from the outer to the inner circle “a” (35 feet, half the length of the chord by symmetry). Call the angle “2A”.

Upon drawing the diagram one sees that the radius of the outer circle is a*cosA and the and the radius of the inner circle is a*sinA. The area of the annulus is therefore

pi*a^2((cosA)^2 - (sinA)^2) = pi*a^2*cos(2A)

The painter does **not** need to find the center of the circles. Read the answer given under “last week’s puzzler” on the home page for more details.

“Upon drawing the diagram one sees that the radius of the outer circle is a*cosA and the and the radius of the inner circle is a*sinA.”

Are you sure you didn’t mean: Upon drawing the diagram one sees that the radius of the outer circle is a***sec**A and the radius of the inner circle is a***tan**A?

Here’s a solution in one sentence:

Because a mathematical Puzzler will have a unique correct answer, we can infer that knowing only the length of the chord is sufficient, therefore the radius of the inner circle may be any size, including zero, in which case the 70’ chord is now the diameter of the outer circle, and the area is pi x the square of 35.

I know, this “proof” would get me laughed out of any geometry class, but for solving a puzzle it is a valid argument. There are many well-known puzzles that rely on the assumption of a unique correct answer for the solution.

My answer was incorrect. David Duffy’s was indeed correct. When I constructed my right triangle identified the wrong angle as the right angle.

I was wrong. See below.

Oops. The “I was wrong” refers to my earlier (1:00) post. The post agreeing with David Duffy’s answer is correct. The pdf is equivalent to his verbal description.

another way to look at it is there is a general solution, using just the chord length, regardless of the size of the inner circle. If you take the extreme, as the inner circle gets smaller and smaller until it is size of radius 0, then the chord becomes a diameter, with radius R, and the area is R^2 pi.

I solved it a different way. It involves the law of intersecting chords, which says if two chords intersect (at any angle) the product of the two pieces of each cord are equal. So one chord is the 70’ one. The other chord is drawn such that it is perpendicular to the 70’ chord and is also the diameter of the larger circle. Then the two pieces of the smaller chord are 35’ each. The two pieces of the larger chord are: (R+r), and (R-r), where R is the radius of the large circle, and r is the radius of the small circle. So the law of intersecting chords says 35^2=(R+r)*(R-r)=R^2-r^2. The area to be painted is pi(R^2-r^2), which = pi*35^2.