# Math Help - Mainly logic with a little math

1. ## Mainly logic with a little math

I got an April Fool's Day set of math problems and I need insight. Here are the problems:

1. A clock strikes six times in five seconds. How many times will it strike 10 seconds? Justify your answer.

2. Let $\gamma$ be a closed non-intersecting curve all of whose points are 1 inch from a given point. Is it possible that $\gamma$ is longer than 1 mile? Justify your answer.

3. Let $\triangle ABC$ be a triangle all of whose sides are shorter than 1 inch. Can the circle which passes through $A$, $B$, and $C$ have circumference which is larger than 1 mile? Justify your answer.

My thoughts:

1. Obvious answer would be 12 but I'm not sure what it's even asking.

2. I can't think of a way this would be even possible.

3. I think the thickness of the circle could make it possible. Make it thick enough to pass through all three sides.

2. ## Problems

I got an April Fool's Day set of math problems and I need insight. Here are the problems:

1. A clock strikes six times in five seconds. How many times will it strike 10 seconds? Justify your answer.
This is like saying: a fence 5 m long has six posts, equally spaced. How many posts will a similar fence have that is 10 m long? Think about it...the posts are 1 m apart.

2. Let $\gamma$ be a closed non-intersecting curve all of whose points are 1 inch from a given point. Is it possible that $\gamma$ is longer than 1 mile? Justify your answer.
No, all the points in a given plane that are 1 inch from a given point lie on a circle, radius 1 inch. And the complete circle only measures $2\pi$ inches all the way round.

3. Let $\triangle ABC$ be a triangle all of whose sides are shorter than 1 inch. Can the circle which passes through $A$, $B$, and $C$ have circumference which is larger than 1 mile? Justify your answer.
Yes it can. You can make the circumference of the circle as big as you like!

How? Well, you can find the centre of a circle by drawing the perpendicular bisectors of two of its chords, and finding where they meet. So if we make, say $AB$ and $BC$, both a half-inch long (so that $BC$ is still less than 1 inch) but very nearly lying in a straight line (in other words $\angle ABC$ is very nearly $180^o$), we can make these perpendicular bisectors very nearly parallel. Hence we can make their point of intersection as far away as we like.

I got an April Fool's Day set of math problems and I need insight. Here are the problems:

1. A clock strikes six times in five seconds. How many times will it strike 10 seconds? Justify your answer.
The clock 'strikes' at the beginning of each second. Therefore it strikes five times in total for the beginning of each of the five seconds, and then once more which marks the end of the fifth/beginning of the sixth.

Therefore for 10 seconds, the clock will strike once for the start of each second, and once to mark the end of the last second. 11 strikes in total.

4. 1)It might not continue chiming through the other seconds. If it is 6'o clock it may only chime 6 times, if it is 8'o clock maybe it chimes 8 times. We dont know what time it is also we dont know the number of chimes per hour (it is usually one, but doesnt have to be!).

2)It depends on how you make a curve. A piece of string can be folded up or wrapped around a sphere to fit the conditions of having all points on the curve 1 inch away from a common point. The limit for length here depends on the string thickness. A thin enough string will be of the required length, in fact given a thin enough string you could make any length.

3)Yes, for the same reasons stated above.