Skip to main content
### Number and algebra

### Geometry and measure

### Probability and statistics

### Working mathematically

### For younger learners

### Advanced mathematics

# Overarch 2

Why do this problem?

This is such a surprising result that one would hope it would intrigue and motivate students to learn more mechanics. It also links to the harmonic series and ideas in pure mathematics. The proof depends on the simple idea of taking moments about a fulcrum.

After the students have done the problem, it is enlightening to discuss the fact that in practice, the building plan has to be made knowing the total of all the overhanging distances first and hence, from the harmonic series, the number of bricks required. This is because, suppose $(n+1)$ bricks are required, the overhang of the first brick to be put in place has to be ${1\over n}$ and the overhang of the topbrick ${1\over 2}$ a brick length.

Possible approach

How about doing some calculations and then building your own structures using dominoes or Jenga pieces? You might challenge the students to build 2 symmetric arches meeting at the top with 10 dominoes and the width at the bottom slightly more than two domino lengths.

As with many problems it is advisable to start with simple cases (here 2 bricks then 3, then 4) and then to make a conjecture about the general result. It is somewhat counter-intuitive to work from the top down but this is inherent in working through these simple cases.

Numbering the bricks from the top down, to achieve the maximum overhang, the $(n-1)$ bricks on top of brick n have centre of mass exactly above the edge of brick $n$, and then we need to take moments about a fulcrum at the edge of the $(n+1)$st brick.

Key question

What forces act on brick $n$ if it balances on a fulcrum at the end of the brick below it?

What is the total number of bricks needed for the top brick to be nowhere vertically above the bottom brick? (Answer: 5 bricks.)

Possible support

Try the problem Overarch 1

Possible extension

Try the problem Harmonically.

Note how the proof that the harmonic series is divergent relates to the second key qestion above. This can be extended to ask how many bricks (roughly) would you use to be certain of getting a total overhang of two brick lengths, then 3, then 4.... giving the number of terms in the partitions of the series in the proof of divergence. However this eventually overestimates the number needed, the divergence proof being based on inequalities. The overarch answers depend on $\log n$ whereas the partial sums in the estimate depend on $\log 2$.

Or search by topic

Age 16 to 18

Challenge Level

- Problem
- Getting Started
- Student Solutions
- Teachers' Resources

Why do this problem?

This is such a surprising result that one would hope it would intrigue and motivate students to learn more mechanics. It also links to the harmonic series and ideas in pure mathematics. The proof depends on the simple idea of taking moments about a fulcrum.

After the students have done the problem, it is enlightening to discuss the fact that in practice, the building plan has to be made knowing the total of all the overhanging distances first and hence, from the harmonic series, the number of bricks required. This is because, suppose $(n+1)$ bricks are required, the overhang of the first brick to be put in place has to be ${1\over n}$ and the overhang of the topbrick ${1\over 2}$ a brick length.

Possible approach

How about doing some calculations and then building your own structures using dominoes or Jenga pieces? You might challenge the students to build 2 symmetric arches meeting at the top with 10 dominoes and the width at the bottom slightly more than two domino lengths.

As with many problems it is advisable to start with simple cases (here 2 bricks then 3, then 4) and then to make a conjecture about the general result. It is somewhat counter-intuitive to work from the top down but this is inherent in working through these simple cases.

Numbering the bricks from the top down, to achieve the maximum overhang, the $(n-1)$ bricks on top of brick n have centre of mass exactly above the edge of brick $n$, and then we need to take moments about a fulcrum at the edge of the $(n+1)$st brick.

Key question

What forces act on brick $n$ if it balances on a fulcrum at the end of the brick below it?

What is the total number of bricks needed for the top brick to be nowhere vertically above the bottom brick? (Answer: 5 bricks.)

Possible support

Try the problem Overarch 1

Possible extension

Try the problem Harmonically.

Note how the proof that the harmonic series is divergent relates to the second key qestion above. This can be extended to ask how many bricks (roughly) would you use to be certain of getting a total overhang of two brick lengths, then 3, then 4.... giving the number of terms in the partitions of the series in the proof of divergence. However this eventually overestimates the number needed, the divergence proof being based on inequalities. The overarch answers depend on $\log n$ whereas the partial sums in the estimate depend on $\log 2$.