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## Monday, 24 February 2014

### Sunday Afternoon Maths I Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### Burning Ropes

Light one rope at both ends and the other at one end. When the first rope finishes burning, light the other end of the second rope. The second rope will finish burning 45 minutes after the start.
##### Extension
What are all the possible times you can measure with two ropes? How about three ropes? Four ropes? n ropes?

#### Pole Position

The two poles can be any distance apart; the distance does not affect the heights.
##### Extension
If the heights of the two poles were 12m and 24m tall, what height would the intersection of the lines be?
If the heights of the two poles were am and bm tall, what height would the intersection of the lines be?

#### Circles

Call the blue area B and the red area R.
R = Area of quarter circle - Area of semicircle - (Area of semicircle - B)
R = ¼π(2r)2 - ½πr2 - (½πr2 - B)
R = πr2 - ½πr2 - ½πr2 + B
R = 0 + B
R = B
##### Extension
Prove that the red and blue areas are equal.

#### Pyramid and Tetrahedron

Let the length of a side of a triangle be 2L (I used 2L and not L to get rid of fractions in the calculations).
By Pythagoras' Theorem, the height of a triangle is L√3.
Using Pythagoras' Theorem again, the height of the square-based pyramid is L√2.
Therefore, the volume of the square-based pyramid is ⅓×(2L)2×L√2.
This simplifies to 4/3√2L3.

Next, we find the area of the tetrahedron.
Call the point on the base of the tetrahedron, directly below the vertex at the top A.
Using cosine in the triangle made by A, the corner of the base and the midpoint of a side of the base, the distance from the corner to A is 2/√3L.
Using Pythagoras' Theorem yet again, we find that the height of the tetrahedron is L2√2/√3
Therefore, the volume of the tetrahedron is ⅓×½×2L×L√3×L2√2/√3.
This simplifies to 2/3√2L3.

Finally, the ratio of volume of the square based pyramid to the tetrahedron is:
4/3√2L3 : 2/3√2L3
2 : 1
##### Extension
What would the ratio be if they were isosceles triangles?

#### 8! minutes

8×7×6×5×4×3×2×1 minutes
= 8×7×4×3×1 hours (dividing by 60)
= 7×4×1 days (dividing by 24)
= 4×1 weeks (dividing by 7)
= 4 weeks
##### Extension
8 is the smallest number n such that n! minutes is a whole number of weeks.
What is the smallest number m such that m! seconds is a whole number of weeks?

## Sunday, 23 February 2014

### Sunday Afternoon Maths I

Following some relatively popular Twitter posting of maths problems, I've decided to start posting a weekly collection of interesting puzzles I have encountered. I'll be posting solutions on the following Monday.
Here's this week's collection, including puzzles from this month's MathsJam:

#### Burning Ropes

You have two ropes and some matches. Each rope, if lit at its end, will burn for 60 minutes. But the rate of burning is not regular, so cutting a rope in half doesn’t result in a burn time of 30 minutes. How can you use the ropes to time exactly 45 minutes?

#### Pole Position

Two poles stand vertically on level ground. One is 10 feet tall, the other 15 feet tall. If a line is drawn from the top of each pole to the bottom of the other, the two lines intersect at a point 6 feet above the ground. What’s the distance between the poles?

#### Circles

There is a quarter circle with radius 2r and centre A and two semi circles with radius r and centres B and C.
Prove that the red area is equal to the blue area.

#### Pyramid and Tetrahedron

If four equal equilateral triangles form the sides of a square-based pyramid, what is the ratio of the volume of the pyramid to the volume of the tetrahedron whose sides are the four triangles?

#### 8! minutes

How many weeks are there in 8! (8×7×6×5×4×3×2×1) minutes?