Maths Item of the Month Archive 2014

December 2014
A Christmas Star

An eight pointed Christmas star is made with a gold layer and a silver layer.

The star for the gold layer is formed from a regular octagon ABCDEFGH by finding the points of intersection of the squares ACEG and BDFH.

The silver layer is formed from the same regular octagon ABCDEFGH by creating the star ADGBEHCF.

What fraction of the gold layer is covered by the silver layer?

Solution

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November 2014
Three's a Crowd

There is just one form of the graph of a quadratic needed to produce the graph of any quadratic using vertical stretches and horizontal/vertical translations: i.e. the graph of y = x² can be transformed into the graph of any quadratic y = ax² + bx + c.

The graph of y = x³ cannot be transformed into the graph of all cubics y = ax³ + bx² + cx + d using horizontal/vertical stretches and horizontal/vertical translations. How many basic forms of a cubic are needed so that all cubics can be obtained and can you prove that these are sufficient to do this?

Solution

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October 2014
FMSP Favourite Problem Posters

Triangle ABC has a right-angle at A.
Semi-circles are drawn with BA, AC and BC as diameters as shown.

Given that AC = 8 and BC = 10, write down the value of one third of the total shaded area.

This is the first of a series of six problems chosen by the FMSP Central Team available as posters. The aim of the posters is to engage students in problem solving and extend the mathematical ideas and approaches they encounter at GCSE.
The full set of problems and the solution to this problem can be seen at: www.furthermaths.org.uk/favourite

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September 2014
Square strings

Each number in the sequence 49, 4489, 444889, 44448889 is a perfect square.
i.e. the first digits are always 4s, the next digits are 8s but always 1 fewer of them, the units digit is always 9.

Prove that the numbers are always perfect squares.

Can you find another sequence (of the form XXXXYYYZ like above but different digits) that behaves in the same way?

Solution

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August 2014
Cubic with Integer Points

Find a cubic equation, with three distinct roots, such that the roots and the stationary points all lie on points with integer coordinates.

If the coefficient of x³ is equal to 1, what is the cubic with this property where the sum of the absolute value of the roots is as small as possible?

Solution

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July 2014
Summer All Over

Take the set of the reciprocals of all the positive integers from 1 to n.

Show that the sum of the products of the possible subsets (including the complete set) is equal to n.

e.g. for n = 3:

Solution

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June 2014
APGP

An arithmetic progression (AP) is a sequence of numbers with a common difference between them: e.g. 3, 7, 11, 15, …

A geometric progression (GP) is a sequence of numbers with a common ratio between them: e.g. 2, 6, 18, 54, …

If the 1^st, 2^nd and 6^th terms of an AP form a GP what is the common ratio?

If the 1^st, 2^nd and n^th terms of an AP form a GP what is the common ratio?

Solution

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May 2014
How many circles?

Circles plotted on coordinate axes can be categorised based on the following criteria (ignoring the scales on the axes):

• Position of the centre
• Positions of the points of intersection with the x-axis
• Positions of the points of intersection with the y-axis

For example the circles above can be categorised as follows:

Circle A

• Centre in the top-right quadrant
• One intersection with the positive x-axis and one intersection with the negative x-axis
• One intersection with the positive y-axis and one intersection with the negative y-axis

Circle B

• Centre on the positive x-axis
• Two intersections with the positive x-axis
• No intersections with the y-axis

Circle C

• Centre in the bottom-right quadrant
• One (repeated) intersection with the positive x-axis
• Two intersections with the negative y-axis

Using these criteria how many different categories of circles are there?

Solution

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April 2014
Divisibility of consecutive integers

Let N = (n+1)(n+2)...(2n), i.e. the product of n consecutive integers from n+1 to 2n.

Prove that, for any positive integer n, N is divisible by 2ⁿ but not a higher power of 2.

For example, for n = 3:
N = 4×5×6
    = 120.
120 is divisible by 8 but not divisible by 16.

Solution

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March 2014
Inner Circle

In the diagram various regular polygons, P, have been drawn whose sides are tangents to a circle, C.

Show that for any regular polygon drawn in this way:

Solution

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February 2014
Cones from a Circle

An angle θ is cut out of a circle of card to create two sectors: a major sector and a minor sector. The two sectors are then folded to make cones.

What angle θ is required to obtain the largest value for the sum of the volumes of the two cones?

Solution

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January 2014
An Unexpected Answer

Mr Student sets his class the following problem:

A committee of 3 students is to be chosen from a group of 13 students of which 8 are girls and 5 are boys. The students are selected at random, without replacement. What is the expected number of girls on the committee?

Anne Student immediately responds that the answer is .
She gives the reason that there are 3 students to be chosen and the proportion of girls is
so she calculated

Is she correct?
If the number of students was different would her method work?

Solution