Junior Mentoring Oct 2010 - May 2012

amtejcTextBoxSupported by the Man Group PLC Charitable Trust

UKMT Junior Mentoring Scheme Notes

A. Even, odd and prime numbers

If a sum or difference of a pair of numbers is even, then both numbers are even or both areodd. To get an odd sum, one must be even and the other odd. If a product of two numbers isodd, both numbers must be odd.

That that 2 is a prime number. It is the only even prime number. All the other prime numbersare odd. Not every odd number is a prime number!

The number 1 is not a prime number. There would be problems if we included it, because everynumber is either prime or can be written as a product of a unique finite list of prime numbers.To consider 1 as a prime number would mean that we could keep on dividing out prime factorsfrom a number ad infinitum.

B. Useful algebraic factorisations

You may be familiar with multiplying out brackets for squares:

(a+ b)2 = a2 + 2ab+ b2 and (a b)2 = a2 2ab+ b2.A third result which is very useful is:

(a+ b)(a b) = a2 b2,which is sometimes referred to as the (factorisation of) the difference of two squares.

Since a square number is always non-negative, the second result leads to a2 2ab+ b2 > 0.This can be written in various ways, for example:

a2 + b2 > 2ab; a2 + b2

2> ab; a

2 + b2

2>a2b2.

If we now replace a2 with x and b2 with y, we deduce that

x+ y

2> xy for x, y > 0.

Equality only occurs when x = y, and this is known as the arithmetic meangeometric meaninequality (or the AMGM inequality for short).

We can use this, for example, to verify that

15 < 12(3 + 5).

C. Adding series of numbers

Adding a set of numbers going up in regular steps, for example 3 + 8 + 13 + + 98, is easywith the right trick. Add 3 to 98, 8 to 93, 13 to 88 and so on, until you have gone all theway through to adding 98 to 3. You now have 20 pairs of numbers each adding to 101. (Whyare there 20 pairs, and not (98 3)/5 = 19 pairs?) Adding all of these pairs together gives20 101 = 2020, so the original series adds up to half of this, which is 1010. (Why do we haveto halve our sum?)

A series of numbers going up in regular steps like this is called arithmetic.

1

2 UKMT JUNIOR MENTORING SCHEME NOTES

D. Similar and congruent triangles

Two triangles are called similar if both have the same set of angles. The corresponding sidesare then in a fixed ratio to each other. For example, consider these two triangles:

AB

C

P

Q

R

If ABC = PQR and CAB = RPQ, so that by the sum of angles in a triangle, BCA =QRP , the triangle ABC and PQR are similar, so we have, for example,

AB

PQ=BC

QR,

by comparing the sizes of the triangles, or

AB

BC=PQ

QR,

which amounts to the same thing, but says that the ratio of corresponding pairs of sides isequal.

Triangles are called congruent if they have the same set of angles and also their correspondingsides are equal.

E. Right angles in a circle

A

B C

T

It is left to the reader to show that if O is the centre of the circle, then there are two isoscelestriangles in the figure, which leads to the conclusion that BAC = 90.If AT is a tangent to the circle, then OAT = 90.

F. Centroid of a triangle

Many solvers of geometry problems like to draw decent diagrams using geometrical instruments,redrawing them as some ideas become irrelevant and other ideas take on increasing importance.When trying to avoid a symmetrical or right-angled look, use angles near these:

45 60

75

UKMT JUNIOR MENTORING SCHEME NOTES 3

Triangles have many interesting properties, including those of their centres. Here is just oneof them:

There are three lines joining the vertices of the triangle to the mid-points of theopposite sides. These are called the medians of the triangle.

The three medians meet at the centroid (the balancing point of the triangle).The centroid divides each median in the ratio 2 : 1.

Here is a proof of this result using areas of triangles.

Let |ABC| stand for the area of triangle ABC. Let B and C be the midpoints of CA and ABrespectively, and let the medians BB and CC meet at G. We draw the line AG and let AGmeet BC at X. We are going to prove that X is the midpoint of BC. Draw this diagram outfor yourself to see what is happening.

Using common bases or common heights, and using the fact that BB is a median, it followsthat

|ABB| = |CBB| and |ABG| = |CBG| |AGB| = |CGB|Similarly, |CGA| = |CGB| because CC is a median. |AGB| = |AGC|.Now let BX = k.CX. By a similar argument, |AGB| = k.|AGC|.Hence k = 1 and X is the midpoint of BC.

Furthermore, we have

|BGX| = |CGX| |BGX| = 1

2|CGB| = 1

2|ABG|

GX = 12 AG.

You are advised to measure the lengths in your diagram to check this works.

G. Percentage changes

Regular polls and surveys have shown that these are one of the least understood arithmeticaltopics that have to be used in commercial activities by the general population. In the case ofValue Added Tax,1 a percentage is added by the store to the price they want to charge.VAT is currently 15%, but will be rising once again to 17.5% from January. Suppose they wantto collect 12.99. They add 15% of this, namely 1.95, charge the customer 14.94 and paythe government the 1.95. Note that we can also understand this calculation as multiplyingthe original price that the shop wishes to charge by 115

100to determine the actual selling price.

A useful diagram for understanding this is as follows:

Original price Final price 115

100

In the reverse case, suppose they advertise an article at a price of 19.99. To find the pricethat the store collects, we have to reverse the process and divide by 115

100, giving 17.38, so that

19.9917.38 = 2.61 goes to the government. Another way of thinking about this is that1A VAT registered trader claims back from the government the VAT on everything bought, but then pays

the government the VAT on everything sold. In this way, the trader only pays the extra VAT on the markupmade, i.e., the increase in value of the product. This is the reason for the name Value Added Tax.


out of every 115 charged to the customer, 100 goes to the shop-keeper and 15 goes to thegovernment. Note that you cannot multiply by 85

100to find the amount the store-keeper gets.

There is also a term percentage point, and this is used to describe arithmetic differences betweenpercentages. For example, if we think of an item as costing 115 percentage points in a shop, theshop-keeper gets 100 percentage points while the government gets 15. As another example, if thepercentage of Year 6 boys gaining Level 4 or above in their Key Stage 2 SATs in 2002 was 35% andin 2009 it was 70%, we say that there has been a 35 percentage point improvement (being 70 35),even though the actual percentage rise was 100%. This distinction causes much confusion, especiallyin the media.

It is also wise to be suspicious of percentage statistics when applied to small numbers, and alwaysquestion the base on which statistics are applied. Many users try to bamboozle their audience byquoting percentages which are, at best, irrelevant to the argument they are trying to make.

H. Surds

Numbers like

5 are called surds. Their appearance can be off-putting to say the least! How-ever, the best way to treat them is like letters in algebra: they are what they do. In fact, itcan help to replace a surd by a letter, as it may make the algebra less gruesome-looking.

Suppose you put c instead of

5. What happens when you find c2? It comes to exactly 5, soyou can put a number back instead of c2. What about c3? That is c2 c = 5c = 55. Notethat c4 = c2 c2 = 5 5 = 25 exactly.Consider an equilateral triangle with side length 2, as shown below. (Take the dimensions asshown and angles as right angles where they look like 90 in the diagram.) By PythagorasTheorem applied to have the equilateral triangle, you can see the height of the triangle must beexactly

3. Applying Pythagoras Theorem again, you can then see that x must be exactly 4.

Changing into decimals can never prove this result.

22

2

x

13

I. Circle geometry

Why should a tangent be perpendicular to the radius of the circle at the point of contact?

We first show that a radius which bisects a chord is perpendicular to the chord. Consider thisfigure:

O

P Q

CT

Let M be the midpoint of a chord PQ of the circle.


OP = OQ as both are radii of the circle, so OPQ = OMQ.Also, PM = QM , therefore 4OPM is congruent to 4OQM , so OMP = OMQ.Since OMP +OMQ = 180, it follows that OMP = OMQ = 90, so OM is perpendic-ular to PQ.

We can also prove the converse, namely, if OM is perpendicular to PQ, then M is the midpointof PQ.

Now we can consider tangents and radii. Let OM meet the circle at C, as in the diagramabove. Construct TC parallel to PQ and therefore perpendicular to OC. Take T be the pointwhere OP meets the tangent TC. Since OC > OM , we must have OT > OP as 4OPM issimilar to 4OTC. Therefore T must lie outside the circle.This means that no point of the tangent TC can lie inside the circle, as we can draw a linefrom O to any point on the tangent, calling P the point where the line meets the circle andchoosing Q on the circle such the PQ is parallel to CP .

Finally, as the tangent TC meets the circle at C, the tangent touches the circle at only thisone point.

From radius, chord and tangent properties, we can deduce a number of significant theoremsabout angles in a circle:

(a) Angle at the centre is twice that at the circumference

A

B

C

O

BOC = 2BAC

This can be proven by drawing in the line OA and considering isosceles triangles.

(b) Angles on a chord in the same segment are equal

A

B

C

A

BAC = BAC

This can be proven by using the previous theorem: if O is the centre of the circle, thenBOC = 2BAC = 2BAC.

(c) Opposite angles in a cyclic quadrilateral are supplementary


A

B C

C

E

DAB + BCD = 180 DAB + DCE = 180.

This can be proven either by using angle at the centre of a circle is twice the angle at thecircumference. Alternatively, draw radii from the centre of the circle to the vertices of thequadrilateral, and notice that there are four isosceles triangles formed. Noting that theangles of a quadrilateral sum to 360, the result then follows.

(d) The alternate segment theorem

A

B

CT

BCT = BACThis is easiest to show in the case that AC is perpendicular to the tangent, for then theangle ABC is a right angle (see note E above). If it is true in this case, then it is alwaystrue, as if B and C are fixed, the angles BAC and BCT are also fixed using the angleson a chord theorem.

For a more detailed account of geometry with many problems, consult Plane Euclidean Ge-ometry: Theory and Problems by A. D. Gardiner and C. J. Bradley, published by the UKMathematics Trust, http://www.ukmt.org.uk/.

J. Subscript, index or suffix notation

If we have a sequence, we could name the terms of the sequence with letters in alphabeticalorder as a, b, c, . . . , but we would soon run out of letters. Instead, it is useful to use one letterfor the whole sequence and to attach a small number next to it, written just below the line,called a subscript, index or suffix. This number tells us which position in the sequence we areat.

Here are some examples using the Fibonacci sequence, which is defined by adding two consec-utive terms to yield the next term, starting with the two terms 1 and 1:

F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5, F6 = 8, F7 = 13, F8 = 21, etc.

It is useful to take F0 = 0, so that F0 + F1 = F2.

(i) F8 is five positions along from F3.(ii) F 28 means the square of F8, and has the value 21

2 = 441.


(iii) Fn means a term in a general position n. Then Fn1 is the term before Fn, Fn+1 is theterm after Fn, and Fn+2 is the term after Fn+1.

(iv) F2n means the 2n-th term, which is the term in the nth even position; F2n1 is similarlythe term in the nth odd position.

(v) We can see that F 25 is one more than F4 F6. We normally leave out the times sign andwrite F 25 = F4F6 + 1. Check that it is true that F

2n is one more or less than Fn1Fn+1 for

some other values of n.

K. The Greek alphabet

This is often used in mathematics when writers run out of (Latin) letters or wish to usea different style of letter. (There are occasionally times when mathematicians venture intoRussian or Greek in addition, but these are fairly rare, thankfully!)

There are twenty-four characters in the Greek alphabet, usually written in this order:

A alpha aB beta b gamma g delta dE , epsilon eZ zeta zH eta h , theta q

I iota iK , kappa k lambda lM mu mN nu n xi xO o omicron o pi, $ pi p

P , % rho r , sigma sT tau t upsilon u , phi fX chi c psi y omega w

In this table, the first column is the capital letter and the second column shows the lower caseletters. (Seven letters are shown with common variant forms, and they are as follows: the twovariants of the letters epsilon, theta and phi are both commonly found in mathematics booksand papers; the second variant of the letters kappa, pi and rho are found, but are significantlyrarer than the first form; finally, the variant form of sigma is very rare in mathematics books,and is used in Greek when a sigma appears at the end of a word.)

The third column is the name of the letter, with the bold-faced letter being the English pseudo-equivalents. While many pronounce eta with the sound ea as in least, classical scholarsoften use an e sound as in air; the vowels in beta, zeta and theta are pronounced as in eta,and the o in omega is as in bold.

The fourth column shows the keys which correspond to these letters in the standard keyboardmapping for the Symbol font on Windows.

L. Rules of inequalities

The symbols and > are often mistrusted; many students often avoid them duringproblems, substituting = instead, and then try to convince themselves later on that they havegot the direction (< or >) the right way round. There are some simple rules that will keep youon track.

As a general guideline, substitute in simple numbers as you go to check that you are on track.

The rules are as follows:

(i) a < b is the same as b > a.For example, 3 < 5 is equivalent to 5 > 3.

(ii) If a < b, then a+ x < b+ x for any number x.For example, since 3 < 5, it follows that 4 < 2 by taking x = 7. The same is true

for subtraction: if a < b, then a x < b x for any number x.


(iii) If a < b, then ax < bx for any positive number x, but ax > bx for any negative number x.For example, since 3 < 5, it follows that 12 < 20 by taking x = 4, and 12 > 20 by

taking x = 4.(iv) If a < b, then a > b.

For example, since 3 < 5, it follows that 3 > 5. Note that this is equivalent tomultiplying (or dividing) both sides by 1, or by adding (a + b) to both sides (orsubtracting a+ b from both sides.

(v) If a < b, then 1a> 1

bprovided a and b both have the same sign (i.e., ab > 0).

For example, 3 < 5 gives 13> 1

5and 5 < 3 gives 1

5< 1

3. This result follows by

dividing both sides by ab. Note that 3 < 5 but 13 1

5, as in this case we are dividing

by 3 5 < 0, so the inequality switches direction.In summary, you may add or subtract to both sides, just as with an equation, and you maymultiply or divide both sides by a positive number. But if you multiply or divide by a negativenumber, the symbol changes from < to > or vice versa.

There are some standard gotchas which are likely to catch you out. The most frequent casesare those which involve multiplying or dividing by numbers which are (or might be) negative:

(a) Consider the case that a < b, and then the relation of xa to xb or ax

to bx. If we do not

know the sign of x, the outcome will need to be split into two cases in your working, onewith x positive, the other with x negative, and perhaps also the case x = 0.

(b) Consider the situation x < 3. Is it necessarily true that x2 < 9? Perhaps; if x = 1, thenx2 < 9, whereas if x = 5, then x2 > 9.

(c) Conversely, if x2 > 9, then we might again have x = 5, so it does not follow that x > 3.This is very similar to the trap of saying: x2 = 9 means x = 3. We ignore the casex = 3 at our peril!!

To summarise, watch for hidden cases involving multiplying or dividing by negative numbers,taking squares or taking square roots.

JMO mentoring schemeNovember 2010 paper

A questions in each paper are meant to be straightforward - B and C questions more difficult.Hints are included upside down at the bottom of the page. Fold this back and look at them when you need.

A1 Recently we passed the point when the date and time could be recorded as 10/10/10 10h 10m 10s. Howmany seconds from then is it until we reach the time 11/11/11 11h 11m 11s?

A2 A bus leaves Rushton for Portborough at 10:00 am just as another bus leaves Portborough for Rushtonalong the same route. The bus from Rushton goes twice as fast as the bus from Portborough andarrives at Rushton at 10:48 am. When do they meet?

A3 Find the positive integer which when multiplied by 9 gives an integer between 1100 and 1220, andwhich multiplied by 13 gives an integer between 1500 and 1600.[You must show it is the only answer.]

B4 In the diagram AD and BC intersect at X. ABX and CDX are similar.AB = 4, AX = 5, AD = 20 and ABX = 90.AB and CD extended intersect at Y. (not shown)What is the area of BXDY? [not to scale]

B5 24 points are arranged in a regular rectangular grid of 4 rows and 6 columns.How many rectangles or squares can be formed by joining four of thepoints so that the edges are not parallel to the rows or columns of the grid?

C6 Three circles of radius 1 unit touch each other externally.A small circle is placed in the middle so that it touches eachof these circles. What is its radius?

C7 A square board is made up of 25 unit squares which we will call tiles, coloured alternately black andwhite like a chess board with a black tile in the centre.A triomino is a shape made up of 3 unit squares, like the one shown in the diagram. Eight triominoeslaid on the board can cover the board except for one tile.Which of the 25 tiles can be left uncovered given a suitable arrangement of the 8 triominoes?[It is assumed that the edges of triominoes run along the edges of the tiles.]

Hints :

6.Use the results of notes F and H.

7.A triomino will cover either 2 white tiles and 1 black tile or vice versa.

Divide cases into leaving a black tile uncovered or a white tile uncovered.

A B

X

C

D

amtejcTextBoxsupported by

JMO mentoring schemeDecember 2010 paper


A1 You travel the first 48 miles of a journey at an average of 36 mph and the second 48 miles at anaverage of v mph. What must v be if you want to average 45 mph over the 96 mile journey?

A2 Without using decimal equivalents (or a calculator) prove that these three fractions are arranged inorder of size, largest first :

6 7 8 11 13 15

B3 Beth has 4 of her friends round for a sleepover. She puts two airbeds down in her own bedroom; theother bedroom has two beds. In how many ways can they split themselves between the two bedroomsif she decides not to sleep in her own bed?

B4 Its the last month in which to play with the year number 2010.(i) Which values from one to ten can you form with the digits of 2010, each digit used exactlyonce in the order in which it appears, if you are allowed to use + ( ) or putting digits together toform numbers?(ii) If you use the digits in any order, can you get any extra values?(iii) Can you obtain all the values from one to ten using the digits in order if you are allowed to addthe use of powers, square roots and factorials. How?Note that 1! = 1, 2! = 2 1, 3! = 3 2 1, 4! = 4 3 2 1, etc. It is defined that that 0! = 1; it follows from the rules offactorials in a way similar to 50 = 1 follows from the rules of indices. You probably have not met this before.

B5 When we write 789 in decimal, we mean 7 hundreds plus 8 tens plus 9 units.If abc represents a decimal number (not algebra), we would have to use

100a + 10b + cfor its value. Show that if a + b + c is divisible by 9, then so is abcdivisible by 9 and if abc is divisible by 9, then a + b + c is divisible by 9.

B6 ABC is right-angled at A and AB < AC. BCDE is a square.G is on AC so that AGE = 90.F is on EG so that BFE = 90.Prove that ABFG is a square.

C7 In the square, the two equal arms of the black saltire cross extend a distance x from the corners of thesquare as shown in the diagram.. The centre length is 2a. The black cross covers the same area as thefour white triangles. Prove that x = a (2 1).[Note that it is unnecessary to expand / multiply out brackets.]

If the width of each arm of the cross (perpendicular to the edges of the arms)is 1 unit, find the area of the square.

Hints :

6.Some lines for you to fill in. Look for a triangle congruent to ABC and prove it is so.

7.Divide the square into eight triangles using the axes of symmetry. For the second part, note that the hypotenuse of an

isosceles right-angled triangle with other sides both 1 is

2.

A

B C

G

E D

F

x 2a x

Supported by

Charitable Trust

JMO mentoring schemeJanuary 2011 paper


A1 2011 + p is a power of 2 where p is a prime number. What is the minimum value of 27p ?

A2 How many triangles are there in this diagram? If the diagram was based on half a 5 by 5 squareinstead of half a 4 by 4 square, how many extra triangles would there be?

A3 The sum of two numbers is 80. One is four times the other. What is their product?

B4 New Years Day in 2011 is on a Saturday. On what day will it be in 2111?

B5 Assume : the flesh of fresh peaches has a water content of 75% by weight;when left in the sun to dry, the flesh loses 80% of its water content;the stone accounts for 10% of the weight of the original peach.

How many grams of fresh peaches are needed to make up a 180 g pack of dried peaches with theirstones removed?

B6 Each of the four triangular faces of a regular tetrahedron is to be painted either red or green. Howmany different (distinguishable) ways are there of painting the tetrahedron? We allow the tetrahedronto be turned and viewed from any direction, so, for example, painting the front face red and the restgreen is considered the same as painting the bottom face red and the rest green.

C7 You are given that 2 + y = x2 . Put x = 2 + a and y = 2 + b where both a, b are between 0 and 1.Show that if y is close to 2, then x is closer.What about the case where x is a little less than 2?

C8 ABCDEF is a regular hexagon, the lengths of each of the sides being 2 units.DF and DB cut EC at X and Y respectively.M is where CE cuts DA.What are the lengths of DM and CE?Find the area of BDF.Hence find the area of the trapezium XYBF.

Hints :

4.Dont forget leap years.

7.It is instructive to study a graph y = x

2

2 and look at how x and y change near the point (2, 2).

8.This is all to do with

3. See notes F and H.

C B

A

FE

DY

XM

Supported by

Charitable Trust

JMO mentoring schemeFebruary 2011 paper


A1 In the country of Hargravia they use six kinds of coins as money : 1, 5, 10, 50, 100, 500.What is the smallest number of coins you need in order to pay exactly 777, assuming that you havesufficiently many coins of each kind and that you do not necessarily need to use all the kinds of coin ?

A2. A road surfacing gang is made up of a certain number of workers and machinery. Three gangs haveasphalted 20 km of road in 10 days. How many additional gangs should be brought in if all the work isto be finished in another 15 days and there is still 50 km of road to be asphalted ?[Assuming, of course, that they do not get in each others way !]

B3 In how many ways can one choose two different numbers from the set {1, 2, 3, ..., 2011} such thattheir sum is an even number ?

B4 You are given that n is a positive integer with the property that when we add n and the sum of itsdigits, we get the number 313. What are possible values of n?

B5 Let a and b be positive integers so that b = (a (a a)). Find the least value of a + b.

C6 Is it possible to draw a triangle in which the sum of any two angles is less than 120 ? Prove youranswer.

C7 ABC is isosceles with AB = AC and BAC = 120 . D is on BC and BAD = 90 .Show that DC = 1/3 BC. ADC has a circle inscribed in it. The radius of this circle is 1 cm. What is the area of ABC ?

Hints :

5.Square and square and square again. If a and b are the powers of a common number, what is that number?

6.Assume each pair add to less than 120. What happens?

7.Write down the ratio of the sides in a 30, 60, 90 triangle. Let I be the centre of the circle.

Draw lines from I to meet AC and CD at 90 and mark the intersections as X and Y respectively.

IYD is a 30, 60, 90 triangle and IX = IY = 1 cm which gives you XD, etc.

Supported by

Charitable Trust

A

B CD

Not to scale

JMO mentoring schemeMarch 2011 paper


A1 A short time back my plumber mentioned a mathematical trick which seems to be circulating as somerather curious and amazing fact. If you add the two digit version of your year of birth to your age (onyour birthday this year) you get 111. Is this true?

B2 In the figure ACB = ADB = 90 .Let M be the mid-point of AB (not shown).Find five isosceles triangles in this figure using these pointsand prove that MCD is one of them.[See note E attached to October 2010 paper]

B3. Last month a date occurred, namely 11th Feb 2011,which can be written as an 8 digit palindrome 11022011.How many more such dates are there before the year 3000?[Allow a leading zero for the day number if necessary.]

B4 In a parallelogram ABCD where BC < BD, E is on the line BC so that BE + BC = BD. Prove that thecircle through A with centre D touches the circle through E with centre B.

B5 The Fibonnacci sequence can be defined by the relation :F

n+1 = F

n + F

n1where you think of F

n as the current term, F

n1 as the previous term, F

n+1 as the next term and, for

example, Fn+5

as the term 5 places on from the current term.F

1 = 1, F

2 = 1 and it is convenient to define F

0 = 0.

Show that Fn+5

= 8 . Fn + 5 . F

n1 .

[See note J attached to October 2010 paper for more on suffix notation.]

C6 Two circles overlap so that the centre of each lies on the circumference of the other (like a partialeclipse of the sun by the moon). The centres are S and M.Take each circle to have radius 6. Find the areas of :(i) SMY;(ii) the sector formed by the radii SM and SX

and the arc MX;(iii) the part of the sun still showing.

C7 Prove that every positive integer may be representedby a sum of Fibonacci numbers, no two of whichare consecutive in the sequence.[See B5 for the definition of the Fibonacci numbers.]

Hints :

5.Work out F

n+2

, F

n+3

, ... using particular numbers in the sequence for F

n

and F

n1

, keeping track of the values of F

n

and F

n1

as you proceed. Then replace your numbers with the symbols. You may find it easier to use a, b, c, d, ... instead of Fs.

6.Refer back to C9 in the October paper and also to note H attached to that paper.

7.The difficulty is in writing down the argument clearly though you may spot why the definition of the Fibonacci sequence

(in question B5) is crucial.

A

B

CD

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

1234567890123456789012345678

S M

X

Y

Supported by

Charitable Trust

JMO mentoring schemeApril 2011 paper


A1 A powerful sports car accelerates from 0 to 60 mph in 5 seconds so that the speed-time graph is astraight line. Using distance = average speed time, find in metres the distance it travels in 5 seconds.1 mile = 1609 m, 1 hour = 3600 seconds.

B2. Show that 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + ... + 1/15 + 1/16) > 3.B3 Let A, B, C and P be distinct points on the circumference of a circle,

with P lying between B and C, as shown in the diagram. A straight lineis drawn from P meeting BC at L so that PL is perpendicular to BC.[This is called the perpendicular from P to BC and L is the foot of the perpendicular.]Likewise M and N are the feet of the perpendiculars from P to AB and ACrespectively The line AB or AC is extended as necessary.Find, giving reasons, three quadrilaterals apart from ABCPwhose vertices lie on circles.

B4 Show that (2 1)(2 + 1) = 1.If x = 1 / (2 + 1), prove that 2 = 1 + x and that x = 1 / (2 + x).

B5 Prove that the minimum number of trianglesinto which a convex n-sided polygon can be dissected is n 2.

C6 a, b and c are integers with no common factor higher than 1 satisfying the equation c2 = a2 + b2 .Prove that c must be an odd number.

C7 Pascals triangle can be laid out :columns 0 1 2 3 4 5 6 7

row 0 1 0 0 0 0 0 0 0row 1 1 1 0 0 0 0 0 0row 2 1 2 1 0 0 0 0 0row 3 1 3 3 1 0 0 0 0row 4 1 4 6 + 4 1 0 0 0row 5 1 5 + 10 10 5 1 0 0row 6 1 + 6 15 20 15 6 1 0The advantage of this method is that each element can be referred to by its position. Thus we write

6 = 15 (example shown in green/large bold). 4 It has the property that two adjacent elements in a row add to the number in the next row underneaththe right element (shown in red/bold italic). Write this example using the position notation.Explain why adding along a diagonal from the left upwards (as example shown in blue/bold upright)produces a Fibonacci number.

** Reminder : notes published with the October paper can be found separately on the UKMT webpage following the links :www.ukmt.org.uk > Mentoring > Junior > Ideas in number theory, ... (pdf file).

Hints :

3.Its all to do with right angles formed on a diameter in a semi-circle : see note E.

4.Remember a

2

b

2

= (a b)(a + b) ? [See note B]

5.Consider cutting up the polygon into triangles using an extra point inside the polygon.

6.Consider why c can not be an even number. Can two odd square numbers add to an even square number?

7.Careful study will show why it is sensible to start counting from row 0 and column 0. Writing out more examples like the

one with the red numbers will give you confidence in using the notation.

A

CB

PM

NL

Supported by

Charitable Trust

JMO mentoring schemeMay 2011 paper


A1 Make the shape shown by cutting and folding a sheet of paper 4 units by 2 units.No rejoining is allowed. The tab standing up is 2 units by 1 unit in size.

B2 A pilot has to eject from his fighter aircraft. In order to clear thetail fin he must rise to a height at least the height of the tailfin in the time the tail fin moves to where he was when he pressedthe ejector button. Assuming that the aircraft is travelling at 350 mph,that the tail of the aircraft is 20 metres behind him and that the height of the tail fin is 1.5 metres,find the minimum average acceleration that the rockets in the ejector seat have to provide.[See A1 of April paper.] Average acceleration = change of speed time.

B3. Zebedee is out training on his bicycle.He averages u mph on the outward leg of his journey and thenv mph on the return journey, which is of equal distance. Find anexpression for Zebedees average speed for his whole journeyin terms of u and v.

B4 p, q and r are prime numbers so that p < q < r and p/q + q/r = s/qr .Prove that if s is prime, then p = 2.

B5 The rectangle shown is dissected into 11 squares of differentsizes. (Diagram not to scale.) The smallest square (shaded) is 9 by 9.What are the dimensions of the rectangle ?

You are not expected to be able to write out proofs for C6 and C7 but to give some account of what is going on.

C6 Find the values of the expressions 1 + 1 / 2, 1 + 1 / (2 + 1 / 2), 1 + 1 / (2 + 1 / (2 + 1 / 2)) (see * below)and 1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / 2))) in the form a / b where a, b are integers with a > b.Each expression is the previous one with the innermost 2 replaced by (2 + 1 / 2). Can you see a way toget each simplified fraction from the previous simplified fraction? With the aid of April question B4,suggest what number these fractions approach as this happens.

C7 Explain why a sequence of consecutive Fibonacci numbers can be added in different proportionsusing Pascals triangle (binomial coefficients shown in red/bold italic) to produce another Fibonaccinumber,

e.g. 1 . 5 + 3 . 8 + 3 . 13 + 1 . 21 = 89.

Hints :

2.These numbers are approximate and your answer will be an estimate.

3.Assume distance for one leg is d, find the times taken for each leg to add together. d should not appear in the answer.

4.Multiply out. This is all about odd and even numbers.

5.You can let one of the squares adjacent to the black square have a side x and express all the rest in terms of x.

6.With these fractions you must work from inside outwards. Consider replacing 1 / (2 + 1 / 2) with x .

7.Start by replacing each number on the left hand side of the example by the sum of the two previous Fibonacci numbers and

see what happens.

1 * The third of these written out fully would look like : 1 + 1

2 +2 + 12

Supported by

Charitable Trust

JMO mentoring scheme

June 2011 paper

This is the last paper of this academic year and has a bit more material.Hints are included upside down at the bottom of the page. Fold this back and look at them when you need.

1 Find all the two digit positive integers n such that the remainder of the division of 2011 by n is 13.

2 How many positive divisors does 6! have including 6! and 1? [6! = 6 5 4 3 2 1.]

3 In a rectangle ABCD, let E be the midpoint of the side BC and F the midpoint of the side CD.If AFE = 90 and BC = 2, find CF.

4 A sequence of numbers of which the 1st is -5 and the 26th is +2 is defined by the rule that eachnumber is the sum of the one before it and the one after it. Find the 12th number.

5 A circle has a diameter AB and centre O. P is placed on the circle and a perpendicular is dropped fromP onto AB to meet AB at Q. Show that D APQ, D PBQ and D ABP are all similar.If PQ = g, AQ = x and BQ = y, show that g = (xy).By condering a suitable length on the diagram to represent a = (x + y), show that a g.[a is the ARITHMETIC MEAN of x and y and g is the GEOMETRIC MEAN of x and y .]

6 Recall question 1 in December 2010. Consider replacing 48 miles by distance d, 36 mph by u and keepthe second speed as v. Prove that the average speed for the whole journey of 2d is :

h = 2/(1/u + 1/v)[Note that this does not depend on the value of d: h is the HARMONIC MEAN of u and v .]

7 A circle has a centre O and radius a. A line passes through O and extends outside the circle. P is apoint on the circle. A perpendicular is dropped from P to a line and meets at U. The tangent to thecircle at P (a tangent must be at 90 to the radius OP) meets at X.If OU = u, by using similar triangles prove that OX = a/u.

8 (i) Assume a > 0, b > 0. Explain why if a b, then 1/a 1/b. (ii) Referring to questions 5 and 6, and by putting x = 1/u and y = 1/v,

prove that h (uv) and hence h (u + v).[This shows why 36 mph then 60 mph each maintained over equal distances only averages 45 mph and not 48 mph.]

9 A child's puzzle consists of four square cards as shown below, each of which has two tabs and twoslots, each differently shaded (on one side only). When two cards are placed next to each other, the tabmust be the same shade as the slot over which it fits as in the example on the right. Ignoring rotatedversions, show that there are only two arrangments in which they can form a larger (2 by 2) square.(Unused tabs may project outside the larger square.)

1/2. Note that the number of divisors ofp

n

wherep is prime andn is an integer isn + 1.

3 Write down some expressions using Pythagoras's rule.

4 Let the first number bex, the second bey and work out what the next few are in terms ofx andy.

5/7 There are similarities in the proofs.

8 Part (i) is needed in the proof of part (ii).

9 Note how many of each shade are tabs and how many are slots to prove which cards can NOT fit.

Supported by

Charitable Trust

A B C

CB

D


October 2011 paper

This is the introductory paper of the season - they will be produced every month up to June 2012 inclusive.Some questions are devised to help you learn aspects of mathematics which you may not meet in school.

Hints are upside down at the bottom of the page; fold the page back to view them when needed.There are also a set of notes on various topics included with the mailing of this paper.

1 To make short pastry one uses flour to fat (butter, margarine, etc.) in the ratio 2 : 1. To make flakypastry requires a ratio 4 : 3. I have 3 kg of flour and 2 kg of fat and wish to use all the ingredientsmaking some of each type of pastry. How much flaky pastry do I make?

2 What is the 2011th character in the sequence ABCDEDCBABCDEDCBABCDEDCBAB?

3 A polygon of 100 sides has all possible diagonals drawn from a chosen vertex to the others. Howmany triangles are formed? What is the sum of the interior angles of the 100 sided polygon?

4 Write down the units digits of all the square numbers from 1 to at least 20. How can you tell that 573 is not a square number?

5 The exterior angles of a triangle are in the ratio 4 : 5 : 6. What is the least (interior) angle of the triangle?

6 You tell your friend that there is the same chance of obtaining an even number of heads when tossingfour coins as of a head showing when tossing just one coin.

She doesnt believe you; how do you convince her?

7 In a quadrilateral ABCD, side AB = side BC = diagonal CA = 2 and side AD = 1. D is located so that acircle using diagonal BD as a diameter will pass through A. Prove that BD = 5.

8 p and q are two prime numbers chosen so that p + q and p - q are also prime. Prove that p - q is a prime number.

9 Work out the day of the week for 14 Sep 1752, the first day of the new calendar in England when anAct of Parliament brought the calendar into alignment with the Gregorian calendar used in most ofWestern Europe by that time.

[Leap years occur every 4 years with the exception that every 100th year is not a leap year but every 400th year is a leapyear. Look up Gregorian calendar on Wikipedia.]

10 Construct a sequence of five single digit positive integers a, b, c, d, e so that :b divides a + b but a does not do so;c divides a + b + c but a and b do not do so;d divides a + b + c + d but a, b and c do not do so;e divides a + b + c + d + e but a, b, c and d do not do so;

[Taken from Enigma in New Scientist, 23 Jul 2011.]

2/4/9 It all depends on remainders after division. The methods are similar, qu 4 counting in 10s and qu 9 counting in 7s. For more notes go to http://www.ukmt.org.uk/ : follow links Mentoring>Junior>Modular Arithmetic at NRICH3 You may know a rule for this but use this question to prove it for yourself.5 The exterior angle of a triangle is that between a side and its neighbour extended externally.6 Work out all the ways you can arrange 4 coins in a row, each showing a head or a tail. Remember 0 is an even number.7 Always make a drawing which is fairly accurate when doing geometry. NB: Is all the data needed to solve the problem?8 You can work out the values of p and q but you should justify your answers.10 There are a number of possibilities for a and b but these get reduced as you try to fit successively c, d and e.

Supported by

Charitable Trust


November 2011 paper

Earlier questions are generally easier and later questions more difficult.Some questions are devised to help you learn aspects of mathematics which you may not meet in school.

Hints are upside down at the bottom of the page; fold the page back to view them when needed.

1 At most, how many Mondays can there be in 45 consecutive days?

2 In the dark, Emily picks some socks from a drawer which contains 6 black socks, 14 blue socks and8 green socks. To be sure to pick two of the same colour, what is the minimum number of socks shemust pick?

3 Lets write all the natural numbers from 1 to 2011 inclusive one after the other so that they form a new(rather long) natural number. How many digits does this number have?

4 You choose a three digit number a such that when the digits of a are reversed, they form a differentthree digit number b. [0 can not be a leading digit in either the numbers a or b.]Let c be the (non-negative) difference between a and b. How many possible values of c are there?

5 Katrina is standing in a rectangular garden. Her distances from the corners of the garden are 6 m, 7 m,9 m and d m where d is an integer. Find d.

6 We choose six (different) numbers out of the ten integers 1, 2, 3, , 10. Prove that their product isdivisible by a perfect square greater than 1.

7 An equilateral triangle ABC has sides of length 6.D, E and F are on BC, CA and AB respectively so that CD = AE = 2 (so that BD = CE = 4) andso that AEF = 90.What is the area of DDEF?

4 You may find it helpful to consider a couple of specific cases then generalise with letters for each digit.

5 Set up some equations using Pythagoras's rule. You will have to consider possible different orders for the data.

6 What is the maximum number of integers that can be chosen so that you cannot divide by a perfect square > 1?

7 You may need to use Pythagoras's rule applied to a 30

, 60

, 90

triangle. See note H available on UKMT website

http://www.ukmt.org.uk/ and follow links Mentoring > Junior > Mentoring scheme resources pdf file

Supported by

Charitable Trust


December 2011 paper

Generally earlier questions are easier and later questions more difficult.Some questions are devised to help you learn aspects of mathematics which you may not meet in school.


1 (a) Luke writes all the consecutive even numbers from 2 to 2012 (inclusive) on a board. Louise thenerases all the numbers that are multiples of three. How many numbers are left?

(b) How many digits does the answer to 111222333444555666777888999 37 have?

2 DABC is scalene and strictly acute angled, that is, all sides are of different lengths and it has no rightor obtuse angles.D is a point on BA (extended past A if necessary). It is also on the perpendicular bisector of BC.

Prove that ADC = 2 ABC or that ADC + 2 ABC = 180.

3 Show that the tens digit of every power of 3 is even.

4 Major Tom has landed on a planet populated by purple cats, who always tell the truth, and black cats,who always lie. In the pitch blackness, he meets 5 cats. The first cat says: I am purple. The secondcat says: At least 3 of us are purple. The third cat says: The first cat is black. The fourth cat says:At least 3 of us are black. The fifth cat says: Were all black. How many of the 5 cats are purple?

5 How many pairs of integers (x, y), x > 1 and y > 1, are there such that x2 + y = xy + 1 ?

6 A blacksmith is building a fence consisting of uprights 18 cm apart. Instead of welding a baracross the top (like the dotted line), he makes individual arcs of circles like the one shown inthe diagram. (He continues this pattern along the fence.) The highest point of the arc is 33cm above the dotted line. All the pieces of metal lie in the same vertical plane.How long is the piece of metal used to make one of the circular arcs?

7 (i) Prove that if n is even, an+1 + 1 = (a + 1)(an - an-1 + an-2 - an-3 + + 1). (ii) Which numbers of the form mm+1 + 1 are prime if m is a positive integer? You should prove your result.

8 We are given a 50 by 50 chessboard, divided into 2500 squares which can be coloured black or white.Initially they are all white. At any step, a whole row or column of squares can reverse colour. Thus if32 of the squares in a row are white and the other 18 are black, a step will change them to 32 black and18 white. We aim to perform a sequence of steps so that we finish with exactly y black squares wherey is the year number. Show that it was possible to do this last year but that it is not possible this year ornext year.

2 The result you prove will depend on whether AB > AC or AC > AB.

3 Consider the remainders after a few powers are divided by 20.

4 Draw lines from the centre of the circle to the ends and top of the arc. You should find some simple angles.

5 Rearrange the equation. Remember that(x- 1)(x + 1) =x

2

- 1.

6 Spot the statements that contradict each other.

7 Multiply out in part (i) and use it in part (ii).

8 What happens if you do just a column step and a row step? Use some algebra.

Supported by

Charitable Trust


January 2012 paper



1 What is the sum of 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5 + + 34 + 34 + 35 + 35 + 36?

2 How many positive integers n are there such that 8n + 50 is a multiple of 2n + 1?

3 (i) What is the area of a square whose diagonal is d? (ii) What is the area of a regular octagon whose side is d?

4 Let E, F and G be the mid-points of the sides AB, BC and CD of a rectangle ABCD. The lines DE andBG intersect the line AF in the points M and N (respectively). Find the ratio MN : AF.

5 How many different pairs of numbers can you choose from the set {1, 2, 3, , 2012} such that theirsum is an even number?

6 ABCD is a regular tetrahedron. A plane passes through A parallel to D BCD (which is opposite A).There are three more planes through B, C and D defined in a similar way. These four planes intersectin pairs to give the edges of a larger tetrahedron; each vertex is the intersection of three of theseplanes. What is the ratio of the volume of ABCD to the volume of the new tetrahedron?

7 In this, and some subsequent papers, define a RAT to be a right angled triangle with integer value lengths of sides. Notethat if the sides have fractional values for their lengths, we can always obtain integer values by using a smaller unit. Thusthe set (0.15, 0.2, 0.25) using a particular unit becomes the set (3, 4, 5) if we use a unit twenty times smaller. You cannever do this in cases such as a right angled triangle with sides (1, 1, 2).

(i) You are given a RAT with lengths a, b and c such that a + b = c and such there is no commonfactor between all three of a, b and c. Prove that just one of a and b is even and that c is odd.

(ii) If m and n are positive integers (m > n), prove that the set of values (m - n, 2mn, m + n) can beused to contruct a RAT.

In questions in subsequent papers we will call m and n key numbers for the triangle (abbreviated to KN).

8 A triangle has two vertices A (-3, 1) and B (3, -1). The third vertex C is at (2, k). Find all the valuesof k such that the triangle is isosceles.

1 Letn be the number of tests in total anda the average before the last two tests. Form andsimplify equations.

2 What are the positive integersa andb such that 8n + 50 =a(2n + 1) +b?

3 Dissect the octagon into 4 triangles, 4 rectangles and a central square. What is the smaller side of one of the rectangles?

4. Extend the figure by rotating the rectangle and its elements about F through 180.

5 Try writing out the pairs for a small set, e.g. {1, 2, 3, , 10}.

6 Draw the larger tetrahedron first and study how ABCD fits inside it. Use note F on medians in the notes on the website.

7 (i) Note that the square of an even number must be a multiple of 4. What about squares of odd numbers?

(ii) Look at note B in the notes (www.ukmt.org.uk > Mentoring > Junior > Ideas in ) to help you with some algebra.

8 You are advised to draw an accurate diagram on graph paper to help you analyse this problem.

Supported by

Charitable Trust


February 2012 paper



1 What is the maximum number of sides that a regular polygon can have so that its interior angles are aninteger value measured in degrees?

2 You have two jugs with capacities of 7 litres and 5 litres and a tap for a supply of water.Using these jugs you have to fill a bottle (capacity about 1 litres) with exactly 1 litre of water.How do you do it?

3 Arrange 10, 3 and 2 + 3 in increasing order of magnitude without using decimal approximations.

4 DABC is equilateral; D, E and F are the mid-points of its sides BC, CA and AB. How many non-congruent triangles can be obtained by choosing 3 points from A, B, C, D, E and F?

A triangle like BDC does not count because the points are in a straight line; the triangle is called degenerate.

5 On the second to last test of the school year, Barbara scored 98 and her average score so far thenincreased by 1. On the last test she scored 70 at which her average score then decreased by 2. Howmany tests has she taken through the school year?

6 Refer back to January question 7. What are the KNs (key numbers) for these RATs with lengths of sidesas given by the three numbers within each pair of brackets?

(a) [ 3, 4, 5 ], (b) [ 5, 12, 13 ], (c) [ 8, 15, 17 ], (d) [ 21, 20, 29 ], (e) [ 15, 20, 25 ] .. careful! Which two RATs are the same shape? Can you identify this from their KNs? [More on this next month.]

7 If N = 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 and M 3 is the maximum cube number which divides into N, what is M?

8 This question is designed to encourage you to look at computer geometry packages and to expand your knowledge ofgeometry. Look at the next page for details.

You are given three points A, B and C on a circle. The tangents to the circle at A and B intersect at T.The line through T parallel to AC meets the line through B and C at D. Prove that DDAC is isosceles.

See section I (about circles) in the notes found under Junior Mentoring on the UKMT website, particularly paragraphs(b) angles on a chord, (c) opposite angles of a cyclic quadrilateral and (d) angles relating to tangents to a circle.

3 Square the numbers and then subtract a suitable number before squaring again.

5 Form some equations usingn = the number of tests anda = the average before the last two tests.

7Note that if a numbern divides intox, then the nearest numbers tox which can be multiples ofn arex-n andx +n.

8 Keeping the circle and A, B and T fixed, vary the position of C.

Look at howTDC is related to bothTAB andACB and what this tells you about where D may lie.

Supported by

Charitable Trust

You can use Geogebra which can be downloaded off the web without charge. See their website www.geogebra.org for version 4.0 to do this on your operating system.

Geogebra is dynamic software allowing you to draw diagrams which can be adjusted continuously justby grabbing a starting point. These become controls for the shape and size of your diagram.

You will see a row of icons at the top of the sheet below the menu bar. (For reference they have beennumbered from 1 to 12 on this sheet.) You can get rid of the algebra view and the axes by clicking onthe drop-down list from the view menu. Each icon will reveal several options by clicking down amenu from the marker at the bottom right hand corner of an icon. The following are sufficient fordrawing the diagram required.

icon 1 Pointer. icon 2 For inserting points, including points of intersection. icon 3 For drawing lines or segments (parts of lines). icon 4 For drawing parallel lines and tangents. icon 6 For drawing circles. icon 12 For hiding or revealing objects.

There are several ways to draw the diagram. Here is one suggestion.

a) Select New Point from icon 2. Click in two positions. The points will be labelled A and B. Right-clickon B to bring up a property box and change its name to O as the centre of the circle.

b) Select Circle with Centre through Point from icon 6, click on O and then on A.The circle centre Owill appear.

You can hide the point O either by using icon 12 or the hide option in the property box. Whenever icon 12 Hide/Show isactive, all hidden objects appear so you can always recover them. Try it now to find out how it works and what you can doby right-clicking an object.

c) Select Point on Object from icon 2. Click on the circle twice more to give points B and C.

d) Select Segment between Two Points from icon 3. Click on A and then on C. The line between Aand C will be drawn. You may also wish to do this for AB.

e) Select Tangents from icon 4. Click on point A and then on the circle. The tangent will appear. Do thesame for B. You may have to grab B and C and move each round the circle to get the tangents to meetat a convenient place within the window.

f) Select Intersect Two Objects from icon 2, click on each tangent line in turn and a point will bemarked where they meet. Re-label it T by the same method as step (a).

g) Select Parallel Line from icon 4. Click on T and then on the line AC. An infinite parallel line willappear through T.

h) Select Line through Two Points from icon 3. Click on C and then on B.

i) Select Intersect Two Objects from icon 2, click on each of these last two lines to obtain point D.

You can now join any other lines you like. Try moving C round the circle without disturbing A and Band see what happens, particularly watching how D moves.

j) The following may be revealing. Select Locus from icon 4, click on D and then click on C. Now grabC again and move it round the circle.

If you found this useful and would like more geometrical material like this, please tell us by emailingthe UKMT office using [email protected] .


March 2012 paper



1 The product of 4 consecutive odd numbers is 23961009. What is the smallest of these odd numbers?

2 Simplify (164x2).

3 An equilateral triangle whose edges measure 20 cm is dissected and rearranged to form a squarewithout overlap or gaps. How long is the side of the square?

To see how this can be done, refer to www.mathworld.wolfram.com/Dissection.html

4 A 10 digit number abcdefghij includes all the digits 0 to 9. How many such numbers are there givena + j = b + i = c + h = d + g = e + f = 9?

5 In a magic forest there are talking foxes, snakes and turtles. Turtles always speak the truth and foxesalways lie. Snakes only lie on rainy days. Hogwarts student Hermione Granger talks to four of theseanimals who tell her one by one: (1st) its raining today; (2nd) that animal has just lied;(3rd) the weathers fair today; (4th) that animal has just lied or I'm a snake!At most, how many turtles did Hermione talk to?

6 Let x = m - n, y = 2mn and z = m + n where m and n are integer KNs. [See January question 7 and February question 6.]

Show that x - y = (m - n) - 2n. If x - y = 1 or y - x = 1, then [x, y, z] represents a Nearly Isosceles Right Angled Triangle [NIRAT],

namely a right angled triangle where the two shorter sides are nearly equal. Check using the KNs thatthe right angled triangles with sides [3, 4, 5] and [20, 21, 29] fit these conditions.Find the next largest NIRAT.

7 DOAB is equilateral. M is the mid-point of OA. A circle C1 has centre O and radius OA; only the arcbetween A and B needs to be drawn. A second circle C2 has centre A and radius AM; only a quartercircle needs to be drawn from M so that it cuts C1. Finally a circle C3 is formed with its centre N onBM so that it touches both C1 and C2. Why does AN pass through the touching point of C2 and C3?What line passes through the touching point of C1 and C3?

Let OA = 4. Show that the radius of C3 is 1. If P is the intersection of C3 with NB and Q is the intersection of C3 with NM, show that MP = 5 + 1

and that MP MQ = 4.

3 Remember how to find the height of an equilateral triangle?

4 Remember that, once you've decided whata andj are, the other options are more limited.

5 You don't know what sort of day it was so consider the two cases of weather separately and count the possible turtles.

6 Note that [5, 12, 13] is not aNIRAT since one of the nearly equal sizes isz, the length of the hypotenuse.

Also be aware thatx,y andz might not be in ascending order of magnitude.

7 You may find some insight by reflectingC

1

in BM to produce another arc of radius 4 with centre A.

Draw the figure accurately though you need to know that the radius ofC

3

is 1 to construct N when you attempt it first time.

Supported by

Charitable Trust


May 2012 paper



1 A runner covers 5 km in 16 minutes 40 seconds. Along the course, feeling confident that she is fit, therunner increases speed so that each kilometre is run in a time 5 seconds shorted than the previouskilometre. How long does she take over the last kilometre?

2 How many four digit positive integers are there such that the units digit is the sum of the tens digit andthe hundreds digit?

3 Matthew sends an encrypted message to Mark by using an agreed key word which has no lettersrepeated, e.g. XIHLBW. He replaces an original a with X, b with I, c with H, d with L, e withB, f with W, then he continues through the remaining letters of the alphabet; g is replaced with Y(the next unused letter after W), h with Z and then he continues from the start of the alphabet so thati is replaced with A and so on until the substitution cypher is complete.What letter does he replace with R?Luke knows their system and is trying to evesdrop on their messages. He knows that they are using asix letter key word and he is trying to break the code. What is the maximum number of keywordsLuke would have to check before being able to decrypt the message by brute force?

4 A triangle has two angles which measure 30 and 105.The side between these angles has length 2 cm. What is the perimeter of the triangle?

5 Part of a circle is drawn between diagonally opposite vertices of a rectangle comprising square tilesarranged in m rows of n tiles each. If the circle lies entirely within the rectangle and does not passthrough any of the corners of the tiles except the end points, how many tiles does the circle cross?Will this still work if the tiles are rectangular and not square?

6 A square ABCD has 1 m sides. Another square EFGH is constructed inside ABCD so that trianglesCDE, DAF, ABG and BCH are equilateral. What is the area of EFGH?

7 This is the last question continuing the sequence starting with January question 7.

Write down the KNs you obtained from April question 8 in a table including an extra column for rwhere r = m + n. Check that r2 = 2 m2 1. If the numbers in a chosen row are denoted by r and m andin the next row by R and M, find R and M in terms of r and m. Show that R2 2 M2 = (r2 2 m2).Hence prove that r2 = 2 m2 1.

Show that as m increases, the isosceles triangle with sides [m, m, r] approaches a right-angled triangle. The values m and r are solutions to Pell's equation, an equation dating back to late Greek mathematical investigations.

8 Using the pairs r and m from question 7, investigate the values of the fractions r/m in relation to thevalue of 2. Can you explain how the value 2 emerges from this equation?

4 Split the triangle into two right-angled triangles.

5 Consider how the curve crosses the boundaries of the tiles.

6 Work out the diagonal length of the square EFGH.

7 It makes sense to work with the table of numbers in order to get your ideas straight and test your formulae.

Note that means that one of the + or signs apply.

8 What is the hypotenuse of an isosceles right-angled triangle with other sides of lengthm?

Supported by

Charitable Trust

2010-10_Junior_mentoring_paper_and_notes_2010-11_Junior_mentoring_paper2010-12_Junior_mentoring_paper2011-01_Junior_mentoring_paper2011-02_Junior_mentoring_paper[1]2011-03_Junior_mentoring_paper2011-04_Junior_mentoring_paper2011-05_Junior_mentoring_paper2011-06_Junior_mentoring_paper2011-10_Junior_mentoring_paper2011-11_Junior_mentoring_paper2011-12_Junior_mentoring_paper2012-01_Junior_mentoring_paper2012-02_Junior_mentoring_paper2012-03_Junior_mentoring_paper[1]2012-05_Junior_mentoring_paper

Documents

Junior Mentoring Oct 2010 - May 2012