Jul 5, 2011

Why parents can't do maths today

Long division and long multiplication have been replaced in schools by chunking and gridding. While the new methods are meant to make maths easier, parents have been left scratching their heads, writes Rob Eastaway.

The emphasis has moved away from blindly following rules to techniques a child can understand”
Rob Eastaway
I used to think I had a good understanding of maths - until my daughter started going to primary school. That's when I discovered a revolution had taken place in the way arithmetic is taught, and there were techniques and terminology that meant nothing to me.
Let me give you a flavour. In most primary schools, maths lessons are called numeracy. Children work using number lines and learn their number bonds, they fill in Carroll Diagrams, and they calculate using the grid method and something that carries the peculiar name of "chunking".
Like most parents - numerate or otherwise - my first reaction to this was annoyance. Why have they changed it? Now my child gets cross when I try to explain using my methods. Is this why some people reckon the country's maths is going to the dogs?
I decided to find out more, and ended up writing a book aimed at parents, like me, who wanted to have a better understanding of how young children learn maths these days.

Researching the book was a revelation.


What became clear is that at school I was one of the lucky ones. Being strong with numbers, I had no problem learning the black-box techniques of long multiplication and long division, and usually got the right answer.

But for a huge proportion of children, these techniques were a meaningless chore. Ask most adults today to carry out a long multiplication or division sum and they will look blankly at you.

They may have, sort of, got it once, but they can't remember how to do it. And anyway, we have calculators now, don't we?


The point about calculators is important. Many of the techniques we were taught at school date back to Victorian times, when the country needed vast numbers of clerks to perform calculations every day. Today, calculators and spreadsheets can do these tasks far quicker, so the need for everybody to be able to do big calculations by hand has largely disappeared.


That's not to say we don't need strong number skills.
We are inundated by numbers all the time, whether it's somebody flogging us a mobile phone package or a politician trying to convince us about a particular policy. As a society we have to make sense of these numbers if we are to successfully manage our lives.


Do we all need to be able to work out 27 x 43 precisely with a pen and paper? Probably not. But we do need to know that 27 x 43 is roughly 30 x 40, and that this is roughly 1,200. It's partly the need to have a good feel for numbers that is behind the modern methods.

The revolution in the teaching of maths at primary school kicked in with the National Numeracy Strategy in 1999. The emphasis moved away from blindly following rules (remember borrowing one from the next column and paying back?) towards techniques a child understood.

One of the methods that has been adopted widely is the "grid method" for multiplication, which links to a visual method that many children find easier to understand.

Use the step-by-step guide below for a quick refresher on long multiplication, then an introduction to the grid method.
 Long multiplycation: 27 multiply 43?












Another important method, used for division, is "chunking". To understand chunking, you need to think about what division actually means. Division is usually introduced through the idea of sharing. You want to divide 18 sweets fairly between six children. How many sweets do they each get? 18 / 6 = 3.

But what if the problem is this: you need to put 18 sweets into bags of six. How many bags do you need?

This isn't about sharing, it's taking away sweets in chunks of six until there are none left, and then counting the bags. Here, "division" is really repeated subtraction, but calculated in the same way, 18 / 6 = 3.


Chunking is a method based around repeated subtraction and many people find it an easier way to tackle division problems. Ever wondered why six divided by ½ is 12? Think of it as "how many times can I take ½ a pizza away from six pizzas?" and it becomes clear that the answer is indeed 12.

See the grid method and chunking in action


So is the nation's maths better thanks to these new methods? Certainly the horror stories of children being punished or humiliated for getting things wrong have all but disappeared, as have the tedious lessons of endless sums. There is also some evidence that children do have a better understanding of the methods they're using, and make fewer mistakes when they use them.

But that isn't the full story. To become fully numerate you need to know when to use these methods, you need to practise, and you also need to be able to estimate, which means knowing your times tables off by heart.



My own experience, and the feedback I get from others, is that many children are missing out on these basics. Is too much energy being diverted into taking Sats tests? Does the problem lie with teachers who don't have enough maths knowledge? Or is too much emphasis being placed on enjoyment at the expense of rigour?

Perhaps it's all of these things. But we shouldn't be relying just on schools to impart all this knowledge in any case. Children learn maths at home too, whether it's helping with cooking, playing board games or helping mum and dad to measure wallpaper.

Forcing a child to learn the methods we were taught can result in frustration and tantrums. For the sake of harmony at home if nothing else, it's not a bad idea to get familiar with chunking, number lines and the rest.

Maths for Mums and Dads by Rob Eastaway & Mike Askew is published by Square Peg.

Source: BBC

Jun 22, 2011

Patriots vs. Redcoats: An Outdoor Social Studies Game

This outdoor game, a modification of "Capture the Flag", uses ball, baskets, and foam to re-enact the fight between the Redcoats and the Patriots in Massachusetts in 1775.
Can your foam-armed "Patriots" stop the "Redcoats" from capturing Concord's armory? You'll find out!

What You Need:

  • 10-20 tennis balls, or other soft, rubber balls, at least two per person
  • Three empty baskets—laundry baskets work great
  • 2-3 short sticks of foam, such as a section from an old pool “noodle”, about 12-18”
  • 8 sport cones or other markers to place on the field
  • Optional: red, white, and blue streamers; duct tape
Set-Up:
  1. Assemble the props and clear the field area. Collect tennis balls (or any rubber balls of any size) that you have on hand or that you can borrow from friends and family. Place the balls in one of the baskets, and take a quick count. These balls will be the "cannonballs" in the armory for the game.
  2. Gather the foam sticks which will act as the “bayonet muskets” for the patriots. These foam sticks will help young kids extend their reach, so they should be soft to avoid injury. To add more color to the game, have the kids decorate their foam sticks by wrapping red, blue, and white streamers around them, so everyone can be very clear who and where the “Patriots” are!
  3. Set up the field. Find an open space that will accommodate all the kids. At one end of the space, place one empty basket and put four cones around it. This basket represents the British headquarters in "Boston". At the other end, place the basket full of balls and put four cones around it. This basket represents the "Concord Armory." In the middle of the field, place the third basket. This middle basket will represent the patriots' "secret stash" basket that was the property of the Continental Army.

What You Do:

  1. Choose the teams. Depending on the size of the group, select three or so “Patriots” (more, if the group is large). Have the patriots stand in the middle of the field, and give each of them a “bayonet musket” to hold. Everyone else will be a Redcoat.
  2. The Redcoats' goal is to run across the field without being tagged, grab no more than two "cannonballs" from the armory, and make it back to the other end of the field, or "Boston". As long as they have not been tagged, the Redcoats can keep running back and forth until they have brought all their arms back to their side.
  3. The Patriot's goal is to tag the Redcoats with their "bayonet muskets". If a Patriot tags a Redcoat, the Redcoat must sit down where he was tagged on the field until the round is over. Any cannonballs the tagged Redcoat was carrying will be confiscated by the Continental Army and placed them in the "secret stash" basket in the middle of the field.
  4. Each round ends either when all the Redcoats are tagged, or when all the cannonballs have been removed from the "Concord Armory".
  5. The final cannonball count signals the victory: if the majority of the balls have returned to "Boston", then the Redcoats have won, but if the majority of the balls are in the Patriots' stash or in the armory, the Patriots win!

Did You Know?

What really happened in Massachusetts on April 19, 1775?

On that day, American colonists in the Massachusetts towns of Lexington and Concord entered a decisive moment in history. Early that morning, British General Thomas Gage sent 700 “Redcoat” soldiers from Boston to capture the Patriots Samuel Adams and John Hancock in Lexington, and to destroy the Patriots’ armory in Concord.

But thanks to the famous warning from Paul Revere (with help from Dr. Samuel Prescott), Samuel Adams and John Hancock were able to escape the night before the battle. When the British troops did make it to Concord on the morning of April 19, 1775, they began seizing gunpowder and arms until the colonists stood their ground at the Old North Bridge at the edge of the town. Astonished by the colonists' resistance, the British tried to retreat back to Boston. However, by that time, colonists from local towns had poured out of their homes and lined the road to fight.

By the end of the day, the British had lost nearly 300 men, and a clear message had been sent by the colonists: the Americans wanted their independence, and they weren’t going to give up the fight. They didn’t, and today, centuries later, we can still thank them for their gift.

Julie Williams, M.A. Education, taught middle and high school History and English for seventeen years. Since then, she has volunteered in elementary classrooms while raising her two sons and earning a master's in school administration. She has also been a leader in her local PTA. 

Source: http://www.education.com

Create a Two-Sided Opposites Painting

Create a colorful painting—twice! Transform a single sheet of card stock into a double-sided painting of opposites: on one side, a bright daytime landscape, on the other, a dusky evening scene. Not only is this a great way to let your kids dive into art and explore their artistic side, it also provides an excellent lesson on the concept of opposites and the differences between night and day. And as always, any hands-on art project like this also helps strengthen little hand muscles for writing.

What You Need:

  • Card stock or other thick paper
  • Pencil
  • Tempera paints
  • Paint brush
  • Water cup (make sure it's one you won't reuse for food or drinks)
  • Paint palette or washable art tray
  • Hole punch
  • Scissors
  • Yarn or thin ribbon

What To Do:

  1. Sit by a window (or outside if possible) and have your child make a pencil sketch of what he sees on the card stock. This is a great time to introduce him to art terminology such as landscape, horizon line (where the earth meets the sky), foreground, middle ground, and background. Ask him to really look at the area he's drawing. Do trees that are further back look bigger or smaller than the ones that are closer to him? What colors and shapes does he see? Is it sunny, rainy, or cloudy?
  2. Take the drawing to a suitable workspace such as a flat table or an easel. Lay small puddles of paint onto the palette or tray. Try using just the primary colors (blue, red, and yellow) and white. Use this opportunity to talk about the color wheel: encourage him to mix the primary colors together to get new colors, and use the white paint to make lighter shades.
  3. Now start the painting. Invite him to color in his landscape sketch by painting over it with the tempera paint. Once he's finished, set the painting aside to dry.
  4. At dusk, have him sit in the same spot he was in for step 1, and ask him to sketch what he sees on the reverse side of the painting (if you started the activity in the afternoon, wait until the following day to do this step so the front side has time to dry completely). Talk about the landscape: does it look different now than it did earlier in the day? Do the shapes still look as sharp? Is there more or less to see? What does the sky look like?
  5. Bring the drawing to the workspace and encourage him to paint his new nighttime landscape. Talk about the colors he'll need for this painting. Once he's done painting, set the work aside to dry.
  6. Once the painting is completely dry, punch a hole in the top of the painting with the hole punch. Cut a piece of ribbon or yarn, thread it through the hole, and tie a knot just above the edge of the painting.
  7. Hang the painting mobile style so both sides are displayed.
Explore other "opposite" concepts for more art activities. Try themes as simple as inside/outside or beginning/end or as complicated as Impressionism/Realism or classical art/pop art.
Erica Loop has a MS in Applied Developmental Psychology from the University of Pittsburgh's School of Education. She has many years of teaching experience working in early childhood education, and as an arts educator at the Carnegie Museum of Art in Pittsburgh. 

Source: http://www.education.com

2 Grade Math Facts Secret Codes Actvities

When it comes to those basic math facts, there's nothing like practice, practice, practice. But making them interesting is quite another matter. Here's an irresistible activity you can use with your second grader to send messages in “secret code.” Next time you pack a school lunch, for example, try putting in a special note with your love and good wishes … in a code that your child and her friends can decipher over their meal.

What You Need:

What to Do:

  1. Download our Math Fact code sheet, or make your own based on your child's level.
  2. Write your message on the decoder key, by making a small line for each letter of the alphabet in the message. See our sample message as an example.
  3. Write out your message lines, and then let your child take a whack at the math facts to decode your words. The sum of each pair of numbers appears under each line of the message, and matches up with a letter. For a child who is making good progress on math facts, this is a fairly fast activity … but one that will bring a sense of accomplishment that lasts all day and beyond!
Julie Williams, M.A. Education, taught middle and high school History and English for seventeen years. Since then, she has volunteered in elementary classrooms while raising her two sons and earning a master's in school administration. She has also been a leader in her local PTA. 

Source: http://www.education.com

Feb 13, 2011

42nd IMO 2001 shortlisted problems and solutions

Algebra


A1.  Let T be the set of all triples (a, b, c) where a, b, c are non-negative integers. Find all real-valued functions f on T such that f(a, b, c) = 0 if any of a, b, c are zero and f(a, b, c) = 1 + 1/6 f(a+1, b-1, c) + 1/6 f(a-1, b+1, c) + 1/6 f(a+1, b, c-1) + 1/6 f(a-1, b, c+1) + 1/6 f(a, b+1, c-1) + 1/6 f(a, b-1, c+1) for a, b, c all positive.

A2.  Show that any infinite sequence an of positive numbers satisfies an > an-121/n - 1 for infinitely many n.

A3.  Show that x1/(1 + x12) + x2/(1 + x12 + x22) + ... + xn/(1 + x22 + ... + xn2) < √n for all real numbers xi.

A4.  Find all real-valued functions f on the reals such that f(xy) ( f(x) - f(y) ) = (x - y) f(x) f(y) for all x, y.

A5.  Find all positive integers a0 = 1, a1, a2, ... , an such that a0/a1 + a1/a2 + ... + an-1/an = 99/100 and (ak+1 - 1)ak-1 ≥ ak3 - ak2 for k = 1, 2, ... , n-1.

Combinatorics


C1.  What is the largest number of subsequences of the form n, n+1, n+2 that a sequence of 2001 positive integers can have? For example, the sequence 1, 2, 2, 3, 3, of 5 terms has 4 such subsequences.

C3.  A finite graph is such that there are no subsets of 5 points with all 15 edges and every two triangles have at least one point in common. Show that there are at most two points X such that removing X leaves no triangles.

C4.  Let P be the set of all sets of three integers of the form {n, n+1776, n + 3777} or {n, n+2001, n+3777}. Show that we can write the set {0, 1, 2, 3, ... } as a union of disjoint members of P.

C5.  Find all finite sequences a0, a1, a2, ... , an such that am equals the number of times that m appears in the sequence.

C6.  Show that there is a set P of (2n)!/(n! (n+1)!) sequences of 2n terms, half 1s and half 0s, such that any sequence of 2n terms, half 1s and half 0s, is either in P or can be derived from a member of P by moving one term. Moving a term means changing a1, a2, ... , a2n to a1, a2, ... , ai-1, ai+1, ... , ajaiaj+1 ... an for some i, j. For example, by moving the initial 0 we can change 0110 to 1010 or 1100, or by moving the first 1 we can change 0110 to 1010 or 0101.

C7.  There are n piles of stones in a line. If a pile contains at least two stones more than the pile on its right, then a stone can be moved to the pile on its right. Initially, all the piles are empty except the leftmost pile which has n stones. If more than one move is possible, then a possible move is chosen arbitrarily. Show that after a finite number of moves, no more moves are possible and that the final configuration is independent of the moves made. Describe the final configuration. For example, we n = 6, one might get successively: 5 1, 4 2, 4 1 1, 3 2 1.

Geometry


G1.  ABC is an acute-angled triangle. A' is the center of the square with two vertices on BC, one on AB and one on AC. Similarly, B' is the center of the square with two vertices on CA, one on AB and one on BC, and C' is the center of the square with two vertices on AB, one on BC and one on CA. Show that AA', BB' and CC' are concurrent.

G3.  ABC is a triangle with centroid G and side lengths a = BC, b = CA, c = AB. Find the point P in the plane which minimises AP.AG + BP.BG + CP.CG and find the minimum in terms of a, b, c.

G4.  P is a point inside the triangle ABC. The feet of the perpendiculars to the sides are D, E, F. Find the point P which maximises PD.PE.PF/(PA.PB.PC). Which triangles give the largest maximum value?

G5.  ABC is an acute-angled triangle. B' is a point on the perpendicular bisector of AC on the opposite side of AC to B such that angle AB'C = 2A. A' and C' are defined similarly (with angle CA'B = 2C, angle BC'A = 2B). The lines AA' and B'C' meet at A". The points B" and C" are defined similarly. Find AA'/A"A' + BB'/B"B' + CC'/C"C'.

G6.  P is a point outside the triangle ABC. The perpendiculars from P meet the lines BC, CA, AB at D, E, F, respectively. The triangles PAF, PBD, PCE all have equal area. Show that their area must equal that of ABC.

G7.  P is a point inside acute-angled the triangle ABC. The perpendiculars from P meet the sides BC, CA, AB at D, E, F, respectively. Show that P is the circumcenter iff each of the triangles AEF, BDF, CDE has perimeter not exceeding that of DEF.

Number theory


N1.  Prove that we cannot find consecutive factorials with first digits 1, 2, ... , 9.

N2.  Find the largest real k such that if a, b, c, d are positive integers such that a + b = c + d, 2ab = cd and a ≥ b, then a/b ≥ k.

N3.  The sequence an is defined by a1= 1111, a2 = 1212, a3 = 1313, and an+3 = |an+2 - an+1| + |an+1 - an|. Find an, where n = 1414.

N4.  Let p > 3 be a prime. Show that there is a positive integer n < p-1 such that np-1 - 1 and (n+1)p-1 - 1 are not divisible by p2.

N6.  Do there exist 100 positive integers not exceeding 25000 such that the sum of every pair is distinct? 

Note: problems A6, C2, C8, G2, G8 and N5 were used in the Olympiad and are not shown here.

Solutions

Problem A3
Show that x1/(1 + x12) + x2/(1 + x12 + x22) + ... + xn/(1 + x22 + ... + xn2) < √n for all real numbers xi.
 
Solution
Using Cauchy ∑ aibi ≤ √(∑ ai2) √(&sum bi2) with ai = 1, bi = xi/(1+x12+x22+...+xi2) we have lhs ≤ rhs √(∑ bi2).
Now use b12 ≤ x12/(1+x12) = 1 - 1/(1+x12

Problem C1
What is the largest number of subsequences of the form n, n+1, n+2 that a sequence of 2001 positive integers can have? For example, the sequence 1, 2, 2, 3, 3, of 5 terms has 4 such subsequences.
 
Solution
Answer: 6673.
We show that 1, ... , 1, 2, ... , 2, 3, ... , 3 (with 667 1s, 667 2s and 667 3s) is maximal. There are several stages. Let the sequence be a1, a2, ... , a2001. We show first that there is a maximal sequence with a1 ≤ a2 ≤ ... ≤ a2001. For suppose there is a maximal sequence with ai > ai+1. Swapping these two terms cannot decrease the number of subsequences n, n+1, n+2. By a series of such swaps we get another maximal sequence with a1 ≤ a2 ≤ ... ≤ a2001.
Next we show that there is a maximal sequence with ai = ai+1 or ai+1 + 1 for each i. For suppose a maximal sequence has ai ≥ ai+1 + d+1. Then by adding d to ai and all earlier terms we do not decrease the number of subsequences n, n+1, n+2. So by a series of such changes we can get another maximal sequence with ai = ai+1 or ai+1 + 1 for each i.
Thus there is a maximal sequence with b0 terms k, followed by b1 terms k+1, followed by b2 terms k+2 and so on up to bh terms k+h. The number of sequences n, n+1, n+2 is evidently N = b0b1b2 + b1b2b3 + b2b3b4 + ... + bn-2bn-1bn. But if n ≥ 2, then changing from b0, b1, ... , bn to b1, b2, (b0 + b3, b4, ... bn increases N by b0b2b4 + b0b4b5 (or leaves it unchanged if n = 3). So we may take n = 2. (Obviously a maximal sequence cannot have n < 2, because then it has no subsequences of the required form.) Only differences matter, so we may take the first term to be 1. Thus there is a maximal sequence 1, ... , 1, 2, ... , 2, 3, ... , 3 with a 1s b 2s and c 3s, where a + b + c = 2001.
Such a sequence obviously has abc subsequences of the required form. But by the arithmetic/geometric mean inequality, abc is maximised by taking a = b = c. 

Problem C5
Find all finite sequences a0, a1, a2, ... , an such that am equals the number of times that m appears in the sequence.
 
Solution
Answer: (1) 1, 2, 1, 0; (2) 2, 0, 2, 0; (3) 2, 1, 2, 0, 0; (4) N, 2, 1, 0, ... , 0 with N+4 terms, except that aN = 1.
Note that we cannot have a0 = 0. Suppose there are k other non-zero terms. The number of 1s is a1, the number of 2s is a2 etc, so the number of non-zero terms is a1 + a2 + ... + an. Hence a1 + a2 + ... + an = k+1. But if there are k+1 non-zero terms, then k of a1, a2, ... , an are at least 1. So k-1 of them must be 1, one of them 2 and the rest 0. In particular, none of them are 3 or more (*).
If a0 is 1 or 2, then no member of the sequence is 3 or more, so a3 = ... = an = 0.
Suppose a0 = 1. Then a1 cannot be 1 (or there would be two 1s). It cannot exceed 2, so it must be 2. But the only other term that can be 1 is a2, so a2 = 1. We need a 0, so a3 = 0. There cannot be any further terms, because they would have to be 0s and we have enough already. Thus we have the solution 1, 2, 1, 0.
Suppose a0 = 2. We know that a1 cannot exceed 2. Suppose it is 2. Then a2 is at least 2. Like all the terms, it is at most 2. But if it is 2, then there are three 2s. Contradiction. So a1 must be 0 or 1. Suppose it is 0. Then a2 is at least 1 because a0 is 2. But it cannot be 1, because there are no 1s. So it must be 2. So all remaining terms must be 0. Now there are two 0s, so there must be just one more term, and we have the solution 2, 0, 2, 0.
So suppose a1 = 1 (and a0 = 2). Again there are no more 1s. So a2 must be 2. Hence all remaining terms must be 0. We need two more 0s (to get a0 correct), so we have the solution 2, 1, 2, 0, 0.
That exhausts the possibilities for a0 = 1 or 2. So suppose a0 = N > 2. Hence aN = 1. So a1 is at least 1. It cannot be 1 (or there would be two 1s). But by (*) it cannot exceed 2, so it must be 2. Hence, by (*), a2 must be 1. Since the only term exceeding 2 is a0 (by (*) ), all of a3, a4, ... , an except aN must be 0. We need N zeros, so n = N+3. It is easy to check that a0 = N (> 2), a1 = 2, a2 = 1, aN = 1 and the rest of a3, a4, ... , aN+3 is indeed a solution.

Problem C6
Show that there is a set P of (2n)!/(n! (n+1)!) sequences of 2n terms, half 1s and half 0s, such that any sequence of 2n terms, half 1s and half 0s, is either in P or can be derived from a member of P by moving one term. Moving a term means changing a1, a2, ... , a2n to a1, a2, ... , ai-1, ai+1, ... , ajaiaj+1 ... an for some i, j. For example, by moving the initial 0 we can change 0110 to 1010 or 1100, or by moving the first 1 we can change 0110 to 1010 or 0101.
 
Solution
Given a sequence x (of length 2n), let f(x) be the sum of the positions of the 1s mod n+1. For example, x = 0110 has 1s in positions 2 and 3, so f(x) = 2 + 3 = 5 = 2 mod 3. Take Pk to be all sequences with f(x) = k. Now suppose a sequence x has 0 in the first position. If we move it to just after the mth 1 to get the sequence y, then we shift each of those 1s one place to the left, so f(y) = f(x) - m mod n+1. So by taking a suitable m (between 1 and n) - or by making no move at all - we can get y into Pk. Similarly, if x has 1 in the first position and we move it to just after the mth 0 to get a sequence y, then if we moved it past h 1s, each of those 1s has its position reduced by 1, but the 1 we are moving has its position increased by m+h, so f(y) = f(x) + m. So we can again get y into Pk. So every sequence either belongs to Pk or can be derived from a member of Pk by a move. But the n+1 sets Pk form a partition the (2n)!/(n! n!) possible sequences, so at least one of them has at most (2n)!/(n! (n+1)!) members. [If it has less then we can add any sequences to bring it up to (2n)!/(n! (n+1)!) members]. 

Problem G1
ABC is an acute-angled triangle. A' is the center of the square with two vertices on BC, one on AB and one on AC. Similarly, B' is the center of the square with two vertices on CA, one on AB and one on BC, and C' is the center of the square with two vertices on AB, one on BC and one on CA. Show that AA', BB' and CC' are concurrent.
 
Solution
Let the ray AA' meet BC at X. Since AX passes through the center of the square, we have VX = q, WX = p, as shown in the diagram. By similar triangles BX/XC = p/q. Hence BX/XC = (BX + p)/(XC + q) = BW/VC = (BV + s)/(CW + s), where s is the side of the square. But BV = s cot B, CW = s cot C, so BX/XC = (cot B + 1)/(cot C + 1). We are now home by Ceva's theorem. 


Problem G3
ABC is a triangle with centroid G and side lengths a = BC, b = CA, c = AB. Find the point P in the plane which minimises AP.AG + BP.BG + CP.CG and find the minimum in terms of a, b, c.
 
Solution
Applying the sine rule to triangle BGL, we have BG/sin BLG = BL/sin BGK. Similarly applying the sine rule to triangle CGL, we have CG/sin CLG = CL/sin CGK. But L is the midpoint of BC and sin BLG = sin CLG, so BG/CG = sin CGK/sin BGK.
We have BK = 2 R sin BGK, where R is the radius of the circle (because if O is its center then ∠BOK = 2 ∠BGK). Similarly, CK = 2 R sin CGK, so CK/BK = BG/CG, and hence BG/CK = CG/BK.
Similarly, AG/BG = sin BGN/sin AGN = sin BGN/sin CGK (opposite angles). Also BC = 2 R sin BKC = 2 R sin BGN (since BGCK is cyclic, so angle BKC = angle BGN). Hence BC/CK = sin BGN/sin CGK = AG/BG, so BG/CK = AG/BC and hence BC : CK : BK = AG : BG : CG.
We now use Ptolemy's theorem on the quadrilateral PBKC: PK.BC <= BP.CK + CP.BK. Hence PK.AG <= BP.BG + CP.CG. Hence (AP + PK)AG <= AP.AG + BP.BG + CP.CG and so finally AK.AG <= AP.AG + BP.BG + CP.CG with equality iff (1) P lies on the circle between B and C (for equality in Ptolemy) and (2) P lies on AK (for equality in the triangle inequality). Thus the unique minimum is when P = G.
Using the cosine formula, we have AB2 = AL2 + BL2 - 2 BL.AL cos ALB, AC2 = AL2 + CL2 - 2 CL.AL cos ALC. Adding, AB2 + AC2 = 2 AL2 + BC2/2, so AG2 =(4/9) AL2 = (2 AB2 + 2 AC2 - BC2)/9. Hence AG2 + BG2 + CG2 = (AB2 + BC2 + CA2)/3. 

Problem G4
P is a point inside the triangle ABC. The feet of the perpendiculars to the sides are D, E, F. Find the point P which maximises PD.PE.PF/(PA.PB.PC). Which triangles give the largest maximum value?
 
Solution
Let n = PD.PE.PF/(PA.PB.PC). We have PF.PE/PA2 = sin x sin x' = 1/2 cos(x - x') - 1/2 cos(x + x') = 1/2 cos(x - x') - 1/2 cos A ≤ 1/2 - 1/2 cos A = sin2(A/2), with equality iff P lies on the bisector of angle A. Hence n ≤ sin(A/2) sin(B/2) sin(C/2) with equality iff P is the incenter.
But sin(A/2) sin(B/2) = 1/2 cos(A/2 - B/2) - 1/2 cos(A/2 + B/2) = 1/2 cos(A/2 - B/2) - 1/2 sin C/2. So sin(A/2) sin(B/2) sin(C/2) can only be maximal if A/2 = B/2. Hence it can only be maximal if A = B = C. So the unique maximum is if the triangle is equilateral.

Problem G5
ABC is an acute-angled triangle. B' is a point on the perpendicular bisector of AC on the opposite side of AC to B such that angle AB'C = 2A. A' and C' are defined similarly (with ∠CA'B = 2∠C, ∠BC'A = 2∠B). The lines AA' and B'C' meet at A". The points B" and C" are defined similarly. Find AA'/A"A' + BB'/B"B' + CC'/C"C'.
 
Solution
Answer: 4.
Note that ∠B' is not twice ∠B, but twice ∠A, so ∠BAB' = ∠CBC' = ∠ACA' = 90o. Also ∠B'AC = 90o + (90o - ∠B) < 180o. Note also that the reason for ABC being acute-angled is to ensure that ∠AB'C < 180o (so that B' does indeed lie on the opposite side of AC to B). Thus AB'CA'BC' is convex.
Let the circle center B' through A and C meet the circle center A' radius B and C at O. Then ∠AOC = 180o - 1/2 ∠AB'C = 180o - ∠A. Similarly, ∠COB = 180o - ∠C. Hence ∠AOB = 360o - ∠AOC - ∠COB = ∠A + ∠C = 180o - ∠B. Hence O also lies on the circle center C' through A and B. Now since the circles center A' and B' meet at O and C, the points O and C are reflections in A'B'. Similarly for O and A, and O and B.
AA'/A"A' = 1 + AA"/A'A" = 1 + area AB'C'/area A'B'C'. Hence AA'/A"A' + BB'/B"B' + CC'/C"C' = 3 + (area AB'C' + area BC'A' + area CA'B')/area A'B'C' = 3 + (area OB'C' + area OC'A' + area OA'B')/area A'B'C' = 4. 

Problem G6
P is a point outside the triangle ABC. The perpendiculars from P meet the lines BC, CA, AB at D, E, F, respectively. The triangles PAF, PBD, PCE all have equal area. Show that their area must equal that of ABC.
 
Solution
In principle, there are two possible configurations: (1) P can lie between the two rays BA and BC (on the opposite side of AC to B), or similarly between the rays AB and AC or between the rays CA and CB; or (2) P can lie between the rays AB and CB (on the infinite side, opposite to A and C), or the two similar arrangemetns. But the second case implies that one of the triangles PAF, PBD, PCE is a proper subset of another, so it is not possible. Thus, without loss of generality, we have the configuration shown below. Take area PAF = area PBD = area PCE = x. Also put area APE = u, area ABE = v and area CBE = w.
BD/DC = area ADB/area ADC = area PDB/area PDC. So (x - u - v)/(x - v + w) = x/(2x + w). Hence x2 - x(2u + v) - uw - vw = 0 (*). Similarly, CE/EA = area CBE/area CBA = area CPE/area EPA, so w/u = x/v. Substituting in (*) gives x2 - x(2u + ux/w) - ux - uw = 0, or x2(w - u) = uw(3x + w). Similarly, AF/FB = area AFP/area FPB = area AFC/area FCB, so x/(x + u + v) = (2x + v)/(2x + u + v + w) (**). Again, eliminating v, we get x = w(w2 - uw - u2)/(u2 + 2uw). Now eliminating x, we get (w - u)(w + u)(w3 - 3uw2 - 4u2w - u3) = 0. Obviously we cannot have w + u = 0 (for that means area ABC = 0, but then ABC is not a triangle). If w - u = 0, then v = x, so (**) gives x/(2x + u) = 3x/(3x + 2u), so 3x + u = 0, so x = 0, which is impossible. So we must have w3 = 3uw2 + 4u22 + u3 = u(3w + u)(w + u).
Hence x = (w3 - uw(u+w) )/(u2 + 2uw) = (u+w) ( u(3w+u) - uw)/(u2 + 2uw) = u+w, which is the required result. 

Problem G7
P is a point inside acute-angled the triangle ABC. The perpendiculars from P meet the sides BC, CA, AB at D, E, F, respectively. Show that P is the circumcenter iff each of the triangles AEF, BDF, CDE has perimeter not exceeding that of DEF.
 
Solution
One way is obvious: if P is the circumcenter, then D, E, F are the midpoints of the sides and so AEF, BDF, CDE, DEF all have the same area. So assume area AEF ≤ area DEF, area BDF ≤ area DEF and area CDE ≤ area DEF, but P is not the circumcenter. We will obtain a contradiction.
Let the angles in the triangles be as shown in the diagram (there was not room to mark y" = angle DEC.) Take B' so that BDB'F is a parallelogram. Then area BDF = area B'DF, so E cannot lie inside the triangle B'DF (except possibly at B' itself), because then we would have area DEF < area BDF. Suppose E = B'. Then EF is parallel to BC, and DE is parallel to AB. Hence PD is perpendicular to EF, and PF is perpendicular to DE. Hence P is the orthocenter of DEF and so PE is perpendicular to DF, so DF is parallel to AC. Hence BDEF and AFDE are parallelograms, so BF = DE = AF, in other words F is the midpoint of AB. Similarly, D and E are also midpoints. Hence P is the circumcenter. Contradiction.
So we may assume that E is not any point of B'DF. Hence either x' < y or z" < y. Similarly, y' < z or x" < z, and z' < x or y" < x. Without loss of generality we may assume y' < z (there is symmetry between y' and y"). So looking at triangle AEF, z" > y. Hence x' < y. So looking at CDE, y" > x. Hence z' < x. Looking at BDF, x" > z. So x' < y < z", y' < z < x", and z' < x < y".
Now AEPF is cyclic, so ∠AEF = ∠APF, so AP = PF/cos y'. Similarly, BDPF is cyclic, so ∠BDF = ∠BPF, so BP= PF/cos x" > PF/cos y' = AP. Similarly, CP > BP and AP > CP. Contradiction.
 
Problem N1
Prove that we cannot find consecutive factorials with first digits 1, 2, ... , 9.
 
Solution
We can write any integer n in scientific notation as n = n x 10k, where 1 ≤ n < 10 is a real number and k is a positive integer. For example, 356 = 3.56 x 102. Suppose that (N+i)! has first digit i for i = 1, 2, ... , 9. Then (N+i)! is a real number between i and i+1. So for i = 2, ... , 9, we have that N+i is a real number such that 1 < N+i = (N+i)!/(N+i-1)! < (i+1)/(i-1). So certainly N+i < 3 and hence N+i < N+i+1. Thus we must have 1 < N+2 < N+3 < ... < N+9 < 5/4.
But ab = a b unless a b > 10, in which case ab = (a b)/10. So in any case aba b. Hence (N+4)! = (N+1)! N+2 N+3 N+4 < 2 (5/4) (5/4) (5/4) = 250/64 < 4. Contradiction (because it is supposed to have first digit 4). 

Problem N2
Find the largest real k such that if a, b, c, d are positive integers such that a + b = c + d, 2ab = cd and a ≥ b, then a/b ≥ k.
 
Solution
Answer: k = 3 + 2√2.
A few examples of such a, b, c, d are:
2 + 12 = 6 + 8, 2.2.12 = 6.8, so a/b = 6
2 + 15 = 5 + 12, 2.2.15 = 5.12, so a/b = 7.5
3 + 20 = 8 + 15, 2.3.20 = 8.15, so a/b = 6.67
We have (a+b)2 - 8ab = (c+d)2 - 4cd, so (a/b)2 - 6(a/b) + 1 = ((c-d)/b)2 ≥ 0. The plot of x2 - 6x + 1 is shown below. It has zeros at x = 3 - 2√2 < 1 and 3 + 2√2 = 5.83. So since a/b ≥ 1, we must have a/b > 3 + 2√2.
The harder part is showing that we can get as close as we like to 3 + 2√2 and hence that k = 3 + 2√2. The idea is to show that we can make (c-d)/b as small as we like.
Note first that (a+b)2 - 4ab = (c+d)2 - 2cd, so (a-b)2 = c2 + d2. In other words, a-b, c, d is a Pythagorean triple. It is well-known that r2 + s2, r2 - s2, 2rs are such triples for any integers r, s. Then we have a+b = c+d = r2 + 2rs - s2. Hence a = r2 + rs, b = rs - s2. It is now easy to check that a = r2 + rs, b = rs - s2, c = r2 - s2, d = 2rs satisfy a + b = c + d, 2ab = cd.
Suppose we specialise to c - d = 1 (then if we could get arbitrarily large b we would be home). That requires r2 - s2 - 2rs = 1, or (r - s)2 - 2s2 = 1. We recognise that as a Pell equation. It has a solution r - s = 3, s = 2. But if (R, S) is a solution to x2 - 2y2 = 1, so is (3R+4S, 2R+3S). So we can get arbitrarily large r - s and s such that c - d = 1. Then b = rs - s2 = (r - s)s, which can therefore be made arbitrarily large as required.
For example, r - s = 3, s = 2, gives r = 5, so a = 35, b = 6, c = 21, d = 20. Then a/b = 5.8333. The next solution is r - s = 17, s = 12, so r = 29, so a = 1189, b = 204, c = 697, d = 696 and a/b = 5.828431 (cf 3 + 2√2 = 5.828427).

Problem N3
The sequence an is defined by a1= 1111, a2 = 1212, a3 = 1313, and an+3 = |an+2 - an+1| + |an+1 - an|. Find an, where n = 1414.
 
Solution
Answer: 1.
This looks so horrendous that appearances must be deceiving. Suppose we take smaller initial values to try to get a sense of what is going on.
term 1  3  7  6  5  2  4  5  3  3  2  1  2  2  1  1  1  0  1  2  2  1
diff  2  4  1  1  3  2  1  2  0  1  1  1  0  1  0  0  1  1  1  0  1
We now have a repetition of 1, 2, 2, so the sequence has period 7 and just repeats the cycle 1, 2, 2, 1, 1, 1, 0. Take another example, starting with 3, 21, 25. After a longer initial lead-in we again get a sequence with period 7:
3  21  25  22   7  18  26  19  15  11   8   7   4   4   3   1   3   4   3   2   2   1
 18   4   3  15  11   8   7   4   4   3   1   3   0   1   2   2   1   1   1   0
Finally, we try something less random. It is evidently periodic right from the start.
k    k    k   0   k  2k  2k   k   k   k
  0   0   k   k   k   0   k   0   0
The last case shows that the terms do not necessarily become small, but we hope that they do unless the starting values are special.
Note that 100, 0, 100 is followed by 200, so terms can go up. Studying the examples above, the differences look easier to deal with. The examples suggest that if three successive differences are a, b, c, then the next is |c - a|. If three successive differences are a, b, c, then the corresponding terms must be A, B, C, a + b. The next term will be b + c. So the next difference will be |b+c - (a+b)| = |c - a|. That is certainly less than max(a, c) and hence less than max(a, b, c). In particular, the differences are certainly bounded.
Suppose max(a, b, c) = M > 1. The following differences are all <= M, so moving forward one or two places in the sequence if necessary, we can assume that a = M. Thus the differences are M, b, c, d, e, f, g, ... . We claim that max(e, f, g) < M unless b and c are both 0 or M. If not, then max(b, c, d) = max(c, d, e) = max(d, e, f) = max(e, f, g) = M (because once the max dips down it can never get back up). But d = |M - c| = M - c. So at least one of b, c, M-c must be M.
Suppose b = M. Then e = |M - c - M| = c. But max(c, d, e) = max(c, M-c, c) = M, so c = 0 or M. Thus both b and c are 0 or M. Contradiction.
Suppose c = M. Then d = 0, e = |0 - b| = b, f = |b - M| = M - b, so max(d, e, f) = max(b, M-b) = M, so b = 0 or M. Thus both b and c are 0 or M. Contradiction.
Finally, suppose c = 0. Then d = M, e = M-b, f = M-b, g = b, so max(e, f, g) = max(M-b, b) = M, so b = 0 or M. Contradiction.
Now suppose three successive differences are all 0 or M. Then two successive terms (not differences) are 0, M or 2M. Working backwards we find that all terms must be multiples of M:
...  A   B   C
... z   a   b 
a and b are multiples of M, so C = a+b must be a multiple of M. But B = C±b, so B must be a multiple of M. But B = a+z, so z must be a multiple of M. Also A = B±a, so A must be a multiple of M, and so on. However, for the particular values we have been given, the greatest common divisor of the first two terms is 1, so M must be 1. Contradiction.
Thus we have established that if three successive differences have maximum M > 1, then we can find three successive terms starting at most 6 places further on which have a strictly smaller maximum. Continuing in this way, we find that the three successive differences starting at most 6M places further on have maximum 1 or less. In other words every difference is 0 or 1 once one gets sufficiently far. Hence all terms an with n > 6M are 0, 1 or 2. Since N = 1414 is obviously much larger than M for the first few differences, we conclude that aN = 0, 1 or 2.
We now work mod 2. Note that a1, a2, a3 = 1, 0, 1 mod 2. We can now calculate successively:
n 1  2  3  4  5  6  7     8  9 10
an 1  0  1  0  0  1  1     1  0  1
Since a8, a9, a10 are the same as a1,, a2, a3 the sequence must be periodic with period 7. But N = 1414 = 0 mod 7, so aN = a7 = 1 mod 2. Hence aN = 1.

Problem N4
Let p > 3 be a prime. Show that there is a positive integer n < p-1 such that np-1 - 1 and (n+1)p-1 - 1 are not divisible by p2.
 
Solution
By the binomial theorem we have (p - a)p-1 - ap-1 = (p-1) p ap-1 mod p2. For 0 < a < p, (p-1) p a = p mod p2, so (p - a)p-1 and ap-1 are unequal mod p2. Hence at most one of them is 0 mod p2. In particular, since 12 = 1 mod p2, we must have (p-1)p-1 ≠ 1 mod p2. Also, at least (p-3)/2 of 2, 3, ... , p-2 must have p-1th powers ≠ 1 mod p2. So either one of the pairs (2, 3), (4, 5), ... , (p-3, p-2) has both members with p-1th powers ≠ 1 mod p2, in which case we are done, or exactly one member of each pair has p-1th power ≠ 1 mod p2.
Now if p-2 has p-1th power ≠ 1 mod p2 then we are done (because of p-1). If not, then p-3 has p-1th power ≠ 1 mod p2. Now suppose (p-4)p-1 = 1 mod p2. Then (expanding) 1 = 4p-1 - (p-1)p 4p-2 = 4p-1 + p 4p-2 mod p2 (*). But (p-2)p-1 = 1 mod p2, so (expanding) 1 = 2p-1 + p 2p-2 mod p2. Squaring, 1 = 4p-1 + p 4p-1 mod p2. Subtracting from (*), 0 = p(4p-1 - 4p-2) = 3 p 4p-2 mod p2, but that is clearly false. So we must have (p-4)p-1 ≠ 1 mod p2 and we are home (with p-4 and p-3). 

Comment. This is a weak result. There are relatively few a for which ap-1 = 1 mod p2.

Problem N6
Do there exist 100 positive integers not exceeding 25000 such that the sum of every pair is distinct?
 
Solution
Answer: yes.
We show how to find p integers <= 2p2 for any prime p. Since 2.1012 < 25000, that is sufficient.
Let n denote the residue of n2 mod p. We claim that the integers 2p + 1, 4p + 2, 6p + 3, ... , 2p p + p work. If (2ap + a) + (2bp + b) = (2cp + c) + (2dp + d), then 2(a + b)p + (a + b) = 2(c + d)p + (c + d) and hence a + b = c + d and a + b = c + d (since n < p). Hence a + b = c + d mod p and a2 + b2 = c2 + d2 mod p. So a = c + d - b mod p. Substituting in the next equation, (c + d - b)2 + b2 = c2 + d2 mod p, so 2(b2 - bc - bd - cd) = 0 mod p, or 2(b - c)(b - d) = 0 mod p. Hence b = c or d. So the sum of every pair is distinct.

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