IB Maths and GCSE Maths Resources from British International School Phuket. Theory of Knowledge (ToK). Maths explorations and investigations.  Real life maths. Maths careers. Maths videos. Maths puzzles and Maths lesson resources.

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British International School Phuket

Welcome to the British International School Phuket’s maths website. My name is Andrew Chambers and I am currently working at BISP.  I am running my site as the school’s maths resources website for both our students and students around the world.

We are a British international school located on the tropical island of Phuket in Southern Thailand. We offer a number of scholarships each year, catering for a number of national and international standard sports stars as well as for academic excellence. You can find out more about our school here.

There are a huge amount of resources to explore – especially for students doing their IAs and for students looking for revision videos.  You may also like to try our school code breaking site – where you can compete with over 10,000 students from around the world who have made it onto our school leaderboard.


Circular Motion: Modelling a ferris wheel

This is a nice simple example of how the Tracker software can be used to demonstrate the circular motion of a Ferris wheel.  This is sometimes asked in IB maths exams – so it’s nice to get a visual representation of what is happening.

First I took a video from youtube of a Ferris wheel, loaded it into Tracker, and then used the program to track the position of a single carriage as it moved around the circle.  I then used Tracker’s graphing capabilities to plot the height of the carriage (y) against time (t).  This produces the following graph:

As we can see this is a pretty good fit for a sine curve. So let’s use the regression tool to find what curve fits this:

The pink curve with the equation:

y = -116.1sin(0.6718t+2.19)

fits reasonably well.  If we had the original dimensions of the wheel we could scale this so the y scale represented the metres off the ground of the carriage.

There we go!  Short and simple, but a nice starting point for an investigation on circular motion.

The Folium of Descartes

The folium of Descartes is a famous curve named after the French philosopher and mathematician Rene Descartes (pictured top right).  As well as significant contributions to philosophy (“I think therefore I am”) he was also the father of modern geometry through the development of the x,y coordinate system of plotting algebraic curves.  As such the Cartesian plane (as we call the x,y coordinate system) is named after him.

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Pascal and Descartes

Descartes was studying what is now known as the folium of Descartes (folium coming from the Latin for leaf) in the first half of the 1600s.  Prior to the invention of calculus, the ability to calculate the gradient at a given point was a real challenge.  He placed a wager with Pierre de Fermat, a contemporary French mathematician (of Fermat’s Last Theorem fame) that Fermat would be unable to find the gradient of the curve – a challenge that Fermat took up and succeeded with.

Calculus – implicit differentiation:

Today, armed with calculus and the method of implicit differentiation, finding the gradient at a point for the folium of Descartes is more straightforward.  The original Cartesian equation is:

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which can be differentiated implicitly to give:

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Therefore if we take (say) a =1 and the coordinate (1.5, 1.5) then we will have a gradient of -1.

Parametric equations

It’s sometimes easier to express a curve in a different way to the usual Cartesian equation.  Two alternatives are polar coordinates and parametric coordinates.  The parametric equations for the folium are given by:

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In order to use parametric equations we simply choose a value of t (say t =1) and put this into both equations in order to arrive at a coordinate pair in the x,y plane.  If we choose t = 1 and have set a = 1 as well then this gives:

x(1) = 3/2

y(1) = 3/2

therefore the point (1.5, 1.5) is on the curve.

You can read a lot more about famous curves and explore the maths behind them with the excellent “50 famous curves” from Bloomsburg University.

Project Euler: Coding to Solve Maths Problems

Project Euler, named after one of the greatest mathematicians of all time, has been designed to bring together the twin disciplines of mathematics and coding.  Computers are now become ever more integral in the field of mathematics – and now creative coding can be a method of solving mathematics problems just as much as creative mathematics has always been.

The first problem on the Project Euler Page is as follows:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

This is a reasonably straight forward maths problem which we can solve using the summation of arithmetic sequences (I’ll solve it below!) but more interestingly is how a computer code can be written to solve this same problem.  Given that I am something of a coding novice, I went to the Project Nayuki website which has an archive of solutions.  Here is a slightly modified version of the solution given on Project Nayki, designed to run in JAVA:


The original file can be copied from here, I then pasted this into an online JAVA site jdoodle. The only modification necessary was to replace:

public final class p001 implements EulerSolution with public class p001

Then after hitting execute you get the following result:

i.e the solution is returned as 233,168. Amazing!

But before we get carried away, let’s check the answer using some more old fashioned maths. We can break down the question into simply trying to find the sum of multiples of 3 under 1000, the sum of the multiples of 5 under 1000 and then subtracting the multiples of 15 under 1000 (as these will have been double counted). i.e:

(3 + 6 + 9 + … 999)  +  (5 + 10 + 15 + … 995)  – (15 + 30 + 45 + …990)

This gives:

S_333 = 333/2 (2(3)+ 332(3)) = 166,833
S_199 = 199/2 (2(5) + 198(5)) = 99, 500

S_66 = 66/2 (2(15) +65(15) = 33, 165.

166,833 + 99, 500 – 33, 165 = 233, 168 as required.

Now that we have seen that this works we can modify the original code.  For example if we replace:

if (i % 3 == 0 || i % 3 == 0)
if (i % 5 == 0 || i % 7 == 0)

This will find the sum of all the multiples of 5 or 7 below 1000.  Which returns the answer 156,361.

Replacing the same line with:

if (i % 5 == 0 || i % 7 == 0 || i % 3 == 0)

will find the sum of all the multiples of 3 or 5 or 7 below 1000, which returns the answer 271,066.  To find this using the previous method we would have to do:

Sum of 3s + Sum of 5s – Sum of 15s + Sum of 7s – Sum of 21s – Sum 35s – Sum of 105s. Which starts to show why using a computer makes life easier.

This would be a nice addition to any investigation on Number Theory – or indeed a good project for anyone interested in Computer Science as a possible future career.

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Spotting Asset Bubbles

Asset bubbles are formed when a service, product or company becomes massively over-valued only to crash, taking with it most of its investors’ money.  There are many examples of asset bubbles in history – the Dutch tulip bulb mania and the South Sea bubble are two of the most famous historical examples.  In the tulip mania bubble of 1636-37, the price of tulip bulbs became astronomically high – as people speculated that the rising prices would keep rising yet further.  At its peak a single tulip bulb was changing hands for around 10 times the annual wage of a skilled artisan, before crashing to become virtually worthless.

More recent bubble include the Dotcom crash of the early 2000s – where investors piled in trying to spot in what ways the internet would revolutionise businesses.  Huge numbers of internet companies tried to ride this wave by going public with share offerings.  This led to massive overvaluation and a crash when investors realised that many of these companies were worthless.  Pets.com is often given as an example of this exuberance – its stock collapsed from $11 to $0.19 in just 6 months, taking with it $300 million of venture capital.

Therefore spotting the next bubble is something which economists take very seriously.  You want to spot the next bubble, but equally not to miss out on the next big thing – a difficult balancing act!  The graph at the top of the page is given as a classic bubble.  It contains all the key phases – an initial slow take-off, a steady increase as institutional investors like banks and hedge funds get involved, an exponential growth phase as the public get involved, followed by a crash and a return to its long term mean value.

Comparing the Bitcoin graph to an asset bubble

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The above graph is charting the last year of Bitcoin growth.  We can see several similarities – so let’s try and plot this on the same axis as the model.  The orange dots represent data points for the initial model – and then I’ve fitted the Bitcoin graph over the top:

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It’s not a bad fit – if this was going to follow the asset bubble model then it would be about to crash rapidly before returning to the long term mean of around $4000.  Whether that happens or it continues to rise, you can guarantee that there will be thousands of economists and stock market analysts around the world doing this sort of analysis (albeit somewhat more sophisticated!) to decide whether Bitcoin really will become the future of money – or yet another example of an asset bubble to be studied in economics textbooks of the future.


Measuring the Distance to the Stars

This is a very nice example of some very simple mathematics achieving something which  for centuries appeared impossible – measuring the distance to the stars.  Before we start we need a few definitions:

  • 1  Astronomical Unit (AU) is the average distance from the Sun to the Earth.  This is around 150,000,000km.
  • 1 Light Year is the distance that light travels in one year.  This is around 9,500,000,000,000km.  We have around 63000AU = 1 Light Year.
  • 1 arc second is measurement for very small angles and is 1/3600 of one degree.
  • Parallax is the angular difference in measurement when viewing an object from different locations.  In astronomy parallax is used to mean the half the angle formed when a star is viewed from opposite sides of the Earth’s solar orbit (marked on the diagram below).Screen Shot 2017-12-09 at 8.28.33 PM

With those definitions it is easy to then find the distance to stars.  The parallax method requires that you take a measurement of the angle to a given star, and then wait until 6 months later and take the same measurement.  The two angles will be slightly different – divide this difference by 2 and you have the parallax.

Let’s take 61 Cyngi – which Friedrick Bessel first used this method on in the early 1800s.  This has a parallax of 287/1000 arc seconds.  This is equivalent to 287/1000 x 1/3600 degree or approximately 0.000080 degrees.  So now we can simply use trigonometry – we have a right angled triangle with opposite side = 1 AU and angle = 0.0000080.  Therefore the distance is given by:

tanΦ = opp/adj

tan(0.000080) = 1/d

d = 1/tan(0.000080)

d = 720000 AU

which is approximately 720000/63000 = 11 light years away.

That’s pretty incredible!  Using this method and armed with nothing more than a telescope and knowledge of the Earth’s orbital diameter,  astronomers were able to judge the distance of stars in faraway parts of the universe – indeed they used this method to prove that other galaxies apart from our own also existed.

Orion’s Belt

The constellation of Orion is one of the most striking in the Northern Hemisphere.  It contains the “belt” of 3 stars in a line, along with the brightly shining Rigel and the red super giant Betelgeuse.  The following 2 graphics are taken from the great student resource from the Royal Observatory Greenwich:

The angles marked in the picture are in arc seconds – so to convert them into degrees we need to multiply by 1/3600.  For example, Betelgeuse the red giant has a parallax of 0.0051 x 1/3600 = 0.0000014 (2sf) degrees.  Therefore the distance to Betelgeuse is:

tanΦ = opp/adj

tan(0.0000014) = 1/d

d = 1/tan(0.0000014)

d = 41,000,000 AU

which is approximately 41,000,000/63000 = 651 light years away.  If we were more accurate with our rounding we would get 643 light years.  That means that when we look into the sky we are seeing Betelgeuse as it was 643 years ago.

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The Remarkable Dirac Delta Function

This is a brief introduction to the Dirac Delta function – named after the legendary Nobel prize winning physicist Paul Dirac. Dirac was one of the founding fathers of the mathematics of quantum mechanics, and is widely regarded as one of the most influential physicists of the 20th Century.  This topic is only recommended for students confident with the idea of limits and was inspired by a Quora post by Arsh Khan.

Dirac defined the delta function as having the following 2 properties:

Screen Shot 2017-12-02 at 9.22.27 PMThe first property as defined above is that the delta function is 0 for all values of t, except for t = 0, when it is infinite.

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The second property defined above is that the integral of the delta function – and the area of the graph between 2 points (either side of 0) is 1.    We can take the bottom integral where we integrate from negative to positive infinity as this will be more useful later.

The delta function (technically not a function in a normal sense!) can be represented as the following limit:

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Whilst this looks a little intimidating, it just means that we take the limit of the function as epsilon (ε) approaches 0.  Given this definition of the delta function we can check that the 2 properties outlined above hold.

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For the first limit above we set t not equal to 0.  Then, because it is a continuous function when t is not equal to 0, we can effectively replace epsilon with 0 in the first limit above to get a limit of 0.  In the second limit when t = 0 we get a limit of infinity.  Therefore the first property holds.

To show that the second property holds, we start with the following integral identity from HL Calculus:

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Hopefully this will look similar to the function we are interested in.  Let’s play a little fast and loose with the mathematics and ignore the limit of the function and just consider the following integral:

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Therefore (using the fact that the graph of arctanx has horizontal asymptotes at positive and negative pi/2 for the final part) :

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So we have shown above that the integral of every function of this form will have an integral of 1, regardless of the value of epsilon, thus satisfying our second property.

The use of the Dirac Function

So far so good.  But what is so remarkable about the Dirac function?  Well, it allows objects to be described in terms of a single zero width (and infinitely high) spike, but despite having zero width, this spike still has an area of 1.   This then allows the representation of elementary particles which have zero size but finite mass (and other finite properties such as charge) to be represented mathematically.  With the area under the curve = 1 it can also be thought of in terms of a probability density function – i.e representing the quantum world in terms of probability wave functions.

A graphical representation:

This is easier to understand graphically.  Say for example we choose a value epsilon (ε) and gradually make it smaller (i.e we find the limit as ε approaches 0).  When ε = 5 we have the following:

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When ε = 1 we have the following:

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When ε = 0.1 we have the following:

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When ε = 0.01 we have the following:

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You can see that as  ε approaches 0 we get a function which is close to 0 everywhere except for a spike at zero.  The total area under the function remains at 1 for all ε.

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Therefore we can represent the Dirac Delta function with the above graph.  In it we have a point with zero width but with infinite height – and still with an area under the curve of 1!



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The Rise of Bitcoin

Bitcoin is in the news again as it hits $10,000 a coin – the online crypto-currency has seen huge growth over the past 1 1/2 years, and there are now reports that hedge funds are now investing part of their portfolios in the currency.   So let’s have a look at some regression techniques to predict the future price of the currency.

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Here the graph has been inserted into Desmos and the scales aligned.  1 on the y axis corresponds to $1000 and 1 on the x axis corresponds to 6 months.  2013 is aligned with  (0,0).

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Next, I plot some points to fit the curve through.

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Next, we use Desmos’ regression for y = aebx+d. This gives the line above with equation:

y = 5.10 x 10-7 e1.67x + 0.432.

I included the vertical translation (d) because without it the graph didn’t fit the early data points well.

So, If I want to predict what the price will be in December 2019, I use x = 12

y = 5.10 x 10-7 e1.67(12) + 0.432 = 258

and as my scale has 1 unit on the y axis equal to $1000, this is equal to $258,000.

So what does this show?  Well it shows that Bitcoin is currently in a very steep exponential growth curve – which if sustained even over the next 12 months would result in astronomical returns.  However we also know that exponential growth models are very poor at predicting long term trends – as they become unfeasibly large very quickly.   The two most likely scenarios are:

  1. continued growth following a polynomial rather than exponential model
  2. a price crash

Predicting which of these 2 outcomes are most likely is probably best left to the experts!  If you do choose to buy bitcoins you should be prepared for significant price fluctuations – which could be down as well as up.  I’ll revisit this post in a few months and see what has happened.

If you are interested in some more of the maths behind Bitcoin, you can read about the method that is used to encrypt these currencies (a method called elliptical curve cryptography).


This post is inspired by the Quora thread on interesting functions to plot.

  1. The butterfly

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This is a slightly simpler version of the butterfly curve which is plotted using polar coordinates on Desmos as:

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Polar coordinates are an alternative way of plotting functions – and are explored a little in HL Maths when looking at complex numbers. The theta value specifies an angle of rotation measured anti-clockwise from the x axis, and the r value specifies the distance from the origin. So for example the polar coordinates (90 degrees, 1) would specify a point 90 degrees ant clockwise from the x axis and a distance 1 from the origin (i.e the point (0,1) in our usual Cartesian plane).

2. Fermat’s Spiral

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This is plotted by the polar equation:

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The next 3 were all created by my students.

3.  Chaotic spiral (by Laura Y9)

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I like how this graph grows ever more tangled as it coils in on itself.  This was created by the polar equation:

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4.  The flower (by Felix Y9)

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Some nice rotational symmetries on this one.  Plotted by:

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5. The heart (by Tiffany Y9)

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Simple but effective!  This was plotted using the usual x,y coordinates:

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You can also explore how to draw the Superman and Batman logos using Wolfram Alpha here.

A geometric proof for the Arithmetic and Geometric Mean

There is more than one way to define the mean of a number.  The arithmetic mean is the mean we learn at secondary school – for 2 numbers a and b it is:

(a + b) /2.

The geometric mean on the other hand is defined as:


So for example with the numbers 1,2,3 the geometric mean is (1 x 2 x 3)1/3.

With 2 numbers, a and b, the geometric mean is (ab)1/2.

We can then use the above diagram to prove that (a + b) /2 ≥ (ab)1/2 for all a and b. Indeed this inequality holds more generally and it can be proved that the Arithmetic mean ≥ Geometric mean.

Step (1) We draw a triangle as above, with the line MQ a diameter, and therefore angle MNQ a right angle (from the circle theorems).  Let MP be the length a, and let PQ be the length b.

Step (2) We can find the length of the green line OR, because this is the radius of the circle.  Given that the length a+b was the diameter, then (a+b) /2 is the radius.

Step (3) We then attempt to find an equation for the length of the purple line PN.

We find MN using Pythagoras:  (MN)2 = a2 +x2

We find NQ using Pythagoras:  (NQ)2 = b2 +x2

Therefore the length MQ can also be found by Pythagoras:

(MQ)2 = (MN) + (NQ)2

(MQ) = a2 +x2 + b2 +x2

But MQ = a + b.  Therefore:

(a + b) = a2 +x2 + b2 +x2

a2+ b2 + 2ab = a2 +x2 + b2 +x2

2ab = x2 +x2

ab = x2

x = (ab)1/2

Therefore our green line represents the arithmetic mean of 2 numbers (a+b) /2 and our purple line represents the geometric mean of 2 numbers (ab)1/2. The green line will always be greater than the purple line (except when a = b which gives equality) therefore we have a geometrical proof of our inequality.

There is a more rigorous proof of the general case using induction you may wish to explore as well.

Euler’s 9 Point Circle

This is a nice introduction to some of the beautiful constructions of geometry.  This branch of mathematics goes in and out of favour – back in the days of Euclid, constructions using lines and circles were a cornerstone of mathematical proof, interest was later revived in the 1800s through Poncelot’s projective geometry – later leading to the new field of non Euclidean geometry.  It’s once again somewhat out of fashion – but more accessible than ever due to programs like Geogebra (on which the below diagrams were plotted).  The 9 point circle (or at least the 6 point circle was discovered by the German Karl Wilhelm von Feuerbach in the 1820s.  Unfortunately for Feuerbach it’s often instead called the Euler Circle – after one of the greatest mathematicians of all time, Leonhard Euler.

So, how do you draw Euler’s 9 Point Circle?  It’s a bit involved, so don’t give up!

Step 1: Draw a triangle:

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Step 2: Draw the perpendicular bisectors of the 3 sides, and mark the point where they all intersect (D).

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Step 3: Draw the circle through the point D.

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Step 4: From each line of the triangle, draw the perpendicular line through its third angle.  For example, for the line AC, draw the perpendicular line that goes through both AC and angle B. (The altitudes of the triangle).  Join up the 3 altitudes which will meet at a point (E).

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Step 5:  Join up the mid points of each side of the triangle with the remaining angle.  For example, find the mid point of AC and join this point with angle B.  (The median lines of the triangle).  Label the point where the 3 lines meet as F.

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Step 6:  Remove all the construction lines.  You can now see we have 3 points in a line.  D is the centre of the circle through the points ABC, E is where the altitudes of the triangle meet (the orthoocentre of ABC) and F is where the median lines meet (the centroid of ABC).

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Step 7:  Join up the 3 points – they are collinear (on the same line).

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Step 8:  Enlarge the circle through points A B C by a scale factor of -1/2 centered on point F.

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Step 9: We now have the 9 point circle.  Look at the points where the inner circle intersects the triangle ABC.  You can see that the points M N O show the points where the feet of the altitudes (from step 4) meet the triangle.

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The points P Q R show the points where the perpendicular bisectors of the lines start (i.e the midpoints of the lines AB, AC, BC)

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We also have the points S T U on the circle which show the midpoints of the lines between E and the vertices A, B, C.

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Step 10:  We can drag the vertices of the triangle and the above relationships will still hold.

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In the second case we have both E and D outside the triangle.

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In the third case we have E and F at the same point.

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In the fourth case we have D and E on opposite sides of the triangle.

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So there we go – who says maths isn’t beautiful?

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