Fifthteen statement: balls in a box

Happy Christmas! Luckily you have all received math-related presents. Now image Santa Claus has given you a box with n balls, and each ball has a positive integer painted on it. Santa Claus tells you to take two balls out of the box without looking. Then you add up the two numbers painted on the balls that you took out. Let p be the probability that the sum is an even number and let q be the probability that the sum is an odd number.

The question is…

What are the values of n for which it is possible that p=q?

And happy 11111100010!

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Fourteenth statement: a resoluble equation?

Well, winter is coming and so are math problems. We are back with a problem about a (ir?)resoluble equation. Consider the following:

9m = 4n2 + 1

The question is: can you find any real values of m and n for which the equality above is true?

And manganese is not the aswer…

 

(Mn)

 

Extra: pigeons

You would have never guessed that pigeons are useful in mathematics, but they are. There is even a principle with their name: the pigeonhole principle! It states that if there are k boxes and k+1 pigeons there will be at least one box with two pigeons. This may seem obvious, but it can be very useful when it comes to proofs. Let’s see some examples:

  • Proof that if we take any 8 consecutive numbers, there will be 2 whose difference is divisible by 7.

When dividing by 7 there are 7 possible remainders: 0, 1, 2, 3, 4, 5, 6 and 7 (holes). But we have 8 numbers (pigeons), so there will be two numbers with the same remainder. We can write them as:

a = 7k + n and b = 7q + n

Then their difference is: (7k +m) – (7q + n) = 7k + n -7q – n = 7(k-q) which is clearly divisible by 7. We can generalize this problem and say that if we take n consecutive numbers there will be 2 whose difference is divisible by n-1.

  • If we have an equilateral triange with side 2 and we put 5 points inside it, there will be two points whose distance is less than 1.

We can divide this triangle into 4 small triangles of lenght 1 by uniting the midpoint of each side:

pigeon

Since we have 4 triangles (holes) and 5 points (pigeons), there will be at least two points inside the same triangle. And because its side measures 1, their distance will be less than 1.

  • We have written down the numbers from 1 to 100. Show that if we pick 51 there will be at least 2 numbers such that one divides the other.

Take all the odd numbers of the list: 1, 3, 5, … ,99. For each number we create a hole containing all the powers of 2 times the number. So for example, in the hole of number 1 we have:

1, 2, 4, 8, 16, 32…

In the hole of number 3 we have:

3, 6, 12, 24, 48, 96…

We know that the numbers from 1 to 100 are all contained inside these 50 holes (50 because there are 50 odd numbers between 1 and 100) because all the odd numbers are obviously included and the pair numbers you can keep dividing by 2 until you get an odd number. Then, we have 50 holes and 51 numbers (pigeons). Hence, we will pick 2 numbers of the same hole at least, and this means that one will divide the other (the first odd number is a divisor of all the numbers in the hole).

We can find many other problems similar to these ones, but you can observe the applications of the pigeonhole principle!

Thirteenth solution: Ceva’s theorem

Remember we had to prove that if three lines are congruent inside a triangle then the following relationship is true:

AF/FB · BD/DC · CE/EA = 1

We can write the LHS as:

AFC/BFC · BDA/DCA · CEB/EAB

because triangles AFC and BFC share the same height (from vertex C) and hence the proportion is mantained because we have multiplied by the height and then divided by 2. This operation applies to the other 2 fractions. We can do the same but with the triangles that are formed with point O, that is:

AFO/BFO · BDO/DCO · CEO/EAO

Their proportions are also manteined because triangles AFO and BFO have the same height (from vertex O). Then we can write AF/FB as

AF/FB = (AFC-AFO)/(BFC-BFO) = CAO/BOC

We can see on the drawing that if we substract the area of the AFO triangle from the area of the AFC triangle we get the triangle DCA, and the same process applies to the denominator. If we then apply it to the two other fractions we get:


BD/DC = (BDA-BDO)/(DCA-DCO) = BOA/CAO
CE/EA = (CEB-CEO)/(EAB-EAO) = BOC/BOA

So if we write the equality again:

CAO/BOC · BOA/CAO · BOC/BOA

we see that all terms cancel and the expression is equal to 1. This is what we wanted to prove.


triangle1

(Note: on the drawing the heights have been drawn, but this is just a particular case)

Thirteenth statement: Ceva’s theorem

Now we are back to geometry! There is one very important relationship in a triangle called Ceva’s Theorem. Imagine we have a triangle ABC and we draw the lines AO, BO and CO from the vertices to the opposed sides (D, E and F respectively), all of them going through the common point O. The following Picture describes this:

triangle1

Then if AD, BE and CF are concurrent (they all go through the point O), then the following relationship is present:

AF/FB * BD/DC * CE/EA = 1

Prove why it is true.

Twelfth solution: trigonometric functions (STOP reading if you want to solve the problem by yourself)

The first part of the problem consisted to proof the two basic trigonometric functions which are:

sin(x+y) = sinx*cosy + siny*cosx

cos(x+y) = cosx*cosy – sinx*siny

To do so we will use vector rotation. Imagine we have a vector v and we want to rotate it an alpha angle. How can we do that?

blogvector

Take into account the sum and the product of vectors. Let 3v be a vector and let Aalpha be the rotated vector (an angle alpha).

Aalpha(3v) = 3*Aalpha(v)

Likewise, if we add two vectors (AalphaV is the rotation of vetor v) using the parallelogram law we have:

u(v+Aalphav) = uv + uAalphav

blogvector2

Now let’s imagine the complex plane. The x-axis represents the real one, whereas the y-axis represents the imaginary one. For instance, the representation of 2+3i would be 2 units horizontally and 3 units vertically.

Recall we have unit vectors. Let’s call them l1 and l2.

l1 is placed on the x-axis (1,0) and l2 on the y-axis (0,1). These is called the canonic base. We can express all vectors as a combination of l1 and l2:

v = (v1,v2) = v1(1,0) + v2(0,1) = v1*l1 + v2*l2

We can represent l1 and l2 in the goniometric circumference. If l1=(1,0) we can express the rotated vector Aalphal1 as (cosα, sinα). Similarly, if l2 = (0,1) we can say that the rotated vector Aaplhal2 is (-sinα, cosα). Here you can see this representation:

blogvector3.png

l1

blogvector4.png

l2

Aαv = Aα(v1l1 + v2l2) = v1*Aα(l1) + v2*Aα(l2) = v1(cosα, sinα) + v2(-sinα, cosα) = (v1cosα – v2sinα, v1sinα + v2cosα)

This is the expression of the rotation of a vector an angle α.

Now note that the rotation of the angle α+β equals the rotation of first the angle α and then the angle β.

Aα+β = AβAαv

Let’s look at the RHS and let’s apply the last expression of the rotation vector.

AβAαv = ((v1cosα – v2sinα)cosβ – (v1sinα + v2cosα)sinβ, (v1cosα – vsinα)sinβ + (v1sinα + v2cosα)cosβ) = (v1cosαcosβ – v2sinαcosβ – v1sinαsinβ -v2cosαsinβ, v1cosαsinβ – vsinαsinβ + v1sinαcosβ + v2cosαcosβ)

Let’s look at the LHS:

Aα+β = (v1cos(α+β) – v2sin(α+β), v1sin(α+β) + v2cos(α+β))

As it has been said, the two expressions (RHS and LHS) must be equal. Let’s look closer at the first part of the vector. If we equal v1 from RHS with v1 from LHS and we equal v2 from RHS and v2 from LHS we get:

cos(α+β) = cosα*cosβ – sinα*sinβ

sin(α+β) = sinα*cosβ + cosα*sinβ

We can also proof these two expressions by using the complex plane.

Eleventh solution: number 12 (STOP reading if you want to solve the problem by yourself)

We first start by answering the hint questions:

a) Which is the probability of getting a 12 at the end?

In order to get number 12 we need the 6 operations to cancel between themselves. We only have 4 ways to to this:

(2*2*2)/(2/2/2)

(2*2*3)/(2*2*3)

(2*3*3)/(2*3*3)

(3*3*3)/(3*3*3)

Now we look at how can we obtain each of the fractions in different ways. To do so we image 6 boxes which represent the 6 operations. We take the first fraction, that is:

(2*2*2)/(2/2/2)

There are only two possible operations: x2 and /2. Imagining the 6 boxes, the x2 operation can be put in the first 6 boxes, the second x2 in 5 boxes and the third x2 in 4 boxes. That is 6*5*4 different ways of putting the three operations x2. But we have to divide this by the permutations: 3*2*1. Thus we have 20 different ways of obtaining the first fraction. The same happens with the fraction:

(3*3*3)/(3*3*3)

because they are symmetric. Now let’s calculate how many different ways do we have to obtain the second fraction, that is:

(2*2*3)/(2*2*3)

We apply the same method:

x2: 6*5 / 2*1

/2: 4*3 / 2*1

x3: *2

/3: *1

Therefore we have 180 different ways of obtaining it. The same occurs with the third fraction due to the fact that they are also symmetric. In total we have 20+20+180+180 = 400 different ways of obtaining number 12 in the blackboard. If we divide this by the number of total possible operations, which is 46, we can conclude that there is a 9,765% probability.

b) Which is the probability of getting a 18 at the end?

In order to obtain a 18 from 12 we have to multiply it by 3/2. This operation is compulsory, and the 4 operations left have to cancel. We can obtain it in three different ways:

(3*2*2)/(2*2*2)

(3*3*2)/((2*3*2)

(3*3*3)/(2*3*3)

We apply the aforementioned method in order to know in how many ways can we obtain the first fraction:

x2: 6*5 / 2*1

/2: 4*3*2 / 3*2*1

x3: *1

So 60 ways in total, which is the same for the last fraction (symmetry). We apply the same method for the second one:

x2: *6

/2: 5*4 / 2*1

x3: 3*2 / 2*1

/3: *1

Hence we get 180 different ways. In total we have 60+180+60 ways of obtaining 18 in the blackboard after the 6 operations. If we divide this by the number of total operations  (46), we can conclude that there is a 7,324% probability.

c) Which is the probabilty of getting a 36 at the end?

12 has to be multiplied by 3 in order to get 36. Thus we have 5 operations left which have to cancel each other. But that is impossible for that is an odd number of operations. Hence we conclude there is no way of getting number 36 at the end.

Now we get to the real problem.

d) Which is the probability that after these 6 operations there will be an integer number on the blackboard?

A number will be integer if and only if we do not divide it by a number which is not one of its divisors. As number 12 is decomposed as 2*2*3, we can divide it maximum by 2 twice and by 3 once. If we divide it more than this, it is crucial we multiply it by the same number, for if they did not cancel we would obtain a fractionary numer. Now we can write all possible integers: only multiplying, only dividing by 2, only dividing by 3, and dividing by 2 and 3.

If we operate the fractions we obtain all possible integers. Nevertheless there are some numbers which are repeated and hence are more likely to be on the blackboard. Applying the method in the previous questions we can find out in how many ways we can obtaing each integer:

192

2*2*2*2*2 / 2 = 6 ways

2*2*2*2*3 / 3 = 30 ways

972

3*3*3*3*3 / 3 = 6 ways

2*3*3*3*3 / 2 = 30 ways

108

3*3*3*3 / 3*3 = 15 ways

2*3*3*3 / 2*3 = 120 ways

2*2*3*3 / 2*2 = 90 ways

12

2*2*2 / 2*2*2 = 20 ways

3*3*3 / 3*3*3 = 20 ways

2*2*3 / 2*2*3 = 180 ways

2*3*3 / 2*3*3 = 180 ways

48

2*2*2*2 / 2*2 = 15 ways

3*3*2*2 / 3*3 = 90 ways

2*2*2*3 / 2*3 = 120 ways

288

2*2*2*3*3 / 3 = 60 ways

2*2*2*2*3 / 2 = 30 ways

432

2*2*3*3*3 / 3 = 60 ways

2*2*2*3*3 / 2 = 60 ways

648

2*3*3*3*3 / 3 = 30 ways

2*2*3*3*3 / 2 = 60 ways

1458

3*3*3*3*3 / 2 = 6 ways

72

3*3*3*2 / 3*3 = 60 ways

2*2*3*3 / 2*3 = 180 ways

2*2*2*3 / 2*2 = 60 ways

162

2*3*3*3 / 2*2 = 60 ways

3*3*3*3 / 2*3 = 15 ways

243

3*3*3*3 / 2*2 = 15 ways

18

3*2*2 / 2*2*2 = 60 ways

3*2*3 / 3*2*2 = 180 ways

3*3*3 / 2*3*3 = 60 ways

27

3*3*3 / 2*2*3 = 60 ways

2*3*3 / 2*2*2 = 60 ways

128

2*2*2*2*2 / 3 = 6 ways

32

3*2*2*2 / 3*3 = 60 ways

2*2*2*2 / 2*3 = 30 ways

8

3*3*2 / 3*3*3 = 60 ways

2*2*2 / 2*2*3 = 60 ways

3*2*2 / 2*3*3 = 180 ways

3

2*2 / 2*2*2*2 = 15 ways

2*3 / 2*2*2*3 = 120 ways

3*3 / 2*2*3*3 = 90 ways

2

3*2 / 2*2*3*3 = 180 ways

2*2 / 2*2*2*3 = 60 ways

3*3 / 3*3*3*2 = 60 ways

768

2*2*2*2*2*2 = 1 way

1152

2*2*2*2*2*3 = 6 ways

1728

2*2*2*2*3*3 = 15 ways

2592

2*2*2*3*3*3 = 20 ways

3888

2*2*3*3*3*3 = 15 ways

5832

2*3*3*3*3*3 = 6 ways

8748

3*3*3*3*3*3 = 1 way

In conclusion we can see that we can obtain up to 26 integer numbers. If we add all the different ways we see there are 3038. If we divide this by all the possible results (46), we get that there is a 73,803% probabilty of getting an integer number.

From this results we can draw the following graph:

blogcomb

The points represent all integers which can be left in the blackboard after the 6 operations. The number which is written is the total number of ways in which we can obtain that number. The y-axis represents 3 and the x-axis represents 2. Therefore, one unit to the right means «multiply by 2», one unit to the left means «divide by 2», one unit upwards means «multiply by 3» and on unit downwards means «divide by 3». The point in the middle represents number 12 (from 0,0: two units right and one unit upwards = 2*2*3 = 12). If we would like to count in how many ways we can obtaing number 18, the minimum movements are up and left. We can also do: right-left-right-left, up-down-up-down, up–down-right-left and all the possible ways of combining this, that is 300.

Twelfth statement: trigonometric functions

Now that we are back from holidays it’s time to re-think maths and learn how to demonstrate what we believe is unquestionable. In this problem we will talk about a very important topic in mathematics: trigonometry. We can define trigonometric functions which help us demonstrate identities and solve equations. Besides the Pythagorean identities, the most famous ones are:

sin(x+y) = sinx*cosy + siny*cosx

cos(x+y) = cosx*cosy – sinx*siny

Hence, proof these two formulas.

By the way, the solution of the eleventh problem will be ready soon, so think about number 12!

My father

Today is the birthday of the man who has taught me everything I know and who has inspired my passion for maths and science: my father. This blog was created in the first place to keep safely all the problems he explained to me.

He is a exeptional mathematician and, most importantly, a wonderful person. He has always taken exceptional care of me and has backed me up in all the problems I faced. I really look up to him, and I just want to thank him for being my father. I couldn’t be more proud of him.

I have kept this blog secret for a year and a half so that I could improve it, and today’s the day I will tell him its existence. I hope it represents everything we have discussed and thought together. I would have never been able to solve these problems without all the passion he put in explaining me the magic of maths since I was a child.

Simplement gràcies i felicitats. T’estimo!

 

 

 

 

Extra: journey to Mars

As aforementioned, this blog also aims to show the applications of maths in other disciplines. This time we will talk about space – in particular about Mars. We all know that the Red Planet is a present target for mankind. There will certainly be a trip to Mars before 2030. But how far is this planet from us? How long would it take? Which would be the medium velocity of the rocket? We can answer all these questions with simple calculus.

The main question is: what trajectory should the rocket follow? The easy answer is «a straight line». But this is not feasible. It would require an enormous amount of energy because the rocket would tend to follow an elipse due to Sun’s gravity. Therefore we have to take advantage of this gravity. This is what Hohmann’s trajectory does. It was designed by the engineer and mathematician Walter Hohmann in 1925. By applying it, the travel wouldn’t be the shortest one, but it would be the one using less energy. It would only use fuel in two occasions: when leaving Earth’s orbit and when entering Mars’ orbit. The rest of the time it would follow en elipse around the Sun.

To start our maths, we will consider that the orbits of Earth and Mars around the sun are circular, and that they have a constant velocity. But in fact this is not true, as stated by Kepler’s three laws of planetary motion:

  1. The orbit of a planet is an ellipse with the Sun at one of the two foci.
  2. A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
  3. The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

Another term we have to take into consideration is the conjunction. This means the Sun, Mars and Earth are aligned with the Sun inbetween. Our rocket needs this position to get to Mars:

conjunction

Therefore, where has to be Mars when the rocket is thrown? To investigate so, first we will calculate the angular velocity (ω) of each planet. It is calculated by dividing the angle (ϕ) by the time. We just need to know how long does it take for the Earth and for Mars to go all around the Sun (365 and 687 days respectively):

ω Earth = 360° / 365 days = 0,986°/day

ω Mars = 360° / 687 days = 0,524°/day

We can see that Mars’ angular velocity is higher. Now we will calculate the linear velocity (v) of both planets. It is done by dividing the space (x) by the time (t). We need to know the distance with each planet with the Sun (the radium) so that we can calulate the perimeter (x). The distance between the Earth and the Sun is 149.600.000 km and between Mars and the Sun it is 227.900.000 km. We multiply the diameter by π and hence:

v Tierra = (149.600.000 * 2 * π) / (365 * 24 * 3600) =  = 29,8 km/s

v Marte = (227.900.000 * 2 * π) /  = (365 * 24 * 3600) = 24,1 km/s

This is que trajectory our rocket must follow:

blogm1

The blue circle represents the orbit of Earth, the red one represents the orbit of Mars, the yellow point is the Sun, the blue one is the Earth and the red one is Mars. The green ellipse represents the trajectory of the rocket. Here we can see when does the rocket need an impulse (when changing the orbit) and hence the use of fuel:

blogm2

But we can no longer consider that the trajectory followed by the rocket is circular. Therefore, we will have to use Keplers’ laws, as it is an ellipse with the Sun as the focus. Remember that the third law stated that the square of the orbital period (T) of a planet is proportional to the cube of the semi-major axis of its orbit (a). We can write it by:

T2 / a3 = k

where k is a proportional constant. It is the same constant for all the planets that orbit around the same star. So we need to calculate the value of k in our Solar System, as it also includes our rocket. We will use the Earth and the units will be UA instead of kilometers (1 UA = distance Earth-Sun = 149.600.000 km) and years instead of days. Therefore:

12 / 13 = 1

and hence k = 1 in the Solar Sistem. For so, we can say that T2 = a3For our rocket, a equals the major axis of the ellipse. It is calculed by adding the distances Earth-Sun and Mars-Sun and dividing by 2. The distance between Mars and the Sun in UA is:

227.900.000 / 149.600.000 = 1,524 UA

Thus:

a = (1 + 1,524) / 2 = 1,262 UA

 T2 = a3

 T2 = 1,2523

T2 = 2,009916728

T = 1,4177153

but to get to Mars, the rocket only travels through half the ellipse:

1,4177153 : 2 = 0,70885 years

0,70885 years = 8,5 months = 258 days

Now we ask ourselves, where should Mars be when the rocket is launched so that after 258 days it will be precisely in front of (with the Sun inbetween) where the Earth was when the rocket was launched? It is completely necessary, as the Red Planet and Mars must coincide.

To do so we divide the number of days the rocket spends to get to Mars by how many days this planet needs to go all around the Sun, and we multiply it by 360°:

360 * (258 / 687) = 135,7°

So when the rocket is launched Mars has to be at 180 – 135,7 = 44,3° on the right of the Earth.

mart1

And where will the Earth be when the rocket arrives on Mars? We make the same opperation:

360 * (258 / 365) = 254,4°

Hence this will be its position:

mart2

Before we continue, we should review the parts of an ellipse:

  • Major axis: the longer radium (segment from the center)
  • Minor axis: the shorter radium (segment from the center)
  • Focis: two points in the major axis which are equidistant from the center point. The sum of the distances from any point P on the ellipse to those two foci is constant and equal to the major axis.

ellipse

Now we have to find out the medium velocity of our rocket when travelling around the Sun (without taking into account the impulse when leaving the orbit of the Earth). We will do so by dividing the distance by the time. But we just know the time (258 days), so we will calculate the distance with the perimeter of the ellipse. Actually with half of the perimeter.

P = 2pi * sqrt((r*r + s*s) / 2)

where r is the major axis and s is the minor axis. This is not an exact formula, because to calculate the exact value of the perimeter we would need an infinite number of calculus.

We know the value of the major axis (the average of 149.600.000 km y 227.900.000 km = 188.750.000 km), but not of the minor axis. However, we do know that the Sun is one of the focal points of the ellipse (as stated by Kepler’s laws) and that in any point of the ellipse the sum of the distances to the two foci is constant and equal to the major axis.

mart3

In the representation we can appreciate the two focal points (the Sun and a point which is 1 UA from Mars’ orbit), the center of the ellipse, the major axis (purple), which mesurates 1,262 UA (the distance between the two focus is 2,524 – 2 = 0,524; 0,524 : 2 = 0,262) and the minor axis (pink), whose value is unknown. But the sum of the segments Sun – A and Foci – A must be equal to the major axis (1UA + 1,524 UA = 2,524 UA).

 

 

mart4

The segment Sun – A must be equal to 1,262 UA (2,524 UA : 2). The distance Sun – center of the ellipse is:

(2,524 – 2*1) / 2 = (2,524 – 2) / 2 = 0,524 / 2 = 0,262 UA

We apply Pythagoras theorem (s is the semi-minor aixs):

1,2622 = s2 + 0,2622

1,5926 = s2 + 0,0686

s2 = 1,5926 – 0,0686

s2 = 1,524

s = 1,234 UA

s = 184.645.169,72 km

Now we apply the formula for the perimeter of the ellipse:

P = 2pi * sqrt(r*r + s*s)

P = 2pi * sqrt(1,262*1,262 + 1,234*1,234)

P = 2pi * sqrt(1,5926 + 1,5227)

P = 2pi * sqrt(3,11530 / 2)

P = 2pi * 1,55765

P = 2pi * 1,248

P = 7,8417 UA

But the rocket only travels through half the ellipse, which is 7,8417: 2 = 3,921 UA = 586.563.204,927 km. Therefore, it will have a medium velocity of:

v = x / t

v = 586.563.204 km / 258*24*3600 s = 26,31 km

Lastly, we will calculate how many days have to go by so that a conjuction takes places (this is the necessary condition for the lauch of the rocket). The difference of the angular velocities is:

ω Earth – ω Mars = 360° / 365 days – 360° / 687 days = 360° (1 / 365 – 2 / 687)

To find out the number of days between two conjuctions, we divide 360° by the difference of the angular velocities:

360° / 360° * (1/365 – 1/687) = (365*687) / (687-365) = 778,74 days

Once all the calculus are completes, we can conclude that:

  • The rocket must be launched when Mars is positioned 44,3° on the right of Earth.
  • It will take the rocket 258 to get to Mars and it will have a medium velocity of 26,31 km/s. It will travel through 586.563.204 km.
  • When the rocket gets to Mars, the Earth will have travelled 254,4°.
  • The necessary conditions for a Hohmanns’ transference take place every 778 days.