Monthly Archives: November 2014

Proof of SAS Congruency Test of two triangles

SAS Test math blogTheorem: If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are equal in all respects.

Construction: Let ABC, DEF be two triangles in which AB=DE and

AC=DF and the included angle \angle BAC is equal to included angle \angle EDF.

It is required to prove that the

\Delta ABC = \Delta DEF in all respects.

Proof: Apply the \Delta ABC to the \Delta DEF so that the point A falls on the point D; and the side AB along the side DE. Then, because AB=DE so the point B must coincide with the point E. And, because AB falls along DE, and the

\angle BAC=\angle EDF, so AC must fall along DF. And, because

AC=DF, the point C must coincide with the point F. Then, since B coincides with E, and C with F, hence, the side BC must coincide with the side EF. Hence, the \Delta ABC coincides with the \Delta DEF, and is therefore equal to it in all respects. QED.

More later…please post your questions, comments, and I will gladly answer them asap

Nalin Pithwa

Congruency Tests of two triangles — for PreRMO, RMO, IITJEE Foundation Math

Uptil now, we have discussed quite a bit of algebra, number theory and calculus for RMO, IITJEE Main and Advanced Mathematics. Let us switch back to plane geometry for a while.

In school, we all learn about tests of congruency of two triangles. But, have you wondered what is the exact meaning of “two plane figures are congruent”? Try to Google and find the answer for yourself. You can also post your answers as comments to this blog article.

The reasons I discuss these are two fold: (1) do you know the proofs of these congruency tests? In schools, students are just taught the meaning of these and how to apply them. Well, I will give you proofs of these in my blog at some later time. (2) There are some nuances to  these tests; when they cannot be applied, or rather, which *tests* fail to apply.

Here we go…

Two triangles are equal in all respects when the following three parts in each are severally equal:

1)Two sides, and the included angle. (the SAS Test for Congruency of 2 triangles).

2) The three sides. (the SSS Test for congruency of two triangles.

3) Two angles and one side, the side given in one triangle CORRESPONDING to that given in the other (ASA and AAS Test for Congruency of two triangles).

The two triangles are not, however, necessarily equal in all respects, when any three parts of one are equal to the corresponding parts of the other.

For example:

1) When the three angles of one are are equal to  the three angles of the  other, each to each, the fig 1 (attached, please download) shows  that the triangles need not be equal in all respects. (If you recall, such triangles are called similar, not  congruent because congruent means exactly same; two twins are similar, but not same).

2) When two sides and one angle in one are equal to two sides and one angle of the other, the given angles being opposite to equal sides, the  fig 2 (attached, please download) shows that the triangles need not be equal in all respects.

For, if $AB=DE$ and $AC=DF$ and the $\angle ABC = \angle DEF$, it will be seen that the shorter of the given sides in the triangle DEF may lie in either of  the positions DF or DF’.

Important Note: From these data, it may be shown that the angle opposite to the equal sides AB, DE are either equal (as for instance, the $\angle ACB$ and

$\angle DF’E$) or supplementary as (the $\angle ACB and \angle DFE$); and, that in the former case the triangles are equal in all respects. This is called the ambiguous case in the congruence of triangles. 

Note: ambiguous means that which has some uncertainty. In other words, there could be more than one possibilities.

If the given angles at B and E are right angles, the ambiguity disappears.

In  the next few blog articles, we will discuss the proofs of congruency tests of two triangles.

More later,

Nalin PIthwablogfigcongruencetests

 

Part I : Solutions to Pre-RMO (Regional Mathematical Olympiad) Oct 2014

Question Paper Set A:

Qs 1) A natural number is such that k^{2}<2014<(k+1)^{2}. What is the largest prime factor of k?

Solution 1:

HInt: Think of perfect square numbers near 2014.

Hence, 44^{2}<2014<(44+1)^{2}. And, the largest prime factor of 44 is 11. Hence, ans is 11.

Qs 2) The first term of a sequence is 2014. Each succeeding term is the sum of cubes of the digits of the previous term. What is the 2014th term of the sequence?

Solution 2:

Hint: Try the simplest solution…just keep calculating…:-)

a_{1}=2014

a_{2}=2^{3}+1^{3}+4^{3}=73

a_{3}=7^{3}+3^{3}=343+27=370

Clearly, a_{2014}=370, which is the answer.

Qs 3) Let ABCD be a convex quadrilateral with perpendicular diagonals. If AB=20, BC=70, and CD=90, then what is the value of DA?

Solution 3. let the diagonals meet at 0. Let OB=a,OC=b, OD=d and OA=c. Then, we can apply Pythagoras’s theorem to the 4 right angled triangles AOB, BOC, COD, DOA. We get

a^{2}+b^{2}=70^{2}

b^{2}+d^{2}=90^{2}

d^{2}+c^{2}=DA^{2}

a^{2}+c^{2}=20^{2}

Hence, we get, a^{2}+b^{2}+c^{2}+d^{2}=90^{2}+20^{2}. In the RHS, substitute for a^{2}+b^{2}=70^{2} and you will get

DA^{2}=c^{2}+d^{2}.

Question 4: In a triangle with integer side lengths, one side is three times as long as a second side and the length of the third side is 17. What is the greatest possible perimeter of the triangle?

Solution 4:AB=c;AC=b; BC=a; b=3a; c=17

Hint: use triangle inequalities:

a+b>c; b+c>a; c+a>b. Hence, 4a>17 and

a>17/4.. Also, a+17>3a and hence, a<17/2. So,

17/4<a<17/2. So, we get a=8, b=3a=24

Ans. 8+24+17=49.

Question 5. if real numbers a,b,c,d,e satisfy

a+1=b+2=c+3=d+4=c+5=a+b+c+d+e+3, what is the value of

a^{2}+b^{2}+c^{3}+d^{4}+e^{5} ?

Solution 5. b=a-1, c=b-1=a-2, d=c-1=a-3,

e=d-1=a-4 so we get

a+1=a+a-1+a-2+a-3+a-4+3. Hence, a=2.

Ans. 10.

Question 6: What is the smallest possible natural number n for which the equation x^{2}-ax+2014=0 has integer roots?

Solution 6:

Hint: Use the discriminant formula.

x=(n \pm \sqrt (n^{2}-8056))/2.

So, just think of plugging in some values 🙂 and note that n has to be smallest.

Also, n has to be even and also \sqrt (n^{2}-8056) has to be even and a perfect square root.

Ans. 91

Question 7: If x^{(x)^{4}}=4, what is the value of

x^{(x)^{2}}+x^{(x)^{8}} ?

Solution 7:

Warning: Do not use logarithms as there is no formula for log(M+N).

Hint: Try to input some numbers so that the original equation is satisfied.

x=\sqrt (2).Ans.

Question 8: Let S be set of real numbers with mean M. If the means of the sets S \cup {15} and S \cup {15,1} are M+2 and M+1 respectively, then how many elements does S have?

Solution: Use the definition of mean or average value followed by a little manipulation.

Ans. 4

Question 9: Natural numbers k,l,p,q are such that if a and b are roots of

x^{2}-kx+l=0, then a+(1/b) and b+(1/a) are the roots of x^{2}-px+q=0. What is the sum of all possible values of q?

Solution 9: Hint: Use the relationships between roots and coefficients:

a+b=k — equation I

ab=l — equation II

a+(1/b)+b+(1/a)=p — equation III

(a+1/b)(b+1/a)=q — equation IV

From equation IV, we get ab+1+1+(1/ab)=q, that is, l+2+(1/l)=q But q is a natural number. Hence, 1/l is also a natural number and the only way this could be possible is l=1.

Hence, ans q=4.

Question 10 In a triangle ABC, X and Y are points on the segments AB and AC respectively, such that AX:XB=1:2 and AY:YC=2:1. If the area of the triangle AXY is 10, then what is the area of the triangle ABC?

Solution 10: Let AX=k, BX=2k, AY=2m, CY=m where k and m are constants of proportionality.

Use: From trigonometry, (1/2)bc \sin A = area of triangle ABC .

Here, (1/2)k.2m.\sin A=10 and hence, mk \sin A=10.

But, area of triangle ABC is (1/2)(3k)(3m) \sin A=(9/2) \times 10=45. Ans.

Question 11: For natural numbers x and y, let (x,y) denote the greatest common divisor of x and y. How many pairs of natural numbers x and y with

x \leq y satisfy the equation xy=x+y+(x,y)?

To be discussed in the next blog.

Question 12:Let ABCD be a convex quadrilaterall with \angle DAB = \angle BDC = 90 \deg. Let the incircles of triangles ABD and BCD touch BD at P and Q, respectively, with P lying in between B and Q. If AD=999 and PQ=200, then what is the sum of the radii of the incircles of triangles ABD and BDC?

To be discussed in the next blog.

Question 13:For how many natural numbers n between 1 and 2014 (both inclusive) is 8n/(9999-n) an integer?

To be discussed in the next blog.

Question 14: One morning, each member of Manjul’s family drank an 8-ounce mixture of coffee and milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Manjul drank 1/7 th of the total amount  of milk and 2/17 th of  the total amount of coffee. How many people are there in Manjul’s family?

Solution 14: Key Idea: When two different fluids are mixed, they are in a certain proportion or ratio; this ratio of the two fluids remains the same when the mixture is poured out in different amounts.

Now, ratio of milk to coffee in Manjul’s mixture is 17/14.

Let x be the constant of proportionality. Total mixture is 8 ounce. Hence,

17x + 14x = 8 and therefore, x=8/31. But, total milk is

17x=(17/31) \times 8 and total coffee is 14x = (14/31) \times 8.

Hence, ans. 8.

Question 15: Let XOY be a triangle with \angle XOY=90 \deg. Let M and N be the midpoints of legs OX and OY, respectively. Suppose that XN=19 and YM=22. What is XY?$

Solution 15. Let XM=a=MO and ON=b=NY.

Apply Pythagoras’s theorem to right angled triangle MOY:

a^{2}+(2b)^{2}=22 \times 22.

Similarly, from right angled triangle XON, we get:

(2a)^{2}+ b^{2}=19 \times 19

Adding the above two equations, 5(a^{2}+b^{2})=845 and hence,

2(a^{2}+b^{2})=26, which is the desired answer.

Question 16: In a triangle ABC, let I denote the incenter. Let the lines AI, BI, and CI intersect the incircle at P, Q and R, respectively. If \angle BAC=40 \deg, what is the value of \angle QPR in degrees?

To be discussed in the next blog.

Question 17: For a natural number b, let N(b) denote the number of natural numbers a for which the equation x^{2}+ax+b=0 has integer roots. What is the smallest value of b for which N(b)=6?

Question 18: Let f be a one-one function from the set of natural numbers to itself such that f(mn)=f(m)f(n) for all natural numbers m and n. What is the least possible value of f(999)?

Question 19: Let x_{1}, x_{2}, x_{3}, \ldots x_{2014} be real numbers different from 1, such that x_{1}+x_{2}+x_{3}+ \ldots + x_{2014}=1 and

\frac {x_{1}}{1-x_{1}} + \frac {x_{2}}{1-x_{2}}+ \frac {x_{3}}{1-x_{3}}+ \ldots +\frac {x_{2014}}{1-x_{2014}}=1.

What is the value of \frac {x_{1}^{2}}{1-x_{1}}+\frac {x_{2}^{2}}{1-x_{2}}+\frac {x_{3}^{2}}{1-x_{3}}+\ldots + \frac {x_{2014}^{2}}{1-x_{2014}} ?

Solution. 

\frac {2x_{1}}{1-x_{1}}+\frac {2x_{2}}{1-x_{2}}+\frac {2x_{3}}{1-x_{3}}+\ldots+\frac {2x_{2014}}{1-x_{2014}}=2

Let S= \frac {x_{1}^{2}}{1-x_{1}}+\frac {x_{2}^{2}}{1-x_{2}}+\frac {x_{3}^{2}}{1-x_{3}}+\ldots+\frac {1-x_{2014}^{2}}{1-x_{2014}}

hence, S-2=\frac {x_{1}^{2}}{1-x_{1}}-\frac {2x_{1}}{1-x_{1}}+    \frac {x_{2}^{2}}{1-x_{2}}-\frac {2x_{2}}{1-x_{2}}+\ldots+\frac{x_{2014}^{2}}{1-x_{2014}}-\frac {2x_{2014}}{1-x_{2014}}

But, x_{1}+x_{2}+x_{3}+\ldots+x_{2014}=1 and also given that

\frac{x_{1}}{1-x_{1}}+\frac {x_{2}}{1-x{2}}+\frac {x_{3}}{1-x_{3}}+\ldots+\frac {x_{2014}}{1-x_{2014}}=1

(\frac {x_{1}}{1-x{1}}+1)+(\frac {x_{2}}{1-x{2}}+1)+(\frac {x_{3}}{1-x_{3}} +1)+\ldots +(\frac{x_{2014}}{1-x{2014}}+1)=2015.

Hence, S-2+2015=\frac {(1-x_{1})^{2}}{1-x_{1}} + \frac {(1-x_{2})^{2}}{1-x_{2}} +\ldots + \frac {(1-x_{2014})^{2}}{1-x_{2014}}

which  equals 1-x_{1}+1-x_{2}+\ldots +1-x_{2014}=2014 -1=2013.

Hence, S=0.

Question 20: What is the number of ordered pairs (A,B) where A and B are subsets of {1,2,\ldots 5} such that neither A \subseteq B nor

B \subseteq A?

 

 

 

Alexander Grothendieck math enigma dies at 86

Europe

The NY Times on the web.

By BRUCE WEBER and JULIE REHMEYER

NOV. 14, 2014.

Alexander Grothendieck, whose gift for deep abstraction excavated new ground in the field known as algebraic geometry and supplied a theoretical foundation for the solving of some of the most vexing conundrums of modern mathematics, died on Thursday in Ariège, in the French Pyrenees. He was 86.

A vexing character himself, Mr. Grothendieck (pronounced GROAT-en-deek) turned away from mathematics at the height of his powers in the early 1970s and had lived in seclusion since the early 1990s. His death was widely reported in France, where the newspaper Le Monde described him as “the greatest mathematician of the 20th century.” In a statement on Friday, President François Hollande praised him as “one of our greatest mathematicians” and “an out-of-the-ordinary personality in the philosophy of life.”

Algebraic geometry is a field of pure mathematics that studies the relationships between equations and geometric spaces. Mr. Grothendieck was able to answer concrete questions about these relationships by finding universal mathematical principles that could shed unexpected light on them. Applications of his work are evident in fields as diverse as genetics, cryptography and robotics.

“He had an extremely powerful, almost otherworldly ability of abstraction that allowed him to see problems in a highly general context, and he used this ability with exquisite precision,” Allyn Jackson wrote in a 2004 biographical essay about Mr. Grothendieck for Notices of the AMS, a journal of the American Mathematical Society. “Indeed, the trend toward increased generality and abstraction, which can be seen across the whole field since the middle of the 20th century, is due in no small part to Grothendieck’s influence.”

His background and early life were tangled and harrowing. His father, whose name is usually reported as Alexander Schapiro, was a Jewish anarchist who fought against the Russian czarist government. He was captured by the Bolsheviks during the Russian Revolution and eventually escaped to Western Europe. Along the way he lost an arm.

Schapiro made a living as a street photographer and met Johanna Grothendieck, known as Hanka, an aspiring writer who was married to a man named Alf Raddatz, in Berlin, in the mid-1920s. By then Schapiro had changed his name to Alexander Tanaroff, according to Mr. Grothendieck’s biographer, Winfried Scharlau. Introducing himself to Raddatz, Tanaroff said, “I will steal your wife,” and proceeded to do so.

Alexander Grothendieck, who for an unknown reason was named Raddatz at birth (not Schapiro, Tanaroff or Grothendieck), was born in Berlin on March 28, 1928.

Young Alexander’s parents left Germany as the Nazis took power — they participated in the Spanish Civil War — leaving him in the care of foster parents in Hamburg, where he first went to school. In 1939, he reunited with his mother and father in France, but his father was arrested, sent to an internment camp at Le Vernet and eventually moved to Auschwitz, where he died in 1942.

 

With his mother, Alexander lived in Le Chambon, where he finished primary school, and after the war attended college in Montpellier and later in Nancy, beginning his mathematical education in earnest. By the late 1940s he had entered the society of elite European mathematicians.

During the 1950s he taught in São Paulo, Brazil, and at the University of Kansas and lectured at Harvard. In 1958 he joined the faculty of the fledgling Institut des Hautes Études Scientifiques (IHES), which became a leading institute for the support of advanced research in mathematics and physics. (It is now in Bures-sur-Yvette, south of Paris.)

In 1966, Mr. Grothendieck was given the Fields Medal, an international award considered among the highest in mathematics.

He is widely credited for advances that made possible long-sought proofs for befuddling problems, including Fermat’s Last Theorem, the 17th-century conjecture by the French mathematician Pierre de Fermat that no three positive integers — a, b and c — exist that will satisfy the equation an + bn = cn for any integer value of n greater than 2. The first successful proof was published by an Englishman, Andrew Wiles, in the 1990s.

Mr. Grothendieck’s work was also a steppingstone to solutions of enigmas famous among mathematicians — the Poincaré conjecture, for instance — but far more arcane.

Working together with Pierre Deligne, Mr. Grothendieck was responsible for proving an especially thorny set of hypotheses known as the Weil conjectures. But characteristically he did not attack the problem directly. Instead, he built a superstructure of theory around the problem. The solution then emerged easily and naturally, in a way that made mathematicians see how the conjectures had to be true.

He avoided clever tricks that proved the theorem but did not develop insight. He likened his approach to softening a walnut in water so that, as he wrote, it can be peeled open “like a perfectly ripened avocado.”

“If there is one thing in mathematics that fascinates me more than anything else (and doubtless always has), it is neither ‘number’ nor ‘size,’ but always form,” he wrote in a long memoir in the 1980s, “Reapings and Sowings.” “And among the thousand-and-one faces whereby form chooses to reveal itself to us, the one that fascinates me more than any other and continues to fascinate me, is the structure hidden in mathematical things.”

Mr. Grothendieck had long held pacifist views, and by the late 1960s he had also become consumed by environmentalism. In 1966, he refused to travel to Moscow to receive the Fields Medal as a protest against the imprisonment of Soviet writers. He traveled to Hanoi at the height of the Vietnam War and lectured in Paris about the trip. He resigned from IHES, at least in part because some of its funding came from the French Defense Ministry, though he was also feuding with the institute’s founder. And he helped found an organization, Survivre, that promoted environmental activism and opposed the proliferation of nuclear weapons. He studied Buddhism and mysticism.

Over the next two decades, though he taught mathematics for a time at the University of Montpellier, he gradually withdrew from society and, according to his biographer, began devoting himself obsessively to writing what he called his “meditations.”

Mr. Grothendieck was married at least once, to Mireille Dufour. They had three children. He had two other sons with other women. Information about his survivors was not available.

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Hope  the applications of algebraic geometry to cryptography and coding theory etc. inspire you to take up pure mathematics or applied Mathematics…

More later…

Nalin Pithwa

 

 

A variant of mathematical induction — for IITJEE Math and RMO Math

As is generally known, the principal of mathematical induction is used in the following form: We are supposed to prove a proposition for natural numbers. We check whether it is true for the natural number 1, and then assume it to be true for another natural number m. Then, we got to  show  that the proposition also holds for the natural number m+1.

Here is a variation of the principle of mathematical induction: 

Assume the proposition to be true for a first natural number k (you have to  find the particular value of k under the conditions of the given problem). Then, assume it to be true for the natural number m. Finally, we gotta show that the proposition is true for the natural number m+1.

Example. Show that n! > 3^{n} if n is large enough.

Solution. How large is large? We have to find that “first” value for which the above proposition is true. We actually need to check whether the inequality holds for n=1, 2, 3, 4, 5, 6, 7, 8 \ldots

So, if we tabulate the values of LHS and RHS, we find that the first natural number for which this holds is n=7..

Hence, the proposition n! > 3^{n} is true for n=7.

Assume the proposition to be true for some natural number m where m > 7.

That is, m! > 3^{m} holds true. Multiplying both sides by m+1, we get

(m+1)m! > 3^{m} (m+1) that is,

(m+1)! > m \times 3^{m} + 3^{m}, but m>7. Hence, we get

(m+1)! > 7 \times 3^{m} + 3^{m} which equals 8 \times 3^{m}.

Hence, obviously, (m+1)! > 3^{m} \times 3 that is,

(m+1)! > 3^{m+1}. That is, the proposition holds for all natural numbers. Thus, the proof.

More later,

Nalin Pithwa

Divisibility and the GCD — Pre-RMO, RMO and IITJEE foundation Math

We switch back to basic number theory for the express purpose of Pre-RMO, RMO and IITJEE Foundation Mathematics.

As we have already seen in a previous blog on Pythagorean triples, the notions of divisibility and factorizations are important tools in number theory. In this blog, we will look at these ideas more closely.

Suppose that m and n are integers with m \neq 0. We say that m divides n, if n is a multiple of m, that is, if there is an integer k so that n=mk. If m divides n, we write m|n. Similarly, if m does not divide n, then we write

m \not | n. For example, 3|6 and 12|132 since 6=3.2 and 132=12.11

The divisors of 6 are 1,2,3 and 6. On the other hand, 5 \not | 7 since no integer multiple of 5 is equal to 7. A number that divides n is called a divisor of n.

If we are given two numbers, we can look for common divisors, that is, numbers that divide both of them. For example, 4 is a common divisor of 12 and 20, since 4|12 and 4|20. Notice that 4 is the largest common divisor of 12 and 20. Similarly, 3 is a common divisor of 18 and 30, but it is not the largest, since 6 is also a common divisor. The largest common divisor of two numbers is an extremely important quantity that will frequently appear during our number theoretic excursions.

The greatest common divisor of two numbers a and b (not both zero) is the largest number that divides both of them. It is denoted by gcd(a,b). If

gcd(a,b)=1, we say that a and b are relatively prime or co-prime.

Two examples that we mentioned above are

gcd(12,20)=4 and gcd(18,30)=6.

Another example is gcd(225,120)=15

We can check that this answer is correct by factoring 225=3.3.5.5 and

120=2.2.2.3.5 but, in general, factoring a and b is not an efficient way to compute their GCD or HCF.

The most efficient method known for finding the GCD of two numbers is called the Euclidean algorithm. It consists of doing a sequence of divisions with remainder until the remainder is zero. We will illustrate with an example before describing the general method.

As our example, we will compute the gcd(36,132) The  first step is to divide 132 by 36, which gives a quotient of 3 and a remainder of 24. We write this as

132=3 \times 36 +24

The next step is to 36 and divide it by the remainder 24 from the previous step. This gives

36=1 \times 24 +12

Next divide 24 by 12, and we find a remainder of 0,

24=2 \times 12 +0

The Euclidean algorithm says that when you get a remainder of 0 then the remainder from the previous step is the GCD of the original two numbers. So, in  this case we find that the gcd(132,36)=12.

Notice how at each step we replace our old A and B with the numbers B and R and continue the process until we get a remainder of 0. At that point, the remainder R from the previous step is the GCD of our original two numbers.

There is one more practical matter to be mentioned before we undertake a theoretical analysis of the Euclidean algorithm. If we are given A and B, how can we find a quotient Q and the remainder R? Of course, you can always use long division, but that can be time consuming and subject to arithmetic errors if A and B are large. A pleasant alternative is to find a calculator or computer program that will  automatically compute Q and R for you. However, even if you are only equipped with an inexpensive calculator, there is an easy three-step method to find Q and R:

Method to compute Q and R on a Calculator so that A=B \times Q +R

1) Use the calculator to divide A by B. You get a number with decimals.

2) Discard all the digits to the right of the decimal point. This gives Q.

3) To find R, use the formula R=A-B \times Q.

For example, suppose  that A=12345 and B=417. Then,

A/B=29.6043 \ldots so Q=29 and

R=12345 - 417 \times 29=252

We are now ready to analyze the Euclidean algorithm. The general method looks like:

a=q_{1} \times b + r_{1}

b=q_{2} \times r_{1} + r_{2}

r_{1}=q_{3} \times r_{2} + r_{3}

r_{2}=q_{4} \times r_{3} + r_{4}

\vdots

r_{n-3}=q_{n-1} \times r_{n-2} + r_{n-1}

r_{n-2}=q_{n} \times r_{n-1} + r_{n}

r_{n-1}=q_{n+1} \times r_{n} + 0

If we let r_{0}=b and r_{-1}=a, then every line looks like :

r_{t-1} = q_{r+1} \times r_{t} + r_{t+1}.

Why is the last non zero remainder r_{n} a common divisor of a and b?

We start from the bottom and work our way up. The last line $r_{n-1}=r_{n]$. Now, looking at the line above that, we already know that $r_{n}$ divides both $r_{n-1}$ and $r_{n-2}$ so we find that $r_{n}$ also divides $r_{n-3}$. Moving up line by line… finally, we move up to the top line and use fact that $r_{n}$ divides both $r_{1}$ and b to conclude that $r_{n}$ also divides a. This completes our verification that the last nonzero remainder $r_{n}$ is a common divisor of a and b.

But, why is r_{n} the greatest common divisor of a and b? 

Suppose that d is any common divisor of a and b. We will work our way back down the list of equations. So, from the first equation a=q_{1} \times b + r_{1} and the fact that d divides both a and b, we see that d also divides r_{1}. Then, the second equation b=q_{2}r_{1}+r_{2} shows us that d must divide r_{2}. Counting down line by line…therefore, r_{n} must be the greatest common divisor of a and b.

This completes our verification that the Euclidean algorithm actually computes the greatest common divisor, a fact of sufficient importance to be officially recorded.

Theorem (Euclidean Algorithm) To compute the greatest common divisor of two numbers a and b, let r_{-1}=a, let $;atex r_{0}=b$, and compute successive quotients and reminders r_{i-1}=q_{i+1} \times r_{i} + r_{i+1}

for i=0,1,2, \ldots and some remainder r_{n+1} is 0. The last nonzero remainder r_{n} is then the greatest common divisor of a and b.

Fun with computer programming: you can write a computer program to compute the gcd of large numbers ! Try its very easy and real fun 🙂

 

The man and his math by Sangeetha Devi Dundoo

The Hindu, 22 March 2011.

“[Srinivasa Ramanujan] was a genius far ahead of his times and math was his creative expression of numbers and not his means to make money. His was an extraordinary, inspiring and emotional journey,” says film director Dev Benegal. The filmmaker is interviewed about a film he plans to make on the life of mathematician Ramanujan, who grew up in poverty in Kumbakonam, India. Benegal, who has been researching Ramanujan’s life for four years, says “the film will explore his life at an emotional level–the struggle of his parents, particularly his mother; Ramanujan’s relationship with his wife, which is one of the greatest love stories of our times; the sacrifices that the wife had to make which are unknown and unheard of; and the bond that Ramanujan and G.H. Hardy shared.”

I like to inspire budding young minds towards Math…of course, you can say that I am biased, but …some people need inspiration, others have intrinsic motivation…I will share more such stories with you all later…

More later…

Nalin Pithwa

Some problems for Pre-RMO practice

Below are a few problems, which I think might boost a student’s confidence for Pre-RMO.

1) Factorize (show a proof also): a^{3}+b^{3}+c^{3}-3abc

2) Bad calibration.

My bathroom scale is set incorrectly but otherwise it works fine. It shows 10 kilograms when Dan stands on it, 14 kilograms when Sarah stands on it, but 22.5 kilograms Dan and Sarah are both on it. Is the scale set too high?

3) Getting inside a brick.

You  have 3 identical rectangular bricks and a ruler. Must you use a formula, such as the Pythagorean theorem, to find the length of the brick’s diagonal?

4) In the running.

In a cross-country run, Sven placed exactly in the middle among all participants. Dan placed lower, in tenth place, and Lars placed sixteenth. Is it possible to figure out how many runners took part in the race?

5) The sands of time.

Maia bought two unusual sandglasses. One measures a nine-minute interval, and the other measures a thirteen-minute interval. A certain love potion needs to boil for exactly thirty minutes. is it possible to measure such a time interval with these sandglasses under the additional stipulation that you turn over the glass(es) for the first time just as the potion starts to boil?

6) Anyone for tennis?

Each child in a school plays either soccer or tennis. One-seventh of the soccer players also play tennis, and one ninth of the tennis players play soccer. Do more than half the children play tennis?

7) Tiresome parrot.

We recently bought a parrot. The first day it said “O” and the second day “OK”. The third day the parrot said “OKKO” and the day after that “OKKOKOOK”. If this doubling pattern continues and the parrot squawks every second, will it get through squawking on the sixteenth day?

8) A wire cube.

A 12 inch long wire is to be divided into a number of parts and from these we want to construct the frame (that is, the edges) of a cube of 1 inch on a side. Can you do it with three pieces?

9) Paper folding.

You have a strip of paper that is two thirds of a meter long. However, you need a strip exactly half a meter long. Must you have a ruler to cut off such a length?

10) No one twice as rich.

One hundred children in a school are counting their money. Each child has between 1 and 100 cents, and no child has the same amount as any other child. Is it possible to divide the children into two groups so that no child in either group will have twice as much money as any other child in the same group?

Good luck 🙂 Happy problem solving…keep on trucking…

More later…

Nalin Pithwa

What is a catenary or a chain curve? A nice little application of basic calculus

Let us continue our exploration of basic calculus and its application. As you will discover, the invention of calculus is a triumph of the human intellect, it is a fountain head of many ideas in pure mathematics as well as mine of applications to sciences and engineering and even economics and humanities!

Hanging cables

Imagine a cable, like a telephone line or TV cable, strung from one support to another and hanging freely. The cable’s weight per unit length is w and horizontal tension at its lowest part is a vector of length H. If we choose a coordinate system for the plane of the cable in which the x-axis is horizontal, the force of gravity is straight down, the positive y-axis points straight up, and the lowest point of the cable lies at the point y=H/w on the y-axis. (Fig 1), then it can be shown that the cable lies along the graph of the hyperbolic cosine

y=(H/w)\cosh (wx/H).

Such a curve is sometimes called a chain curve or a catenary, the latter deriving from the Latin catena meaning ‘chain’.

a) Let P(x,y) denote an arbitrary point on the cable. Fig 2 displays the tension at P as a vector of  length (magnitude) T, as well as the tension at H at the lowest point A. Show that the cable’s slope at P is

\tan \phi =dy/dx=\sinh (wx/H)

b) Using the result from part (a) and the fact that the tension at P must equal H (the cable is not moving or swinging), show that T=wy. This means that the magnitude of the tension at P(x,y) is exactly equal to the weight of y units of cable.catenarycurveblog

2) (Continuation of above problem). The length of arc AP in Fig 2 is

s=(1/a)\sinh (ax) where a=w/H. Show that the coordinates of P may  be expressed in terms of s as

x=(1/a) (\sinh)^{-1} (as) and y=\sqrt (s^{2}+1/a^{2}).

3) The sag and horizontal tension in a cable. The ends of a cable 32 feet long and weighing 2 pounds per foot are fastened at the same level in posts 30 feet apart.

i) Model the cable with the equation

y=(1/a)\cosh (ax) for -15 \leq x \leq 15

Use information from problem 2  above to show that a satisfies the equation

16a=\sinh 15a

ii) Estimate the horizontal tension in the cable at the cable’s lowest point.

I hope you enjoy a lot 🙂

More later…

Nalin Pithwa