Virtual Math Tutor

Math problem #281

2014-10-08T12:55:00.000-03:00

Math problem: Solve the following irrational equation:

$$\sqrt{14-x}-\sqrt{x-4}=\sqrt{x-1}.$$

Discussions and hints: Consider the following two functions $$f(x) = \sqrt{14-x}$$ and $$g(x) = \sqrt{x-4}+\sqrt{x-1}.$$

Solution

Soap Bubbles. The Beauty of Mathematics

2014-05-31T13:20:00.000-03:00

Yet another proof of the Pythagorean theorem

2014-05-09T07:43:00.001-03:00

Doing math before Powerpoint

2014-02-10T13:46:00.004-04:00

Blackboard mathematics is being done at NASA (1960s). Photo by J.R. Eyerman.

Math problem #280 (Solution)

2014-02-10T13:34:00.002-04:00

Solution to Math Problem #280:

Any odd number $k = 2n+ 1 >3$ can be presented in the following form $k = n + (n+1)$. Any even number which is a multiple of 4 ($k = 4n$) can be presented as the sum $(2n+1) + (2n-1)$. The last number is greater than 1 for $k \ge 8$. Finally, an even number $k = 4n+2$ can be presented as the sum $(2n+3) + (2n-1)$. These numbers are mutually prime, since their difference is equal to 4, while 4 is mutually prime with any odd number. The last number is greater than 1 when $k \ge 10$. Therefore we have presented in this form all natural numbers except 1, 2, 3, 4 and 6. It is easy to see that none of these numbers can be presented in the form specified in the statement of the problem.

Islam is mathematics and love in South Africa

2014-01-09T15:48:00.003-04:00

L-R: Zweli Miyataza, Principal Cass Sehloho, Kahlil Muhammad, and Sister Bashiira at Ingqayizivele School. Photos: Kahlil Muhammad

JOHANNESBURG, South Africa (FinalCall.com) - The Original Mathematics Initiative was conceived in October of 2011 and presented to the Department of Education in South Africa. At the time, the need to improve results in mathematics, science, and technology was being discussed in every corner of the country. Continue

Math problem #280

2014-01-09T15:40:00.002-04:00

Math problem: Find all such natural numbers $k$ that can be presented as a sum of two mutually prime numbers different from 1.

Discussion and hints: Start with odd numbers, for example. Then note that any such number $k = 2n+1 > 3$ can be easily presented as $k = n + (n+1)$.

Solution

Math problem #279 (Solution)

2013-10-25T14:35:00.000-03:00

Solution to Math problem #279:

We note first that the side length of the largest square is equal to the sum of the side lengths of two other squares, namely - the next one going clockwise and the smallest one. Let $x$ be the side length of the largest square. Then the sides of the other squares are $x-1$, $x-2$, $x-3$ and $x-3$ (see the diagram below).

Now we note that the the length of the upper side of the rectangular is $x + (x-1)$, while the length of the lower one is $(x-2) + (x-3) + (x-3)$, but they must be equal. Therefore we arrive at the algebraic equation $$x + (x-1) = (x-2) + (x-3) + (x-3),$$ from which it follows that $2x - 1 = 3x - 8$, hence, $x= 7$. The problem now is completely solved.

Math integral joke

2013-10-23T15:51:00.002-03:00

Dear math enthusiasts: Is this joke really THAT funny?

Math problem #279

2013-10-23T15:38:00.000-03:00

Math problem: A rectangular consists of 6 squares (see below). Find the side length of the greatest square, if the side length the smallest one is equal to 1.

Discussion and hints: The side length of the greatest square is equal to the sum of side lengths of two other squares.

Solution

Martin Gardner -- The Best Friend Mathematics Ever Had

2013-10-23T15:20:00.004-03:00

What do L. Ron Hubbard, H.G. Wells, G.K. Chesterton, Lord Dunsany, Alice in Wonderland, M.C. Escher, John H. Conway, Roger Penrose and Oprah Winfrey have in common? The same thing as Isaac Asimov, Vladimir Nabokov, and Salvador Dali. And the vase illusion shown, which was designed by Atlanta-based magician and illusion artist Victoria Skye.
They're all connected in meaningful ways to Martin Gardner, the best friend mathematics ever had. His wonderful posthumous memoirs Undiluted Hocus-Pocus: The Autobiography of Martin Gardner (Princeton University Press) just appeared. Gardner had a long and productive publishing career, spanned 80 years (from 1930 to 2010) and spawning over 100 books. As Stanford statistician and magician Persi Diaconis says in the foreword to those memoirs, "Pick up anything he wrote. You'll smile and learn something." It's Gardner's profile that frames the vase image above. He continues to have a profile on Twitter today, as followers of @WWMGT (What Would Martin Gardner Tweet?) can attest ... (continue)

Math problem #278 (Solution)

2013-03-28T16:32:00.000-03:00

Solution to Math problem #278:

Let $x$ be the first term, then $18-x$ is the second term. Accordingly, we have to find the minimum value of the following quadratic polynomial: $$f(x) = x^2 + (18-x)^2 = 2x^2 - 36x + 18^2.$$

Taking the derivative and equating the result to zero, we get $$f'(x) = 4x - 36 = 0.$$ Therefore $x=9$ is a critical value. When $x<9$, $f'(x)<0$, while $x>9$, $f'(x)>0$. Therefore $x=9$ is the maximum value.

Mathematics in Forensic Science

2013-03-26T17:57:00.000-03:00

by Leila Schneps and Coralie Colmez

Mathematics is fast becoming one of the most important techniques in crime detection. Where once a Sherlock Holmes would have had to be content with a magnifying glass, or a jury with gut instinct and rational discussion, now a range of methods from probability and statistics are available to help. Today, mathematics lies behind expert conclusions on a hundred forensic matters from fingerprints to DNA. Statistics can be a precious tool when identifying the patterns behind confusing or misleading phenomena.

The University of California at Berkeley was sued for gender bias when it was observed that just 35 percent of female applicants to graduate school were being accepted, versus 44 percent of males. The investigators began by narrowing the problem down to six major departments for which, combined, the inequality shifted to an even more incriminating 46 percent of males versus just 30 percent of females. But then, a department-by-department analysis showed the exact contrary of a bias against women: in four of the six departments, they were actually accepted at a higher percentage rate than males, and in the other two, the male-female ratio was 37-34 percent and 28-24 percent, discrepancies too small to have caused the overall appearance of inequality. Continue

Math problem #278

2013-03-26T17:23:00.000-03:00

Math problem: Express the number 18 as a sum of two numbers $a$ and $b$: $18 = a+b$ so that the sum $a^2 + b^2$ is minimal.

Discussions and hints: Let $x = a$, then $b = 18- x$, which means we have to minimize the following function: $$f(x) = x^2 + (18-x)^2,$$
which is a quadratic polynomial.

Solution

Virtual Math Tutor: Top 10 posts of February 2013

2013-03-01T17:13:00.000-04:00

Below please find February's top ten most popular posts published on Virtual Math Tutor blog - as per total # of pageviews (in that order):

Math problem #277 (Solution)

2013-02-20T17:08:00.000-04:00

Solution to Math problem #277: According to the statement of the problem, we have $$a_1 + a_2 + \cdots + a_{10} > 0,$$ $$a_2 + a_3 + \cdots + a_{11} > 0,$$ $$a_3 + a_4 + \cdots + a_{12} > 0,$$ $$\cdots$$ $$a_{2013} + a_1 + \cdots + a_9 > 0.$$ Adding up these inequalities, we arrive at $$10(a_1 + a_2 + \cdots + a_{2013} > 0$$ and the result follows.

Virtual Math Tutor: Top 10 posts of January 2013

2013-02-01T16:58:00.000-04:00

Below please find January's top ten most popular posts published on Virtual Math Tutor blog - as per total # of pageviews (in that order):

Evelyn Lamb: Setting mathematics in verse

2013-01-16T15:24:00.000-04:00

I am home from the Joint Mathematics Meetings, and I’m still trying to process everything I learned and got fired up about there. Most of my time was spent in formal, serious lectures, many of them quite technical, but on Friday night, I went to something completely different: a mathematical poetry reading, facilitated by the Journal of Humanistic Mathematics. Some of the featured poets are mathematicians or math teachers, while some are full-time writers with an interest in math.

Mathematics is often portrayed as a sterile, robotic discipline, but continue reading

Math problem #277

2013-01-16T15:06:00.002-04:00

Math problem: Let there be given $2013$ numbers. It is known that the sum of any $10$ is them is positive. Show that the total sum is positive too.

Discussion and hints: For example, try to estimate first the number of negative numbers in the set.

Solution

Math problem #276 (Solution)

2013-01-15T15:31:00.000-04:00

Solution to Math problem #276:

Of course it is possible to solve this problem by introducing the substitution $3^k= t$ and then studying the resulting quadratic polynomial. However, it much easier to start with the function $f(k) = 4\cdot 3^{2k+1} + 3^k$, which is obviously a continuous, increasing function.

Next, we note that $f(0) > 1$, $f(-1) > 1$, but $f(-2) < 1$. Hence, the answer is $-2$.

Amazing math multiplication trick

2013-01-14T15:26:00.001-04:00

Math problem #276

2013-01-10T17:10:00.000-04:00

Math problem:

Find the greatest integer number $k$ such that $$4\cdot 3^{2k+1} + 3^k < 1.$$

Discussions and hints: Investigate the function $f(k) = 4\cdot 3^{k+1} + 3^k$ first.

Solution

Math problem #275 (Solution)

2013-01-05T18:46:00.000-04:00

Solution to Math problem #275:

Using the basic trigonometric formula TF1, rewrite the equation in question as follows: $$\sin^{2013}x + \cos^{2013} = \sin^2 x + \cos^2 x,$$ or, $$\sin^2 x(\sin^{2011}x - 1) = \cos^2 x (1 - \cos^{2011}x).$$ The left hand side of the last equation is less or equal to 1, while the right hand side is greater or equal to 0. Hence, we conclude that equality is only possible when both sides of the equation are equal to 0: $$\begin{cases}\sin^2 x (\sin^{2011} - 1) = 0, \\ \cos^2(1 - \cos^{2011}x) = 0.\end{cases}$$ If $\sin x = 0$, then $\cos x \not=0$, which means that $1- \cos^{2011} = 0$. If $\sin x = 1$, then $\cos x = 0$. Finally, we arrive at the following two cases: $$\begin{cases}\sin x = 0, \\ \cos x = 1, \end{cases}$$ or, $$\begin{cases}\sin x = 1, \\ \cos x = 0. \end{cases}$$ Therefore we arrive at the following two families of solutions: $\{2\pi k, \pi/2 + 2\pi k\}$, where $k \in \mathbb{Z}$.

Fast Math Tricks - How to multiply 2 digit numbers up to 100 - the fast way!

2013-01-05T18:18:00.001-04:00

by tecmath

Math problem #275

2013-01-04T17:45:00.002-04:00

Math problem: Solve the following trigonometric equation
$$\sin^{2013}x + \cos^{2013}x = 1.$$

Discussion and hints: Use the basic trigonometric identity TF1.

Solution