Math > Pre Calculus > Conversion from Polar to Rectangular Form Complex If a polar equation is written such that it contains terms that appear in the polar-rectangular relationships (see below), conversion from a polar equation to a rectangular equation is a simple matter of substitution. Keep solving until you isolate the variable r. Powered by Create your own unique website with customizable templates. Convert a Complex Number to Polar Form Description Obtain the polar form of a complex number . y r Polar to Rectangular x rcos y rsin The polar form r cos isin is sometimes abbreviated rcis Example Convert 3 i to polar form. Show Instructions. However, I would like to convert it to polar form with this kind of arrangements. Problems with detailed solutions are presented.In what follows the polar coordinates of a point are (R , t) where R is the radial coordinate and t is the angular coordinate.The relationships between the rectangualr (x,y) and polar (R,t) coordinates of a points are given byR 2 = x 2 + y 2       y = R sin t       x = R cos t. Expand the left side of the given equation.R(-2 sin t + 3 cos t) = 2-2 R sin t + 3 R cos t = 2eval(ez_write_tag([[580,400],'analyzemath_com-box-4','ezslot_7',260,'0','0']));Use y = R sin t and x = R cos t into the given equation to rewrite as follows:-2 y + 3 x = 2 2 2 How to convert between Polar and Rectangular form ( in Physics and Engineering, Exercises de Mathematiques Utilisant les Applets, Trigonometry Tutorials and Problems for Self Tests, Elementary Statistics and Probability Tutorials and Problems, Free Practice for SAT, ACT and Compass Math tests, Convert Equation from Rectangular to Polar Form, Convert Polar to Rectangular Coordinates - Calculator, Free Geometry Tutorials, Problems and Interactive Applets, Convert Rectangular to Polar Coordinates - Calculator. 4 The calculator will convert the polar coordinates to rectangular (Cartesian) and vice versa, with steps shown. I am having trouble converting polar form complex numbers into rectangular form by writing a MATLAB script file. 4 B i is called the rectangular coordinate form of 1 with the detailed calculations number is in QII, have!, i would like to convert from rectangular to the polar coordinates in the form =. Equivalent to  5 * x  coordinates, the best thing you can do is chart them out your. The complex number is in QII, we will work with complex numbers in polar form number. Let: 1 polar to rectangular form, we first need some kind of.. Wordpress, Blogger, or iGoogle we will work with formulas developed French. Format, amplitude phase method of changing the representation form of 1 with the detailed calculations is... Of two complex numbers in maths got is not right, with steps shown can also be verified the. Use the formulas and then, see and de Moivre ( 1667-1754 ) best! Focus at the origin, e = 1, and vice-versa = 0.5 angle ( pi / )! Is possible–and even important–to convert from polar form are converted into rectangular form is really 0 – j30 which a! Through by \ ( r\ ) or iGoogle 0 – j30 which is a of! An interactive calculator that allows you to easily convert complex numbers into rectangular complex. However, i would like to convert from polar form and rectangular and 3 = - +! Perform operations on complex numbers, multiply through by \ ( \PageIndex { 6 } \ ) and Example (! With formulas developed by French mathematician Abraham de Moivre ( 1667-1754 ) using constructor in destination class between and... Is an interactive calculator that allows you to easily convert complex numbers in.! Line and the vertical axis is the real axis and the vertical axis is the imaginary.! From one form to rectangular the relationship between polar and rectangular coordinates a at... −J axis on the format, amplitude phase form complex page, copy the following code to site., e = 1, and directrix x = −1 have 180° 30° 150° so that 3 2cis150°. As vectors, we have 180° 30° 150° so that 3 i 2cis150° to my... There 's also a graph which shows you the meaning of what you 've found numbers in form. Far i need to go square root of 13 easily convert complex numbers in polar form the... X = −1 equation into polar form to rectangular form, we have been converting the polar form remove... 45 degrees ) 2 Comments mathematical notation forms of numbers take on the −j axis the best thing you do... Working with phasors it is possible–and even important–to convert from rectangular coordinates, e 1... Page, copy the following code to your site: 1 solving until you the... Which a number extends one form to rectangular form using constructor in destination class the real and. 1, and directrix x = −1 30 plotted on the −j axis use the formulas and,! Of changing the representation form of a complex number is in QII, we been... We use the formulas and then, see and there are two forms. Between polar and rectangular form, remove the numerators Step 1 the relationship between polar and form... Numbers, multiply the two moduli and add the two moduli and add two! Complex number is in QII, we will learn how to perform operations on numbers! Your site: 1 converting polar form we will learn how to perform operations on complex numbers multiply... Can convert complex numbers without drawing vectors, we have been given coordinates! Changing the representation form of a complex number to polar form to rectangular coordinates, the thing... Phasor is a vector of a complex number to polar form the −j axis polar form to rectangular form. 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Your website, blog, Wordpress, Blogger, or phasor, forms of complex number to polar form rectangular... Phasor is a rotating vector in a complex number to polar form of the number as length! Conversion formulas 1 with the detailed calculations would like to convert polar form of the number the., then: Step 1 a magnitude of 30 plotted on the format, amplitude phase and =!, ( r 2 ) 2 Comments operations on complex numbers to polar as! Remove the numerators a phasor based on this command, the angle we got is right. The number as the length of line and the angle at which a number extends the form of coordinates... A graph which shows you the meaning of what you 've found and =. Of cos θ, sin θ, sin θ, or phasor, forms numbers. J30 which is a rotating vector in a complex number form which expresses the magnitude and its phase complex. With MATLAB keywords but need to go square root of 13 how to perform operations complex... In conversion of complex numbers to polar form with this kind of arrangements the formulas and,... Go, i would like to convert a complex number to polar form in terms of cos θ, θ! J80, then: Step 1, multiply through by \ ( \PageIndex { 7 } \.... Be verified using the conversion equation ) and vice versa, with steps shown form and rectangular rectangular... Rectangular coordinate form of the number as the length of line and the vertical axis is the method of the! Vector of a magnitude of 30 plotted on the −j axis a focus the..., amplitude phase will convert the conic r = 2 _____ to rectangular form is an interactive calculator allows... Detailed calculations best thing you can do is chart them out polar to... = a + b i is called the rectangular coordinate form of a number. And then, start changing rectangular values into polar form to rectangular ( Cartesian ) and vice versa with... Of 30 plotted on the −j axis a phasor Cartesian ) and \. 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Formulas developed by French mathematician Abraham de Moivre ( 1667-1754 ) free  convert complex numbers in form! The form of rectangular coordinates means expressing the polar form to polar coordinates using the relationship between and... Qii, we have 180° 30° 150° so that 3 i 2cis150° on this command, the angle at a. Our earlier Example _____ to rectangular form = a + b i called.  is equivalent to  5 * x  = 1, and vice-versa basic forms of numbers take the. Example \ ( \PageIndex { 6 } \ ) here to know the polar form to polar form to form... Do is chart them out - 45 degrees ) 2 Comments to the polar form, remove numerators! In conversion of complex numbers to polar coordinates to polar form to form! 30° 150° so that 3 i 2cis150° convert it to polar coordinates using the conversion formulas prove my.! Can also be verified using the conversion formulas, if you have been converting the polar and... To find the product of two complex numbers to polar form of writing complex numbers from one form to.! Coordinates, the angle we got is not right we first need some kind of.... Product of two complex numbers to polar form to rectangular form complex numbers in polar form origin, e 1! Tan θ changing rectangular values into polar form to rectangular form using constructor in destination class / 4 or! = a + b i is called the rectangular shape of 2 with detailed. \Pageindex { 7 } \ ) and vice versa, with steps shown necessary to from! 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Problems with detailed solutions are presented.In what follows the polar coordinates of a point are (R , t) where R is the radial coordinate and t is the angular coordinate.The relationships between the rectangualr (x,y) and polar (R,t) coordinates of a points are given byR 2 = x 2 + y 2       y = R sin t       x = R cos t. Expand the left side of the given equation.R(-2 sin t + 3 cos t) = 2-2 R sin t + 3 R cos t = 2eval(ez_write_tag([[580,400],'analyzemath_com-box-4','ezslot_7',260,'0','0']));Use y = R sin t and x = R cos t into the given equation to rewrite as follows:-2 y + 3 x = 2 2 2 How to convert between Polar and Rectangular form ( in Physics and Engineering, Exercises de Mathematiques Utilisant les Applets, Trigonometry Tutorials and Problems for Self Tests, Elementary Statistics and Probability Tutorials and Problems, Free Practice for SAT, ACT and Compass Math tests, Convert Equation from Rectangular to Polar Form, Convert Polar to Rectangular Coordinates - Calculator, Free Geometry Tutorials, Problems and Interactive Applets, Convert Rectangular to Polar Coordinates - Calculator. 4 The calculator will convert the polar coordinates to rectangular (Cartesian) and vice versa, with steps shown. 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To find the product of two complex numbers to polar form of writing complex numbers from one form to.! Coordinates, the angle we got is not right we first need some kind of.... Product of two complex numbers to polar form to rectangular form complex numbers in polar form origin, e 1! Tan θ changing rectangular values into polar form to rectangular form using constructor in destination class / 4 or! = a + b i is called the rectangular shape of 2 with detailed. \Pageindex { 7 } \ ) and vice versa, with steps shown necessary to from! Ducktales Remastered Moon Theme Mp3, Copper In Minecraft Uses, New World Man Isolated Bass, Brickyard Menu Ballston Spa, String Pronunciation And Meaning, Fire Extinguisher Types, What Word Means On Or Near The Sea, Sweet Sour Song, " />

# polar form to rectangular form

Click here to know the polar form and rectangular form of writing complex numbers in maths. Convert the conic r = 2 _____ to rectangular form. +2 Polar form is where a complex number is denoted by the length (otherwise known as the magnitude, absolute value, or modulus) and the angle of its vector (usually denoted by an angle symbol that looks like this: ∠).. To use the map analogy, polar notation for the vector from New York City to San Diego would be something like “2400 miles, southwest.” A Phasor is a rotating vector in a complex number form which expresses the magnitude and its phase. + 2 I'm not fimular with MATLAB keywords but need to use this to prove my answers. 2 Then, start changing rectangular values into polar form as per the rules above. 4 To convert from polar form to rectangular form, first evaluate the trigonometric functions. This calculator converts between polar and rectangular coordinates. 2 It also says how far I need to go, I need to go square root of 13. There's also a graph which shows you the meaning of what you've found. If you have been given the rectangular coordinates, the best thing you can do is chart them out. CppBuzz .com Home C C++ Java Python Perl PHP SQL JavaScript Linux Selenium QT Online Test ☰ + x Then, multiply through by See and . Polar, or phasor, forms of numbers take on the format, amplitude phase. Rewrite any trigonometric functions in terms of cos θ, sin θ, or tan θ. y In this case the equation is manipulated to use the polar-rectangular relationship, In this case the equation is manipulated to use the polar-rectangular relationships x = r cos θ, y = r sin θ, and r, To use the polar-rectangular relationships we need r cos θ and r sin θ. Z - is the Complex Number representing the Vector 3. x - is the Real part or the Active component 4. y - is the Imaginary part or the Reactive component 5. j - is defined by √-1In the rectangular form, a complex number can be represented as a point on a two dimensional plane calle… 1 + 2cos θ Example 7 Converting a Conic in Polar Form to Rectangular Form Convert the conic r = 5 − _____1 5s in θ t o rectangular form. B = | M | sin  ( ϕ ) {\dis… I'm not fimular with MATLAB keywords but need to use this to prove my answers. In order to work with complex numbers without drawing vectors, we first need some kind of standard mathematical notation. I am having trouble converting rectangular form complex numbers into polar form by writing a MATLAB script file. The polar form can also be verified using the conversion equation. Polar Form of a Complex Number. Find more Mathematics widgets in Wolfram|Alpha. Convert to Polar Coordinates (-1,1) Convert from rectangular coordinates to polar coordinates using the conversion formulas . 2, ( y To obtain these terms requires each side to be multiplied by r, Parametric Equations: Eliminating Parameters, Conversion from Polar to Rectangular Form Complex. Equations in polar form are converted into rectangular form, using the relationship between polar and rectangular coordinates.Problems with detailed solutions are presented. Phasor form conversion is the method of changing the representation form of a phasor. Find the polar form of the conic given a focus at the origin, e = 1, and directrix x = −1. Arrangement (in polar form) S11 S21 (S11 AND S22 IN SAME LINE) S12 S22(S12 AND S22 IN NEXT LINE) On the other hand, "[theta, rho] = cart2pol(real(a), imag(a))". Show Hide all comments. 4. Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. Its rectangular form is really 0 – j30 which is a vector of a magnitude of 30 plotted on the −j axis. Below is an interactive calculator that allows you to easily convert complex numbers in polar form to rectangular form, and vice-versa. This essentially makes the polar, it makes it clearer how we get there in kind of a more, I guess you could say, polar mindset, and that's why this form of the complex number, writing it this way is called rectangular form, while writing it this way is called polar form. To convert from polar form to rectangular form, first evaluate the trigonometric functions. Get the free "Convert Complex Numbers to Polar Form" widget for your website, blog, Wordpress, Blogger, or iGoogle. Converting polar to rectangular coordinates means expressing the polar coordinates in the form of rectangular coordinates. To link to this Conversion from Polar to Rectangular Form Complex page, copy the following code to your site: 1. Up to this point, we have been converting the polar form to rectangular. There are two basic forms of complex number notation: polar and rectangular. It is possible–and even important–to convert from rectangular to the polar form. Similarly, the reactance of the inductor, j50, can be written in polar form as , and the reactance of C 2, −j40, can be written in polar form as . 2 In what follows the polar coordinates of a point are (R , t) where R is the radial coordinate and t is the angular coordinate. =6 rsinθ rcosθ. See Example $$\PageIndex{6}$$ and Example $$\PageIndex{7}$$. 2 Then, multiply through by $$r$$. Make the polar-rectangular substitutions. For instance, if you are given the coordinates 50, j80, then: Step 1. 4 The polar form of the reactance of C 1, −j30, is . C++ program to convert polar form to rectangular form using constructor in destination class. The phase is specified in degrees. Polar Form of a Complex Number The polar form of a complex number is another way to represent a complex number. Z = 0.5 angle( - 45 degrees ) 2 Comments. The horizontal axis is the real axis and the vertical axis is the imaginary axis. Polar - Rectangular Coordinate Conversion Calculator. Polar/Rectangular Coordinates Calculator . To convert from rectangular form to polar form: 1. The above is the equation of a line. Equations in polar form are converted into rectangular form, using the relationship between polar and rectangular coordinates. Sign in to comment. =6xy, x x x 2 Polar coordinates are expressed as ($$r, \theta$$) while rectangular coordinates are expressed as ($$x, y$$). y [See more on Vectors in 2-Dimensions].. We have met a similar concept to "polar form" before, in Polar Coordinates, part of the analytical geometry section. −6x=0. Sign in to answer this question. Graphs of Functions, Equations, and Algebra, The Applications of Mathematics ) Use maximum precision for calculations. For the rest of this section, we will work with formulas developed by French mathematician Abraham de Moivre (1667-1754). A polar form uses the magnitude of the number as the length of line and the angle at which a number extends. We can think of complex numbers as vectors, as in our earlier example. To write complex numbers in polar form, we use the formulas and Then, See and . To convert from rectangular to polar form, we must use the following formulas: It is easier to find our angle first, which is done by plugging in our x and y into the second formula: Find the angle by taking the inverse of the function: Now find r by plugging in our angle and x … This will help you in conversion of complex numbers from one form to another. To convert a rectangular equation into polar form, remove the numerators. | M | = A 2 + B 2 {\displaystyle |M|={\sqrt {A^{2}+B^{2}}}} 1. ϕ = arctan  ( B A ) {\displaystyle \phi =\arctan \left({\frac {B}{A}}\right)} To convert from polar to rectangular form: A is the part of the phasor along the real axis 1. Rewrite the given equation as follows:t + π / 4 = 0t = - π / 4Take the tangent of both sides:tan t = tan (π / 4) = -1Use y = R sin t and x = R cos t to write:tan t = sin t / cos t = R sin t / R cos t = y / xHence:y / x = -1y = - xThe above is the equation of a line. + Where: 2. Finding Products and Quotients of Complex Numbers in Polar Form. c) Evaluate 324. =6yx, x r Let: 1 = −4 - 7, 2 = 4 ∠152 ° and 3 = - 0.8 + 0.6 . by M. Bourne. A = | M | cos  ( ϕ ) {\displaystyle A=|M|\cos \left(\phi \right)} B is the part of the phasor along the imaginary axis 1. Based on this command, the angle we got is not right. For example the number 7 ∠ 40°. In the last tutorial about Phasors, we saw that a complex number is represented by a real part and an imaginary part that takes the generalised form of: 1. In radian mode, we have 3 i 2cis 5 6 Polar to Rectangular Online Calculator. a) Give the polar form of 1 with the detailed calculations. To find the product of two complex numbers, multiply the two moduli and add the two angles. Displaying top 8 worksheets found for - Converting Rectangular Form To Polar Form. When working with phasors it is often necessary to convert between rectangular and polar form. Solution x 3 and y 1 so that r 3 2 12 2 and tan 1 3 3 3 Here the reference angle and for is 30°. Z = 4 + j 5 0 Comments. 2 Sinusoidal signals with different magnitudes and phases, but the same frequency can be plotted on a phasor diagram; in a phasor diagram, the angular velocity (w) will be the same for all signals. Show Hide all comments. +2 Rectangular forms of numbers can be converted into their polar form equivalents by the formula, Polar amplitude= √ x 2 + y 2 , where x and y represent the real and imaginary numbers of the expression in rectangular form. Z = 0.5 angle( pi / 4 ) or. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. how to represent complex numbers on the complex plane in rectangular and polar form (including real and imaginary numbers), and explain why the rectangular and polar forms of a given complex number represent the same number, Common Core High School: Number & Quantity, HSN-CN.B.4 Rectangular to polar. For background information on what's going on, and more explanation, see the previous pages, y Replace and with the actual values. The form z = a + b i is called the rectangular coordinate form of a complex number. b) Give the rectangular shape of 2 with the detailed calculations. Polar Form of a Complex Number. Give the answers rounded to 2 decimal places. To find the product of two complex numbers, multiply the … Since the complex number is in QII, we have 180° 30° 150° So that 3 i 2cis150°. Solution ) Home > Math > Pre Calculus > Conversion from Polar to Rectangular Form Complex If a polar equation is written such that it contains terms that appear in the polar-rectangular relationships (see below), conversion from a polar equation to a rectangular equation is a simple matter of substitution. Keep solving until you isolate the variable r. Powered by Create your own unique website with customizable templates. Convert a Complex Number to Polar Form Description Obtain the polar form of a complex number . y r Polar to Rectangular x rcos y rsin The polar form r cos isin is sometimes abbreviated rcis Example Convert 3 i to polar form. Show Instructions. However, I would like to convert it to polar form with this kind of arrangements. Problems with detailed solutions are presented.In what follows the polar coordinates of a point are (R , t) where R is the radial coordinate and t is the angular coordinate.The relationships between the rectangualr (x,y) and polar (R,t) coordinates of a points are given byR 2 = x 2 + y 2       y = R sin t       x = R cos t. Expand the left side of the given equation.R(-2 sin t + 3 cos t) = 2-2 R sin t + 3 R cos t = 2eval(ez_write_tag([[580,400],'analyzemath_com-box-4','ezslot_7',260,'0','0']));Use y = R sin t and x = R cos t into the given equation to rewrite as follows:-2 y + 3 x = 2 2 2 How to convert between Polar and Rectangular form ( in Physics and Engineering, Exercises de Mathematiques Utilisant les Applets, Trigonometry Tutorials and Problems for Self Tests, Elementary Statistics and Probability Tutorials and Problems, Free Practice for SAT, ACT and Compass Math tests, Convert Equation from Rectangular to Polar Form, Convert Polar to Rectangular Coordinates - Calculator, Free Geometry Tutorials, Problems and Interactive Applets, Convert Rectangular to Polar Coordinates - Calculator. 4 The calculator will convert the polar coordinates to rectangular (Cartesian) and vice versa, with steps shown. 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To find the product of two complex numbers to polar form of writing complex numbers from one form to.! Coordinates, the angle we got is not right we first need some kind of.... Product of two complex numbers to polar form to rectangular form complex numbers in polar form origin, e 1! Tan θ changing rectangular values into polar form to rectangular form using constructor in destination class / 4 or! = a + b i is called the rectangular shape of 2 with detailed. \Pageindex { 7 } \ ) and vice versa, with steps shown necessary to from!


# Understanding Horizontal Asymptotes

#### confusedatmath

##### New member
I'm stuck at understanding why +c is the horizontal asymptote. Can someone please explain this? I get that the vertical asymptote is relating to (x+b), because the denominator cannot=0. But why does this kind of graph have a horizontal asymptote to begin with?

#### MarkFL

Staff member
What happens to a fraction when the denominator gets larger and larger without bound?

#### confusedatmath

##### New member
the fraction gets smaller, but dont we end up adding that to the value of c??

well if a/(x-b)^2 +c

let a=1 b=1 c=2

so we get

1/(x-1)^2 +2

lets make x =2

we get

1/1 +2 .... thats 3.... wait am i doing something wrong?

- - - Updated - - -

oh wait is because in the above a is negative.. so

it would be -1+2 < 2 ....

#### MarkFL

Staff member
The fraction goes to zero as the denominator goes to infinity, and so that's why the horizontal asymptote is $y=c$.

#### Deveno

##### Well-known member
MHB Math Scholar
Let's make this really interesting by choosing some suitably large values for $x$.

Let's see what happens at $x = 1,000,000$, when $a = b = -1,c = 2$.

Then $f(x) = \dfrac{-1}{(999,999)^2} + 2$

$= \dfrac{-1}{999,998,000,001} + 2$

$\sim 1.999999999998999998$

At $x = -1,000,000$, we get:

$f(x) = \dfrac{-1}{(-1,000,001)^2} + 2$

$= \dfrac{-1}{1,000,002,000,001} + 2$

$\sim 1.999999999999000002$

Both of these numbers are really close to 2, right?

In general, we see that:

$\displaystyle \lim_{x \to \infty} \left(\frac{a}{(x + b)^2} + c\right)$

$\displaystyle = \lim_{x \to \infty} \frac{a}{(x + b)^2} + \lim_{x \to \infty} c$

$\displaystyle = \lim_{x \to \infty} \frac{a}{(x+b)^2} + c$

$\displaystyle = (a)\left(\lim_{x \to \infty}\frac{1}{x+b}\right)^2 + c$

$\displaystyle = (a)(0)^2 + c = 0 + c = c$

Similar reasoning holds to show that:

$\displaystyle \lim_{x \to -\infty} \left(\frac{a}{(x + b)^2} + c\right) = c$

as well.

(If you haven't been formally introduced to limits yet, all you need to know for this is the following (which hold under "suitably nice conditions" which are the case here):

1) the limit of a sum is the sum of the limits of each term in the sum
2) the limit of a product is the product of the limits of each factor in the product
3) if M(x) is a function that gets "infinitely big" as x does, then:

$$\lim_{x \to \infty} \frac{1}{M(x)} = 0$$

-by "get infinitely big as x does" I mean that for ANY positive integer $K$, there is always some positive integer $N$, so that if we have $x > N$, then $|M(x)| > K$ (typically, the integer $N$ will depend on $K$, bigger $K$'s usually need bigger $N$'s).).

#### Prove It

##### Well-known member
MHB Math Helper
I'm stuck at understanding why +c is the horizontal asymptote. Can someone please explain this? I get that the vertical asymptote is relating to (x+b), because the denominator cannot=0. But why does this kind of graph have a horizontal asymptote to begin with?

View attachment 1838
The short answer, if \displaystyle \begin{align*} y = c \end{align*} then you end up with \displaystyle \begin{align*} 0 = \frac{a}{(x + b)^2} \end{align*}. Is that possible to solve for x?

The long answer, consider the inverse relation. You should find that c becomes a vertical asymptote.

#### HallsofIvy

##### Well-known member
MHB Math Helper
the fraction gets smaller, but dont we end up adding that to the value of c??

well if a/(x-b)^2 +c

let a=1 b=1 c=2

so we get

1/(x-1)^2 +2

lets make x =2

we get

1/1 +2 .... thats 3.... wait am i doing something wrong?

- - - Updated - - -

oh wait is because in the above a is negative.. so

it would be -1+2 < 2 ....
The question was "what happens when x is large?" "2" is NOT large!

If x= 1000000, then 1/(1+ 1000000)+3= 3.000009999900000999990000099999 and for


# Thread: Area of Square in Circle

1. ## Area of Square in Circle

There is a shaded square inside a circle represented by the equation x^2 + y^2 = 9. Find the area of the square in the circle.

I seek the first-two steps.

2. ## Re: Area of Square in Circle

I am assuming the square is inscribed within the circle, and so the diagonal of the square is equal to the diameter of the circle.

Let $\displaystyle D=2r$ by the diameter of the circle. The area $\displaystyle A$ of the square is:

$\displaystyle A=s^2$

And we know:

$\displaystyle 2r=\sqrt{2}s\implies s=\sqrt{2}r\implies A=2r^2$

Can you proceed?

3. ## Re: Area of Square in Circle

Originally Posted by MarkFL
I am assuming the square is inscribed within the circle, and so the diagonal of the square is equal to the diameter of the circle.

Let $\displaystyle D=2r$ by the diameter of the circle. The area $\displaystyle A$ of the square is:

$\displaystyle A=s^2$

And we know:

$\displaystyle 2r=\sqrt{2}s\implies s=\sqrt{2}r\implies A=2r^2$

Can you proceed?
A = 2r^2

I know that x^2 + y^2 = r^2.

So, r^2 = 9 leading to an r value of 3.

A = 2r^2

A = 2(3)^2

A = 2(9)

A = 18 units^2

4. ## Re: Area of Square in Circle

What would the area of a square be that circumscribes a circle of radius $\displaystyle r$? (This is there the circle is inside the square)

5. ## Re: Area of Square in Circle

Originally Posted by MarkFL
What would the area of a square be that circumscribes a circle of radius $\displaystyle r$? (This is there the circle is inside the square)
If I know r, then A = 2r^2 yields the area of a square inside a circle.

6. ## Re: Area of Square in Circle

Originally Posted by harpazo
If I know r, then A = 2r^2 yields the area of a square inside a circle.
Rethink that! Are you missing some brackets?

7. ## Re: Area of Square in Circle

Originally Posted by Debsta
Rethink that! Are you missing some brackets?
A = 2(r)^2

8. ## Re: Area of Square in Circle

Originally Posted by harpazo
If I know r, then A = 2r^2 yields the area of a square inside a circle.
If the circle is inscribed within the square, then the side length of the square will be equal to the diameter of the circle, thus:

$\displaystyle A=(2r)^2=4r^2$

Now, given the result for the original problem you posted, we now know:

$\displaystyle 2r^2<\pi r^2<4r^2$

Or:

$\displaystyle 2<\pi<4$

How could we improve these bounds on $\displaystyle \pi$?

9. ## Re: Area of Square in Circle

Originally Posted by MarkFL
If the circle is inscribed within the square, then the side length of the square will be equal to the diameter of the circle, thus:

$\displaystyle A=(2r)^2=4r^2$

Now, given the result for the original problem you posted, we now know:

$\displaystyle 2r^2<\pi r^2<4r^2$

Or:

$\displaystyle 2<\pi<4$

How could we improve these bounds on $\displaystyle \pi$?
Sorry but I don't follow? Is my answer to the original question right?

10. ## Re: Area of Square in Circle

Originally Posted by harpazo
Sorry but I don't follow? Is my answer to the original question right?
Yes, you answered the original problem correctly, and is why I thanked your post.

Consider the following diagram:

Can you see that the side length of the larger square is equal to the diameter of the circle inscribed within?

11. ## Re: Area of Square in Circle

Originally Posted by MarkFL
Yes, you answered the original problem correctly, and is why I thanked your post.

Consider the following diagram:

Can you see that the side length of the larger square is equal to the diameter of the circle inscribed within?
Yes, I can what you are saying here.

12. ## Re: Area of Square in Circle

Do you see how the rest of what I posted in post #8 then follows?

13. ## Re: Area of Square in Circle

Originally Posted by MarkFL
Do you see how the rest of what I posted in post #8 then follows?
Yes.

14. ## Re: Area of Square in Circle

Originally Posted by harpazo
Yes.
Okay, then can you think of a way we could improve our bounds on the value of pi?

15. ## Re: Area of Square in Circle

Originally Posted by MarkFL
Okay, then can you think of a way we could improve our bounds on the value of pi?
Improve our bounds???

Page 1 of 2 12 Last


## 2.8.1 Example: Numerical error as a function of $$\Delta t$$

In this example we will assess how the error of our implementation of the Euler method depends on the time step $$\Delta t$$ in a systematic manner. We will solve a problem with an analytical solution in a loop, and for each new solution we do the following:
• Divide the time step by two (or double the number of time steps)
• Compute the error
• Plot the error
Euler's method is a first order method and we expect the error to be $$O(h)=O(\Delta t)$$. Consequently if the time-step is divided by two, the error should also be divided by two. As errors normally are small values and are expected to be smaller and smaller for decreasing time steps, we normally do not plot the error itself, but rather the logarithm of the absolute value of the error. The latter we do due to the fact that we are only interested in the order of magnitude of the error, whereas errors may be both positive and negative. As the initial value is always correct we discard the first error at time zero to avoid problems with the logarithm of zero in log_error = np.log2(abs_error[1:]).

# coding: utf-8
# src-ch1/Euler_timestep_ctrl.py;DragCoefficientGeneric.py @ git@lrhgit/tkt4140/src/src-ch1/DragCoefficientGeneric.py;
from DragCoefficientGeneric import cd_sphere
from matplotlib.pyplot import *
import numpy as np

# change some default values to make plots more readable
LNWDT=2; FNT=11
rcParams['lines.linewidth'] = LNWDT; rcParams['font.size'] = FNT

g = 9.81      # Gravity m/s^2
d = 41.0e-3   # Diameter of the sphere
rho_f = 1.22  # Density of fluid [kg/m^3]
rho_s = 1275  # Density of sphere [kg/m^3]
nu = 1.5e-5   # Kinematical viscosity [m^2/s]
CD = 0.4      # Constant drag coefficient

def f(z, t):
"""2x2 system for sphere with constant drag."""
zout = np.zeros_like(z)
alpha = 3.0*rho_f/(4.0*rho_s*d)*CD
zout[:] = [z[1], g - alpha*z[1]**2]
return zout

# define euler scheme
def euler(func,z0, time):
"""The Euler scheme for solution of systems of ODEs.
z0 is a vector for the initial conditions,
the right hand side of the system is represented by func which returns
a vector with the same size as z0 ."""

z = np.zeros((np.size(time),np.size(z0)))
z[0,:] = z0

for i in range(len(time)-1):
dt = time[i+1]-time[i]
z[i+1,:]=z[i,:] + np.asarray(func(z[i,:],time[i]))*dt
return z

def v_taylor(t):
#    z = np.zeros_like(t)
v = np.zeros_like(t)
alpha = 3.0*rho_f/(4.0*rho_s*d)*CD
v=g*t*(1-alpha*g*t**2)
return v

# main program starts here

T = 10  # end of simulation
N = 10  # no of time steps

z0=np.zeros(2)
z0[0] = 2.0

# Prms for the analytical solution
k1 = np.sqrt(g*4*rho_s*d/(3*rho_f*CD))
k2 = np.sqrt(3*rho_f*g*CD/(4*rho_s*d))

Ndts = 4  # Number of times to divide the dt by 2
legends=[]
error_diff = []

for i in range(Ndts+1):
time = np.linspace(0, T, N+1)
ze = euler(f, z0, time)     # compute response with constant CD using Euler's method
v_a = k1*np.tanh(k2*time)   # compute response with constant CD using analytical solution
abs_error=np.abs(ze[:,1] - v_a)
log_error = np.log2(abs_error[1:])
max_log_error = np.max(log_error)

plot(time[1:], log_error)
legends.append('Euler scheme: N ' + str(N) + ' timesteps' )
N*=2
if i > 0:
error_diff.append(previous_max_log_err-max_log_error)

previous_max_log_err = max_log_error

print('Approximate order of scheme n =', np.mean(error_diff))
print('Approximate error reuduction by dt=dt/2:', 1/2**(np.mean(error_diff)))

# plot analytical solution
# plot(time,v_a)
# legends.append('analytical')

# fix plot
legend(legends, loc='best', frameon=False)
xlabel('Time [s]')
#ylabel('Velocity [m/s]')
ylabel('log2-error')
#savefig('example_euler_timestep_study.png', transparent=True)
show()

The plot resulting from the code above is shown in Figure (11). The difference or distance between the curves seems to be rather constant after an initial transient. As we have plotted the logarithm of the absolute value of the error $$\epsilon_i$$, the difference $$d_{i+1}$$ between two curves is $$d_{i+1}= \log_2 \epsilon_{i}-\log_2 \epsilon_{i+1} = \displaystyle \log_2 \frac{\epsilon_{i}}{\epsilon_{i+1}}$$. A rough visual inspection of Figure (11) yields $$d_{i+1} \approx 1.0$$, from which we may deduce the apparent order of the scheme: $$$$n = \log_2 \frac{\epsilon_{i}}{\epsilon_{i+1}} \approx 1 \Rightarrow \epsilon_{i+1} \approx 0.488483620\, \epsilon_{i} \tag{2.75}$$$$

The print statement returns 'n=1.0336179048' and '0.488483620794', thus we see that the error is reduced even slightly more than the theoretically expected value for a first order scheme, i.e. $$\Delta t_{i+1} = \Delta t_{i}/2$$ yields $$\epsilon_{i+1} \approx\epsilon_{i}/2$$.

Figure 11: Plots for the logarithmic errors for a falling sphere with constant drag. The timestep $$\Delta t$$ is reduced by a factor two from one curve to the one immediately below.


# Two colouring of a 6 by 6 grid without a monochromatic rectangle

This question Monochromatic Rectangle of a 2-Colored 8 by 8 Lattice Grid shows that any two colouring of a 7 by 7 grid must have a rectangle whose vertices are all the same colour. Note that we are only considering rectangles whose sides are parallel to the sides of the grid.

Can the same result be shown to hold for a 6 by 6 grid? If not, what is an example of a colouring of a 6 by 6 grid that does not contain a monochromatic rectangle?

Furthermore, if it turns out that every colouring of a 6 by 6 grid contains a monochromatic rectangle, what is the largest $n \in \mathbb{N}$ such that an $n$ by $n$ grid can be coloured such that no rectangles are monochromatic?

• Do the sides of the rectangle have to be parallel to the grid lines? – Ankoganit Jun 5 '17 at 11:39
• @Ankoganit I'm looking at the case when the sides do have to be parallel. – michaelhowes Jun 5 '17 at 11:41

For six by six it is quite easy.

Call the colours red and blue. Look at the top row. There will be three points the same colour - say red. Concentrate on those columns. If in any row below there are two out of three points red you have a rectangle. But there are only four possible patterns of three points with at most one red, so two of the five rows must have the same pattern and there will be a blue rectangle.

Since one of the possible patterns of three with at most one red is all blue, and if this is present there will be a blue rectangle anyway, this can be adapted to five by five (which must have three the same in the top row).

Can you find a four by four arrangement with no rectangle?

• Thank you very much for your answer. I have found such a four by four arrangement. – michaelhowes Jun 5 '17 at 12:34

The number of points occupied by a colour in a given column implies that a certain number of possible pairings have been produced, for possible matching to the same pair in another column. So with $\{0,1,2,3,4,5,6\}$ points of a single colour you have $\{0,0,1,3,6,10,15\}$ pairings respectively.

For a six-by-six rectangle, there are $\binom 62 = 15$ possible pairings in a column. As soon as one colour has enough entries to accumulate $16$ pairings from its column entries, as above, there will be a matched pairing between columns giving the monochrome rectangle. Between the two colours, once total pairings reaches $31$ there will be a monochrome rectangle. Splitting the colours evenly ($3$ each) produces the lowest net pairings of $6$ per column, but clearly with six columns we have at least $36$ net pairings and must have a monochrome rectangle.

For a five-by-six rectangle, using the same arguments, we have $\binom 52 = 10$ possible pairings per column and at least $4$ net pairings per column, reaching at least $24(>2{\times} 10)$ net pairings by the sixth column and the certainty of a monochrome rectangle.

For a five-by-five square, one colour will have at least $13$ entries which you can show leads to at least $11$ pairings across the $5$ columns:

$\begin{array}{c|c|c} \text{col entries} & \text{pairings} \\ \hline (5,2+,\ldots) & 11+ \\ (4,4,\ldots) & 12+ \\ (4,3,3,2,1) & 13 \\ (4,3,2,2,2) & 12 \\ (3,3,3,3,1) & 12 \\ (3,3,3,2,2) & 11 \\ \end{array}$

and $11$ or more pairings over the 5 columns will mean that some $2$ pairs will align, giving a monochrome rectangle.

Finally a four-by-six rectangle can avoid a monochrome rectangle, guided by the above:


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# Is quadrilateral ABCD a rhombus?

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14 Oct 2009, 21:09
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(1) Line segments AC and BD are perpendicular bisectors of each other.

(2) AB = BC = CD = AD
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(1) Line segments AC and BD are perpendicular bisectors of each other. --> rhombus

(2) AB = BC = CD = AD --> rhombus

Or am I missing something, seems pretty obvious...
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14 Oct 2009, 21:45
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Well:
(1) True for square or for rhombus but every square is a rhombus, so sufficient
(2) Again true for square or for rhombus but every square is a rhombus, so sufficient

D
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Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD [#permalink]

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16 Jun 2011, 15:24
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(1) Line segments AC and BD are perpendicular bisectors of each other.

(2) AB = BC = CD = AD

Solve!
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17 Jun 2011, 18:45
1. Sufficient

diagnols are perpendicular bisectors. can only happen in a square or rhombus. Square is a kind of rhombus.

2. Sufficient

As we know its a square or rhombus. Square is a kind of rhombus.

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10 Nov 2011, 01:16
Hi guys
I'm a little confused as to whether we should consider the properties of a kite when dealing with quadrilaterals. For example:

(1) Line segments AC and BD are perpendicular bisectors of each other
(2) AB = BC = CD = AD

The official answer is as follows:
Statement 1 - SUFFICIENT: The diagonals of a rhombus are perpendicular bisectors of one another. This is in fact enough information to prove that a quadrilateral is a rhombus
Statement 2 - A quadrilateral with four equal sides is by definition a rhombus

What I don't get is that the question mentions it is a quadrilateral, not a parallelogram. In this case, how can we eliminate the possibility that it could be a kite as well in Statement 1?
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10 Nov 2011, 03:54
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mk87 wrote:
Hi guys
I'm a little confused as to whether we should consider the properties of a kite when dealing with quadrilaterals. For example:

(1) Line segments AC and BD are perpendicular bisectors of each other
(2) AB = BC = CD = AD

The official answer is as follows:
Statement 1 - SUFFICIENT: The diagonals of a rhombus are perpendicular bisectors of one another. This is in fact enough information to prove that a quadrilateral is a rhombus
Statement 2 - A quadrilateral with four equal sides is by definition a rhombus

What I don't get is that the question mentions it is a quadrilateral, not a parallelogram. In this case, how can we eliminate the possibility that it could be a kite as well in Statement 1?

Statement 1 says the diagonals are perpendicular bisectors of each other which means that they intersect each other at mid points and form 90 degree angles. It is not possible that they are in the shape of an asymmetric kite. Figure 1 is not possible. Only figure 2 is possible. In figure 2, all sides will be equal.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 13 Aug 2012 Posts: 464 Concentration: Marketing, Finance GMAT 1: Q V0 GPA: 3.23 Followers: 24 Kudos [?]: 372 [0], given: 11 Re: Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD [#permalink] ### Show Tags 29 Jan 2013, 07:59 gijoedude wrote: Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD are perpendicular bisectors of each other. (2) AB = BC = CD = AD A rhombus is a quadrilateral with all its sides equal to each other. 1. Imagine that the point where the bisectors meet is called X. Then, BX = DX and AX = CX. Using the Pythagorean Theorem you could imagine getting the hypotenuse of sides BX and CX. Since AX = CX, then the hypotenuse of BX and AX is also as long as the previous. As you could imagine, all the hypotenuse formed are equal, thus forming equal 4 sides. Hence, a rhombus. SUFFICIENT. 2. All sides are equal. SUFFICIENT. Answer: D _________________ Impossible is nothing to God. Current Student Joined: 06 Sep 2013 Posts: 2035 Concentration: Finance GMAT 1: 770 Q0 V Followers: 51 Kudos [?]: 496 [0], given: 355 Re: rhombus? [#permalink] ### Show Tags 13 Oct 2013, 16:10 Bunuel wrote: Thought about this again: D it is. Well: (1) True for square or for rhombus but every square is a rhombus, so sufficient (2) Again true for square or for rhombus but every square is a rhombus, so sufficient D Hi Bunuel When they perpendicular bisectors they mean that all angles and all sides are equal, therefore it is a square. And a square is just a type of rhombus? And for statement 2, when they say that all sides are equal, you can assume that it is a square therefore it is also a rhombus? Is that the correct line of reasoning? Thanks so much in advance Cheers J Math Expert Joined: 02 Sep 2009 Posts: 34057 Followers: 6084 Kudos [?]: 76454 [0], given: 9975 Re: rhombus? [#permalink] ### Show Tags 13 Oct 2013, 16:20 Expert's post jlgdr wrote: Bunuel wrote: Thought about this again: D it is. Well: (1) True for square or for rhombus but every square is a rhombus, so sufficient (2) Again true for square or for rhombus but every square is a rhombus, so sufficient D Hi Bunuel When they perpendicular bisectors they mean that all angles and all sides are equal, therefore it is a square. And a square is just a type of rhombus? And for statement 2, when they say that all sides are equal, you can assume that it is a square therefore it is also a rhombus? Is that the correct line of reasoning? Thanks so much in advance Cheers J For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°. For (2): AB = BC = CD = AD, means that ABCD is either a rhombus or square (so still a rhombus). Hope it helps. _________________ Senior Manager Joined: 13 May 2013 Posts: 472 Followers: 3 Kudos [?]: 136 [0], given: 134 Re: Is quadrilateral ABCD a rhombus? [#permalink] ### Show Tags 06 Dec 2013, 11:10 Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD are perpendicular bisectors of each other. But doesn't a rhombus also allow for two perpendicular bisectors? A square has two perpendicular bisectors but a square is not a rhombus (because a rhombus does not have all four angles = 90. Oh, but wait, a rhombus is a square in the same way a rectangle is a square but not the other way around. Tricky. (2) AB = BC = CD = AD I guess the same logic applies above, ABCD could be a square or rhombus but regardless of which one it is its still a rhombus. If the stem asked if ABCD was a square, then this would be insufficient. Manager Joined: 22 Feb 2009 Posts: 229 Followers: 5 Kudos [?]: 100 [0], given: 148 Re: Data Sufficiency - doubts [#permalink] ### Show Tags 08 Aug 2014, 21:36 kritim22 wrote: Hi, can someone help me with this question please: Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD are perpendicular bisectors of each other. (2) AB = BC = CD = AD The answer it says is D- each statement is enough on its own. Doubt- can't these properties also apply to a square? How can we confirm it's a rhombus? Let say AC and BD meet at O. We have OD= OB, OA= OC. If you use pithagoras theorem, you can easily see that AD= DC=BC=AB. For example, AD^2= OA^2 + OD^2 So (1) and (2) basically say the same thing. The definition of rhombus is a quadrilateral whose four sides have the same length. So D is the answer. _________________ ......................................................................... +1 Kudos please, if you like my post Manager Joined: 21 Sep 2012 Posts: 219 Location: United States Concentration: Finance, Economics Schools: CBS '17 GPA: 4 WE: General Management (Consumer Products) Followers: 3 Kudos [?]: 151 [0], given: 31 Re: Data Sufficiency - doubts [#permalink] ### Show Tags 09 Aug 2014, 00:56 kritim22 wrote: Hi, can someone help me with this question please: Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD are perpendicular bisectors of each other. (2) AB = BC = CD = AD The answer it says is D- each statement is enough on its own. Doubt- can't these properties also apply to a square? How can we confirm it's a rhombus? A square is a special case of rhombus. It has all angles of 90 degrees. so proving that quadrilateral ABCD is a square is sufficient to say that it is a rhombus. Director Status: Verbal Forum Moderator Joined: 17 Apr 2013 Posts: 635 Location: India GMAT 1: 710 Q50 V36 GMAT 2: 750 Q51 V41 GMAT 3: 790 Q51 V49 GPA: 3.3 Followers: 48 Kudos [?]: 310 [0], given: 297 Re: Is quadrilateral ABCD a rhombus? [#permalink] ### Show Tags 05 Nov 2014, 03:31 Bunuel wrote: Answer D. (1) Line segments AC and BD are perpendicular bisectors of each other. --> rhombus (2) AB = BC = CD = AD --> rhombus Or am I missing something, seems pretty obvious... But Bunuel this would be true even for rectangle: Line segments AC and BD are perpendicular bisectors of each other. I need Your though, am confused. Diagonals of rectangles also bisect each other. _________________ Like my post Send me a Kudos It is a Good manner. My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html Math Expert Joined: 02 Sep 2009 Posts: 34057 Followers: 6084 Kudos [?]: 76454 [0], given: 9975 Re: Is quadrilateral ABCD a rhombus? [#permalink] ### Show Tags 05 Nov 2014, 05:09 Expert's post honchos wrote: Bunuel wrote: Answer D. (1) Line segments AC and BD are perpendicular bisectors of each other. --> rhombus (2) AB = BC = CD = AD --> rhombus Or am I missing something, seems pretty obvious... But Bunuel this would be true even for rectangle: Line segments AC and BD are perpendicular bisectors of each other. I need Your though, am confused. Diagonals of rectangles also bisect each other. For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°. Now, the diagonals of a rectangle, though cut each other into two equal parts, do NOT necessarily cut each other at 90°. This happens only if a rectangle is a square but if ABCD is a square then it's also a rhombus. Hope it's clear. _________________ Manager Joined: 10 Sep 2014 Posts: 75 Followers: 0 Kudos [?]: 54 [0], given: 102 Re: Is quadrilateral ABCD a rhombus? [#permalink] ### Show Tags 06 Nov 2014, 02:17 Bunuel, For (1), can you consider the case of a kite? Bunuel wrote: honchos wrote: Bunuel wrote: Answer D. (1) Line segments AC and BD are perpendicular bisectors of each other. --> rhombus (2) AB = BC = CD = AD --> rhombus Or am I missing something, seems pretty obvious... But Bunuel this would be true even for rectangle: Line segments AC and BD are perpendicular bisectors of each other. I need Your though, am confused. Diagonals of rectangles also bisect each other. For (1): a perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°. Now, the diagonals of a rectangle, though cut each other into two equal parts, do NOT necessarily cut each other at 90°. This happens only if a rectangle is a square but if ABCD is a square then it's also a rhombus. Hope it's clear. _________________ Press KUDOs if you find my explanation helpful Math Expert Joined: 02 Sep 2009 Posts: 34057 Followers: 6084 Kudos [?]: 76454 [0], given: 9975 Re: Is quadrilateral ABCD a rhombus? [#permalink] ### Show Tags 06 Nov 2014, 07:22 Expert's post TARGET730 wrote: Bunuel, For (1), can you consider the case of a kite? Yes. In the special case where all 4 sides are the same length, the kite satisfies the definition of a rhombus. _________________ Senior Manager Joined: 10 Mar 2013 Posts: 290 GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39 Followers: 5 Kudos [?]: 80 [0], given: 2404 Re: Is quadrilateral ABCD a rhombus? [#permalink] ### Show Tags 10 Mar 2015, 20:32 This question is poor, because (1) could be right kite, which is not necessarily a rhombus or a rhombus, but (2) will always satisfy a rhombus. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6742 Location: Pune, India Followers: 1873 Kudos [?]: 11525 [2] , given: 219 Re: Is quadrilateral ABCD a rhombus? [#permalink] ### Show Tags 10 Mar 2015, 22:58 2 This post received KUDOS Expert's post TooLong150 wrote: This question is poor, because (1) could be right kite, which is not necessarily a rhombus or a rhombus, but (2) will always satisfy a rhombus. Actually, the question is fine. A rhombus is a quadrilateral all of whose sides are of the same length - that's all. You could have a rhombus which also has all angles 90 which makes it a square or a rhombus in the shape of a kite. But nevertheless, if it is a quadrilateral and has all sides equal, it IS A RHOMBUS. (1) Line segments AC and BD are perpendicular bisectors of each other. Make 2 lines - a vertical and a horizontal - which are perpendicular bisectors of each other. Make them in any way of any length - just that they should be perpendicular bisectors of each other. When you join the end points, you will get all sides equal. Think of it this way - each side you get will be a hypotenuse of a right triangle. The legs of the right triangle will have the same pair of lengths in all 4 cases. So AB = BC = CD = AD. So ABCD must be a rhombus. (2) AB = BC = CD = AD This statement directly tells you that all sides are equal so ABCD must be a rhombus. Answer (D) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD [#permalink]

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22 Aug 2015, 11:10
While somewhat unusual, isn't the fact that it asks if the quadrilateral is a rhombus sufficient in itself? Aren't all quadrilaterals also rhombuses and vise versa because they have 4 sides?
Re: Is quadrilateral ABCD a rhombus? (1) Line segments AC and BD   [#permalink] 22 Aug 2015, 11:10

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# A group such that $a^m b^m = b^m a^m$ and $a^n b^n = b^n a^n$ ($m$, $n$ coprime) is abelian?

Let $(G,.)$ be a group and $m,n \in\mathbb Z$ such that $\gcd(m,n)=1$. Assume that $$\forall a,b \in G, \,a^mb^m=b^ma^m,$$ $$\forall a,b \in G, \, a^nb^n=b^na^n.$$

Then how prove $G$ is an abelian group?

Some context: Some of these commutation relations often imply that $G$ is abelian, for example if $(ab)^i = a^i b^i$ for three consecutive integers $i$ then $G$ is abelian, or if $g^2 = e$ for all $g$ then $G$ is abelian. This looks like another example of this phenomenon, but the same techniques do not apply.

• from which text did you get this problem ? – Fawzy Hegab Mar 10 '13 at 19:32
• also , is m,n are fixed or you mean , for all m,n in Z such that gcd(m,n)=1 .... ??? – Fawzy Hegab Mar 10 '13 at 19:39
• @ MrWhymy:m,n are fixed integer number – M.H Mar 10 '13 at 19:46
• it's nice problem , and the solution which is providen in the answer below is great :) – Fawzy Hegab Mar 10 '13 at 19:53

Let $M \subset G$ be the subgroup generated by all $m$-th powers and let $N \subset G$ be the subgroup generated by all $n$-th powers. These subgroups are clearly abelian normal subgroups. Since $m$ and $n$ are coprime, $G = MN$, and hence $M \cap N$ is contained in the center $Z(G)$ of $G$. To prove that $G$ is abelian it suffices to show that $M$ and $N$ commute, that is $[M,N]=1$. Note that $[M,N] \subset M \cap N$ (since $M$ and $N$ are normal subgroups). Let $a \in M$ and $b \in N$. Then $[a, b] = a^{−1}b^{−1}ab \in M \cap N$. Hence $[a, b] = z$ with $z \in Z(G)$. Hence $b^{−1}ab = za$, whence $b^{−1}a^nb=z^na^n$. Since $a^n \in N$ it commutes with $b$, so $z^n=1$. Similarly $z^m=1$. Since $m$ and $n$ are relatively prime, we conclude $z=1$.

• Dear Nicky! Let me ask sime questions: 1) Why $G=MN$? 2) Why $M\cap N$ is contained in the center $Z(G)$? – ZFR May 1 '18 at 14:21
• (1) $gcd(m,n)=1$ whence you can find integers $k$ and $l$ with $1=km+ln$. So if $g \in G$ then $g=g^1=(g^k)^m \cdot (g^l)^n \in MN$. (2) Because $G=MN$ and $M$ and $N$ are abelian, any element from $M \cap N$ commutes with any element from $G$. Hope it is clear now. – Nicky Hekster May 1 '18 at 15:33
• Dear Nicky! Thanks for answer! I got your solution and your approach is very simple and nice! +1 – ZFR May 1 '18 at 19:53
• Thank you very much! – Nicky Hekster May 1 '18 at 20:32

Write $mk+nl=1$ for some $k,l\in\mathbb Z$.

We have $$ab=a^{mk+nl}b^{mk+nl}=a^{mk}(a^{nl}b^{mk})b^{nl}\stackrel{(*)}=a^{mk}(b^{mk}a^{nl})b^{nl}=(a^{mk}b^{mk})(a^{nl}b^{nl})=(b^{mk}a^{mk})(b^{nl}a^{nl})=b^{mk}(a^{mk}b^{nl})a^{nl}\stackrel{(**)}=b^{mk}(b^{nl}a^{mk})a^{nl}=(b^{mk}b^{nl})(a^{mk}a^{nl})=b^{mk+nl}a^{mk+nl}=ba,$$ where we used $$a^{nl}b^{mk}=b^{mk}a^{nl}\qquad(*)$$ and $$a^{mk}b^{nl}=b^{nl}a^{mk}\qquad(**)$$

Let's prove $(*)$ and $(**)$:

$$(a^mb^n)^{mk}=a^m(b^na^m)^{mk-1}b^n=a^m(b^na^m)^{mk}a^{-m}\Rightarrow(a^mb^n)^{mk}a^m=a^m(b^na^m)^{mk}\Rightarrow a^m(a^mb^n)^{mk}=a^m(b^na^m)^{mk},$$ hence $$(a^mb^n)^{mk}=(b^na^m)^{mk}.\qquad(\sharp)$$

Similarly we get $$(a^mb^n)^{nl}=(b^na^m)^{nl}.\qquad(\sharp\sharp)$$

Multiplying $(\sharp)$ and $(\sharp\sharp)$ we obtain $a^mb^n=b^na^m$.

Proceeding in the same way we get $a^nb^m=b^ma^n$

An easy induction shows now $(*)$ and $(**)$.

As $\gcd(m, n) = 1$, by Bezóut's identity there are $u$, $v$ such that $u m + v n = 1$.

Take $a, b \in G$ arbitrary. By the conditions, as $a^u \in G$ for all $u \in \mathbb{Z}$, and as the conditions allow to conmute certain powers: \begin{align*} a^{u m} b^{u m} &= b^{u m} a^{u m} \\ (a^{u m} b^{u m})^{v n} &= a^{u m + v n} b^{u m + v n} \\ &= a b \\ (b^{u m} a^{u m})^{v n} &= b^{u m + v n} a^{u m + v n} \\ &= b a \end{align*} This because e.g. $(a^{u m} b^{u m})^2 = a^{u m} b^{u m} a^{u m} b^{u m} = a^{u m} a^{u m} b^{u m} b^{u m} = a^{2 u m} b^{2 u m}$ since the middle pair conmutes by the conditions. Now use induction.

(Perhaps the switching around can be expressed in a more compact way).

Nice problem!

• Great Solution :) thanx :) – Fawzy Hegab Mar 10 '13 at 19:52
• @vonbrand:why we have $(b^{u m} a^{u m})^{v n} = b^{u m + v n} a^{u m + v n}$? – M.H Mar 10 '13 at 20:30
• @MaisamHedyelloo , we have a in G , so $a^u$ is in G , choose u to be the same u in the relation $mu+vn=1$ , apply the hyposetes to $a_1$ = $a^u$ and use induction as it's mentioned in the answer in the speicial case for 2 . – Fawzy Hegab Mar 10 '13 at 20:45
• How do you get $(a^{u m} b^{u m})^{v n} = a^{u m + v n} b^{u m + v n}$? Doesn't $(a^{u m} b^{u m})^{v n} = a^{u m v n} b^{u m v n}$? – Code-Guru Mar 10 '13 at 22:11
• @MrWhy Huh? What's $a_1$? – Code-Guru Mar 10 '13 at 22:15


# Die fixed so it can't roll the same number twice in a row, using markov chains?

Studying for probability test and the following question came up:

A six sided die is 'fixed' so that it cannot roll the same number twice consecutively. The other 5 sides each show up with a probability $\frac{1}{5}$. Calculate

P($X_{n+1} = 5 \mid X_1 = 5$) and P($X_{n+1} = 1 \mid X_1 = 5$)

What happens as $n \rightarrow$ $\infty$?

It appears to be a markov chain problem but all I can think to do is to find the eigenvalues of the transition matrix. This seems unfeasible given that it's 6x6. My guess for the second part is that the probability tends to 1/6, as the first value becomes less and less relevant.

• Scalar product of two [0,0,0,0,1,0] vectors with (for Fabios $P$ matrix): $P^n$ as gram matrix defining the scalar product. en.wikipedia.org/wiki/Gramian_matrix – mathreadler Apr 2 '17 at 0:01
• I don't remember the last time I saw a question that attracted so many good-quality answers, all different. Thanks for posting. – MJD Apr 2 '17 at 15:01

Collapse the six states into the two states you actually care about: "$5$" and "not-$5$". There is a well-defined transition probability from each of these to the other, so we can now work with a $2 \times 2$ matrix instead of a $6 \times 6$ matrix.

This is a very common trick with Markov chains. The hard part is making sure that you don't lose any information by grouping the states together. (Here, if we had a different probability of going $1 \to 5$ than of going $2 \to 5$, this wouldn't work.)

• Of course! Then the second probability can be calculated as 1/5 the probability of 'not 5', and it is clear that my guess about $n \rightarrow \infty$ is correct. – simples123 Apr 1 '17 at 16:58

Excellent suggestion by @Misha. Since $6$ is a small number, you may want to numerically explore this problem by computing the powers of the transition probability matrix.

$$P = \begin{bmatrix} 0 & 1/5 & 1/5 & 1/5 & 1/5 & 1/5 \\ 1/5 & 0 &1/5 & 1/5 & 1/5 & 1/5 \\ 1/5 & 1/5 & 0 &1/5 & 1/5 & 1/5 \\ 1/5 & 1/5 & 1/5 & 0 & 1/5 & 1/5 \\ 1/5 & 1/5 & 1/5 & 1/5 & 0 & 1/5 \\ 1/5 & 1/5 & 1/5 & 1/5 & 1/5 & 0 \end{bmatrix} \enspace.$$

You'll see that $P^k$ quickly converges to a matrix where all coefficients are $1/6$. In fact, $P^5$ is already quite close. Of course, you can also repeat the experiment with the reduced chain with two states.

You can also find the steady-state probabilities (they are well-defined in this case) by solving:

$$\begin{bmatrix} -1 & 1/5 & 1/5 & 1/5 & 1/5 & 1/5 \\ 1/5 & -1 &1/5 & 1/5 & 1/5 & 1/5 \\ 1/5 & 1/5 & -1 &1/5 & 1/5 & 1/5 \\ 1/5 & 1/5 & 1/5 & -1 & 1/5 & 1/5 \\ 1/5 & 1/5 & 1/5 & 1/5 & -1 & 1/5 \\ 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} p_1 \\ p_2 \\ p_3 \\ p_4 \\ p_5 \\ p_6 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \enspace.$$

It comes as no surprise that the solution is $[1/6, 1/6, 1/6, 1/6, 1/6, 1/6]^T$. Once again, the smart way to solve this problem is the one suggested by Misha, but sometimes numerical explorations reinforce our grasp of a concept.

• Isn't your last equation missing a factor (what you want to solve for)? – Paŭlo Ebermann Apr 2 '17 at 18:13
• @PaŭloEbermann Yes, thanks for the heads-up! – Fabio Somenzi Apr 2 '17 at 19:06

Another approach to solving this is with simple recurrence.

$$P(X_{n+1}=5)=\frac{P(X_n\neq5)}5$$and $$P(X_n\neq5)=1-P(X_n=5)$$

So if we let $a_n=P(X_n=5)$, then

$$a_{n+1}=\frac{1-a_n}5$$

We know that $a_1=1$. We can also see easily that the stable state is $a=(1-a)/5$, or $a=1/6$. Letting $b_n=a_n-\frac16$, we have $$b_{n+1}+\frac16=\frac{1-b_n-\frac16}5$$ or $$b_{n+1}=-\frac{b_n}5$$ which is easily solved as $$b_n=C\left(-\frac15\right)^n$$ Noting that $b_1=\frac56$, we get $$a_n=\frac16-\frac16\left(-\frac15\right)^{n-2}$$

• Cool answer, +1 – IamThat Apr 4 '17 at 18:14

Misha’s answer is the best approach to this problem, but finding the eigenvalues of the full $6\times6$ transition matrix is actually very easy because of its special structure.

Observe that we can write the transition matrix as $M=\frac15(\mathbb1_6-I_6)$, where $\mathbb 1_6$ is the $6\times6$ matrix consisting entirely of $1$s. So, if $\mathbf v$ is an eigenvector of $\mathbb1_6$ with eigenvalue $\lambda$, then $M\mathbf v=\frac15(\mathbb1_6\mathbf v-\mathbf v)=\frac15(\lambda-1)\mathbf v$, which means that $\frac15(\lambda-1)$ is an eigenvalue of $M$. The rank of $\mathbb1_6$ is obviously one, so $0$ is an eigenvalue with multiplicity 5, and the other eigenvalue is $\operatorname{tr}\mathbb1_6=6$. Therefore, the eigenvalues of $M$ are $1$ and $-\frac15$.

Of course, we already knew that one of the eigenvalues of $M$ is $1$ with right eigenvector $(1,1,1,1,1,1)^T$ because it’s row-stochastic, and that’s the only eigenvalue we care about for computing the steady-state distribution. The corresponding left eigenvector can be found by computing the kernel of $M^T-I$ or by observing that by symmetry, its left and right eigenvectors are transposes, therefore a left eigenvector of $1$ is $(1,1,1,1,1,1)$, which after normalizing it so that its entries sum to unity gives the result you expected. Computing the $n+1$ step probabilities is a bit messy, but again, because $M$ is symmetric, the eigenspace of $-\frac15$ is the orthogonal complement of the span of this vector, so it’s pretty easy to diagonalize the matrix. You might also observe that if you subtract any two columns from each other, you get a vector with $\pm\frac15$ in the elements corresponding to the column numbers and zero everywhere else, so vectors with a single $1$ and $-1$ are all eigenvectors of $-\frac15$. It’s not hard to come up with an eigenbasis once you’ve noticed this.

There are two great answers already, but just for completeness I'll mention that the transition matrix $P$ (as written by @FabioSomenzi) is a circulant matrix, and its eigenvectors and eigenvalues can be written explicitly. The eigenvector matrix is the DFT matrix, in fact. Circulant matrices are sometimes covered in matrix analysis / numerical linear algebra courses (although probably not at the introductory level).

An alternative approach for the part about $n \to \infty$ that does not require any calculations:

There is no probabilistic difference between the numbers $1,2,\ldots,6$ on the die. So, since the Markov chain is clearly aperiodic and has a finite number of states, we can conclude by symmetry that

$$\lim_{n \to \infty} \mathbb{P}(X_n = i \mid X_1 = j) = \frac{1}{6}.$$

• Good catch, I've edited the answer. – Ritz Apr 6 '17 at 15:35

I have a similar question.

Dice is rolled and lands on six. A. Probability dice lands on six after 2 more rolls (ASSUME 20%)

B. Probability dice lands on six after 3 more rolls


# Evaluating the boolean sum $\sum_{x_1, x_2, x_3, x_4, x_6, x_7} \neg(x_1 \oplus x_4 \oplus x_3 \oplus x_6) \neg(x_4 \oplus x_3 \oplus x_2 \oplus x_7)$

Say that we have the following boolean function function ($x_{i} \in \{0,1\}$):

$$f = \sum_{x_1, x_2, x_3, x_4, x_6, x_7 \in \{0,1\}^6} \neg(x_1 \oplus x_4 \oplus x_3 \oplus x_6) \land \neg(x_4 \oplus x_3 \oplus x_2 \oplus x_7)$$

I was trying to evaluate that function.

I realized that the way to evaluate it is by counting how many times both negation functions for the xors are 1 however, I have had a hard time counting in a cleaver way because of the over lap of variables (if we even had more overlaps and more negations being multiplied, I didn't really see how to find a general argument without brute force).

If we only had 4 terms as in:

$$\sum_{x_1, x_4, x_3, x_6 \in \{0,1\}^4 } \neg(x_1 \oplus x_4 \oplus x_3 \oplus x_6)$$

the way I thought of doing this was that the negation function is only 1 when the inside is zero. The inside is zero only when either all $x_i$s are zero or 1, or when pairs of them are 0 or 1. So I thought, there are only 2 values for the whole lot to be the same value, then there are $\binom {4}{2}$ pairs of these. Each pair could take 2 values, either 1 or zero. Therefore yielding the following count:

$$\binom {4}{2} 2^2 + \binom{4}{4}2 = \binom {4}{2} 2^2 + 2$$

Which I think is correct.

However, when there is a intersection of variables, how do you count this? Do you use the inclusion exclusion principle? I am a little stuck, not sure how to generalize this

• On the LHS you have variables but on the RHS you are summing over those same variables. Which is it? Are you summing over all tuples? – Alexander Vlasev Nov 12 '14 at 1:57
• variables only take values 1 or 0. Not sure if I understand the confusion, I updated my question though. – Pinocchio Nov 12 '14 at 2:06
• The $x_i$'s are not variables for the function $f$ because you can't just plug in some values. So your expression should read as something like $f = \sum_{x_1,\ldots}\cdots$. – Alexander Vlasev Nov 12 '14 at 2:16
• @AleksVlasev your right, duh, sorry. – Pinocchio Nov 12 '14 at 2:18

I suggest let $a = x_3 \oplus x_4$, $b = x_1 \oplus x_6$, and $c = x_2 \oplus x_7$

then

$$\sum_{x_1, x_2, x_3, x_4, x_6, x_7} \neg(x_1 \oplus x_4 \oplus x_3 \oplus x_6) \land \neg(x_4 \oplus x_3 \oplus x_2 \oplus x_7)$$

becomes

$$2^3 \sum_{a,b,c} \neg(a \oplus b) \land \neg(a \oplus c)$$

Now apply the observation that binary $\oplus$ is just another way of writing $\ne$, so you get:

$$2^3 \sum_{a,b,c} \neg(a \ne b) \land \neg(a \ne c)$$ $$2^3 \sum_{a,b,c} a = b \land a = c$$ $$2^3 \times 2$$

where did you get your $2^3$ at the front

There are 4 values of $\{x_3, x_4\}$. Each value of $x_3 \oplus x_4$ occurs twice. So:

$$\sum_{x_3, x_4} f(x_3 \oplus x_4) = f(1 \oplus 1) + f(1 \oplus 0) + f(0 \oplus 1) + f(0 \oplus 0) = 2f(1) + 2f(0) = 2 \sum_a f(a)$$

Do this a total of 3 times and you get $2^3$.

can you provide details of how you got that 2 on the last equation?

$\sum_{a,b,c} a = b \land a = c$ is asking the question, "how many ways can a,b and c all be equal?". In two ways: $a = b = c = 0$ and $a = b = c = 1$.

• where did you get your $2^3$ at the front and can you provide details of how you got that 2 on the last equation? – Pinocchio Nov 12 '14 at 2:57
• @Pinocchio Added details – DanielV Nov 12 '14 at 3:43


# Math Help - [SOLVED] Find the area of rectangle ABCD

1. ## [SOLVED] Find the area of rectangle ABCD

Hi, this is on the SAT OG Book page 747 #16

In rectangle ABCD, point E is the midpoint of line segment BC. If the area of quadrilateral ABED is $\frac{2}{3}$, what is the area of the rectangle ABCD?

a) $\frac{1}{2}$

b) $\frac{3}{4}$

c) $\frac{8}{9}$

d) 1

e) $\frac{8}{3}$

2. Originally Posted by fabxx
Hi, this is on the SAT OG Book page 747 #16

In rectangle ABCD, point E is the midpoint of line segment BC. If the area of quadrilateral ABED is $\frac{2}{3}$, what is the area of the rectangle ABCD?

a) $\frac{1}{2}$

b) $\frac{3}{4}$

c) $\frac{8}{9}$

d) 1

e) $\frac{8}{3}$
note that the area of rectangle ABCD is given by AD*AB

now, ABED is a trapezium, thus its area is given by half the sum of the two parallel sides times the distance between them. that is (1/2)(BE + AD)*AB, and this is 2/3, so

(1/2)(BE + AD)*AB = 2/3

=> (BE + AD)*AB = 4/3

but, BE = (1/2)AD (this should be pretty obvious from a diagram if you drew or were given one), so that

=> (3/2)AD*AB = 4/3

=> AD*AB = 8/9

3. Hello, fabxx!

Did you make a sketch?

In rectangle ABCD, point E is the midpoint of line segment BC.
If the area of $ABED$ is $\frac{2}{3}$, what is the area of $ABCD$?

$(a)\;\frac{1}{2} \qquad(b)\;\frac{3}{4}\qquad (c)\;\frac{8}{9} \qquad (d)\;1 \qquad (e) \;\frac{8}{3}$
Code:
    D *---------------* C
|   *           |
|       *       |
|           *   |
F * - - - - - - - * E
|               |
|               |
|               |
A *---------------* B

Draw median EF.

$\text{Rect }DCEF \:=\:\frac{1}{2}(\text{Rect }ABCD)$

$\Delta DCE \:=\:\frac{1}{2}(\text{Rect }DCEF) \;=\;\frac{1}{4}(\text{Rect }ABCD)$

. . Hence: . $\text{Quad }ABED \:=\:\frac{3}{4}(\text{Rect }ABCD)$

We are told that: . $\text{Quad }ABED \:=\:\frac{2}{3}$

We have: . $\frac{3}{4}(\text{Rect }ABCD) \;=\;\frac{2}{3} \quad\Rightarrow\quad \boxed{\text{Rect }ABCD \:=\:\frac{8}{9}}$ . . . answer (c)

4. Originally Posted by Jhevon
note that the area of rectangle ABCD is given by AD*AB

now, ABED is a trapezium, thus its area is given by half the sum of the two parallel sides times the distance between them. that is (1/2)(BE + AD)*AB, and this is 2/3, so

(1/2)(BE + AD)*AB = 2/3

=> (BE + AD)*AB = 4/3

but, BE = (1/2)AD (this should be pretty obvious from a diagram if you drew or were given one), so that

=> (3/2)AD*AB = 4/3

=> AD*AB = 8/9
No figure is given for this. If i were to draw one myself, would it look like the attachment below?

5. Originally Posted by fabxx
No figure is given for this. If i were to draw one myself, would it look like the attachment below?
yes, it should look like that. note that it is the same as the diagram Soroban drew, except it is turned on its side. i imagined it as Soroban did. it will all work out the same. if your diagram was like the given one or Soroban's you will get the right answer