# Is the limit of this infinite step construction an equilateral triangle?

Just for fun (inspired by sub-problem described and answered here):

Let's pick three points on a circle, say $A,B,C$. Move one point ($A$ for example) until the triangle becomes isosceles ($A'BC$) with all angles acute:

Now we have triangle with sides $AB$ and $AC$ equal. Pick any of the two, say $AC$ and move $B$ until the triangle becomes isosceles again, with all angles acute:

Now we have a triangle with sides $AB$ and $BC$ equal. Pick any of the two, say $BC$ and move $A$ until the triangle becomes isosceles again, with all angles acute:

Repeate the same process infinite number of times.

Can we prove that the end result is always an equilateral triangle? It looks so but I might be wrong. I have checked several initial configurations and always ended up with something looking like an equilateral triangle.

• Regarding the title question, here's a more complicated procedure: 1. Prove whether P=NP or not. 2. Draw an equilateral triangle with compass and straightedge (or any other method). – user253751 Sep 18 '18 at 1:32
• @immibis Very funny :) – Oldboy Sep 18 '18 at 5:08
• @user202729 Title changed as suggested. Thanks! – Oldboy Sep 18 '18 at 5:09
• @immibis On second thought, your procedure is not more complicated. Assuming P=NP is true or not the proof will always take finite time. And my “construction” takes infinite number of steps, each step taking constant time. Just kidding :) – Oldboy Sep 18 '18 at 21:39
• However the steps are all the same, which make it pretty simple :) – user253751 Sep 18 '18 at 22:57

Think about what happens to the maximum difference between angles over time.

For simplicity, let's start with an isoceles triangle with angles $x,y,y$. This triangle has "maximum angle difference" $\vert y-x\vert$. Then when we move one of the $y$-angled points, our new triangle will have angles

$$y, {x+y\over 2}, {x+y\over 2}$$

since the angle of the point being moved doesn't change.The maximum difference of angles in this new triangle is

$$\left\vert {y\over 2}-{x\over 2}\right\vert={1\over 2}\vert y-x\vert.$$

So each time we perform this transformation, the maximum angle difference goes down by a factor of two. Whatever the initial value $\vert y-x\vert$ was, this means that the maximum angle difference goes to zero,$^*$ which in turn means that in the limit the angles are equal.

$^*$This is because it's a geometric sequence with ratio in $(-1,1)$ (namely, ${1\over 2}$): if $r\in(-1,1)$ then for any $a$ we have

$$\lim_{n\rightarrow\infty}ar^n=0.$$

Note that it would not have been enough to simply know that the maximum angle difference decreases, since not every decreasing sequence goes to zero!

• I have also noticed that the difference between angles decreases. But having a decreasing sequence does not mean that the sequence converges to zero. You have proved the key point! – Oldboy Sep 18 '18 at 5:18
• Angles $x$ and $y$? Aaah mine eyes! – user332714 Sep 18 '18 at 7:08

By the Inscribed Angle Theorem, moving a vertex around the circle preserves the angle at that vertex. Now, suppose that, at stage $i$, the apex angle is $\theta_i$, so that the base angles are $\frac12(\pi - \theta_i)$. But this apex angle was the base angle of the previous step, giving this recurrence $\theta_{i} = \frac12(\pi-\theta_{i-1})$. Thus, \begin{align}\theta_n &= -\frac12\theta_{n-1} + \frac12\pi \\[6pt] &=\frac12\left(-\frac12(\pi-\theta_{n-2})+\pi\right) = \frac14\theta_{n-2}+\frac12\pi-\frac14\pi \\[6pt] &= \cdots \\[6pt] &= \left(-\frac12\right)^{n}\theta_0 \;-\; \sum_{i=1}^n\left(-\frac12\right)^{n}\pi \\[6pt] \lim_{n\to\infty}\theta_n &= 0\cdot\theta_0 \;-\; \frac{(-1/2)}{1-(-1/2)}\pi \\ &=\frac{\pi}{3} \end{align}

Thus, in the limit, the triangle becomes equilateral. $\square$

Assume WLOG that the initial triangle is isoceles. Let $\alpha$ be the apical angle, and let $\beta$ be a remaining angle. Then the transformation in question sends

$$\begin{bmatrix}\alpha \\ \beta \end{bmatrix}\mapsto \begin{bmatrix}0 & 1 \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix} \begin{bmatrix}\alpha \\ \beta \end{bmatrix}\text{.}$$ Let $\mathsf{X}$ be the $2\times 2$ transformation matrix on the rhs. $\mathsf{X}$ has characteristic polynomial $x^2-\tfrac{1}{2}x-\tfrac{1}{2}=0.$ By the Cayley–Hamilton theorem, $$\mathsf{X}^2=\tfrac{1}{2}\mathsf{X}+\tfrac{1}{2}\text{.}$$ Therefore we have a Sylvester formula $$f(\mathsf{X})=f(1)\left(\frac{1+2\mathsf{X}}{3}\right)+f(-\tfrac{1}{2})\left(\frac{2-2\mathsf{X}}{3}\right)$$ for any polynomial $f$ (where the matrices in brackets are the Frobenius covariants). Thus, $$\mathsf{X}^n=\frac{1+2\mathsf{X}}{3}+(-\tfrac{1}{2})^n\left(\frac{2-2\mathsf{X}}{3}\right)\text{.}$$ The second term converges to zero, so $$\begin{split} \lim_{n\to\infty}\mathsf{X}^n&=\frac{1+2\mathsf{X}}{3}\\ &=\frac{1}{3}\begin{bmatrix}1 & 2 \\ 1 & 2\end{bmatrix}\\ &=\frac{1}{3}\begin{bmatrix} 1\\ 1\end{bmatrix}\begin{bmatrix}1&2\end{bmatrix}\text{,} \end{split}$$ $$\lim_{n\to\infty} \begin{bmatrix}0 & 1 \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix}^n \begin{bmatrix}\alpha \\ \beta \end{bmatrix}=\begin{bmatrix}\tfrac{\alpha+2\beta}{3}\\ \tfrac{\alpha+2\beta}{3}\end{bmatrix}\text{.}$$ i.e., the apical and side angles approach equality as the operation is repeated.


# Homework Help: Definite integral of step function

Tags:
1. Jun 18, 2015

### Byeonggon Lee

I need to prove whether this expression is true or false:

$\sum\limits_{k=1}^{n}\int_{k-1}^{k}[x]dx = \frac{n(n-1)}{2}$

I'm so confused because as I know, definite integral is possible only when the target function is continuous in closed interval.

In this case, function $[x]$ should be continuous in interval $[k-1,k]$
but actually [x] is not continuous in $[k,k-1]$
$[x]=k-1$ when $k-1\leq x<k$, $[x]=k$ when $x=k$

So I thought that this expression is false,
but in my book's answer, it is true:

"""
Because$[x]=k-1$ when $k-1\leq x<k$, $[x]=k$ when $x=k$
$\sum\limits_{k=1}^{n}\int_{k-1}^{k}[x]dx$
$=\int_{0}^{1}[x]dx+\int_{1}^{2}[x]dx+\int_{2}^{3}[x]dx+...+\int_{n-1}^{n}[x]dx$
$=\int_{0}^{1}0dx+\int_{1}^{2}1dx+\int_{2}^{3}2dx+...+\int_{n-1}^{n}(n-1)dx$
$=0+\bigg[x\bigg]_{1}^{2}+\bigg[2x\bigg]_{2}^{3}+...+\bigg[(n-1)x\bigg]_{n-1}^{n}$
$=0+1+2+...+(n-1)$
$=\frac{n(n-1)}{2}$
"""

The book also says that I need to think interval as $k\leq x<k+1$
and I don't understand how is it possible to omit equal sign at $x<k+1$

2. Jun 18, 2015

### DEvens

An integral is an area. The troublesome points at the end of the interval mean you are uncertain about the area at a point. What is the difference in the area under a curve if you move one single point up or down by 1?

Another way to approach it is this. Think of the integral as a limit of a sum of areas. It happens that in this case you can very easily construct the limit because the function you are looking at is piece-wise constant.

3. Jun 18, 2015

### Ray Vickson

A definite integral certainly can exist for a discontinuous function. Jump discontinuities in f(x) at an interval's endpoint a or b does not affect the area under the curve y = f(x) from x = a to x = b. In other words,
$$\int_{[a,b]} f(x) \, dx = \int_{(a,b]} f(x) \, dx = \int_{[a,b)} f(x) \, dx = \int_{(a,b)} f(x) \, dx$$
We can denote all four of these by the common symbol $\int_a^b f(x) \, dx$.

4. Jun 18, 2015

### RUber

You can read a little more on this idea at https://en.wikipedia.org/?title=Riemann_integral.
For example, to your point on continuity:
"A function on a compact interval [a, b] is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure)."
Set of measure zero means a finite (or countably infinite) set of distinct points. In the case of the step function is discontinuous at the integers which for a maximum n is finite, and without a maximum is countably infinite. Both sets are measure zero, so you are okay to integrate.

5. Jun 18, 2015

### WWGD

Just to nitpick, or to add a bit: while in this context of step functions discontinuities are finite, you may have in other cases uncountably-infinite sets of measure zero, e.g., the Cantor set.

6. Jun 18, 2015

### SammyS

Staff Emeritus
I believe your function is also called the floor function, $\displaystyle \ \lfloor x\rfloor \,,\$ and the greatest integer function, $\displaystyle \ [\![ x]\!]\ .\$

On the interval $\displaystyle \ [k-1\,,\ k)\,,\$ we have $\displaystyle \ [x]=k-1\ .$

Take the limit: $\displaystyle \ \lim_{a\to k^-} \int_{k-1}^a [x]\,dx \ . \$


# $f: \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2)=5$

The function $f : \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2) = 5$.

• What math contest is this from? Apr 20, 2013 at 17:38
• @Potato: KöMaL, Problem B. 4538. Apr 28, 2013 at 14:44
• Another example of a question that should have been closed before anybody answered it! Apr 30, 2013 at 11:03

Hint 1: $3=(x+1)-(x-2)$

$(x-2)f(x)-(x+1)f(x-1)=3\Leftrightarrow (x-2)f(x)-(x+1)f(x-1)=(x+1)-(x-2)\Leftrightarrow (x-2)(f(x)+1)-(x+1)(f(x-1)+1)=0$

Hint 2: Is there a function closely related to $f$ that would verify a simpler equation?

$g(x)=f(x)+1$

$(x-2)g(x) - (x+1)g(x-1)=0 \Leftrightarrow (x-2)g(x)=(x+1)g(x-1) \Leftrightarrow g(x)=\cfrac{x+1}{x-2}g(x-1)$

$g(x)=\cfrac{x+1}{x-2}g(x-1)=\cfrac{x+1}{x-2}\cfrac{x+1-1}{x-2-1}g(x-1-1) = \left(\prod\limits_{k=0}^{n-1} \cfrac{x+1-k}{x-2-k}\right) g(x-n)$

Now give the good value to $n$, simplify the product and you'll have the expression of $g$ you need.

• Yeah right. Some terms cancel each other >_< Apr 20, 2013 at 20:42
• What is "the good value"? Apr 28, 2013 at 14:31
• The value sot hat you can evaluate $g(x-n)$ Apr 28, 2013 at 14:39
• Big fan of answers with this format +1 May 1, 2013 at 3:26
• Same here, format+1. May 5, 2013 at 1:11

Let $x-1=y$, then, $$(y-1)f(y+1)-(y+2)f(y)=3$$ $$\implies f(y+1)=\frac{3+(y+2)f(y)}{y-1} \;\;\;\;\;(1)$$

Lemma: $\forall \; n \geq 2 \in \mathbb{N}$, $f(n)=n(n-1)(n+1)-1$.

Base Case: If $n=2$ then $f(n)=1 \cdot 2 \cdot 3 -1=5$ which is true by information provided in the question.

Inductive Step: Assume for $n=k$ that $f(k)=k(k-1)(k+1)-1$.

Then by equation $(1)$, $$f(k+1)=\frac{3+(k+2)(k(k-1)(k+1)-1)}{k-1}$$ $$=\frac{k^4+2k^3-k^2-3k+1}{k-1}$$ $$=\frac{(k-1)(k^3+3k^2+2k-1)}{k-1}$$ $$=k^3+3k^2+2k-1$$ $$=k(k^2+3k+2)-1$$ $$=k(k+1)(k+2)-1$$

Thus completing the induction.

Hence our Lemma is true and $f(n)=n(n-1)(n+1)-1 \; \forall \; n \geq 2 \in \mathbb{N}$

In particular, if $n=2013$, $f(2013)=2013 \cdot 2012 \cdot 2014-1=8157014183$ (I confess I used a calculator).

P.S I found the lemma by trying small cases.

The conditions allow you to calculate $f(x+1)$ if you know $f(x)$. Try calculating $f(3), f(4), f(5), f(6)$, and looking for a pattern.

Running the following Mathematica program:

f = DifferenceRoot[Function[{f, x}, {(x - 2) f[x] - (x + 1) f[x - 1] == 3, f[2] == 5}]];
f[2013]

We get the answer = 8157014183.


Which of the following complex numbers is equivalent to $\frac{3-5i}{8+2i}$?

I encountered this question in one of my SAT practice tests.

I know the answer is option C, however the only way I got the answer was by trial and error of trying multiple ways to simplify the equation and I ended up rationalising it to get answer choice C.

Is there any other, perhaps easier or more direct method, that I can use to solve these types of questions?

• Mutiplying top and bottom by the conjugate of the denominator is the standard approach. No need for trial and error -- the standard method will always work. – quasi Mar 29 '17 at 6:36
• I ended up rationalising it Don't know that there is anything more direct than that. You don't even need to carry out all calculations, just figure out the sign of the imaginary part to decide between C and D. (That's assuming you discarded choices A, B upfront, as expected) – dxiv Mar 29 '17 at 6:39
• As a multiple choice question, there will be a minus sign, but A is likely to be wrong – Henry Mar 29 '17 at 6:45
• As a multiple choice question ,you can compare $|z|=|\dfrac{3-5i}{8+2i}|=\dfrac{|3-5i|}{|8+2i|}=\dfrac{\sqrt{2(17)}}{\sqrt{4(17)}}=\dfrac{\sqrt2}{2}$ – Khosrotash Mar 29 '17 at 6:56
• @Khosrotash: interesting proposal, but unfortunately it leads to harder computation than the standard way (because of $(7^2+23^2)/34^2$) :-( – Yves Daoust Mar 29 '17 at 7:10

Two ways to approach this problem. First: As quasi suggests in the comments, multiply by the conjugate of the denominator.

\begin{align} \frac{3-5i}{8+2i} & = \frac{3-5i}{8+2i} \times \frac{8-2i}{8-2i} \\ & = \frac{24-40i-6i-10}{8^2+2^2} \\ & = \frac{14-46i}{68} = \frac{7-23i}{34} \end{align}

The second approach, given that it's a multiple choice problem, is to multiply each of the answers by $8+2i$ and see if you obtain $3-5i$. I think the first approach is simpler, but they'll both work.

Multiplying the top and bottom by the complex conjugate is how you handle this. That gives:

$$\frac{3-5i}{8+2i}\cdot\frac{8-2i}{8-2i}=\frac{(3-5i)(8-2i)}{8^2+2^2}=\frac{7-23i}{34}$$

Notice that at the second step we are guaranteed to have a real denominator because $(a+bi)(a-bi)=a^2+b^2$ is always real.

I think the method you are using easier one.

$$\frac{3-5i}{8+2i}$$

Multiply and divide the fraction by conjugate of denominator,

$$\frac{3-5i}{8+2i} \cdot \frac{8-2i}{8-2i}$$

$$\frac{24-6i-40i+10i^2}{64-4i^2}$$

$$\frac{24-6i-40i-10}{64+4}$$

$$\frac{14-46i}{68}$$

$$\frac{7-23i}{34}$$

Alternatives:

Let the answer be $a+ib$. Rewrite the initial equation as $$(a+ib)(8+i2)=3-i5.$$

Expanding, you get the $2\times2$ system $$\begin{cases} 8a-2b=3,\\ 2a+8b=-5.\end{cases}$$

Then by Cramer or simply adding four times the first equation and the second

$$34a=7$$

and subtracting the first from four times the second,

$$34b=-23.$$

Get rid of the denominators and try the products

$$(3\pm i20)(8+i2)=24\mp40+i(6\pm160)\to64-i154,$$ $$(7\pm i23)(8+i2)=56\mp46+i(14\pm144)\to102-i170.$$

(We select the signs that match those of $3-i5$.)

Then we obtain the identity

$$(7-i23)(8+i2)=34(3-i5).$$

The real way:

Use the division formula

$$\frac{a+ib}{c+id}=\frac{ac+bd}{c^2+d^2}+i\frac{bc-ad}{c^2+d^2},$$

giving

$$\frac{14}{68}-i\frac{46}{68}.$$

Then compare to the proposed answers. As C seems to match but D is similar, double check the signs or try the product, for safety.


# Regular average calculated accumulatively

is it possible to calculate the regular average of a sequence of numbers when i dont know everything of the sequence, but just everytime i get a new number i know the total count of numbers and the average for the numbers - 1.

for example: 2 3 10 the average is of course: 5

but in the last step to calculate i only have access to the previous average of 2 and 3: 2.5 the next number: 10 and the count of numbers: 3

if this is possible, how?

Yes, and you can derive it from the expression for the average. Let the average of the first $n$ numbers be $\mu_n$. The formula for it is

$$\mu_n = \frac{1}{n} \sum_{i=1}^n x_i$$

Then you can derive

$$n \mu_n = \sum_{i=1}^nx_i = x_n + \sum_{i=1}^{n-1} x_i = x_n + (n-1)\mu_{n-1}$$

and hence, dividing by $n$,

$$\mu_n = \frac{(n-1) \mu_{n-1} + x_n}{n}$$

i.e. to calculate the new average after then $n$th number, you multiply the old average by $n-1$, add the new number, and divide the total by $n$.

In your example, you have the old average of 2.5 and the third number is 10. So you multiply 2.5 by 2 (to get 5), add 10 (to get 15) and divide by 3 (to get 5, which is the correct average).

Note that this is functionally equivalent to keeping a running sum of all the numbers you've seen so far, and dividing by $n$ to get the average whenever you want it (although, from an implementation point of view, it may be better to compute the average as you go using the formula I gave above. For example, if the running sum ever gets larger than $10^{308}$ish then it may be too large to represent as a standard floating point number, even though the average can be represented).

• As you're computing the previous sum $(n-1)\mu_{n-1}$ as part of the formula, I think this wouldn't help is the sum gets too large. Jun 23, 2016 at 1:12
• @danijar Completely true - a better approach is to keep running sums of the $x_n$ using a method that is robust to rounding error (e.g. Kahan summation) and a separate running count of $n$, and divide whenever you need the mean. That way, your error is bounded by the accuracy of your floating point type. Jun 23, 2016 at 7:44
• If you are likely to end up with numbers larger than $10^{308}$ then you either need to scale down your inputs, or use a more capacious floating point type. Jun 23, 2016 at 7:46

A very simple thought process results in the same formula for running average. If you have $N$ previous measures (of course the measures could all be different) the average you calculate is exactly the same as if all measures were the same as the computed average value. Then, computing the running average of the $N+1$ is equal to $N$ times the previously computed average plus the $N+1$ measure all divided by $N+1$. I know that this is the same as the formula posted in the other answer but no derivation with sums is needed or more obscure mathematical thinking is needed (OK, maybe not really obscure).

What you are asking for is commonly called sequential estimation. A general approach is described in [Robbins, H. and S. Monro (1951). A stochastic approximation method. Annals of Mathematical Statistics 22, 400–407.]

To add to the derivation of Chris Taylor, I personally like this rewriting as it goes quite intuitively (easy to remember). $$\mu_n = \mu_{n-1} + \frac{1}{n}(x_n - \mu_{n-1})$$

# algorithmically: sequential average computation
avg += (x_n - avg)/n


## Monday, July 27, 2009

### The Broken Stick Experiment

I found several interesting video lectures at the BLOSSOMS Video Library. One in particular that I enjoyed is called The Broken Stick Experiment [Flash]. In it, Professor Richard Larson asks the seemingly simple question,
If you break a stick randomly in two places, what is the probability that you can form a triangle from the three pieces?
If one of the pieces is too long, then the other two can't meet to form the triangle.
The lecture is only 33 minutes, and Professor Larson builds up a very pretty geometric proof, so I won't spoil it by posting the answer here. I will provide code for a simulation, though (yes, this is still a programming blog). See if you can guess the answer before running the simulation or watching the video.
import java.util.Arrays;import java.util.Random;public class BrokenStick {   public static void main(String[] args) {       Random random = new Random();       int trials = 10000;       int triangles = 0;       double[] breaks = new double[2];       double[] sides = new double[3];       for( int i=0; i < trials; ++i )       {           breaks[0] = random.nextDouble();           breaks[1] = random.nextDouble();           Arrays.sort(breaks);           sides[0] = breaks[0];           sides[1] = breaks[1] - breaks[0];           sides[2] = 1.0 - breaks[1];           Arrays.sort(sides);                     if( sides[2] < (sides[0] + sides[1]) )           {               triangles++;           }       }       System.out.println("Triangles: " + triangles);       System.out.println("Trials: " + trials);       double p = (double)triangles / trials;       System.out.println("Proportion: " + p);   }}

The simulation breaks 10,000 sticks each into three random pieces using Java's Random class to come up with random lengths. If the longest piece of broken stick is longer than the two remaining pieces, a triangle can't be formed. Ten-thousand trials should be enough to accurately estimate the proportion of triangles made from the broken sticks to two decimal places. How close was your guess?

Related:
The Broken Stick Revisited

rbonvall said...

Thanks, I had a fun time solving it :)

At least to me, this problem lends itself nicely to a geometric solution. In the a-b-c space, the possible breaks must satisfy a+b+c=1, which is a plane. Then, it reduces to find the fraction of the "positive part" of the plane that satisfies the constraints. And I always have fun drawing intersecting planes :P

Bill the Lizard said...

rbonvall,
"And I always have fun drawing intersecting planes :P"

Your comment reminds me of the first lecture (or maybe the second, I can't quite remember) in MIT's Linear Algebra video series. Gilbert Strang shows several methods for solving systems of equations, and makes a complete mess of drawing intersecting planes. He uses that as an illustration of why solving systems of equations using matrices is far superior. :)

John | Retro Programming said...

Thanks, it's an interesting problem. I think I've got the solution, I just need to run a program to double check my result :-)

Alexey Busygin said...

I think it should be 33,3%.

Bill the Lizard said...

Alexey,
It's not 33%, but your initial guess is closer than mine was.

Trevel said...

I estimate at just under 25% triangle. Trying to figure if there's a flaw in my logic, but I can't pick it out...

tg said...

I agree with Trevel. Independent breaks which both must occur on the same half of the stick. <25%

tg said...

Got the right answer, but not sure if the logic is right:

If two breaks occur on the same half of the stick, then on piece is longer than the sum of the other two. These are independent events. The probability of one break on 50% of the stick is 1/2, the probability of two (independent) breaks on the same half is 1/2*1/2 = 0.25.

Bill the Lizard said...

tg,
Wouldn't that indicate that there's a 25% chance of not forming a triangle?

tg said...

back to the drawing board....

Bill the Lizard said...

tg,
Don't feel bad, you almost had me convinced. It took me ten minutes to spot the flaw, and I've watched the video twice! :)

tg said...

How about one more try using the same concept (except using it correctly to represent all three lengths, as I didn't do the first go round):

Probability of NOT forming P(NF) =
P(a > b+c)*P(b>a+c)*P(c>b+a)

As mentioned earlier P(a>b+c) = probability of two breaks on 1/2 of the stick = 0.25.

P(NF) = 0.25*0.25*0.25 = 0.75
Thus P( NOT NF) = 0.25

Bill the Lizard said...

tg,
0.25 * 0.25 * 0.25 = 0.015625

Back to your original statement, though, you said "If two breaks occur on the same half of the stick, then on piece is longer than the sum of the other two." This is correct, but it's only one way that you could get one piece that's longer than the other two. You could also break the stick on opposite ends of the stick and have the middle piece be too long.

tg said...

B the L, you are a patient individual.

One final attempt to redeem myself. The "*" was supposed to be "+". The risks of writing at work.

We are looking for the Probability of (a>b+c) OR (b>a+c) OR (c>a+b).

a + b + c = L
P(a>b+c) = P(a>L/2)

For a>L/2, two breaks must occur on the right half:
P(of a break occurring on the right half) = 0.5
P(of two independent breaks occurring on the right)=
P(a>L/2) = 0.5*0.5 = 0.25

For b>L/2, two independent breaks must occur on either the left quarter, the right quarter, or both breaks on the left quarter, or both on the right quarter. The probability of ONE break landing on that 50% of the stick = 0.5. The probability of another break independently landing on the same 50%of the stick = 0.5*0.5 = 0.25.

c is the same as a, so
P(c>b+a) = P(a>b+c) = 0.25

For any one length > L/2 = (a>b+c) OR (b>a+c) OR (c>a+b) = 0.25+0.25+0.25.

If this is wrong, spare me the indignity and just delete my comments.

Bill the Lizard said...

tg,
Now you have it! I like the approach you used because you're showing the exact opposite of what the problem asks for, the probability of NOT forming a triangle. This kind of thinking is a very powerful tool in mathematics, and is often overlooked.

Sam152 said...

Very interesting post, my nifty c++ program calculated about 18.9% successes after 10,000 tries. How close is this?

Bill the Lizard said...

Sam152,
That's a little bit low; lower than rounding error would explain. Do you want to share your code? You could either post it here, or as a Stack Overflow question and I'd be happy to take a look.

fiberfiend6891 said...

Just a heads up - the link directly to the video has http doubled and doesn't work when clicked directly (I had to copy/paste and get rid of the double):
http://http//blossoms.mit.edu/video/larson-watch.html

This is my first time seeing your blog, and I absolutely LOVE it!

Bill the Lizard said...

fiberfiend6891,

Tom Raywood said...

Bill the Lizard,

Hi. Great question.

There is one thing I'm not clear on, though, from the description. Do obtuse triangles count here, or is the solution limited to acute?

Bill the Lizard said...

Tom,
You could end up with an acute, obtuse, or (rarely) equilateral triangle. You've got me curious now, so I'll have to modify the code later and run another simulation to see if acute or obtuse triangles are more common. Good question!

Tom Raywood said...

Bill the Lizard,

Good. If you limit outcomes to acute triangles, (including equilaterals), what probability do you get?

Oh, and another question [intended to confirm an assumption on my part]: Is this strictly an ordered set or can any of the three pieces form the base?

Bill the Lizard said...

Tom,
Any of the three sides could form the base (I'm assuming by base you mean the longest side). In the code I gave you can see how I work around this by calculating the length of the three sides, then immediately sorting them (on line 24) so that I always know where the longest side ends up.

I'm working on an entire post to answer your other question. :)

Tom Raywood said...

Bill the Lizard,

Gee, no I hadn't imagined that the base of the triangle had to be the largest of the three pieces. All of these unstated parameters make it very difficult to approach the problem in a systematic way.

So far, yes, the triangle can be of any sort, the pieces can be moved around at will and, finally, (if I understand your last answer), the base of the triangle is expected to be whichever piece [of the stick] that's the longest.

Is this correct? Is there anything else in the way of allowances and/or limitations?

Bill the Lizard said...

Tom,
Disregard my last answer, as I had misunderstood your question. There are no constraints on what kind of triangle you can get (acute, obtuse, right, or no triangle at all) or on the order of the pieces. Any of the pieces can be the longest when you randomly select the two break-points. The only reason I sort the pieces in my code is because I need to know which piece is longest, so that I can tell if the pieces form a triangle or not. This has no impact on the final outcome.

Tom Raywood said...

Bill the Lizard,

Whoa, hey, no, it can't be as simple as that can it?

All it takes for the three pieces to form a triangle is that none of the pieces have a length greater than or equal to 1/2 the length of the stick. And that's a full 50% of the time.

What gives? I thought certainly the question would pose a greater challenge than that.

Bill the Lizard said...

Tom,
Close. "All it takes for the three pieces to form a triangle is that none of the pieces have a length greater than or equal to 1/2 the length of the stick." This much is true, but that doesn't happen a full 50% of the time (which was also my initial guess before I watched the full lecture).

Tom Raywood said...

Bill the Lizard,

So true. (I'm feeling better already.) After 100 simulations with n=1,000,000 I'm seeing slightly less than 50%. More specifically I'm seeing 49.994067% and am definitely curious as to what explains this slight deviation. I think I'd rather contemplate it for a while though, that is, to see if I can figure it out myself.

Tom Raywood said...

Bill the Lizard,

To follow up here, beginnings of a guess are as follows: For some small range [in the form of a difference], as two legs of the triangle both approach 1/2, the small difference between them cannot be bridged by the infinitesimally small remaining leg. Be nice to make a formal statement.

Bill the Lizard said...

Tom,
Check your simulation code against mine. Your answer is off by quite a bit more than just rounding error. You can post your code here, or on Stack Overflow if you'd like me to take a look at it (provided it's in a language I know reasonably well, Java, C, C++, Python, PHP, or any dialect of Basic).

Tom Raywood said...

Bill the Lizard,
Definitely courting tedium at this point.

n R1 R2 R3 T P1 P2 P3 C1

These represent column headings in Excel.

R1, R2 and R3 denote random numbers using the RANDBETWEEN function; I chose for each number the range from 1 to 10,001.

T denotes the sum of R1, R2 and R3 for each row.

P1, P2 and P3 denote which percentage of T is entailed of R1, R2 and R3 respectively.

C1 (for condition 1) represents a check to see whether P1, P2 and P3 all three entail percentages less than 0.5, posting a 1 if they do and a 0 if they don't.

n denotes the number of iterations which, as I said, I took the time to set at 1 million.

Totaling the C1 column provides a sum which, divided into 1 million, quite closely approximates the probability that no 'leg of the triangle' will be greater than or equal to one half.

Additionally I generated a macro which recalculates the worksheet 100 times, posts each result to a table and then finds the average of that set.

You can imagine my surprise to hear that the actual probability varies significantly from what this arrangement points up.

I'll take a look at that lecture this weekend.

Bill the Lizard said...

Tom,
From your explanation it looks like you're generating three random lengths to represent the three pieces of stick. The problem with this approach is that there's nothing to make sure the three pieces all add up to the original length of the stick.

Try it by generating only 2 random numbers (x and y) that represent the break points. If 10,000 is the length of the original stick, and you pick 2 random numbers between 0 and 10,000 then the lengths of the three pieces will be x, y-x, and 10,000-y when x < y, or y, x-y, and 10,000-x when x > y. (It's easier if you just select two random numbers then sort them so x < y, then just always use x, y-x, and 10,000-y for the three lengths, which is essentially what I did in my code.)

I'll try to get my next post up this weekend. I have the answer to your question about acute and obtuse triangles, but proving why I'm getting the number I'm getting is proving to be a little bit complicated.

Tom Raywood said...

Bill the Lizard,
I'll work this up and see what gives. It's definitely counterintuitive though that always having the same stick length would even matter, that is, any more than it would matter that triangle of type Q presents as q1, q2, ...qn for n of any size. Be nice to hear the why behind your assertion. Maybe the lecture will help in this regard.

Bill the Lizard said...

Tom,
The particular value of the stick length doesn't matter as much as making sure the three pieces add up to whatever length you decide on. I used a stick length of 1.0 in my similation, but it doesn't hurt anything if you choose a length of 10,000.

What really makes a difference is how you choose your random lengths. If you choose three random lengths then add them together, you don't really know the length of the stick you started with. If you start with a specific length, then make two random breaks within the boundaries of the stick, you're guaranteed to end up with three random lengths that add up to your original stick length.

It's possible that I just didn't correctly understand the explanation of your spreadsheet. Without seeing the formulas you used I can't be sure that this part of your simulation isn't correct. The errors you're getting could be from another part of the calculation, but this is the most logical place to start looking.

Tom Raywood said...

Bill the Lizard,

I suspect most people would anticipate that having a consistent or particular stick length shouldn't matter here. What's at issue is where the two breaks appear, that is, relative to overall length. A million sticks all of different lengths and all randomly broken in two places should demonstrate the same probabilistic characteristics of a million sticks all of the same length because, again, what's at issue is relative in any event.

This principle can be demonstrated as follows. In my simulation each stick length is represented by T, that is, the sum of R1, R2 and R3. If I wanted/needed to ensure that each and every occurrence of T always comes out to equal a certain value, a few more columns could be inserted which accomplish this simply by adjusting the sizes of R1, R2 and R3 by some definitive amount. For example, if for one row T presented as equal to 16,000 [compared to, say, the value of 21,000 I'd predetermined as a uniform stick length], obviously all I'd have to do is increase each of the original random numbers by 21/16.

So even though now all million stick lengths are the same, the actual percentage of that length occupied by each R1, R2 and R3 hasn't changed at all. The outcome would still match the 50% probability that my simulation currently produces.

In short, Woe to The Foe, heh heh heh.

Here are the formulas I employ for...

...{R1,R2,R3}
................ =RANDBETWEEN(1,10001)

..........{T}
................ =SUM(E2:G2), because R1, R2 and R3 occupy columns E, F and G.

...{P1,P2,P3}
................ =E2/H2; =F2/H2; and =G2/H2, because T occupies column H.

.........{C1}
................ =IF((J2<0.5)*(K2<0.5)*(L2<0.5),1,0), because P1, P2 and P3 occupy columns J, K and L.

And fun was had by all.

Spartacus said...

1. Probability that both cuts are on opposite sides of the midpoint: 1/2
2. Probability that two cuts on opposite sides of the midpoint are less than half the stick length apart: 1/2

Therefore the probability of forming a triangle is 1/2*1/2=1/4

Here is a more difficult problem with an even more surprising result: You break the stick into two pieces, then randomly choose one of the two pieces and break it into two pieces. What is the probability that the resulting three pieces can form a triangle?

You will be amazed by the solution!

(Hint: the answer is not just a simple fraction like the original problem; however there is a very nice closed form)

Bill the Lizard said...

Spartacus,
I've read about half-a-dozen variations on this problem (mostly linked from the notes on the BLOSSOMS page) but I haven't seen that one. This might take a while. :)

Anonymous said...

Sam152's solution is correct; it is the answer the Spartacus's variant.

Bill's original problem statement was under-specified, admitting two interpretations that yield different answers. (Do you choose two cutpoints independently and uniformly on the original stick; or do you make one uniform random cut, and then make another uniform random cut on one of the pieces?)

See http://www.cut-the-knot.org/Curriculum/Probability/TriProbability.shtml

Anonymous said...

This issue might be very old and might already been resolved. but just for clarification your value for second side is wrong and should be taken as follows:

breaks[1] = random.nextDouble()*(1-breaks[0]);

With this correction, the probability comes to 0.193 which is the correct solution.

Bill the Lizard said...

Anonymous,

No, that's incorrect. The second break can take place at any point along the length of the stick. Your change restricts it to only the length of the first piece broken off.

Anonymous said...

Ok yes you are correct, this imposes a restriction and thus reduces the probability. the correct answer is 0.25 only and not 0.193.
thanks.

asha rani said...

Is it ok for each triangle n What is n??


# Modular arithmetic - efficiently calculating the remainders of factorials

When working on this question regarding the divisibility of the sum of factorials, I decided to write some code to test "small values" of the problem using the following code.

f[p_] := Total[Mod[#!, p] & /@ Range[p - 1]];
Table[Mod[f@Prime@i, Prime@i], {i, 1, 500}]

Basically, what the code does is sum up all the factorials $$1!+2!+3!+\dots+(p-1)!$$

and find the remainder modulo $p$, for prime $p$.

Unfortunately, my code as written takes a very long time to run. Checking the first 500 primes takes 88.280966 seconds on my computer, but checking the first 2000 primes took me about 4 hours.

Is there any way to improve the code, or is it already the best we can do?

As for optimizations not involving the code, I used Wilson's Theorem, which states that for all primes $p$,

$$(p-1)!\equiv-1 \bmod p$$

Using the above theorem, we can modify the code as follows.

h[p_] := Total@Flatten[{Mod[#!, p], PowerMod[(# - 1)!*(-1)^(#), -1, p]} & /@ Range[(p - 1)/2]];
Table[Mod[h@Prime@i, Prime@i], {i, 1, 500}]

This is considerably faster than the previous code, since checking the first 500 primes takes only 25.896166 seconds. However, checking the first 2000 primes still takes an inordinately long time.

This is bit faster:

toPrime = 500;
sums = Accumulate@FoldList[Times, 1, Range[2, Prime@toPrime - 1]];
primes = Prime[Range[toPrime]];
Mod[sums[[primes - 1]], primes]

Precompute factorial sums and primes. Mod is fast on lists.

• you are way too humble! calculating your sum for even the first 2000 primes takes less than a second. However, is there a way to get around storing large numbers in sums in memory? It keeps crashing my computer when I try toPrime=5000. – Vincent Tjeng Mar 26 '13 at 6:03
• +1 (that's freaking fast!) Can you explain why Accumulate@FoldList[#1 #2 &, 1, Range[n] is so much quicker to Accumulate@Array[#! &, n] + 1? I really don't get it. – gpap Mar 26 '13 at 11:23
• @gpap Calculating factorial so many times costs a lot. Since we know we want all the factorials up to Prime[toPrime]-1, we ultimately gain a lot keeping the intermediate results with FoldList. – Michael E2 Mar 26 '13 at 12:04
• @MichaelE2 Yes, worked it out myself in the meantime - you multiply the previous result and don't calculate a factorial at every step. Thanks – gpap Mar 26 '13 at 12:11
• Is there any reason why you used #1 #2 & instead of Times? – J. M. is away Jun 15 '15 at 12:53

Let $x \equiv r_1 \bmod p$ and $y \equiv r_2 \bmod p$. Then, $x y \equiv r_1 r_2 \bmod p$. So, we can compute the sum of the factorials mod p using:

f[p_] := Mod[ Total @ FoldList[ Mod[Times[##], p]&, Range[p-1]], p]

Let's compare this to the naive implementation:

t[p_] := Mod[Sum[k!, {k, p-1}], p]

For example:

f[Prime[500]] //AbsoluteTiming
t[Prime[500]] //AbsoluteTiming

{0.00058, 1813}

{0.085628, 1813}

The nice thing about using Mod[Times[##], p] as the FoldList function is that the output should be a machine number unless you are working with very large primes. That means that f can be compiled:

fc = Compile[{{p, _Integer}},
Mod[ Total @ FoldList[ Mod[Times[##], p]&, Range[p-1]], p],
RuntimeAttributes->{Listable}
];

Let's compare for a large prime:

f[Prime[10^6]] //AbsoluteTiming
fc[Prime[10^6]] //AbsoluteTiming

{1.83041, 9308538}

{1.05449, 9308538}

Faster, but the real difference is that fc is Listable. Hence, comparing timings on a list shows a much larger difference. For example:

r1 = f /@ Prime[Range[5000, 5500]]; //AbsoluteTiming
r2 = fc[Prime[Range[5000, 5500]]]; //AbsoluteTiming

r1 === r2

{2.06691, Null}

{0.395402, Null}

True

Finally, since the list of all factorials is not stored anywhere, the memory used is much more manageable. Here is the memory and timing for the first 5000 primes:

r1 = fc[Prime[Range[5000]]]; //MaxMemoryUsed //AbsoluteTiming

{3.03463, 462848}

This is quite a bit smaller than @MichaelE2's answer:

MaxMemoryUsed[
toPrime = 5000;
sums = Accumulate@FoldList[Times, 1, Range[2, Prime@toPrime - 1]];
primes = Prime[Range[toPrime]];
r2 = Mod[sums[[primes - 1]], primes]
] //AbsoluteTiming

r1 === r2

{2.40441, 4064607008}

True

The compiled answer uses .462KB while the approach where all the factorials are precomputed takes 4GB, and the timing is not too different.


# While plotting polar coordinate system graphs, should we include the negative of angle $\theta$?

While plotting polar coordinate system graphs, should we include the negative of angle $\theta$?

For example, while plotting $r=\theta$ for negative values the spiral comes out to be reverse of the graph plotted using positive values. Upon searching in google I find only one spiral. So are both spirals part of the curve?

• There's no reason to omit negative values of $\theta$, other than that the pictures are sometimes less pretty. – Aaron Montgomery Oct 30 '17 at 17:46
• Negative values of $\theta$ are perfectly fine. Often times when doing integrals in polar coordinates, it is convenient to use a parameterization with negative values rather than positive ones. – superckl Oct 30 '17 at 17:49

It's not really about $\theta$, which of course can be both positive or negative. Rather it's about the values of $r$ — should we allow negative values of $r$ or not? And this is not really a mathematical question, but rather a matter of convention. Most textbooks I've seen allow both positive and negative values of $r$. However, some require that $r$ must be non-negative ($r\ge0$).


# How many points uniquely define a square?

I was wondering how many points uniquely define a square. Now it is clear to me that if we have $$n$$ random points on a square then this does not necessarily uniquely define the square (for example they may all lie on the same edge).

My question was more the following: supposing I was able to choose exactly which points of my square to use, what is the minimum number of points I could pick to uniquely define the square, and which points should I pick?

Edit: the points need not lie on the vertices.

• Isn’t it 3? Pick three corner points. Jun 27 at 18:17
• @EDX Which four points are enough? If you pick the four corners, then I can for example draw a different square for which those four points lie on the midpoints of the sides, so that's not enough. Jun 27 at 18:23
• @EDX I think it depends on how we understand the question. If we take a square and pick the four corner points, there will actually be infinitely many squares containing those points, although they won't be in the corners. Jun 27 at 18:24
• @subrosar yes the points need not lie on the corners
– JLB
Jun 27 at 18:24
• I can not think of a way to have four points uniquely define a square, but five points should suffice. so long as exactly one set of the three points are colinear. From the three that are colinear we can tell the orientation of the square as any other edges must come at 90 degree angles to it. From there you should be able to draw lines which the edges will fall upon. You should be able to then find the distance between opposite lines which will be the edge length and you should be able to complete the square from there. Jun 27 at 18:37

Four points can be enough to unambiguously define a square.

Take two of the corners, an additional point from the edge between those corners, and an additional point from the middle of the edge opposite the edge the first three points lied on.

A second observer coming to look at these points having been told they all lie on the same square will learn from the three colinear points the orientation of the square and that these lie on the same edge and additional edges will either be perpendicular or parallel to this.

They will then note that since if drawing a perpendicular from the fourth point to this edge it lands in the middle of the edge formed by our colinear points that this must be on the opposite edge. Next, since they know this opposite edge must be parallel to the original edge we now know the lengths of the edges must be equal to this minimum distance between our fourth point and the edge formed by our three colinear points.

Finally, having noted that this length is equal to the length between our colinear points that these two must have been corners of our square.

• That's easier than my construction; well done! Jun 27 at 18:58

Four points are enough, if chosen correctly.

In particular, I claim that the only square which passes through the four points $$A(0,0)$$, $$B(0,0.3)$$, $$C(0.4,0)$$, and $$D(1,1)$$ is the unit square with corners $$(0,0), (0,1), (1,0), (1,1)$$. The solution lets you find many more similar examples.

To prove this claim, here are some inequalities about the distances between points on a square with side length $$s$$:

• Two points on the same side are at a distance between $$0$$ and $$s$$;
• Two points on adjacent sides are at a distance between $$0$$ and $$\sqrt2 s$$;
• Two points on opposite sides are at a distance between $$s$$ and $$\sqrt 2 s$$.

We can check that if $$A,B,C,D$$ lie on a square, then they can't all lie on two adjacent sides of the square: there's no two lines that contain all four points and intersect at right angles. (It's enough to check that $$AB$$ is not perpendicular to $$CD$$, $$AC$$ is not perpendicular to $$BD$$, and $$AD$$ is not perpendicular to $$BC$$.) So there must be some pair of points on opposite sides.

The six distances between the points are $$AB=0.3$$, $$AC=0.4$$, $$BC=0.5$$, $$CD \approx 1.16$$, $$BD \approx 1.22$$, and $$AD \approx 1.41$$. So we see that if any two of $$A,B,C$$ were on opposite sides, the side length $$s$$ would be at most $$0.5$$, but $$D$$ is more than $$0.5 \sqrt2 \approx 0.71$$ away from all the rest. Therefore $$A,B,C$$ are all on two adjacent sides of the squares.

In particular, one side of the square must contain two of $$A,B,C$$. Can a side of the square contain $$B$$ and $$C$$? No: line $$BC$$ separates $$A$$ from $$D$$, which a side of the square can't do. If we assume $$A,B$$ are on one side of the square and $$C$$ is on an adjacent side, or that $$A,C$$ are on one side of the square and $$B$$ is on an adjacent side, we get the same conclusion: the square has a corner at $$A$$ with one side containing segment $$AB$$ and one side containing segment $$AC$$.

Since $$D$$ does not lie on either line, it must be on one of the other sides of the square: a side parallel to $$AB$$, and a side parallel to $$AC$$. We get the same square and try both.

• If we draw line $$AB$$, line $$AC$$, and a line through $$D$$ parallel to $$AB$$, those lines must contain three sides of the square. The only such square is the unit square.
• If we draw line $$AB$$, line $$AC$$, and a line through $$D$$ parallel to $$AC$$, those lines must contain three sides of the square. The only such square is the unit square.

Three points are never enough.

Take three points $$A,B,C$$. If they form a right triangle, we can draw ever-bigger squares with that right angle as a corner, containing all three. If they form an obtuse triangle, that's even better; suppose without loss of generality that the altitude from $$C$$ lands on line $$AB$$ at a point $$H$$ outside segment $$AB$$. Then we can draw ever-bigger squares with one corner at $$H$$ containing all three points.

For an acute triangle, begin by rotating the points arbitrarily. Let $$x_{\min}, x_{\max}, y_{\min}, y_{\max}$$ be the lowest and highest $$x$$- and $$y$$-coordinates among the three points; exclude the finitely many orientations where there are ties. Draw the four lines $$x = x_{\min}, x=x_{\max}, y = y_{\min}, y = y_{\max}$$; each of them passes through one point and together they contain all three (otherwise we'd get an obtuse angle). One of the points lies on two of the lines.

Without loss of generality, $$A$$ lies on $$x = x_{\max}$$ and $$y = y_{\max}$$ and also $$x_{\max} - x_{\min} \ge y_{\max} - y_{\min}$$. Then erase the line through $$y = y_{\max}$$ and instead draw the line through $$y = y_{\min} + x_{\max} - x_{\min}$$. These four lines define a square containing all three points. Since we get this for all but finitely many ways to rotate $$A,B,C$$ (equivalently, for all but finitely many choices of a horizontal and vertical direction) we get infinitely many squares.

• Ahhhhh now I see why three doesn’t suffice. Jun 27 at 20:19

If you specify what points you chose, then two are sufficient: choose opposite vertices. A diagonal can then be drawn between them, and a perpendicular bisector of the same length can be drawn, with the ends being the other two vertices. Then the sides can be drawn between them.

• Sometimes the simple answer is the correct answer… “supposing I was able to choose exactly which points of my square to use, what is the minimum number of points I could pick”
– Matt
Jun 28 at 7:27
• @Matt There is still some ambiguity. If Alice wants to communicate to Bob what square she has, two points being sufficient requires not only that Alice can choose which points to tell Bob, but that Bob knows that Alice chose those points. If all that Bob knows is that these two points are on the square, but doesn't know that they are opposite vertices, then Bob will not know which square they're from. Jun 29 at 0:55
• I understand your point, but also Bob knows it is a square and not some other completely irregular shape (otherwise the problem would be impossible). So it is reasonable Bob could know the two points are opposite vertices
– Matt
Jun 29 at 6:42
• Acccumulation : you require some communication between two persons must succeed and Matt : you assume such persons are perfectly reasonable in optimizing. But OP only is about one person figuring it out on its own. If that person knows about his own invented conventions (one could claim that to be cheating), two points would be enough. Jul 3 at 22:30

Two points are enough if they are ordered - a corner A, and the corner B clockwise from A. This is the minimum, as a square has 4 degrees of feeedom - center, size, and orientation.

A slighly different question is: whats the minimum number of points such that one and only one square will lie on them. 5 is enough - four corners and somewhere on an edge. Four might be enough.


## anonymous one year ago Does the series $\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+1\}}$converge?

1. anonymous

Using the fact that $$\text{lcm}\{n,n+1\}=\dfrac{n(n+1)}{\text{gcd}\{n,n+1\}}$$, I'm almost tempted to say that $\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+1\}}\sim\sum_{n\ge1}\frac{1}{n^2}$ but I don't think I can jump to that conclusion just yet.

2. mathmate

We agree that lcm(n,n+1) is actually n(n+1). (The gcd is 1, by Euclid's algorithm) Since S=$$\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+1\}}=\sum_{n\ge1}\frac{1}{n(n+1)}<\sum_{n\ge1}\frac{1}{n^2}$$ The strict inequality holds for each and every term, and since $$\sum_{n\ge1}\frac{1}{n^2}$$ is a geometric series that is known to converge (=pi^2/6 prove it), so by comparison, S<pi^2/6 so S converges.

3. mathmate

* series, not geometric

4. anonymous

Right, since $$n$$ and $$n+1$$ are consecutive, that makes their lcm very easy to determine. And we can find the value of the series while we're at it quite easily: $\sum_{n\ge1}\frac{1}{n(n+1)}=\sum_{n\ge1}\left(\frac{1}{n}-\frac{1}{n+1}\right)=1$Neat!

5. ganeshie8

Nice! any two consecutive integers are coprime, gcd(n, n+1) = gcd(n, 1) = 1 so lcm(n, n+1) = n(n+1)

6. anonymous

consider that the least common multiple of $$n,n+1$$ is obviously $$n(n+1)$$ since $$n+1-n=1$$ and thus they must be coprime. this reduces to $$\sum_{n=1}^\infty\frac1{n(n+1)}=\sum_{n=1}^\infty\frac{n+1-n}{n(n+1)}=\sum_{n=1}^\infty\left(\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}\right)\\\quad =\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)$$ which is obviously telescoping and since $$1/(n+1)\to0$$ it converges

7. imqwerty

well i read that a series that converges has got some bounds and that it approaches a specific value. so in this case its going like 1/2 , 1/6, 1/12 , 1/20..... so u mark a line of length 1 on a number line and u plot the point 1/2 u knw 1/2 is still left between 1/2 and 1 and then u add 1/6 still 1/3 is left so u have bounds and u knw that it can never go > 1 so yea it converges... https://en.wikipedia.org/wiki/Convergent_series

8. anonymous

Could we make a more general claim here? Numerically, it would seem that $\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+k\}}$also converges to $$1$$ for $$k\in\mathbb{N}$$, though much more slowly.

9. anonymous

for prime $$k$$ the problem is barely any harder -- you merely have to be careful about the multiples of $$k$$: $$\sum_{n=1}^\infty\frac1{[n,n+k]}=\sum_{m=1}^\infty\sum_{n=mk+1}^{(m+1)k}\frac1{[n,n+k]}\\\quad=\sum_{m=1}^\infty\left(\sum_{n=mk+1}^{mk+k-1}\frac1{n(n+k)}+\frac1{(m+1)(m+2)k}\right)$$

10. ParthKohli

$\gcd (n,n+k) \le k$$\Rightarrow {\rm lcm }(n,n+k) = \frac{n(n+k)}{\gcd(n,n+k)} \ge \frac{n(n+k)}{k}$$\Rightarrow \frac{1}{{\rm lcm} (n,n+k) }\le \frac{k}{n(n+k)}$We can sum the right-side telescopically so I guess it is true that the sum is convergent?

11. anonymous

yeah, showing its convergence along those lines is exactly what I figured

12. anonymous

the sum of the terms for $$n$$ coprime to $$k$$ is accomplished with relatively little effort $$\sum_{n=mk+1}^{(m+1)k-1}\frac1{n(n+k)}=\frac1k\left(\frac1{mk+1}-\frac1{(m+2)k-1}\right)$$ so we get: $$\frac1k\sum_{m=0}^\infty\left(\frac1{mk+1}-\frac1{(m+2)k-1}+\frac1{(m+1)(m+2)}\right)$$now split it up and telescope, I suppose? $$\sum\left(\frac1{mk+1}-\frac1{(m+2)k-1}\right)=1+\frac1{k+1}\\\sum\left(\frac1{m+1}-\frac1{m+2}\right)=1$$ so we get $$\frac1k\left(2+\frac1{k+1}\right)=\frac2k+\frac1{k(k+1)}=\frac3k-\frac1{k+1}$$ for prime $$k$$

13. anonymous

oops, ignore that -- I read $$(m+2)k+1$$ rather than $$(m+2)k-1$$, so that series does not quite telescope so cleanly

14. anonymous

our problem term for prime $$k$$ is this: $$\frac1{mk+1}-\frac1{(m+2)k-1}$$

15. anonymous

Would that I could hand out more medals...

16. thomas5267

How do you show $$\gcd (n,n+k) \leq k$$ without using Bézout's Identity?

17. anonymous

i mean, if (positive) $$d$$ divides (positive) $$n$$ and $$n+k$$ then it also divides their difference, so $$d$$ must divide $$k$$. it follows that the greatest common divisor is at most $$k$$

18. anonymous

you don't necessarily need the full strength of Bezout's identity, just the fact that divisibility is preserved by sums which is very elementary

19. ganeshie8

$$\gcd(n,n+k)=\gcd(n.k)\le k$$

20. thomas5267

\begin{align*} \sum_{n=1}^\infty\frac1{[n,n+k]}&=\sum_{m=0}^\infty\sum_{n=mk+1}^{(m+1)k}\frac1{[n,n+k]}\\ &=\sum_{m=0}^\infty\left(\sum_{n=mk+1}^{mk+k-1}\frac1{n(n+k)}+\frac1{(m+1)(m+2)k}\right)\\ \end{align*} \begin{align*} &\phantom{{}={}}\sum_{m=0}^\infty\sum_{n=mk+1}^{mk+k-1}\frac1{n(n+k)}\\ &=\sum_{m=0}^\infty\left(\frac{1}{mk+1}-\frac{1}{(m+1)k-1}\right)\\ &=\sum_{m=0}^\infty\left(\frac{1}{mk+1}-\frac{1}{(m+1)k+1}+\frac{1}{(m+1)k+1}-\frac{1}{(m+1)k-1}\right)\\ &=\sum_{m=0}^\infty\left(\frac{1}{mk+1}-\frac{1}{(m+1)k+1}-\frac{2}{(m+1)^2k^2-1}\right)\\ &=1-\sum_{m=0}^\infty\frac{2}{(m+1)^2k^2-1} \end{align*} I think the sum converges but I couldn't prove it.

21. anonymous

Comparing to the series $$\sum\frac{1}{n^2}$$ would suffice for convergence. It has a somewhat daunting closed form: $\sum_{m=0}^\infty \frac{2}{(m+1)^2k^2-1}=\frac{k-\pi\cot\dfrac{\pi}{k}}{2k}$

22. anonymous

(closed form courtesy of Mathematica)

23. thomas5267

I thought $$\dfrac{2}{(m+1)k^2-1}\geq\dfrac{1}{n^2}$$ for some reason! $\min(k)=2\\ \frac{2}{4n^2-1}-\frac{1}{n^2}=\frac{2n^2-4n^2+1}{n^2(4n^2-1)}=\frac{-2n^2+1}{n^2(4n^2-1)}\leq0$

24. anonymous

$\sum_{m\ge0}\frac{1}{(m+1)^2k^2-1}=\sum_{m\ge1}\frac{1}{m^2k^2-1}\le\sum_{m\ge1}\frac{1}{m^2}$ since $1\le k^2-1~~\implies~~ m^2\le m^2k^2-1~~\implies~~\frac{1}{m^2k^2-1}\le\frac{1}{m^2}$(The first inequality is true since $$k\ge2$$.)


# Ten people are seated at a rectangular table - Permutations homework

I got the following question for homework.

Ten people are to be seated at a rectangular table for dinner. Tanya will sit at the head of the table. Henry must not sit beside either Wilson or Nancy. In how many ways can the people be seated for dinner?

The approach I have used is to take $9!$ and subtract it from the seats where Henry, Wilson or Nancy can sit. I end up getting the answer $211680$ but for some reason, the back of the book says $201 600$. What am I doing incorrectly?

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This looks like an error in the book: your answer appears to be correct.

We can count the allowed seatings directly. Suppose that Henry sits next to Tanya. Then there is one seat forbidden to Wilson and Nancy, so there are $\binom72$ ways to choose seats for Wilson and Nancy. There are $2$ ways to seat Wilson and Nancy in those $2$ seats and $6!$ ways to seat the unnamed people. Finally, Henry can be on either side of Tanya, so there are altogether $2\cdot2\cdot6!\cdot\binom72$ acceptable seatings with Henry next to Tanya.

Now suppose that Henry is not seated next to Tanya. Then there are $2$ seats forbidden to Wilson and Nancy, so there are $\binom62$ ways to choose their seats. As before there are $2$ ways to seat them in those two seats and $6!$ ways to seat the unnamed people. Finally, there are $7$ possible choices for Henry’s seat, so there are altogether $7\cdot2\cdot6!\cdot\binom62$ acceptable seatings with Henry not next to Tanya. Altogether, then, there are

$$2\cdot6!\left(2\binom72+7\binom62\right)=1440(42+105)=211,680$$

acceptable seatings.

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Using inclusion-exclusion, it's possible to see where the error in the back of the book may have come from.

With Tanya assigned to the head of table, there are $9!$ ways to seat the other people with no restrictions. Among these, there $2\cdot8!$ ways in which Henry sits beside Wilson (the factor of $2$ corresponds to which one sits to the left of the other), and another $2\cdot8!$ ways in which Henry sits beside Nancy. Subtracting these, we get the book's answer

$$9!-2\cdot8!-2\cdot8!=201600$$

But this double-subtracts the $2\cdot7!$ ways in which Henry sits between Wilson and Nancy. So the correct answer is

$$9!-2\cdot8!-2\cdot8!+2\cdot7!=211680$$

-


02
dez

# system of linear equations problems

Many problems lend themselves to being solved with systems of linear equations. Case 2: Parallel Lines The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation. Materials include course notes, lecture video clips, JavaScript Mathlets, a quiz with solutions, practice problems with solutions, a problem solving video, and problem sets with solutions. Once you do that, these linear systems are solvable just like other linear systems. $\begin{cases}5x +2y =1 \\ -3x +3y = 5\end{cases}$ Yes. Solve age word problems with a system of equations. The problem also says that Mia will be twice as old as Shaheena. This example shows one iteration of the gradient descent. When you solve systems with two variables and therefore two equations, the equations can be linear or nonlinear. There can be any combination: 1. But let’s say we have the following situation. Systems of linear equations can be used to model real-world problems. Solving using Matrices by Row Operations. Quiz 3. Just select one of the options below to start upgrading. Setting up a system of linear equations example (weight and price) (Opens a modal) Interpreting points in context of graphs of systems (Opens a modal) Practice. You’re going to the mall with your friends and you have $200 to spend from your recent birthday money. At the first store, he bought some t-shirts and spent half of his money. A system of linear equations is called homogeneous if the constants$b_1, b_2, \dots, b_m$are all zero. “Systems of equations” just means that we are dealing with more than one equation and variable. Wouldn’t it be cl… Consider the nonlinear system of equations Cramer's Rule. The best way to get a grip around these kinds of word problems is through practice, so we will solve a few examples here to get you … Word Problems with Systems of Linear Equations: This worksheet is designed for Algebra students to practice those dreaded word problems that are solved with linear systems of equations. Is the point$(1 ,3)$a solution to the following system of equations… Solution of a non-linear system. No Problem 2. A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. Solving using an Augmented Matrix. Solve simple cases by inspection. Problem 1. Find them out by checking. Systems of Equations - Problems & Answers. Systems of linear equations are a common and applicable subset of systems of equations. When solving linear systems, you have two methods at your disposal, and which one you choose depends on the problem: System of equations word problem: walk & ride, Practice: Systems of equations word problems, System of equations word problem: no solution, System of equations word problem: infinite solutions, Practice: Systems of equations word problems (with zero and infinite solutions), Systems of equations with elimination: TV & DVD, Systems of equations with elimination: apples and oranges, Systems of equations with substitution: coins, Systems of equations with elimination: coffee and croissants. Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. Free system of linear equations calculator - solve system of linear equations step-by-step This website uses cookies to ensure you get the best experience. By … System of Linear Equations Word Problems Calvin went to Chicago's Magnificent Mile to do some Christmas shopping. A system of linear equations is a set of two or more linear equations with the same variables. Is the point$(0 ,\frac{5}{2})$a solution to the following system of equations? SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY. A system of linear equations is a group of two or more linear equations that all contain the same set of variables. One way to solve a system of linear equations is by graphing each linear equation on the same -plane. To solve the system of equations, you need to find the exact values of x and y that will solve both equations. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Below is an example that shows how to use the gradient descent to solve for three unknown variables, x 1, x 2, and x 3. You really, really want to take home 6items of clothing because you “need” that many new things. Systems of Linear Equations. Solve the following system of equations by elimination. For problems 1 – 3 use the Method of Substitution to find the solution to the given system or to determine if the system … 9,000 equations in 567 variables, 4. etc. Gradient descent can also be used to solve a system of nonlinear equations. Looking for fun activities to teach kids about Solving Word Problems Involving Linear Equations and Linear Inequalities? Systems of Linear Equations and Problem Solving. 6 equations in 4 variables, 3. (5.2.3) – Solve mixture problems with a system of linear equations. In mathematics, a system of linear equations (or linear system) is a collection of one or more linear equations involving the same set of variables. ... Systems of equations word problems (with zero and infinite solutions) Get 3 of 4 questions to level up! This is one reason why linear algebra (the study of linear systems and related concepts) is its own branch of mathematics. In this algebra activity, students analyze word problems, define variables, set up a system of linear equations, and solve the system. HIDE SOLUTIONS. 1. Systems of 2 linear equations - problems with solutions Test. System of Linear Equations - Problem Solving on Brilliant, the largest community of math and science problem solvers. Section 8.1, Example 4(a) Solve graphically: Linear systems are usually expressed in the form Ax + By = C, where A, B, and C are real numbers. In your studies, however, you will generally be faced with much simpler problems. When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. For example, the sets in the image below are systems of linear equations. This premium worksheet bundle contains a printable fact file and 10 fun and engaging worksheets to challenge your students and help them learn about Solving Word Problems Involving Linear Equations and Linear Inequalities. Substitution Method. Problem 1 Two of the following systems of equations have solution (1;3). B. Determining the value of k for which the system has no solutions. To use Khan Academy you need to upgrade to another web browser. So a System of Equations could have many equations and many variables. There are 7 questions. Donate or volunteer today! A large pizza at Palanzio’s Pizzeria costs$6.80 plus $0.90 for each topping. Our mission is to provide a free, world-class education to anyone, anywhere. System of NonLinear Equations problem example. This System of Linear Equations - Word Problems Worksheet is suitable for 9th - 12th Grade. A solution is a mixture of two or more different substances like water and salt or vinegar and oil. This section provides materials for a session on solving a system of linear differential equations using elimination. Khan Academy is a 501(c)(3) nonprofit organization. The directions are from TAKS so do all three (variables, equations and solve) no matter what is asked in the problem. The statement above gives the following equation m + 10 = 2 × (s + 10) m + 10 = 2 × s + 2× 10 m + 10 = 2s + 20 You now have a system of linear equation to solve m + s = 40 equation 1 m + 10 = 2s + 20 equation 2 Use equation 1 to solve for m m + s = 40 m + s - s = 40 - s m = 40 - s If you're seeing this message, it means we're having trouble loading external resources on our website. For example, + − = − + = − − + − = is a system of three equations in the three variables x, y, z.A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied. One application of systems of equations are mixture problems. This section covers: Systems of Non-Linear Equations; Non-Linear Equations Application Problems; Systems of Non-Linear Equations (Note that solving trig non-linear equations can be found here).. We learned how to solve linear equations here in the Systems of Linear Equations and Word Problems Section.Sometimes we need solve systems of non-linear equations, such as those we see in conics. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. System of linear equations System of linear equations can arise naturally from many real life examples. Add the second equation to the first equation and solve for x. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. set of two (or more) equations which have two (or more) variables Solving Systems of Linear Equations. It has 6 unique word problems to solve including one mixture problem … If the two lines intersect at a single point, then there is one solution for the system: the point of intersection. Simultaneous equations (Systems of linear equations): Problems with Solutions. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$4x - 7\left( {2 - x} \right) = 3x + 2$$, $$2\left( {w + 3} \right) - 10 = 6\left( {32 - 3w} \right)$$, $$\displaystyle \frac{{4 - 2z}}{3} = \frac{3}{4} - \frac{{5z}}{6}$$, $$\displaystyle \frac{{4t}}{{{t^2} - 25}} = \frac{1}{{5 - t}}$$, $$\displaystyle \frac{{3y + 4}}{{y - 1}} = 2 + \frac{7}{{y - 1}}$$, $$\displaystyle \frac{{5x}}{{3x - 3}} + \frac{6}{{x + 2}} = \frac{5}{3}$$. So far, we’ve basically just played around with the equation for a line, which is . 1. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Solve each of the following equations and check your answer. Section 7-1 : Linear Systems with Two Variables. Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve age word problems with a system of equations. A solution of the system (*) is a sequence of numbers$s_1, s_2, \dots, s_n$such that the substitution$x_1=s_1, x_2=s_2, \dots, x_n=s_n$satisfies all the$m$equations in the system (*). Updated June 08, 2018 In mathematics, a linear equation is one that contains two variables and can be plotted on a graph as a straight line. You discover a store that has all jeans for$25 and all dresses for 50. Improve your math knowledge with free questions in "Solve systems of linear equations" and thousands of other math skills. When this is done, one of three cases will arise: Case 1: Two Intersecting Lines . Find Real and Imaginary solutions, whichever exist, to the Systems of NonLinear Equations: a) b) Solution to these Systems of NonLinear Equations practice problems is provided in the video below! A system of linear equations is a system made up of two linear equations. Solution: Rewrite in order to align the x and y terms. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6 . 2 equations in 3 variables, 2. Generally speaking, those problems come up when there are two unknowns or variables to … Solving systems of equations word problems worksheet For all problems, define variables, write the system of equations and solve for all variables. If you're seeing this message, it means we're having trouble loading external resources on our website. The same rules apply. a)\begin{array}{|l} x + y = 5 \\ 2x - y = 7; \end{array}\$ In "real life", these problems can be incredibly complex. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. Answer: x = .5; y = 1.67. Solving using Matrices by Elimination. We can use the Intersection feature from the Math menu on the Graph screen of the TI-89 to solve a system of two equations in two variables.


Learn faster with a math tutor. 9th - 10th grade. The two pairs of congruent sides may be, but do not have to be, congruent to each other. Get better grades with tutoring from top-rated professional tutors. This is one of the most important properties of parallelogram that is helpful in solving many mathematical problems related to 2-D geometry. 3) Are opposite angles of a parallelogram congruent? Solve for x. (10 Properties of parallelogram: Opposite sides of parallelogram are equal . The diagonals of a parallelogram bisect each other. Sides of A Parallelogram The opposite sides of a parallelogram are congruent. (By definition). The angles of a parallelogram are congruent. Ask yourself which approach looks easier or quicker. Line segments XY and ZW are also congruent. One way all sides of the two parallelograms could be congruent would be if $ABCD$ and $EFGH$ are squares with the same side length: in this case they would be congruent. Properties of parallelograms are as follows: i. Parallelogram Properties DRAFT. Tags: Question 19 . CCSS.MATH.CONTENT.HSG.CO.C.11 Prove theorems about parallelograms. Is the quadrilateral a parallelogram? . Give a reason. Expand Image In fact, one method of proving a quadrilateral a rhombus is by first proving it a parallelogram, and then proving two consecutive sides congruent, diagonals bisecting verticies, etc. A parallelogram is a quadrilateral that has opposite sides that are parallel. If one side is longer than its opposite side, you do not have parallel sides; no parallelogram! Draw the diagonal BD, and we will show that ΔABD and ΔCDB are congruent. Opposite sides are parallel and congruent. It is a quadrilateral with two pairs of parallel, congruent sides. Opposite angles are congruent. High School: Geometry » Congruence » Prove geometric theorems » 11 Print this page. Parallelogram Properties DRAFT. A parallelogram is defined to be a quadrilateral with 2 pairs of opposite sides parallel. Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram bisect each other. Consecutive angles are supplementary (A + D = 180°). The two diagonals of a parallelogram bisect each other. A parallelogram is a quadrilateral with two pairs of parallel sides. 5) Does a diagonal of a parallelogram bisect a pair of opposite angles? A parallelogram also has the following properties: Opposite angles are congruent; Opposite sides are congruent; Adjacent angles are supplementary; The diagonals bisect each other. Yes, if both pairs of opposite angles are congruent, then you have a parallelogram. The diagonals of an isosceles trapezoid are congrent. 0 Why are these two lines not congruent (and other ways to figure out if other shapes are not congruent) Solve for x. HELP ASAP 30 points Part 1 out of 2 To repair a large truck or bus, a mechanic might use a parallelogram lift. 2 years ago. One has to be on the lookout for double negatives. 62% average accuracy. Opposite sides are congruent. You can draw parallelograms. The opposite angles are congruent. 60 seconds . A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . If yes, state how you know. There are two ways to go about this. If one angle is right, then all angles are right. The diagonals of a rectangle are the bisectors of the angles. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. An equilateral quadrilateral is a square. That is true. Here are some important things that … Try to move the vertices A, B, and D and observe how the figure changes. The opposite sides of a parallelogram are congruent so we will need two pairs of congruent segments: Now if we imagine leaving $\overline{AB}$ fixed and ''pushing down'' on side $\overline{CD}$ so that these two sides become closer while side $\overline{AD}$ and $\overline{BC}$ rotate clockwise we get a new parallelogram: There is one right angle in a parallelogram and it is not a rectangle. If a quadrilateral has three angles of equal measure, then the fourth angle must be a right angle. The converse is also true that if opposite sides of a quadrangle are equal then its a parallelogram. 62% average accuracy. The figure is a parallelogram. The diagonals of a trapezoid are perpendicular. For our parallelogram, we will label it WXYZ, but you can use any four letters as long as they are not the same as each other. A rhombus is a parallelogram but with all four sides equal in length; A square is a parallelogram but with all sides equal in length and all interior angles 90° A quadrilateral is a parallelogram if: Both pairs of opposite sides are parallel. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. Tags: Question 19 . The diagonals of a quadrilateral_____bisect each other, If the measures of 2 angles of a quadrilateral are equal, then the quadrilateral is_____a parallelogram, If one pair of opposite sides of a quadrilateral is congruent and parallel, then the quadrilateral is______a parallelogram, If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is_____a parallelogram, To prove a quadrilateral is a parallelogram, it is ________enough to show that one pair of opposite sides is parallel, The diagonals of a rectangle are_____congruent, The diagonals of a parallelogram_______bisect the angles, The diagonals of a parallelogram______bisect the angles of the parallelogram, A quadrilateral with one pair of sides congruent and on pair parallel is_______a parallelogram, The diagonals of a rhombus are_______congruent, A rectangle______has consecutive sides congruent, A rectangle_______has perpendicular diagonals, The diagonals of a rhombus_____bisect each other, The diagonals of a parallelogram are_______perpendicular bisectors of each other, Consecutive angles of a quadrilateral are_______congruent, The diagonals of a rhombus are______perpendicular bisectors of each other, Consecutive angles of a square are______complementary, Diagonals of a non-equilateral rectangle are______never angle bisectors, A quadrilateral with one pair of congruent sides and one pair of parallel sides is_____a parallelogram. B) The diagonals of the parallelogram are congruent. The opposite angles of a parallelogram are supplementary. Studying the video and these instructions, you will learn what a parallelogram is, how it fits into the family of polygons, how to identify its angles and sides, how to prove you have a parallelogram, and what are its identifying properties. Reason for statement 8: If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. Triangles can be used to prove this rule about the opposite sides. Use a straightedge (ruler) to draw a horizontal line segment, then draw another identical (congruent) line segment some distance above and to one side of the first one, so they do not line up vertically. Opposite angles are congruent. Parallelogram definition A quadrilateral with both pairs of opposite sides parallel. 1981 times. Opposite sides are congruent. If both pairs are congruent, you have either a rhombus or a square. SURVEY . 2. A parallelogram is a quadrilateral that has opposite sides that are parallel. Check for any one of these identifying properties: You can use proof theorems about a plane, closed quadrilateral to discover if it is a parallelogram: You have learned that a parallelogram is a closed, plane figure with four sides. You can have almost all of these qualities and still not have a parallelogram. Opposite angles are congruent. The opposite sides of a parallelogram are congruent. Local and online. Go with B. Give a reason. The diagonals of a rectangle bisect eachother. Theorem: If ABCD is a parallelogram then prove that its opposite sides are equal. We already mentioned that their diagonals bisect each other. The diagonals of a kite are the perpendicular bisectors of each other. More generally, a quadrilateral with 4 congruent sides is a rhombus. If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is_____a parallelogram Always To prove a quadrilateral is a parallelogram, it is ________enough to show that one pair of opposite sides is parallel The area of a parallelogram is also equal to the magnitude of the vector cross product of two adjacent sides. , all 4 sides are congruent (definition of a rhombus). Get better grades with tutoring from top-rated private tutors. Check Next Opposite angles are congruent. Want to see the math tutors near you? Property that is characteristic of a parallelogram is that opposite sides are congruent. The diagonals of a quadrilateral are perpendicular and the quadrilateral is not a rhombus. Opposite angels are congruent (D = B). Prove that opposite sides of a parallelogram are congruent. Other shapes, however, are types of parallelograms. Which statement can be used to prove that a given parallelogram is a rectangle? 4) Do the diagonals of a parallelogram bisect each other? Observe that at any time, the opposite sides are parallel and equal. Now consider just the interior angles of parallelograms, ∠W, ∠X, ∠Y, and ∠Z. Theorem 1: Opposite Sides of a Parallelogram Are Equal In a parallelogram, the opposite sides are equal. In this lesson we will prove the basic property of a parallelogram that the opposite sides in a parallelogram are equal. Q. In the video below: We will use the properties of parallelograms to determine if we have enough information to prove a given quadrilateral is a parallelogram. The four line segments making up the parallelogram are WX, XY, YZ, and ZW. Notice that line segments WX and YZ are congruent. Use this to prove that the quadrilateral must be a parallelogram. Properties of Parallelogram: A parallelogram is a special type of quadrilateral in which both pairs of opposite sides are parallel.Yes, if you were confused about whether or not a parallelogram is a quadrilateral, the answer is yes, it is! The first rhombus above is a square while the second one has angles of 60 and 120 degrees. If only one set of opposite sides are congruent, you do not have a parallelogram, you have a trapezoid. Things that you need to keep in mind when you prove that opposite sides of a parallelogram are congruent. These geometric figures are part of the family of parallelograms: For such simple shapes, parallelograms have some interesting properties. Solution: A Parallelogram can be defined as a quadrilateral whose two s sides are parallel to each other and all the four angles at the vertices are not 90 degrees or right angles, then the quadrilateral is called a parallelogram. Give the given and prove and prove this. A parallelogram does not have other names. The interior angles are ∠W, ∠X, ∠Y, and ∠Z. Make sure that second line segment is parallel to (or equidistant from) the first line segment. Yes. The consecutive angles of a parallelogram are supplementary. The opposite sides are equal and parallel; the opposite angles are also equal. 1-to-1 tailored lessons, flexible scheduling. 1981 times. Connecting opposite (non-adjacent) vertices gives you diagonals WY and XZ. A parallelogram is a flat shape with four straight, connected sides so that opposite sides are congruent and parallel. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . 2 years ago. Start at any vertex (corner). The area of a parallelogram is twice the area of a triangle created by one of its diagonals. We will show that in that case, they are also equal to each other. That the other pair of opposite sides are congruent. Opposite sides are congruent (AB = DC). Is the quadrilateral a parallelogram? Take a rectangle and push either its left or ride side so it leans over; you have a parallelogram. In today's lesson, we will show that the opposite sides of a parallelogram are equal to each other. Let’s play with the simulation given below to better understand a parallelogram and its properties. This means a parallelogram is a plane figure, a closed shape, and a quadrilateral. The parallelogram has the following properties: Opposite sides are parallel by definition. An equilateral parallelogram is equiangular. Proving That a Quadrilateral is a Parallelogram, Opposite sides are parallel -- Look at the parallelogram in our drawing. The bad in Answer A is due to your teachers written grammar. Is it possible to prove a quadrilateral a parallelogram with two consecutive and two opposite congruent sides? A parallelogram also has the following properties: Opposite angles are congruent; Opposite sides are congruent; Adjacent angles are supplementary; The diagonals bisect each other. That’s a wrap! A parallelogram is defined as a quadrilateral where the two opposite sides are parallel. In the figure given below, ABCD is a parallelogram. The figure shows a side view of the li … ft. FGKL, GHJK, and FHJL are parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. The Angle-Side-Angle Triangle Congruence Theorem can be used to prove that, in a parallelogram, opposite sides are congruent. There are more than two right angles in a trapezoid. Opposite sides are congruent -- The base side (Y Z Y Z) and the top side (W X W X) of our parallelogram are equal in length (congruent); the left side (XY X Y) and right side (ZW Z W) are also congruent To be a parallelogram, the base and top sides must be parallel and congruent, and … The opposite sides of parallelogram are also equal in length. CCSS.MATH.CONTENT.HSG.CO.C.11 Prove theorems about parallelograms. To show these two triangles are congruent we’ll use the fact that this is a parallelogram, and as a result, the two opposite sides are parallel, and the diagonal acts as a transv… Parallelogram law. Parallelogram Theorem #2: The opposite sides of a parallelogram are congruent. SURVEY . View Untitled document (3).pdf from MATH 100 at Basha High School. If the four sides do not connect at their endpoints, you do not have a closed shape; no parallelogram! Mathematics. Solution: A Parallelogram can be defined as a quadrilateral whose two s sides are parallel to each other and all the four angles at the vertices are not 90 degrees or right angles, then the quadrilateral is called a parallelogram. The bottom (base) side, Opposite sides are congruent -- The base side (. C) The diagonals of the parallelogram bisect the angles. 60 seconds . Get help fast. So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. G 2 T: + 3 5 6 8 17 K 3 Which angles are congruent to 21? <2 2 are congruent to 21. As with any quadrilateral, the interior angles add to 360°, but you can also know more about a parallelogram's angles: Using the properties of diagonals, sides, and angles, you can always identify parallelograms. Opposite sides are congruent. One interesting property of a parallelogram is that its two diagonals bisect each other (cut each other in half). Or: Both pairs of opposite sides are congruent. To explore these rules governing the sides of a parallelogram use Math Warehouse's interactive parallelogram. Triangles can be used to prove this rule about the opposite sides. D) The opposite angles of the parallelogram are congruent. Opposite sides of a … ∴ ∴ AB = CD A B = C D and AD= BC A D = B C Properties of a parallelogram. A parallelogram is a flat shape with four straight, connected sides so that opposite sides are congruent and parallel. So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. Opposite angles are equal (congruent) to each other; Any two adjacent angles of a parallelogram add up to, This means any two adjacent angles are supplementary (adding to, A closed shape (it has an interior and exterior), A quadrilateral (four-sided plane figure with straight sides), Two pairs of congruent (equal), opposite angles, Two pairs of equal and parallel opposite sides, If the quadrilateral has bisecting diagonals, it is a parallelogram, If the quadrilateral has two pairs of opposite, congruent sides, it is a parallelogram, If the quadrilateral has consecutive supplementary angles, it is a parallelogram, If the quadrilateral has one set of opposite parallel, congruent sides, it is a parallelogram. To be on the lookout for double negatives proof 2 here ’ use. 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