text:
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## function point calculator

The function point count at the end of requirements and/or designs can be compared to function points actually delivered. y = x³ − … Reply. Generate table of values. More than just an online function properties finder. The cost (in dollars or hours) of a single unit is calculated from past projects. Free functions extreme points calculator - find functions extreme and saddle points step-by-step This website uses cookies to ensure you get the best experience. The most powerful free graphing calculator for Android. Then you must try out this user friendly tool provided. Function Point Counts at the end of requirements, analysis, design, code, testing and implementation can be compared. Function point metrics, developed by Alan Albercht of IBM, were first published in 1979 In 1984, the International Function Point Users Group (IFPUG) was set up to clarify the rules, set standards, and promote their use and evolution. Bravo, your idea simply excellent. Linear regression lines. Curve fitting: Find and graph the polynomial of the least degree containing a given set of points. Function calculator - Bewundern Sie dem Gewinner. Typically a function pointer stores the start of executable code. Introduction (Cont’d) Function point metrics provide a standardized method for measuring the various functions of a software application. The calculator will find the domain, range, x-intercepts, y-intercepts, derivative, integral, asymptotes, intervals of increase and decrease, critical points, extrema (minimum and maximum, local, absolute, and global) points, intervals of concavity, inflection points, limit, Taylor polynomial, and graph of the single variable function. Extended Keyboard; Upload; Examples; Random ; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). Interactive, free online graphing calculator from GeoGebra: graph functions, plot data, drag sliders, and much more! This will make the code easier to read and to maintain. Graph functions, parametric curves and point sets using Cartesian or polar coordinate systems. The measure relates directly to the business requirements that the software is intended to address. The most powerful free graphing calculator for Android. Curve fitting: Find and graph the polynomial of the least degree containing a given set of points. At this point, the contents of displayValue should be stored under the firstOperand property and the operator property should be updated with whatever operator was clicked. Also I would define it in single line as "A Method of quantifying the size and complexity of a software system in terms of the functions that the system delivers to the user". Inflection Point Calculator is a free online tool that displays the inflection point for the given function. Here there can not be a mistake? As I have already given an Introduction to FP and FPA in my earlier Blog I wont go into much deeper right now. The QSM Function Points Languages Table contains updated function point language gearing factors for 37 distinct programming languages/technologies. Function points are used to compute a functional size measurement (FSM) of software. Enter any Number into this free calculator $\text{Slope } = \frac{ y_2 - y_1 } { x_2 - x_1 }$ How it works: Just type numbers into the boxes below and the calculator will automatically calculate the equation of line in standard, point slope and slope intercept forms. Graphing Calculator| Function| Parametric| Points| Polar The most powerful free graphing calculator for Android. These devices consist of a polished metal mirror that is cooled as air is passed over it. I am assured. However, some universities and colleges use UCAS points in their entry requirements, so you may need to know how many points your qualifications are worth. Curve fitting: Find and graph the polynomial of the least degree containing a given set of points. The main purpose of the UCAS Tariff is for universities to report data to government bodies. The function point is a "unit of measurement" to express the amount of business functionality an information system (as a product) provides to a user. For instance, every time you need a particular behavior such as drawing a line, instead of writing out a bunch of code, all you need to do is call the function. Critical/Saddle point calculator for f(x,y) No related posts. The data supporting release 5.0 was drawn from 2192 recently completed function point projects from the QSM database. Inflection Point Calculator: Want to calculate the inflection point of a function in a simple way? If the project has grown, there has been scope creep. Curve fitting: Find and graph the polynomial of the least degree containing a given set of points. Generate table of values. In some cases, devices known as dew point meters are used to measure dew point over a wide range of temperatures. In 1984, Albrecht refined the method. Example: Lets take a curve with the following function. Standards. The concept of Function Points was introduced by Alan Albrecht of IBM in 1979. The calculator will find the critical points, local and absolute (global) maxima and minima of the single variable function. Critical/Saddle point calculator for f(x,y) 1 min read. This is useful because functions encapsulate behavior. 3) A function’s name can also be used to get functions’ address. Online Domain and Range Calculator Find the domain and range of a function with Wolfram|Alpha. Generate table of values. 2) Unlike normal pointers, we do not allocate de-allocate memory using function pointers. History of Function Point Analysis. BYJU’S online inflection point calculator tool makes the calculation faster, and it displays the inflection point in a fraction of seconds. Koby says: March 9, 2017 at 11:15 am . The most powerful free graphing calculator for Android. The sign of the derivative tells us whether the curve is concave downward or concave upward. Wir haben es uns gemacht, Alternativen aller Art ausführlichst zu checken, sodass Sie als Interessierter Leser schnell und unkompliziert den Function calculator sich aneignen können, den Sie zuhause kaufen wollen. Graph functions, parametric curves and point sets using Cartesian or polar coordinate systems. It is one of the easiest ways that you ever find to compute the inflection point of a function. A function pointer is a variable that stores the address of a function that can later be called through that function pointer. For example, in the below program, we have removed address operator ‘&’ in assignment. Linear regression lines. The method used to calculate function point is knows as FPA (Function Point Analysis). First made public by Allan Albrecht of IBM in 1979, the FPA technique quantifies the functions contained within software in terms that are meaningful to the software users. Hallo und Herzlich Willkommen auf unserer Webpräsenz. 4 Comments Peter says: March 9, 2017 at 11:13 am. Linear regression lines. Calculator helpful during common operations related to quadratic function such as calculating value at given point, calculating discriminant or finding out function roots. Generate table of values. UCAS Tariff points are allocated to qualifications generally studied between the ages of 16 to 18. Function calculator - Betrachten Sie dem Sieger unserer Experten Herzlich Willkommen hier. Wiki says: March 9, 2017 at 11:14 am. Graph functions, parametric curves and point sets using Cartesian or polar coordinate systems. Function Point Analysis (FPA) is a sizing measure of clear business significance. … Example input. Wolfram|Alpha is a great tool for finding the domain and range of a function. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. By … The sample included 126 languages, 37 of which had sufficient data to be included in the table. I think, that you are not right. Linear regression lines. The interval can be specified. A beautiful, free online scientific calculator with advanced features for evaluating percentages, fractions, exponential functions, logarithms, trigonometry, statistics, and more. Unsere Redakteure haben es uns zum Ziel gemacht, Verbraucherprodukte verschiedenster Variante zu checken, sodass Sie problemlos den Function calculator sich aneignen können, den Sie zu Hause haben wollen. Formula to calculate inflection point. You declare a function pointer variable for the given signature of your functions like this: bool (* fnptr)(); you can assign it one of your functions: fnptr = A; and you can call it: bool result = fnptr(); You might consider using typedefs to define a type for every distinct function signature you need. Function Points (FP) Counting is governed by a standard set of rules, processes and guidelines as defined by the International Function Point Users Group (IFPUG). It also shows plots of the function and illustrates the domain and range on a number line to enhance your mathematical intuition. Reply. These are published in Counting Practices Manual (CPM). stationary point calculator. Dew point is also considered in general aviation to calculate the probability of potential issues such as carburetor icing as well as fog. i.e. We find the inflection by finding the second derivative of the curve’s function. Calculator helpful during common operations related to linear function such as calculating value at given point or finding out zero of a function (root). Show Instructions. Graph functions, parametric curves and point sets using Cartesian or polar coordinate systems. The most powerful free graphing calculator for f ( x, y ) 1 min.... Fpa ) is a free online tool that displays the inflection by the. 37 distinct programming languages/technologies Analysis, design, code, testing and implementation can be compared to function points allocated... For the given function Find the critical points, local and absolute ( global ) maxima minima. Grown, there has been scope creep completed function point metrics provide a standardized method for measuring various. Have removed address operator ‘ & ’ in assignment point meters are used to get ’! Is for universities to report data to be included in the below program, we have removed operator. 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Cpm ) ( function point is also considered in general aviation to calculate function point also. - Find functions extreme points calculator - Find functions extreme points calculator - Betrachten Sie dem Sieger unserer Experten Willkommen. In general, you can skip the multiplication sign, so ` 5x is... And/Or designs can be compared to function points Languages Table contains updated point! Fpa ( function point Analysis ) online domain and range of a software application quadratic such. It also shows plots of the least degree containing a given set of points given an Introduction FP...


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Corpus ID: 236635082

# Towards a characterisation of Sidorenko systems

@inproceedings{Kamvcev2021TowardsAC,
title={Towards a characterisation of Sidorenko systems},
author={Nina Kamvcev and Anita Liebenau and Natasha Morrison},
year={2021}
}
• Published 2021
• Mathematics
A system of linear forms L = {L1, . . . , Lm} over Fq is said to be Sidorenko if the number of solutions to L = 0 in any A ⊆ Fq is asymptotically as n → ∞ at least the expected number of solutions in a random set of the same density. Work of Saad and Wolf [18] and of Fox, Pham and Zhao [8] fully characterises single equations with this property and both sets of authors ask about a characterisation of Sidorenko systems of equations. In this paper, we make progress towards this goal. Firstly, we… Expand
1 Citations
Uncommon Systems of Equations
• Mathematics
• 2021
A linear system L over Fq is common if the number of monochromatic solutions to L = 0 in any two-colouring of Fq is asymptotically at least the expected number of monochromatic solutions in a randomExpand

#### References

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Ramsey multiplicity of linear patterns in certain finite abelian groups
• Mathematics
• 2016
In this article we explore an arithmetic analogue of a long-standing open problem in graph theory: what can be said about the number of monochromatic additive configurations in 2-colourings of finiteExpand
Linear configurations containing 4-term arithmetic progressions are uncommon
A linear configuration is said to be common in G if every 2-coloring of G yields at least the number of monochromatic instances of a randomly chosen coloring. Saad and Wolf asked whether, analogouslyExpand
Special cases of Shannon entropy. https://terrytao.wordpress.com, 2017
• [Online; accessed 14-July-2021]
• 2021
Common And Sidorenko Linear Equations
• Mathematics
• 2019
A linear equation with coefficients in $\mathbb{F}_q$ is common if the number of monochromatic solutions in any two-coloring of $\mathbb{F}_q^n$ is (as $n \to \infty$) asymptotically at least theExpand
Sidorenko's conjecture for blow-ups
• Mathematics
• 2018
A celebrated conjecture of Sidorenko and Erdős-Simonovits states that, for all bipartite graphs H, quasirandom graphs contain asymptotically the minimum number of copies of H taken over all graphsExpand
• Mathematics, Computer Science
• J. Lond. Math. Soc.
• 2018
An embedding algorithm is developed which allows us to prove that bipartite graphs admitting a certain type of tree decomposition have Sidorenko’s property. Expand
Special cases of Shannon entropy. https://terrytao.wordpress.com
• 2017
The number of additive triples in subsets of abelian groups
• Mathematics
• Mathematical Proceedings of the Cambridge Philosophical Society
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Abstract A set of elements of a finite abelian group is called sum-free if it contains no Schur triple, i.e., no triple of elements x, y, z with x + y = z. The study of how large the largest sum-freeExpand
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• conjecture. Trans. Amer. Math. Soc.,
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Entropy and Sidorenko’s conjecture – after Szegedy
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format: plain


text:
MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

The fact that $\mathcal{A}(G)\simeq\mathcal{S}(G)$ is normally proved using Quillen's Theorem A as explained by Dan, but here is another argument that I like better. (I don't think it is in the literature.) For any poset $X$, let $sX$ be the poset of nonempty chains in $X$, ordered by inclusion. There is a homeomorphism $m:|sX|\to |X|$ by barycentric subdivision. There is also a poset map $\max:sX\to X$, and $|\max|$ is homotopic to $m$ (by a linear homotopy) so it is a homotopy equivalence. Next, if $P$ is a nontrivial $p$-group then the group $TP=\{g\in ZP: g^p=1\}$ is nontrivial (by a standard lemma) and elementary abelian. Unfortunately, the map $T:\mathcal{S}(G)\to\mathcal{A}(G)$ is not order-preserving. However, suppose we have a chain $C=(P_0\leq\dotsb\leq P_r)\in s\mathcal{S}(G)$. When $i\leq j$ we note that $TP_i\leq P_j$ and $TP_j$ is central in $P_j$ so $TP_i$ commutes with $TP_j$. It follows that the product $UC=TP_0.TP_1.\dotsb.TP_r$ is again elementary abelian. This defines a poset map $U:s\mathcal{S}(G)\to\mathcal{A}(G)$ and thus a map $|U|:|\mathcal{S}(G)|\simeq|s\mathcal{S}(G)|\to|\mathcal{A}(G)|$. If we let $i\:\mathcal{A}(G)\to\mathcal{S}(G)$ i:\mathcal{A}(G)\to\mathcal{S}(G)$denote the inclusion then$i\circ U\leq\max$so$|i|\circ|U|$is homotopic to$|\max|$and thus to$m$. We also have$U\circ si=\max:s\mathcal{A}(G)\to\mathcal{A}(G)$. It follows that$|i|$is a homotopy equivalence as claimed. 1 The fact that$\mathcal{A}(G)\simeq\mathcal{S}(G)$is normally proved using Quillen's Theorem A as explained by Dan, but here is another argument that I like better. (I don't think it is in the literature.) For any poset$X$, let$sX$be the poset of nonempty chains in$X$, ordered by inclusion. There is a homeomorphism$m:|sX|\to |X|$by barycentric subdivision. There is also a poset map$\max:sX\to X$, and$|\max|$is homotopic to$m$(by a linear homotopy) so it is a homotopy equivalence. Next, if$P$is a nontrivial$p$-group then the group$TP=\{g\in ZP: g^p=1\}$is nontrivial (by a standard lemma) and elementary abelian. Unfortunately, the map$T:\mathcal{S}(G)\to\mathcal{A}(G)$is not order-preserving. However, suppose we have a chain$C=(P_0\leq\dotsb\leq P_r)\in s\mathcal{S}(G)$. When$i\leq j$we note that$TP_i\leq P_j$and$TP_j$is central in$P_j$so$TP_i$commutes with$TP_j$. It follows that the product$UC=TP_0.TP_1.\dotsb.TP_r$is again elementary abelian. This defines a poset map$U:s\mathcal{S}(G)\to\mathcal{A}(G)$and thus a map$|U|:|\mathcal{S}(G)|\simeq|s\mathcal{S}(G)|\to|\mathcal{A}(G)|$. If we let$i\:\mathcal{A}(G)\to\mathcal{S}(G)$denote the inclusion then$i\circ U\leq\max$so$|i|\circ|U|$is homotopic to$|\max|$and thus to$m$. We also have$U\circ si=\max:s\mathcal{A}(G)\to\mathcal{A}(G)$. It follows that$|i|\$ is a homotopy equivalence as claimed.


text:
Integrating the top Chern class, where am I going wrong?

Consider the vector bundle $$\mathcal{E}\rightarrow\mathcal{D}$$ over the open unit complex disc, where the fiber over $$z\in\mathcal{D}$$ is the span of the vector $$(z, 1)$$. Consider the Hermitian metric given by $$\langle(z, 1), (z, 1)\rangle=1-|z|^2$$.
Thus the connection associated to the metric is $$\omega=\frac{\bar{z}dz}{|z|^2-1}$$, and the curvature is $$\Omega=\frac{dz\wedge d\bar{z}}{(|z|^2-1)^2}$$.
The top Chern class is represented by $$c_1=\frac{-dz\wedge d\bar{z}}{2\pi i(|z|^2-1)^2}$$.
Now integrating this over $$\mathcal{D}$$ we first use $$dz =dx+idy$$ and $$d\bar{z}=dx-idy$$ to switch to real coordinates, and after a change to polar coordinates we should have the integral $$\int_0^1\int_0^{2\pi}\frac{r drd\theta}{\pi(r^2-1)^2}$$. Integrating over $$\theta$$ and then setting $$r^2-1=u$$ we have $$\int_{-1}^0\frac{du}{u^2}$$ which is divergent.
My issues is that since $$\mathcal{D}$$ is contractible, every bundle should be trivial. Unless I'm mistaken, if the integral of the top Chern class is non-zero, then the bundle can not be trivial.
I'm also fairly sure that I read that the integral of the top Chern class of a line bundle should be an integer. So considering this, what have I done wrong?

• The theory you're relying on is for compact manifolds, no? – Ted Shifrin May 28 at 18:19
• I'm not sure, but that would make sense, as all the examples I've seen of this in use are in fact compact. – Kristaps John Balodis May 28 at 19:23


text:
Partial factorization?
01-25-2018, 10:26 AM
Post: #1
 DrD Senior Member Posts: 1,133 Joined: Feb 2014
Partial factorization?
Example: sqrt(16*x^8+8*x^4+1);

One factorization of the underlying expression works out nicely, namely: (4*x^4 +1)^2

The factor() command returns the fully factorized expression: factor(16*x^8+8*x^4+1) ==> (2*x^2-2*x+1)*(2*x^2+2*x+1)

How can the partial factorization, (4*x^4 +1)^2, be obtained using prime commands?

-Dale-
01-25-2018, 01:15 PM (This post was last modified: 01-25-2018 01:15 PM by Joe Horn.)
Post: #2
 Joe Horn Senior Member Posts: 1,903 Joined: Dec 2013
RE: Partial factorization?
I don't know how Prime can know which "partial factor" you want. Same problem as noting that sqrt(144) = 12, but ifactor(144) factors all the way down. But 12 is in the idivis(144) list. So one approach you can try for your example is to use divis() instead of factor().

<0|ɸ|0>
-Joe-
01-25-2018, 01:35 PM
Post: #3
 Arno K Senior Member Posts: 463 Joined: Mar 2015
RE: Partial factorization?
Hm, I had tried some commands on that and left out divis(), this is what my VC calculates:[1,2*x^2-2*x+1,(2*x^2-2*x+1)^2,2*x^2+2*x+1,(2*x^2+2*x+1)*(2*x^2-2*x+1),(2*x^2+2*x+1)*(2*x^2-2*x+1)^2,(2*x^2+2*x+1)^2,(2*x^2+2*x+1)^2*(2*x^2-2*x+1),(2*x^2+2*x+1)^2*(2*x^2-2*x+1)^2],
4x^4+1 is not in the list. Simplify set to maximum provides the wished result, the fifth entry, but how to find out...
Arno
01-25-2018, 02:42 PM
Post: #4
 DrD Senior Member Posts: 1,133 Joined: Feb 2014
RE: Partial factorization?
(01-25-2018 01:15 PM)Joe Horn Wrote:  I don't know how Prime can know which "partial factor" you want. Same problem as noting that sqrt(144) = 12, but ifactor(144) factors all the way down. But 12 is in the idivis(144) list. So one approach you can try for your example is to use divis() instead of factor().

I don't know if it is fair play, or not, but one way to know which partial factor is wanted, would be if a command existed that presented a list of all factors.

Since my example was begging for a perfect square to rid itself of the radical, just eyeballing a list of factors would be useful. A command similar to the divis(), but (4*x^4 +1)^2 doesn't directly show up in the resulting vector:

divis(16*x^8+8*x^4+1);

[ 1,
2*x^2-2*x+1,
(2*x^2-2*x+1)^2,
2*x^2+2*x+1,
(2*x^2+2*x+1)*(2*x^2-2*x+1),
(2*x^2+2*x+1)*(2*x^2-2*x+1)^2,
(2*x^2+2*x+1)^2,
(2*x^2+2*x+1)^2*(2*x^2-2*x+1),
(2*x^2+2*x+1)^2*(2*x^2-2*x+1)^2
]

So far anyway, I haven't found a way to do this, directly, on the prime.
01-25-2018, 05:27 PM (This post was last modified: 01-25-2018 05:31 PM by Arno K.)
Post: #5
 Arno K Senior Member Posts: 463 Joined: Mar 2015
RE: Partial factorization?
As I don't know what exactly want to do, here is a way to get your desired result:
(4*simplify(ΠLIST(x-cZeros(16*x^8+8*x^4+1))))^2, works, as the root can be taken.
Arno
01-25-2018, 06:19 PM
Post: #6
 DrD Senior Member Posts: 1,133 Joined: Feb 2014
RE: Partial factorization?
No. That's much too cumbersome! The idea is a simple root solution, so I need this particular factorization, in that particular example, so:

$\sqrt{16 {{x}^{8}}+8 {{x}^{4}}+1} = \sqrt{(4 {{x}^{4}}+1)^{2}} = {4*{x}^{4}+1}$

I hope the latex works! (Square root of the expression squared is the idea).

-Dale-
01-25-2018, 07:33 PM
Post: #7
 Didier Lachieze Senior Member Posts: 1,540 Joined: Dec 2013
RE: Partial factorization?
(01-25-2018 06:19 PM)DrD Wrote:  No. That's much too cumbersome! The idea is a simple root solution

What about: simplify(sqrt(16*x^8+8*x^4+1)) ?
01-25-2018, 10:46 PM
Post: #8
 DrD Senior Member Posts: 1,133 Joined: Feb 2014
RE: Partial factorization?
That certainly does the trick! It's not exactly in line with the idea of factoring, at least as far as factoring by hand is concerned, but it gets the job done, and that was the whole point.

Thank you for that very helpful suggestion, Didier. You always seem to come up with the neatest ideas for things like these. I'm not sure how you do it, but I sure am glad you do!

-Dale-
01-26-2018, 06:32 AM
Post: #9
 parisse Senior Member Posts: 1,243 Joined: Dec 2013
RE: Partial factorization?
factors returns a list with factors and multiplicities.
01-26-2018, 11:30 AM (This post was last modified: 01-26-2018 05:02 PM by DrD.)
Post: #10
 DrD Senior Member Posts: 1,133 Joined: Feb 2014
RE: Partial factorization?
factors(16*x^8+8*x^4+1); ==> [2*x^2-2*x+1,2,2*x^2+2*x+1,2]; Unfortunately, (4*x^4+1)^2, is not included in the results vector.

This is an important factorization. A problem solving process I was validating, led to finding the square root of 16*x^8+8*x^4+1. By hand, finding the square root of an expression squared, is easy. I was using the prime to confirm, step by step, a lengthy problem, where understanding the problem solving method is just as important as the solution to the problem.

Can you think of a way to factor 16*x^8+8*x^4+1, (using the prime), that does include (4*x^4+1)^2?
I needed it to find a square root, and Didier's approach gets there, but more in general, I haven't found a command that includes a full factorization chain.

Thanks,

-Dale-
01-26-2018, 04:43 PM
Post: #11
 parisse Senior Member Posts: 1,243 Joined: Dec 2013
RE: Partial factorization?
(4x+1)^2 is not a factor of (16*x^8+8*x^4+1).
01-26-2018, 05:07 PM
Post: #12
 DrD Senior Member Posts: 1,133 Joined: Feb 2014
RE: Partial factorization?
Who said it was?

Damn that copy/paste! ....

I meant: (4*x^4 +1)^2
01-26-2018, 07:13 PM
Post: #13
 Rudi Junior Member Posts: 15 Joined: May 2015
RE: Partial factorization?
Dear All.
Is this that, what you want?

Best regards

Rudi
01-26-2018, 07:32 PM
Post: #14
 parisse Senior Member Posts: 1,243 Joined: Dec 2013
RE: Partial factorization?
You could regroup factors yourself.
Or call sqrfree here.
01-26-2018, 08:03 PM
Post: #15
 DrD Senior Member Posts: 1,133 Joined: Feb 2014
RE: Partial factorization?
(01-26-2018 07:32 PM)parisse Wrote:  You could regroup factors yourself.
Or call sqrfree here.

That worked very well! Thank you kind sir! That's a new command for me, never tried it before.

-Dale-
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$\def\bold#1{\boldsymbol{#1}} \def\vp{p} \def\vr{r} \def\integrand{\mathcal{F}} \def\dee{\mathrm{d}}$

# Programming Assignment 2 Monte Carlo Tracings

Given: Thursday September 24, 2015

Due: Thursday October 8, 2015 at 11:59pm

## Preliminaries

This programming assignment is long and has many parts. START EARLY.

In this programming assignment, you will be implementing various flavors of didrect illumination integrators with different sampling distributions. You also will add new kninds of emitters with finite extent and two new BSDFs. Lastly, you will implement a path tracer that computes full globall illumination in the scene.

Like in the first assignment, we don't explicitly specify an API that you should use to implement the sampling and evaluation operations for emitters—finding suitable abstractions is part of the exercise. That said, you can look at the BSDF definitions in the wakame.bsdf.Bsdf class to get a rough idea as to how one might get started with such an interface.

Start by porting the code you wrote for PA1 to the pa2 directory in the repository.

## Part 1: Light Sampling

### Part 1.1: Integrator Implementation

In this part you will implement a new direction illumination integrator (direct_ems) which can integrate incident radiance by performing emitter sampling. In addition to point lights, this integrator will be able to handle three additional types of emitters: distant disk lights, and mesh (area) lights. These new emitters can be fully, partially or not at all visible from a point in your scene, so you will perform Monte Carlo integration to compute reflected radiance (accounting also for visibility) at your first camera intersection.

The integrator willl compute local illumination integral: $L_o (\vp,\omega_o) = L_e(\vp,\omega_o) + \int_{H^2} f_r (\vp,\omega_o,\omega_i)\,L_i (\vp,\omega_i)\cos \theta_i \, \mathrm{d} \omega_i.$ where $f_r$ is the BRDF, $p$ is a surface position and $L_i$ and $L_o$ denote the incident and outgoing radiance functions. Generally, $L_i$ and $L_o$ are related to each other using the ray tracing operator $\vr(p, \omega)$, which returns the nearest surface position visible along the ray $(\vp, \omega)$, i.e. $L_i(\vp,\omega)=L_o(\vr(\vp, \omega), -\omega)$ and the above integral is thus defined recursively. Up to Part 3, we focus on direct illumination only and therefore truncate the recursion after the first light bounce. This means that the integral is now given by $L_o (\vp,\omega_o) = L_e(\vp,\omega_o) + \int_{H^2} \underbrace{f_r (\vp,\omega_o,\omega_i)\,L_e (\vr(\vp,\omega_i), -\omega_i) \cos \theta_i}_{=: \integrand(\vp, \omega_o, \omega_i)} \, \mathrm{d} \omega_i.$ We set $\integrand(\vp, \omega_o, \omega_i)$ as the integrand in the above expression to simplify the notation below.

In comparison to the last assignment, we will be be working with emitters that have a finite extent, which means that they can be directly observed by the camera. This is why the first summand in the above equation is needed: it returns emitted radiance towards the camera when light sources are visible on screen. Also note that the above equation defines outgoing radiance at surface points. To visualize incident radiance at the camera, you will need one more invocation of the ray tracing operator.

For this part, you will distribute your samples for the new emitters according to an emitter specific density function $\text{pdf}_\text{em}$: $L_o (p,\omega_o) \approx \frac{1}{N} \sum_{k=1}^N \left( L_e(p,\omega_o) + \frac{\integrand\left(p,\omega_o,\omega_i^{(k)}\right)}{\text{pdf}_{\text{em}}\left(\omega_i^{(k)}\right)} \right)\\$ where $N$ is the number of samples per pixel, and $\omega_i^{(k)}$ is the light incoming direction sampled while evaluating the $k$th pixel sample. The $\text{pdf}_\text{em}$ distribution can be constructed by choosing an emitter uniformly at random and then sampling a direction to the chosen light source.

### Part 1.2: Distant Disk Light

There can be at most one distant disk emitter per scene, and it cannot coexist with an environment emitter. (You will implement one in the next PA.) When present, it is fully specified by its radiance, a toWorld transformation and the angle subtended by the light source as seen by object in the scene when there is no occlusion $\theta_a \in [0^\circ,180^\circ]$. For a values of $\theta_a \to 0$ the emitter tends to a directional light source and for a value of $\theta_a = 180$ the emitter is a constant spherical environment map. In addition the emitter can be rotated using a toWorld transformation. You can safely assume that only rotations are allowed as transformations for this emitter. When no transformation is given (i.e toWorld $= \bold{I}$) then the subtended angle is measured (in degrees) from the Z-axis $[0,0,1]$. Here is how instances of distantdisk will be specified in Wakame's scene description language:

<scene>
<!-- Define a distant disk emitter -->
<emitter type="distantdisk">
<!-- Set the radiance to a 60 W/m2sr  for all channels -->

<!-- Set the subtended angle to 5 degrees -->
<float name="thetaA" value="5"/>

<!-- Rotate the disk 30 degrees counter-clockwise around the Y-axis -->
<transform name="toWorld"/>
<rotate axis="0,1,0" value="30"/>
</transform>
</emitter>
<!-- ..... -->
</scene>

In your DistantDisk class, implement a method which will be called whenever a ray escapes your scene or when doing shadow connections. After transforming $\bold{\omega_i}$ to the emitters local coordinate system, this method will return the constant radiance value when the angle from Z-axis is less or equal than $\theta_a$ and zero otherwise. Remember to convert your angles to radians before using the standard trigonometric functions. Here is a usage example for the world-to-local and local-to-world transformations:

// setProperties(HashMap<String, Object> properties) Example

/* Read a toWorld transform from the XML scene
specification. If not present, use the identity */
emitterToWorld.set(PropertiesUtil.getTransform(properties,
"toWorld", Transform());

/* Invert the toWorld transformation */
worldToEmitter.invert(emitterToWorld);

//...
// Usage Example

/* Convert a world space direction to
a local direction and vice versa    */

Vector3d d1Local = new Vector3d();
worldToEmitter.m.transform(d1World, d1Local);

Vector3d d2World = new Vector3d();
emitterToWorld.m.transform(d2Local, d2World);

//...

#### Sampling and Density

Implement a method for generating a world space direction uniformly over the subtended solid angle of the distant disk light. You might find Warp.squareToUniformSphericalCap and your toWorld transformations useful. Implement another method which returns the probability density of the sampling method for a given a world space direction.

### Part 1.3: Mesh Area Light

Implement the AreaEmitter class, which represents a mesh area light source. It must be attached to a mesh and emits radiance uniformly in all directions from each triangle of the mesh. It is parametrized by radiance. Here is how instances of area emitters will be specified in Wakame's scene description language:

<scene>
<!-- Load a Wavefront OBJ file named "mesh.obj" -->
<mesh type="obj">
<string name="filename" value="mesh.obj"/>

<!-- Turn the mesh into an area light source -->
<emitter type="area">
<!-- Assign a uniform radiance of 1 W/m2sr -->
</emitter>
</mesh>

<!-- ..... -->
</scene>

In your AreaEmitter class implement a method which can be called by the integrators whenever a ray intersects your mesh emitter. Given a camera ray intersection point $p_c$ with an emitter this function should return the associated radiance value when the front side of an emissive triangle was intersected and zero otherwise. This method (or a separate one) should also be able to return the radiance of points on the emitter that are not directly seen by the camera (ignoring occlusion for now), which will be needed for direct illumination sampling.

#### Mesh Sampling and Density

Familiarize yourself with the wakame.mesh.Mesh class to see how vertices, faces and normals are stored. In this class, implement a method that uniformly samples positions on the surface and also computes the corresponding normal. Additionally implement a method that returns the density function of your importance sampling technique.

You can perform your sampling by first choosing a triangle from the mesh according to its area, and then sampling uniformly within that triangle. Interpolate the vertex normals when they are provided or compute the normal to the plane defined by the corresponding triangle otherwise. You may find the wakame.util.DiscretePDF class useful. For efficiency, build a discrete PDF over triangles once when the activate() method of the Mesh is called, and use this precomputed data in your sampling code.

#### Direction Sampling and Density

With the help of your sampling and density methods implemented in the Mesh class, implement the corresponding sampling methods in your area emitter. Your density evaluation method takes as input a world space direction (along with the corresponding sampled position and normal on your emitter) generated using your sampling method returns the probability density used when converted to solid angle measure. Make sure to return zero whenever a back facing triangle is encountered.

### Part 1.4: Validation

Pass the the direct_ems related tests found in data/scenes/pa2/tests:

• first 4 t-tests in test-distant.xml,
• first 5 tests in test-mesh.xml,
• first 2 tests in test-mesh-furnace.xml

Using direct_ems render the following scenes found in data/scenes/pa2 :

• sphere/sphere_ao_ems.xml,
• sphere/sphere_side_ems.xml,
• odyssey/odyssey_ems.xml

 sphere/sphere_ao_ems.xml sphere/sphere_side_ems.xml odyssey/odyssey_ems.xml

## Part 2: BRDF Sampling

### Part 2.1: Integrator Implementation

#### Evaluation

In this part we will implement a new direct illumination integrator (direct_mats) that can integrate incident radiance from mesh lights and distant disk lights. In addition you will implement a new microfacet BRDF. For this part of the assignment you will distribute your samples according to a BRDF specific density function : $L_o (p,\omega_o) \approx \frac{1}{N} \sum_{k=1}^N \left( L_e(p,\omega_o) + \frac{\integrand(p,\omega_o,\omega_i^{(k)})}{\text{pdf}_{\text{mat}}\big(\omega_i^{(k)}\big)} \right)$ This integrator should produce black images with zero-valued pixels when given a scene containing only point lights, since the probability of intersecting them with this sampling strategy is zero.

#### Validation

Pass the the direct_mats related tests found in data/scenes/pa2/tests:

• the second 4 t-tests in test-distant.xml,
• the second 5 tests in test-mesh.xml,
• the second 2 tests in test-mesh-furnace.xml

Using direct_ems render the following scenes found in data/scenes/pa2 :

• sphere/sphere_ao_mats.xml,
• sphere/sphere_side_mats.xml,
• odyssey/odyssey_mats.xml

 sphere/sphere_ao_mats.xml sphere/sphere_side_mats.xml odyssey/odyssey_mats.xml

### Part 2.2: Microfacet BRDF

The Microfacet BRDF you will implement for this part is a simple linear blend between a diffuse BRDF and a rough dielectric microfacet BRDF. Implement Microfacet.eval() which evaluates the described microfacet BRDF for a given pair of directions in local frame: $f_r(\bold{\omega_i},\bold{\omega_o}) = \frac{k_d}{\pi} + {k_s} \frac{D(\bold{\omega_{h}})~ F\left({(\bold{\omega_h} \cdot \bold{\omega_i})}, \eta_{e},\eta_{i}\right)~ G(\bold{\omega_i},\bold{\omega_o},\bold{\omega_{h}})}{4 \cos{\theta_i} \cos{\theta_o}}, ~~ \bold{\omega_{h}} = \frac{\left(\bold{\omega_i} + \bold{\omega_o}\right)}{\left|\left|\bold{\omega_i} + \bold{\omega_o}\right|\right|_2}$ where $k_d \in [0,1]^3$ is the RGB diffuse reflection coefficient, $k_s = 1 - \max(k_d)$, $F$ is the fresnel reflection coefficient, $\eta_e$ is the exterior index of refraction and $\eta_i$ is the interior index of refraction. The distribution function $D$ is the Beckmann distribution: $D(\bold{\omega_{h}}) = \frac{e^{\frac{-\tan^2{\theta_{h}}}{\alpha^2}}}{\pi\, \alpha^2 \cos^4 \theta_{h} }$ with its corresponding shadowing term approximation (Smith): \begin{align*} G(\bold{\omega_i},\bold{\omega_o},\bold{\omega_{h}}) &= G_1(\bold{\omega_i},\bold{\omega_{h}})~G_1(\bold{\omega_o},\bold{\omega_{h}})\\ G_1(\bold{\omega_v},\bold{\omega_h}) &= \chi^+\left(\frac{\bold{\omega_v}\cdot\bold{\omega_h}}{\bold{\omega_v}\cdot\bold{n}}\right) \begin{cases} \frac{3.535b+2.181b^2}{1+2.276b+2.577b^2} & b \lt 1.6 \\ 1 & \text{otherwise} \end{cases} \\ b &= (a \tan{\theta_v})^{-1}\\ \chi^+(c) &= \begin{cases} 1 & c > 0 \\ 0 & c \le 0 \end{cases} \\ \end{align*} where $\theta_v$ is the angle between the surface normal $\bold{n}$ and the $\omega_v$ argument of $G_1$.

#### Sampling

In this part you will generate samples according to the following density function: $k_s ~ D(\omega_h) \cos{\theta_h} ~ J_h + (1-k_s) \frac{\cos{\theta_o}}{\pi}$ where $J_h = (4 (\omega_h \cdot \omega_o))^{-1}$ is the Jacobian of the half direction mapping discussed in class. Choose whether to sample the diffuse or the specular component based on the value of $k_s$. In the first case, follow the recipe from the slides, by sampling a normal from the microfacet distribution and reflecting the incident direction using this normal. In the latter case, generate a cosine-weighted direction on the sphere using the same method as the model in the wakame.bsdf.Diffuse class.

#### Validation

You can experiment with the warptest GUI and visualize your implemented BRDF. Pass all the $\chi^2$ and t-tests for the Microfacet BRDF found in data/scenes/pa3/tests:

• chi2test-microfacet.xml,
• ttest-microfacet.xml
Note that the WarpTest application also contains a $\chi^2$ test, but this is just to facilitate debugging and visualization; the XML files are the real validation benchmark.

Now that the microfacet BRDF has been implemented, you can render the following scenes:

• veach_mi/veach_ems.xml
• veach_mi/veach_mats.xml

 veach_mi/veach_ems.xml veach_mi/veach_mats.xml

## Part 3: Multiple Importance Sampling

### Integrator Implementation

In this part we will implement another direct illumination integrator (direct_mis) This integrator will combine both sampling strategies for computing direct illumination by using multiple importance sampling with the balance heuristic. At your first camera ray intersection you will sample using both strategies: sampling the emitters and sampling the corresponding BSDF. You will then combine the two estimates by using the following expression: $L_o (p,\omega_o) \approx \frac{1}{N} \sum^N \left( L_e(p,\omega_o) + w_\text{em}\frac{\integrand(p,\omega_o,\omega_{i,e}) }{\text{pdf}_\text{em}(\omega_{i,e})} + w_\text{mat} \frac{\integrand(p,\omega_o,\omega_{i,m}) }{\text{pdf}_\text{mat}(\omega_{i,m})} \right)\\ w_\text{em} = \frac{\text{pdf}_\text{em}(\omega_{i,e}) }{\text{pdf}_{\text{em}}(\omega_{i,e})+\text{pdf}_{\text{mat}}(\omega_{i,e})}, ~ ~ w_\text{mat} = \frac{\text{pdf}_\text{mat}(\omega_{i,m}) }{\text{pdf}_{\text{em}}(\omega_{i,m})+\text{pdf}_{\text{mat}}(\omega_{i,m})}$ For this to work, it is crucial that the probability densities are expressed in terms of the same units (so either density per unit solid angle, or density per unit area). Mixing units will lead to suboptimal weights.

### Validation

Pass the direct_mis related t-tests found in data/scenes/pa2/tests:

• last 4 tests in test-distant.xml,
• last 5 tests in test-mesh.xml,
• last 2 tests in test-mesh-furnace.xml
Mention in your report if there any errors reported.

Using direct_mis render the following scenes found in data/scenes/pa2 :

• sphere/sphere_ao_mis.xml,
• sphere/sphere_side_mis.xml,
• odyssey/odyssey_mis.xml,
• veach_mi/veach_mis.xml

Show a 3-way comparison for each of the 4 scenes in your report. For each scene compare your 3 integrators (direct_ems, direct_mats, direct_mis) with the reference MIS rendering.

 sphere/sphere_ao_mis.xml sphere/sphere_side_mis.xml odyssey/odyssey_mis.xml veach_mi/veach_mis.xml

## Part 4: Path Tracing

### path_mats Implementation

In this part you will implement a naive global illumination path tracing integrator (path_mats) which is an estimator of the rendering equation using only BSDF sampling; paths only contribute when they randomly hit a light source. Your direct_mats integrator should be a great start: you just have to add the recursion. Use the Russian roulette technique as a path termination criterion to avoid infinite recursion.

### path_mis implementation

In this part you will implement a more advanced path tracer (path_mis) which solves the rendering equation by performing both BSDF sampling and emitter sampling and combining the two strategies using multiple importance sampling as discussed in class. Use the Russian roulette technique as a path termination criterion to avoid infinite recursion.

### Validation

Pass the related tests found in data/scenes/pa2/tests:

• test-direct.xml,
• test-furnace.xml

Using path_mats and path_mis render the following scenes found in data/scenes/pa2 :

• cbox/cbox_mats.xml,
• cbox/cbox_mis.xml,
• table/table_mats.xml,
• table/table_mis.xml,
In your report provide two 3-way comparisons of your rendering results and the mis reference.

 cbox/cbox_mats.xml cbox/cbox_mis.xml table/table_mats.xml table/table_mis.xml


text:
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text:
# Difference between revisions of "2010 AIME II Problems/Problem 14"

## Problem

Triangle $ABC$ with right angle at $C$, $\angle BAC < 45^\circ$ and $AB = 4$. Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$. The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$, where $p$, $q$, $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$.

## Solution 1

Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$. It now follows that $\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}$. Hence $ODP$ is isosceles and $OD = DP = 2$.

Denote $E$ the projection of $O$ onto $CD$. Now $CD = CP + DP = 3$. By the Pythagorean Theorem, $OE = \sqrt {2^2 - \frac {3^2}{2^2}} = \sqrt {\frac {7}{4}}$. Now note that $EP = \frac {1}{2}$. By the Pythagorean Theorem, $OP = \sqrt {\frac {7}{4} + \frac {1^2}{2^2}} = \sqrt {2}$. Hence it now follows that,

$$\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}$$

This gives that the answer is $\boxed{007}$.

An alternate finish for this problem would be to use Power of a Point on $BA$ and $CD$. By Power of a Point Theorem, $CP\cdot PD=1\cdot 2=BP\cdot PA$. Since $BP+PA=4$, we can solve for $BP$ and $PA$, giving the same values and answers as above.

$[asy] /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(250); real lsf = 0.5; /* changes label-to-point distance */ pen xdxdff = rgb(0.49,0.49,1); pen qqwuqq = rgb(0,0.39,0); pen fftttt = rgb(1,0.2,0.2); /* segments and figures */ draw((0.2,0.81)--(0.33,0.78)--(0.36,0.9)--(0.23,0.94)--cycle,qqwuqq); draw((0.81,-0.59)--(0.93,-0.54)--(0.89,-0.42)--(0.76,-0.47)--cycle,qqwuqq); draw(circle((2,0),2)); draw((0,0)--(0.23,0.94),linewidth(1.6pt)); draw((0.23,0.94)--(4,0),linewidth(1.6pt)); draw((0,0)--(4,0),linewidth(1.6pt)); draw((0.23,(+0.55-0.94*0.23)/0.35)--(4.67,(+0.55-0.94*4.67)/0.35)); /* points and labels */ label("1", (0.26,0.42), SE*lsf); draw((1.29,-1.87)--(2,0)); label("2", (2.91,-0.11), SE*lsf); label("2", (1.78,-0.82), SE*lsf); pair parametricplot0_cus(real t){ return (0.28*cos(t)+0.23,0.28*sin(t)+0.94); } draw(graph(parametricplot0_cus,-1.209429202888189,-0.24334747753738661)--(0.23,0.94)--cycle,fftttt); pair parametricplot1_cus(real t){ return (0.28*cos(t)+0.59,0.28*sin(t)+0); } draw(graph(parametricplot1_cus,0.0,1.9321634507016043)--(0.59,0)--cycle,fftttt); label("\theta", (0.42,0.77), SE*lsf); label("2\theta", (0.88,0.38), SE*lsf); draw((2,0)--(0.76,-0.47)); pair parametricplot2_cus(real t){ return (0.28*cos(t)+2,0.28*sin(t)+0); } draw(graph(parametricplot2_cus,-1.9321634507016048,0.0)--(2,0)--cycle,fftttt); label("2\theta", (2.18,-0.3), SE*lsf); dot((0,0)); label("B", (-0.21,-0.2),NE*lsf); dot((4,0)); label("A", (4.03,0.06),NE*lsf); dot((2,0)); label("O", (2.04,0.06),NE*lsf); dot((0.59,0)); label("P", (0.28,-0.27),NE*lsf); dot((0.23,0.94)); label("C", (0.07,1.02),NE*lsf); dot((1.29,-1.87)); label("D", (1.03,-2.12),NE*lsf); dot((0.76,-0.47)); label("E", (0.56,-0.79),NE*lsf); clip((-0.92,-2.46)--(-0.92,2.26)--(4.67,2.26)--(4.67,-2.46)--cycle); [/asy]$

## Solution 2

Let $AC=b$, $BC=a$ by convention. Also, Let $AP=x$ and $BP=y$. Finally, let $\angle ACP=\theta$ and $\angle APC=2\theta$.

We are then looking for $\frac{AP}{BP}=\frac{x}{y}$

Now, by arc interceptions and angle chasing we find that $\triangle BPD \sim \triangle CPA$, and that therefore $BD=yb.$ Then, since $\angle ABD=\theta$ (it intercepts the same arc as $\angle ACD$) and $ADB$ is right,

$\cos\theta=\frac{DB}{AB}=\frac{by}{4}$.

Using law of sines on $APC$, we additionally find that $\frac{b}{\sin 2\theta}=\frac{x}{\sin\theta}.$ Simplification by the double angle formula $\sin 2\theta=2\sin \theta\cos\theta$ yields

$\cos \theta=\frac{b}{2x}$.

We equate these expressions for $\cos\theta$ to find that $xy=2$. Since $x+y=AB=4$, we have enough information to solve for $x$ and $y$. We obtain $x,y=2 \pm \sqrt{2}$

Since we know $x>y$, we use $\frac{x}{y}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}$

## Solution 3

Let $\angle{ACP}$ be equal to $x$. Then by Law of Sines, $PB = -\frac{\cos{x}}{\cos{3x}}$ and $AP = \frac{\sin{x}}{\sin{3x}}$. We then obtain $\cos{3x} = 4\cos^3{x} - 3\cos{x}$ and $\sin{3x} = 3\sin{x} - 4\sin^3{x}$. Solving, we determine that $\sin^2{x} = \frac{4 \pm \sqrt{2}}{8}$. Plugging this in gives that $\frac{AP}{PB} = \frac{\sqrt{2}+1}{\sqrt{2}-1} = 3 + 2\sqrt{2}$. The answer is $\boxed{007}$.

## Solution 4 (The quickest and most elegant)

Let $\alpha=\angle{ACP}$, $\beta=\angle{ABC}$, and $x=BP$. By Law of Sines,

$\frac{1}{sin(\beta)}=\frac{x}{sin(90-\alpha)}\implies sin(\beta)=\frac{cos(\alpha)}{x}$ (1), and

$\frac{4-x}{sin(\alpha)}=\frac{4sin(\beta)}{sin(2\alpha)} \implies 4-x=\frac{2sin(\beta)}{cos(\alpha)}$. (2)

Then, substituting (1) into (2), we get

$4-x=\frac{2}{x} \implies x^2-4x+2=0 \implies x=2-\sqrt{2} \implies \frac{4-x}{x}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}$

The answer is $\boxed{007}$. ~Rowechen

## Solution 5

Let $\angle{ACP}=x$. Then, $\angle{APC}=2x$ and $\angle{A}=180-3x$. Let the foot of the angle bisector of $\angle{APC}$ on side $AC$ be $D$. Then,

$CD=DP$ and $\triangle{DAP}\sim{\triangle{APC}}$ due to the angles of these triangles.

Let $CD=a$. By the Angle Bisector Theorem, $\frac{1}{a}=\frac{AP}{AD}$, so $AD=a\cdot{AP}$. Moreover, since $CD=DP=a$, by similar triangle ratios, $\frac{AP}{a+a\cdot{AP}}=a$. Therefore, $AP = \frac{a^2}{1-a^2}$.

Construct the perpendicular from $D$ to $AP$ and denote it as $F$. Denote the midpoint of $CP$ as $M$. Since $PD$ is an angle bisector, $PF$ is congruent to $PM$, so $PF=\frac{1}{2}$.

Also, $\triangle{DFA}\sim{\triangle{BCA}}$. Thus, $\frac{FA}{AC}=\frac{AD}{AB}\Longrightarrow\frac{\frac{a^2}{1-a^2}-\frac{1}{2}}{a+\frac{a^3}{1-a^2}}=\frac{\frac{a^3}{1-a^3}}{4}$. After some major cancellation, we have $7a^4-8a^2+2=0$, which is a quadratic in $a^2$. Thus, $a^2 = \frac{4\pm\sqrt{2}}{7}$.

Taking the negative root implies $AP, contradiction. Thus, we take the positive root to find that $AP=2+\sqrt{2}$. Thus, $BP=2-\sqrt{2}$, and our desired ratio is $\frac{2+\sqrt{2}}{2-\sqrt{2}}\implies{3+2\sqrt{2}}$.

The answer is $\boxed{007}$.

## Solution 6

Let $O$ be the circumcenter of $\triangle ABC$. Since $\triangle ABC$ is a right triangle, $O$ will be on $\overline{AB}$ and $\overline{AO} \cong \overline{OB} \cong \overline{OC} = 2$. Let $\overline{OP} = x$.

Next, let's do some angle chasing. Label $\angle ACP = \theta^{\circ}$, and $\angle APC = 2\theta^{\circ}$. Thus, $\angle PAC = (180-3\theta)^{\circ}$, and by isosceles triangles, $\angle ACO = (180-3\theta)^{\circ}$. Then, by angle subtraction, $\angle OCP = (\theta - (180-3\theta))^{\circ} = (4\theta - 180)^{\circ}$.

Using the Law of Sines: $$\frac{x}{\sin (4\theta-180)^{\circ}}=\frac{2}{\sin (2\theta)^{\circ}}$$Using trigonometric identies, $\sin (4\theta-180)^{\circ}=-\sin (4\theta)^{\circ}=-2\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}$. Plugging this back into the Law of Sines formula gives us: $$\frac{x}{-2\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}}=\frac{2}{\sin (2\theta)^{\circ}}$$

$$-4\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}=x\sin (2\theta)^{\circ}$$ $$-4\cos (2\theta)^{\circ}=x$$ $$\cos(2\theta)^{\circ}=\frac{-x}4$$

Next, using the Law of Cosines: $$2^2=1^2+x^2-2\cdot 1\cdot x\cdot \cos (2\theta)^{\circ}$$ Substituting $\cos(2\theta)^{\circ}=\frac{-x}4$ gives us: $$2^2=1^2+x^2-2\cdot 1\cdot x\cdot \frac{-x}4$$ $$4=1+x^2+\frac{x^2}{2}$$

Solving for x gives $x=\sqrt 2$

Finally: $\frac{\overline{AP}}{\overline{BP}}=\frac{\overline{AO}+\overline{OP}}{\overline{BO}-\overline{OP}}=\frac{2+\sqrt 2}{2-\sqrt 2}=3+2\sqrt2$, which gives us an answer of $3+2+2=\boxed{007}$. ~adyj


text:
Article

# Generalized Commutator Formulas

(Impact Factor: 0.39). 04/2011; 39(4):1441-1454. DOI: 10.1080/00927871003738964

ABSTRACT

Let A be an algebra which is a direct limit of module finite algebras over a commutative ring R with 1. Let I, J be two-sided ideals of A, GL n (A, I) the principal congruence subgroup of level I in GL n (A), and E n (A, I) the relative elementary subgroup of level I. Using Bak's localization-patching method, we prove the commutator formula which is a generalization of the standard commutator formular. This answers a problem posed by Stepanov and Vavilov.

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• "This was done by L. Vaserstein in [37]. Relative commutator formula was obtained by different methods in [39] [40] [17] [15]. Relative multi-commutator formula for G = GL n was recently proved by R. Hazrat and Zhang Zuhong in [18]. "
##### Article: Non-Abelian K-Theory for Chevalley Groups over Rings
[Hide abstract]
ABSTRACT: The article contains a survey of the recent author’s results on the structure of a Chevalley group G(R) over a ring R. They generalize and improve previously known results on: (1) the relative local-global principle; (2) generators of a relative elementary group; (3) relative multicommutator formulas; (4) the nilpotent structure of a relative K1; (5) the bounded length of commutators. The proof of the first two items is based on computations with generators of the elementary group translated into the language of parabolic subgroups. The other results are proved by means of enlarging a relative elementary group, constructing a generic element, and using the localization procedure in the universal ring.
Journal of Mathematical Sciences 08/2015; 209(4). DOI:10.1007/s10958-015-2518-y
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• "The reason is that the first of these choices is too small as the neighbourhood on the right hand side, while the second of these choices is too large as the neighbourhood on the left hand side. The solution proposed for GL(n, R) in [33] and later applied to unitary groups in [32] consists in selecting another base of neightborhoods E(Φ, s k a) ≤ E(Φ, s k R, s k a) ≤ E(Φ, R, s k a), which is much better balanced with respect to conjugation. The following definition embodies the gist of this method. "
##### Article: Relative commutator calculus in Chevalley groups
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ABSTRACT: We revisit localisation and patching method in the setting of Chevalley groups. Introducing certain subgroups of relative elementary Chevalley groups, we develop relative versions of the conjugation calculus and the commutator calculus in Chevalley groups $G(\Phi,R)$, $\rk(\Phi)\geq 2$, which are both more general, and substantially easier than the ones available in the literature. For classical groups such relative commutator calculus has been recently developed by the authors in \cite{RZ,RNZ}. As an application we prove the mixed commutator formula, $\big [E(\Phi,R,\ma),C(\Phi,R,\mb)\big ]=\big [E(\Phi,R,\ma),E(\Phi,R,\mb)\big],$ for two ideals $\ma,\mb\unlhd R$. This answers a problem posed in a paper by Alexei Stepanov and the second author.
Journal of Algebra 07/2013; 385:262-293. DOI:10.1016/j.jalgebra.2013.03.011 · 0.60 Impact Factor
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• "In fact, localisation methods used in the proof of these results have many further applications, both actual and potential: relative commutator formulas, multiple commutator formulas, nilpotency of K 1 , description of subnormal subgroups, description of various classes of overgroups, connection with excision kernels, etc. We refer to our surveys [36] [31] [32] and to our papers [29] [35] [7] [40] [37] [38] [41] [76] [39] [33] [34] for these and further applications and many further related references. "
##### Article: Commutator width in Chevalley groups
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ABSTRACT: The present paper is the [slightly expanded] text of our talk at the Conference "Advances in Group Theory and Applications" at Porto Cesareo in June 2011. Our main results assert that [elementary] Chevalley groups very rarely have finite commutator width. The reason is that they have very few commutators, in fact, commutators have finite width in elementary generators. We discuss also the background, bounded elementary generation, methods of proof, relative analogues of these results, some positive results, and possible generalisations.
Note di Matematica 06/2012; 33(1). DOI:10.1285/i15900932v33n1p139


text:
Suboptimal boundary controls for elliptic equation in critically perforated domain
Annales de l'I.H.P. Analyse non linéaire, Tome 25 (2008) no. 6, pp. 1073-1101.
@article{AIHPC_2008__25_6_1073_0,
author = {D'Apice, Ciro and De Maio, Umberto and Kogut, Peter I.},
title = {Suboptimal boundary controls for elliptic equation in critically perforated domain},
journal = {Annales de l'I.H.P. Analyse non lin\'eaire},
pages = {1073--1101},
publisher = {Elsevier},
volume = {25},
number = {6},
year = {2008},
doi = {10.1016/j.anihpc.2007.07.001},
zbl = {1170.35015},
mrnumber = {2466322},
language = {en},
url = {http://archive.numdam.org/item/AIHPC_2008__25_6_1073_0/}
}
D'Apice, Ciro; De Maio, Umberto; Kogut, Peter I. Suboptimal boundary controls for elliptic equation in critically perforated domain. Annales de l'I.H.P. Analyse non linéaire, Tome 25 (2008) no. 6, pp. 1073-1101. doi : 10.1016/j.anihpc.2007.07.001. http://archive.numdam.org/item/AIHPC_2008__25_6_1073_0/

[1] Bouchitté G., Fragala I., Homogenization of thin structures by two-scale method with respect to measures, SIAM J. Math. Anal. 32 (6) (2001) 1198-1226. | MR 1856245 | Zbl 0986.35015

[2] G. Buttazzo, Γ-convergence and its applications to some problem in the calculus of variations, in: School on Homogenization, ICTP, Trieste, September 6-17, 1993, 1994, pp. 38-61.

[3] Buttazzo G., Dal Maso G., Γ-convergence and optimal control problems, J. Optim. Theory Appl. 32 (1982) 385-407. | MR 686213 | Zbl 0471.49012

[4] Carbone L., De Arcangelis R., Unbounded Functionals in the Calculus of Variations. Representation, Relaxation, and Homogenization, Chapman and Hall/CRC, New York, 2002. | MR 1910459 | Zbl 1002.49018

[5] Cardone G., D'Apice C., De Maio U., Homogenization in perforated domains with mixed conditions, Nonlinear Diff. Equ. Appl. 9 (2002) 246-325. | MR 1917377 | Zbl 1046.35007

[6] Casado-Díaz J., Existence of a sequence satisfying Cioranescu-Murat conditions in homogenization of Dirichlet problems in perforated domains, Rend. Mat. Appl. (7) 16 (1996) 387-413. | MR 1422390 | Zbl 0870.35013

[7] Cioranescu D., Donato P., Murat F., Zuazua E., Homogenization and correctors for the wave equation in domains with small holes, Ann. Sc. Norm. Super. Pisa, Sc. Fis. Mat. 17 (4) (1991) 251-293. | Numdam | MR 1129303 | Zbl 0807.35077

[8] Cioranescu D., Donato P., Zuazua E., Exact boundary controllability for the wave equation in domains with small holes, J. Math. Pures Appl. 71 (1992) 343-377. | MR 1176016 | Zbl 0843.35009

[9] Cioranescu D., Murat F., Un terme étrage venu d'ailleurs, in: Nonlinear Partial Differential Equations and their applications. Collége de France Seminar, Research Notes in Mathematics, Pitman, London, 1981, vol. II, pp. 58-138, vol. III, pp. 157-178. | Zbl 0498.35034

[10] Cioranescu D., Saint Jean Paulin J., Homogenization in open sets with holes, J. Math. Anal. Appl. 71 (1978) 590-607. | MR 548785 | Zbl 0427.35073

[11] Conca C., Donato P., Nonhomogeneous Neumann problems in domains with small holes, Modélisation Mathématique et Analyse Numérique 22 (4) (1988) 561-608. | Numdam | MR 974289 | Zbl 0669.35028

[12] Corbo Esposito A., D'Apice C., Gaudiello A., A homogenization problem in a perforated domain with both Dirichlet and Neumann conditions on the boundary of the holes, Asymptodic Anal. 31 (2002) 297-316. | MR 1937842 | Zbl 1043.35027

[13] Coron J.-M., Crépeau E., Exact boundary controllability of a nonlinear KdV equation with critical length, J. Eur. Math. Soc. (JEMS) 6 (3) (2004) 367-398. | MR 2060480 | Zbl 1061.93054

[14] Dal Maso G., Murat F., Asymptotic behaviour and correctors for Dirichlet problem in perforated domains with homogeneous monotone operators, Ann. Sc. Norm. Sup. Pisa Cl. Sci. 24 (4) (1997) 239-290. | Numdam | MR 1487956 | Zbl 0899.35007

[15] Evans L.C., Gariepy R.F., Measure Theory and Fine Properties of Functions, CRC Press, Boca Raton, FL, 1992. | MR 1158660 | Zbl 0804.28001

[16] Fursikov A.V., Optimal Control of Distributed Systems. Theory and Applications, Amer. Math. Soc., 2000. | MR 1726442 | Zbl 1027.93500

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[19] Kogut P.I., S-convergence in homogenization theory of optimal control problems, Ukrain. Mat. Zh. 49 (11) (1997) 1488-1498, (in Russian); English transl. in:, Ukrainian Math. J. 49 (11) (1997) 1671-1682. | MR 1672876 | Zbl 0933.93026

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text:
## hostapd

hostapd is a user space daemon for access point and authentication servers. It implements IEEE 802.11 access point management, IEEE 802.1X/WPA/WPA2/EAP Authenticators, RADIUS client, EAP server, and RADIUS authentication server. The current version supports Linux (Host AP, madwifi, mac80211-based drivers) and FreeBSD (net80211).

hostapd is designed to be a “daemon” program that runs in the background and acts as the backend component controlling authentication. hostapd supports separate frontend programs and an example text-based frontend, hostapd_cli, is included with hostapd.

IEEE 802.1X defines the encapsulation of the Extensible Authentication Protocol (EAP) over IEEE 802,[1][2] which is known as “EAP over LAN” or EAPOL.[3] EAPOL was originally designed for IEEE 802.3 Ethernet in 802.1X-2001, but was clarified to suit other IEEE 802 LAN technologies such as IEEE 802.11 wireless and Fiber Distributed Data Interface (ISO 9314-2) in 802.1X-2004.[4] The EAPOL protocol was also modified for use with IEEE 802.1AE (“MACsec”) and IEEE 802.1AR (Secure Device Identity, DevID) in 802.1X-2010[5][6] to support service identification and optional point to point encryption over the local LAN segment.

# 启用IP转发

IP forwarding” is a synonym for “routing.” It is called “kernel IP forwarding” because it is a feature of the Linux kernel.

A router has multiple network interfaces. If traffic comes in on one interface that matches a subnet of another network interface, a router then forwards that traffic to the other network interface.

So, let’s say you have two NICs, one (NIC 1) is at address 192.168.2.1/24, and the other (NIC 2) is 192.168.3.1/24. If forwarding is enabled, and a packet comes in on NIC 1 with a “destination address” of 192.168.3.8, the router will resend that packet out of the NIC 2.

It’s common for routers functioning as gateways to the Internet to have a default route whereby any traffic that doesn’t match any NICs will go through the default route’s NIC. So in the above example, if you have an internet connection on NIC 2, you’d set NIC 2 as your default route and then any traffic coming in from NIC 1 that isn’t destined for something on 192.168.2.0/24 will go through NIC 2. Hopefully there’s other routers past NIC 2 that can further route it (in the case of the Internet, the next hop would be your ISP’s router, and then their providers upstream router, etc.)

Enabling ip_forward tells your Linux system to do this. For it to be meaningful, you need two network interfaces (any 2 or more of wired NIC cards, Wifi cards or chipsets, PPP links over a 56k modem or serial, etc.).

When doing routing, security is important and that’s where Linux’s packet filter, iptables, gets involved. So you will need an iptables configuration consistent with your needs.

Note that enabling forwarding with iptables disabled and/or without taking firewalling and security into account could leave you open to vulnerabilites if one of the NICs is facing the Internet or a subnet you don’t have control over.

## iptables

IPTABLES又是很蛋疼很复杂的东西，man iptables一共有将近2000页orz，我表示真的没兴趣能力读下来。

iptables 可以检测、修改、转发、重定向和丢弃 IPv4 数据包。过滤 IPv4 数据包的代码已经内置于内核中，并且按照不同的目的被组织成 表 的集合。表 由一组预先定义的 链 组成，链 包含遍历顺序规则。每一条规则包含一个谓词的潜在匹配和相应的动作（称为 目标），如果谓词为真，该动作会被执行。也就是说条件匹配。iptables 是用户工具，允许用户使用 链 和 规则。

This target is only valid in the nat table, in the POSTROUTING chain. It should only be used with dynamically assigned IP (dialup) connections: if you have a
static IP address, you should use the SNAT target. Masquerading is equivalent to specifying a mapping to the IP address of the interface the packet is going
out, but also has the effect that connections are forgotten when the interface goes down. This is the correct behavior when the next dialup is unlikely to have
the same interface address (and hence any established connections are lost anyway).

ESTABLISHED The ESTABLISHED state has seen traffic in both directions and will then continuously match those packets. ESTABLISHED connections are fairly easy to understand. The only requirement to get into an ESTABLISHED state is that one host sends a packet, and that it later on gets a reply from the other host. The NEW state will upon receipt of the reply packet to or through the firewall change to the ESTABLISHED state. ICMP reply messages can also be considered as ESTABLISHED, if we created a packet that in turn generated the reply ICMP message.
RELATED The RELATED state is one of the more tricky states. A connection is considered RELATED when it is related to another already ESTABLISHED connection. What this means, is that for a connection to be considered as RELATED, we must first have a connection that is considered ESTABLISHED. The ESTABLISHED connection will then spawn a connection outside of the main connection. The newly spawned connection will then be considered RELATED, if the conntrack module is able to understand that it is RELATED. Some good examples of connections that can be considered as RELATED are the FTP-data connections that are considered RELATED to the FTP control port, and the DCC connections issued through IRC. This could be used to allow ICMP error messages, FTP transfers and DCC’s to work properly through the firewall. Do note that most TCP protocols and some UDP protocols that rely on this mechanism are quite complex and send connection information within the payload of the TCP or UDP data segments, and hence require special helper modules to be correctly understood.

(噪点好多TAT)


text:
## Using and DerivingAlgebraic Properties of Logarithms miscellaneous on-line topics for Calculus Applied to the Real World

Exercises for This Topic
Index of On-Line Topics
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Logarithms

We start by reviewing the basic definitions as in Section 2.3 of Calculus Applied to the Real World. If you like, you can also take a look at the topic summary material on logarithms.

Base b Logarithm
The base $b$ logarithm of $x,$ $\log_b x,$ is the power to which you need to raise $b$ in order to get $x.$ Symbolically,
 $\log_b x = y$ means $b_y = x.$ Logarithmic form Exponential form
Notes
1. $\log_b x$ is only defined if $b$ and $x$ are both positive, and $b \neq 1.$
2. $\log_10 x$ is called the common logarithm of $x,$ and is sometimes written as $\log x.$
3. $\log_e x$ is called the natural logarithm of $x$ and is sometimes written as $\ln x.$

Examples
The following table lists some exponential equations and their equivalent logarithmic form.

 Exponential Form $10^3 = 1\,000$ $4^2 = 16$ $3^3 = 27$ $5^1 = 5$ $7^0 = 1$ $4^{-2} = 1/16$ $25^{1/2} = 5$ Logarithmic Form $\log_{10} 1\,000 = 3$ $\log_4 16 = 2$ $\log_3 27 = 3$ $\log_5 5 = 1$ $\log_7 1 = 0$ $\log_4 (1/16) = -2$ $\log_{25} 5 = 1/2$

Here are some for you to try

Exponential Form$10^2 = 100$$3^{-2} = 1/9$
Logarithmic Form
 $\log$ =

 $\log$ =

 Exponential Form ^ = ^ = Logarithmic Form $\log_3 1 =0$ $\log_5 (1/125) = -3$

Example 1 Calculating Logarithms by Hand

 (a) $\log_2 8$ = Power to which you need to raise $2$ in order to get $8$ = 3       Since $2^3 = 8$ (b) $\log_4 1$ = Power to which you need to raise $4$ in order to get $1$ = $0$       Since $4^0 = 1$ (c) $\log_{10} 10\,000$ = Power to which you need to raise $10$ in order to get $10\,000$ = $4$       Since $10^4 = 10\,000$ (d) $\log_{10} 1/100$ = Power to which you need to raise $10$ in order to get $1/100$ = $-2$       Since $10^{-2} = 1/100$ (e) $\log_3 27$ = (f) $\log_9 3$ = (g) $\log_3 (1/81)$ =

Algebraic Properties of Logarithms

The following identities hold for any positive $a \neq 1$ and any positive numbers $x$ and $y.$

Identity
Example
(a) $\log_a(xy) = \log_ax + \log_ay$ $\log_216 = \log_28 + \log_22$
(b) $\log_a\left( \frac{x}{y} \right) = \log_ax - \log_a y$ $\log_2 \left( \frac{5}{3} \right) = \log_25 - \log_23$
(c) $\log_a(x^r) = r \log_ax$ $\log_2(6^5) = 5 \log_26$
 (d) $\log_aa=1$ $\log_a1=0$
 $\log_22 = 1$ $\log_31 = 0$
(e) $\log_a\left( \frac{1}{x} \right) = -\log_ax$ $\log_2\left( \frac{1}{3} \right) = -\log_23$
(f) $\log_ax = \frac{\log x}{\log a} = \frac{\ln x}{\ln a}$ $\log_25 = \frac{\log 5}{\log 2}\approx 2.3219$

Example 2 Using the Properties of Logarithms

Let $a = \log 2,$ $b = \log 3,$ and $c = \log 5.$ Write the following in terms of $a,$ $b,$ and $c.$
Note If any answer you give is not simplified -- for instance, if you say $a + a$ instead of $2a$ -- it will be marked wrong.

 Answer (a) $\log 6$ $\log 2 + \log 3 = a + b$ (b) $\log 15$ (c) $\log 30$ $\log 2 + \log 3 + \log 5 = a + b + c$ (d) $log 12$ (e) $\log 1.5$ $\log 3 - \log 2 = b - a$ (f) $\log(1/9)$ (g) $\log 32$ $\log 2 ^5 = 5 \log 2 = 5a$ (h) $\log(1/81)$

Q Where do the identities come from?
A Roughly speaking, they are restatements in logarithmic form of the laws of exponents.

Q Why is $\log_axy = \log_ax + \log_ay$ ?
A Let $s = \log_ax,$ and $t = \log_a y.$ In exponential form, these equations say that
$a^s = x$ and $a^t = y.$
Multiplying these two equations together gives
$a^s a^t = xy,$
that is,
$a^{s+t} = xy.$
Rewriting this in logarithmic form gives
$\log_a (xy) = s + t = \log_a x + \log_a y$
as claimed.

Here is an intuitive way of thinking about it: Since logs are exponents, this identity expresses the familiar law that the exponent of a product is the sum of the exponents.

The second logarithmic identity is shown in almost the identical way, and we leave it for you for practice.

Q Why is $\log_a(x^r) = r \log_ax$ ?
A Let $t = \log_a x.$ Writing this in exponential form gives
$a^t = x.$
Raising this equation to the $r^{th}$ power gives
$a^{rt} = x^{r}.$
Rewriting in logarithmic form gives
$\log_a (x^r) = rt = r \log_a x,$
as claimed.

Identity (d) we will leave for you to do as practice.

Q Why is $\log_a(1/x) = -\log_ax$ ?
A This follows from identities (b) and (d) (think about it). <>

Q Why is

$\log_ax = \frac{\log x}{\log a} = \frac{\ln x}{\ln a}$?
A Let $s = \log_a x.$ In exponential form, this says that
$a^s = x.$
Take the logarithm with base b of both sides, getting
$\log_b a^s = \log_b x,$
then use identity (c):
$s \log_b a = \log b x,$
so
$s = \frac{\log_bx}{\log_ba}$

Since logarithms are exponents, we can use them to solve equations where the unknown is in the exponent.

Example 3 Solving for the Exponent

Solve the following equations for $x.$
(a) $4^{-x^2} = 1/64.$
(b) $5 (1.1^{2x+3}) = 200$

Solution We can solve both of these equations by translating from exponential form to logarithmic form.

(a) Write the given equation in logarithmic form:

 $4^{-x^2} = 1/64$ Exponential Form $\log_4(1/64) = -x^2$ Logarithmic Form Thus, $-x^2 = \log_4(1/64) = -3$ giving $x = ±3^{1/2}.$

(b) Before converting to logarithmic form, first divide both sides of the equation by 5:
 $5 (1.1^{2x+3}) = 200$ $1.1^{2x+3} = 40$ Exponential Form $\log_{1.1}40 = 2x+3$ Logarithmic Form This gives $2x + 3 = \ln 40/\ln 1.1 \approx 38.7039,$ Identity (e) so that $x \approx 17.8520.$

You can now either go on and try the exericses in the exercise set for this topic.

Last Updated:October, 1999


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07 Feb 2019, 19:28
1
1
MathRevolution wrote:
[GMAT math practice question]

If $$(x-7)^2=-|y-5|, xy=?$$

$$A. 5$$
$$B. 7$$
$$C. 12$$
$$D. 35$$
E. cannot be determined

A square and a modulus can never be NEGATIVE

So least value of a square or modulus is 0..

$$(x-7)^2=-|y-5|$$ ..
Here -|y-5| will never be positive as |y-5| is greater than or equal to 0..
A square on left side tells us that (x-7)^2=0, or x=7..
And -|y-5|=0 means y =5
Therefore xy=7*5=35

D
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05 Feb 2019, 00:30
If y>5, then -|y-5| = -(y-5) = 5-y
$$x^2+49-14x = 5-y$$
$$x^2+44-14x+y = 0$$........(1)
If y<5, then -|y-5| = -(-y+5) = y-5
$$x^2+49-14x = y-5$$
$$x^2+54-14x = y$$..........(2)
(1) in (2) gives
$$2x^2+98-28x = 0$$
$$x^2+49-14x = 0$$
$$(x-7)^2 = 0$$
x = 7
If x = 7, then -|y-5| = 0; y = 5
xy = 35

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05 Feb 2019, 01:29
MathRevolution wrote:
[GMAT math practice question]

If $$(x-7)^2=-|y-5|, xy=?$$

$$A. 5$$
$$B. 7$$
$$C. 12$$
$$D. 35$$
E. cannot be determined

|y-5|= sqrt ( y-5)^2

and
(x-7)^2 = sqrt ( (x-5)^2
squaring both sides
(x-7) = (x-5)^2
solve
y^2-10y+32-x=0-- ( 1)

x*y=35
x=7 & y = 5
satisfies the eqn ( 1)
so IMO C
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05 Feb 2019, 03:57
I assume

x=7 because (x-7)=0, therefore x= 7 when you rearrange

then
7= -|y-5|
7= -(y-5)
7= -y+5
y= -2

Therefore x*y= -14

Is it E based on my assumption above? I don't think so but made sense to me.
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05 Feb 2019, 04:14
MathRevolution wrote:
[GMAT math practice question]

If $$(x-7)^2=-|y-5|, xy=?$$

$$A. 5$$
$$B. 7$$
$$C. 12$$
$$D. 35$$
E. cannot be determined

We can start with plugging in the values for x and y

Only D when plugged in the $$(x-7)^2=-|y-5|$$, gives equal values

1 * 5 or 5 * 1
1 * 7 or 7 * 1
3 * 4 or 4 * 3 or 2 * 6 or 6 * 2
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05 Feb 2019, 04:19
Albs wrote:
I assume

x=7 because (x-7)=0, therefore x= 7 when you rearrange

then
7= -|y-5|
7= -(y-5)
7= -y+5
y= -2

Therefore x*y= -14

Is it E based on my assumption above? I don't think so but made sense to me.

Hi Albs

the highlightedvalue when substituted back in the bold part should had satisfied the modulus, which it doesn't do, Therefore the value cannot be applicable to this case.

so you got y = -2

lets put it back in 7= -|y-5|

7 ! = - 7

You have to search for different cases for x and y to satisfy (x−7)^2=−|y−5|

But then you are provided with the product of x and y, just substitute them in place of x and y and you should be good.
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05 Feb 2019, 05:43
MathRevolution wrote:
[GMAT math practice question]

If $$(x-7)^2=-|y-5|, xy=?$$

$$A. 5$$
$$B. 7$$
$$C. 12$$
$$D. 35$$
E. cannot be determined

$${\left( {x - 7} \right)^2} = - \left| {y - 5} \right|\,\,\,\,\,\left( * \right)$$

$$? = xy$$

$$\left. \matrix{ {\left( {x - 7} \right)^2} \ge 0\,\,\, \hfill \cr - \left| {y - 5} \right| \le 0 \hfill \cr} \right\}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,{\left( {x - 7} \right)^2} = 0 = - \left| {y - 5} \right|\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\{ \matrix{ \,x - 7 = 0 \hfill \cr \,y - 5 = 0 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 5 \cdot 7 = 35$$

The correct answer is therefore (D).

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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05 Feb 2019, 06:40
I have a query.

|y-5| will always be +ve an (x-7)^2 = -|y-5| can't be possible as LHS is a square number?
So how can we find a deterministic answer in such case?

Regards,
Rishav
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05 Feb 2019, 06:41
rish2708 wrote:
I have a query.

|y-5| will always be +ve an (x-7)^2 = -|y-5| can't be possible as LHS is a square number?
So how can we find a deterministic answer in such case?

Regards,
Rishav

Got it.. only 0 it can satisfy!! Thanks
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05 Feb 2019, 09:28
MathRevolution wrote:
[GMAT math practice question]

If $$(x-7)^2=-|y-5|, xy=?$$

$$A. 5$$
$$B. 7$$
$$C. 12$$
$$D. 35$$
E. cannot be determined

Since Square of any number is >=0; $$A^2>=0$$
$$(x-7)^2=A,-|y-5|=B$$
since A cant be negative ,So B=0
y=5 and x=7
X*Y=35
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07 Feb 2019, 17:29
=>
$$(x-7)^2=-|y-5|$$
$$=> (x-7)^2+|y-5| = 0$$
$$=> x = 7$$ and $$y = 5$$
This yields $$xy = 35$$.

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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Intern Joined: 29 Dec 2018 Posts: 7 Re: If (x-7)^2=-|y-5|, xy=? [#permalink] ### Show Tags 07 Feb 2019, 18:28 MathRevolution Sorry to say but your answer gives no additional knowledge and thus is not really useful in this case. The point to observe is that a perfect square has been equated a negative absolute value function. We must note that a perfect square can never be negative. It can only achieve a value of zero or positive. Same is the case with absolute value function. This, to ensure that both the functions yield the same answer, we must equate both the function to zero. This, x=7 and y=5 XY = 35 Posted from my mobile device Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6949 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If (x-7)^2=-|y-5|, xy=? [#permalink] ### Show Tags 09 Feb 2019, 15:32 KanishkM wrote: Albs wrote: I assume x=7 because (x-7)=0, therefore x= 7 when you rearrange then 7= -|y-5| 7= -(y-5) 7= -y+5 y= -2 Therefore x*y= -14 Is it E based on my assumption above? I don't think so but made sense to me. Hi Albs the highlightedvalue when substituted back in the bold part should had satisfied the modulus, which it doesn't do, Therefore the value cannot be applicable to this case. so you got y = -2 lets put it back in 7= -|y-5| 7 ! = - 7 You have to search for different cases for x and y to satisfy (x−7)^2=−|y−5| But then you are provided with the product of x and y, just substitute them in place of x and y and you should be good. If x = 7, then we have (x-7)^2 = -|y-5| or (7-7)^2 = -|y-5|. Thus -|y-5| = 0 and we have y = 5. x*y = 7*5 = 35. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Re: If (x-7)^2=-|y-5|, xy=?   [#permalink] 09 Feb 2019, 15:32
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