Sunday, January 23, 2022

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange


Is multiplying a constant matrix to a multivariate normal another multivariate normal??

Posted: 23 Jan 2022 12:50 AM PST

Suppose that we have an $K\times1$ OLS estimator: $$\hat{\beta}\xrightarrow{\quad d \quad}N(\beta, V).$$

That is, the estimator is a multivariate normal random vector.

Let $A$ be a $M\times K$ constant matrix (note that the matrix is not a square matrix).

Then, does the following statement hold always?

$$A\hat{\beta}\xrightarrow{\quad d \quad} N(A\beta, AVA').$$

That is, does multiplying a constant matrix always preseve the normality?

Why mean is the balancing point?

Posted: 23 Jan 2022 12:48 AM PST

4,5,7,12,34 are the data points. Median is 7. But, why mean would be the balancing point here? I know the mean distribute the data points evenly among individuals, which means if any individual would've scored any point, he'd have scored the mean . But, I don't understand this balancing point thing.

Why not every negative odd number can be positive

Posted: 23 Jan 2022 12:48 AM PST

I have this equation: $$ -3=\sqrt[3]{-27}=\left(-27\right)^{\frac{1}{3}}=\left(-27\right)^{\frac{2}{6}}=\sqrt[6]{\left(-27\right)^2}=\sqrt[6]{729}=3 $$

But I don't know where the error is that has been made that -3 equals to 3.

Why can more general problems — paradoxically — be easier to solve or prove?

Posted: 23 Jan 2022 12:53 AM PST

These answers list instances of the Inventor's Paradox, but doesn't expatiate WHY this can happen?

Here's an analogy. Let Dialect 1 have two grammars, one Traditional and one Simplified grammar that all D1 speakers know fluently. Let Dialect 2 be Dialect 1 with merely Simplified grammar. By my intentional construction, Dialect 1 is generaller than 2, and Dialect 2 is easier to learn.

Yet the Inventor's Paradox says the opposite! Incontrovertibly, it's counterintuitive that Dialect 1 would be easier to learn. What's wrong with my analogy?

By the bye, if you're an linguist, feel free to instantiate Dialects 1 and 2. Do such Dialects 1 and 2 exist in real life?

Does there always exist an element such that the relative trace equal to zero over the finite fields with even characteristic?

Posted: 23 Jan 2022 12:45 AM PST

Let $k$ be an odd positive integer and $j$ an even positive integer with $1\leq j\leq k-1$. Let $t={\rm gcd}(k,j)$. Let $\mathbb{F}_{2^k}$ be the finite fields. Denote ${\rm Tr}^{k}_{t}(x):=x+x^{2^t}+\cdots+x^{2^{k/t-1}}$ for an element $x$ in $\mathbb{F}_{2^k}$. I guess that there always exist an element in $x\in \mathbb{F}_{2^k}$ such that ${\rm Tr}^{k}_{t}(x^{2^t-1})=0.$

For the case of ${\rm gcd}(k,j)=1$, this assertion clearly holds.

About the definition of the closure of topological space.

Posted: 23 Jan 2022 12:45 AM PST

Let $X$ be a topological space.

For $A\subset X$, one of the definitions of the closure of $ A$ is $$cl(A):=\displaystyle\bigcap_{F \in \mathcal F_A} F , \ \mathrm{where} \ \mathcal F_A=\{ F \subset X \mid A \subset F,\ F : \mathrm{closed}\}.$$

My question ;

Does $\mathcal F_A \neq \emptyset$ hold for $A\neq \emptyset$ ?

If $A$ is closed, $A\in \mathcal F_A \neq \emptyset$ but does $\mathcal F_A\neq \emptyset$ also hold for non-closed set $A$ ?

I think the whole set $X$ is included in $\mathcal F_A$ since $A\subset X$ and $X$ is topological space so $X$ itself is closed, and thus $X \in \mathcal F_A \neq \emptyset.$

Is this idea correct ?

$|f|$ is not integrable but $f$ is integrable

Posted: 23 Jan 2022 12:45 AM PST

Could someone provide me an example of a function $f:[a, +\infty[\rightarrow \mathbb{R}$ such that $|f(x)|$ is not integrable in $[a, +\infty[$ but f is integrable in the same interval?

Find out the wrong number in the given series. 64, 24, 31, –32, –81, –174?

Posted: 23 Jan 2022 12:45 AM PST

Find out the wrong number in the given series $64$, $24$, $31$, $–32$, $–81$, $–174$

(a) $24$ (b) $31$ (c) $–32$ (d) $–81$

$31$ seems to be an answer as it is a prime number, but i'm not sure. how to approach such question, kindly looking for help, thanks.

Computing house edge of Lunar New Year dice game

Posted: 23 Jan 2022 12:46 AM PST

Here is an apt description of the game: https://medium.com/@Rob_TactVP/how-to-play-hoo-hey-how-812532d405a1

I'll repeat the rules of the game here.

Traditionally, Hoo Hey How uses the following pay table:

  • 1:1 for one matching symbol.
  • 2:1 for two matching symbols.
  • 3:1 for three matching symbols.
  • 30:1 for any triple.

For example: Betting \$5 on "Fish" and having one fish symbol turn up nets \$10 (your initial \$5, plus \$5 in winnings). Betting \$5 on "Fish" and having it appear on two dice, returns \$15 (your \$5 wager and \$10 in winnings). Three symbols pays \$20 (your \$5 bet and \$15 profit). However, you're also allowed to wager on a "Triple" option. This pays out at 30 to one, making it look like an enticing option. It's not. The odds of all three dice displaying the same symbol are very slim.

I wanted first to evaluate the expected value of betting on any of the 6 symbol choices, and ignore the Triple bet for now.

I came up with an expected value of \$0.875 for an investment of \$1, which is alarmingly low and significantly worse than the supposed 8% house advantage that I have been reading about (including from this article).

So let me walk you through the logic that I used to calculate this.

First, the probability of getting all 3 dice matching your chosen symbol: $(1/6)^3 = 1/216$

Second, the probability of getting exactly 2 dice matching your chosen symbol: $(1/6)^2 (5/6) = 5/216$

Third, the probability of getting exactly 1 die matching your chosen symbol which I derived by first finding the probability of at least 1 die matching, which should be: $1 - (5/6)^3$

This is the inverse of the probability of all 3 dice not matching (each with a 5/6 chance of that).

Therefore the probability of exactly 1 die matching should be:

$1 - (\frac{5}{6})^3 - \frac{5}{216} - \frac{1}{216} = 1 - \frac{125 + 6}{216} = \frac{216 - 131}{216} = \frac{85}{216}$

And the expected value calculation:

\$1 invested yields $\$4\frac{1}{216} + \$3\frac{5}{216} + \$2\frac{85}{216} = \$\frac{4 + 15 + 170}{216} = \$\frac{189}{216} = \$\frac{7}{8} = \$0.875$

Have I neglected to consider something in this analysis? I avoided counting the cases out but it would not be too difficult to do this.

Based on this, a player should be losing around \$12.50 per \$100 spent, not \$8. I wonder if it's just a simple mistranslation where a 1/8 loss got interpreted as an 8% loss.

Let's revisit the 30:1 for triple. If I interpret this description literally, that'd be 31/36 * \$1 expected value, 0.861111, which is remarkably consistent with (and actually a smidge lower than) 0.875 which is a hint to me that it may actually be correct. This means the expected value is not going to change a whole lot whether the player bets more on individual symbols or on the triple.

Association of Categorical and Qualitative Variables

Posted: 23 Jan 2022 12:22 AM PST

I am trying to look at the association between the cost of goods depending on the area. Group A bought their product in one area and group B bought theirs in another. Group A is larger but group B is small ($n_B \leq$ 20) in comparison. The box plot of both shows group A's cost is clustered around a larger value than group B's. Cost in group A is distributed bimodally while cost is distributed unimodally in group B- which I verified by density plot. Group B's distribution is non-normal as well with a different variance than group A. I was wondering if there was another coefficient or test that might be applicable to this situation to claim an association between group membership/location and cost. Normally I would try a non-parametric test but most that I recall seem to have limits on variances, similarity of distributions, etc, which this data violates.

Why is this domain incorrect?

Posted: 23 Jan 2022 12:19 AM PST

I have to find the domain of this function: $$f(x) = \arcsin\left(\frac{|x|}{x+1} \right).$$

My attempt:

  • because arcsine is the inverse of sine function, then its domain is the image of sine: $[-1, 1]$.
  • the absolute value is defined in $]-\infty, +\infty[$,
  • the denominator is defined in $]-1, +\infty[$.

The domain of $f(x)$ is the intersection of these domain. Therefore, it's $]-1, 1]$. But this is wrong, because the correct domain is $[-1/2, +\infty[$, where does this $-1/2$ come from?

On denseness of a subspace of trigonometric polynomials in $L^1(\mathbb T).$

Posted: 23 Jan 2022 12:22 AM PST

$\mathbf {The \ Problem \ is}:$ Let, $e_n(\theta)=e^{in\theta}$ & $A=span\{e_n | n\geq 0\} \subset L^1(\mathbb T).$ Let $B=\bar A.$ Does $e_n \in B$ for any $n<0?$

$\mathbf {My \ approach}:$ The set of all trigonometric polynomials $\mathcal T$ is dense in $C(\mathbb T)$ under sup-norm(by Stone-Weierstrass) and if the answer to this question is positive, then $A$ becomes dense in $C(\mathbb T).$

But, is this right ? Any small hint ? Thanks in adv.

Volume of Solid Equation Set Up - $y=\sqrt{e^{x}+1}$

Posted: 23 Jan 2022 12:07 AM PST

I just wanted to check I have set up my integral correctly for Volume of Solid.

The region is bound between $y=\sqrt{e^{x}+1}$ and the x-axis on the interval $[0,1]$ around line $y=-1$.

I can choose which method - so, using the washer method, $$ V=\pi \int_{0}^{1}\left(\sqrt{e^{x}+1}+1\right)^{2}-(1)^{2} d x $$

Meaning of summation from + to - index

Posted: 23 Jan 2022 12:48 AM PST

I am currently working through this document describing the calculation of an accurate Geopotential model.

In equation 6.15, there is a double summation. The first summation specifies summation for all values of f (frequencies of waves in this case), and the second indicates that, for every frequency, we should perform a summation of the following expression for + and -. The expression includes multiple sets of +- signs, with one of them being actually a -+ sign.

My question is, does this summation for + and - mean to perform a summation of 2 terms, the first one obtained by taking the first sign in each +- sign (i.e., for the case of the -+ sign, it would be a -), and the second one obtained by taking the second sign in each +- sign (i.e., a + for the -+ sign)?

Editing to leave the specific formula (I am including a simplified to show only the notation I'm having trouble with):

$$\sum_{+}^- (\mathcal{C}^\pm \mp i\mathcal{S}^\pm)e^{\pm i\theta}$$

How to convert decimal number to 256 base numbers?

Posted: 23 Jan 2022 12:33 AM PST

I have rgb leds that takes colors as r , g , b values from 0-255 for each. So there is $\\{256*256*256}$ amount of colors as in decimal how do I convert the colors equally into n number of leds. It would be nice to have a formula to convert the decimal color to r,g,b values for example the first and last led (n led) will have $\\{(0,0,0)}$ and $\\{(255,255,255)}$ colors. If I am not not mistaken it looks like base 256 number I am not Shure how to convert to it thanks in advance.

For example if there are 16,777,216 leds (the maximum number of leds otherwise the colors runout) the led r,g,b values go like,

$\\{(0,0,0),(0,0,1),(0,0,2),..............,(255,255,253),(255,255,254),(255,255,255)}$

Number of homomorphisms from $\mathbb Z_5 \times \mathbb Z_5 \times \mathbb Z_5$ to $S_5$ and backwards

Posted: 23 Jan 2022 12:22 AM PST

I have the following question on group theory:

A. show that there are 745 different homomorphisms from $\mathbb Z_5 \times \mathbb Z_5 \times \mathbb Z_5$ to $S_5$
B. How many different homomorphisms are from the group $S_5$ to $\mathbb Z_5 \times \mathbb Z_5 \times \mathbb Z_5$ ?

I have done a similar question about the number of homomorphisms from $D_4$ to $Q_8$, which uses the generating set of $D_4$. I assume that for A I need to do the same.
However, I don't think I have a generated set for $S_5$, and it appears I need to look at the kernel and image of the homomorphism.
I don't really have a strong grasp on group theory, so I could use some help with this. Thank you for answering!

6 coins' box, an expected value of the number of tosses which heads was tossed.

Posted: 23 Jan 2022 12:04 AM PST

In a box there are 6 coins, three ordinary coins, two with heads on both sides and one with tails on both sides. A coin is selected randomly from the box and is tossed 100 times.

  1. What is the expectation value of the number of tosses of heads?
  2. What is the expectation value of the number of tosses of heads, if it's given that in the first toss we got heads?

As for 1. I am not so sure but the expectation value of one toss of heads is: $3\times (1/2)+2\times 1$, so in our case the expectation value should be $(\frac{7}{2})^{100}$, am I correct?

As for 2. I think you should replace 100 with 99, am I correct?

Thanks in advance!

Taylor-expansion of numerator and denominator separately

Posted: 23 Jan 2022 12:03 AM PST

To Taylor-expand $f(x) = \frac{\ln{(x+1)}}{\sin{x}}$, can you expand the numerator and denominator separately?

So if $g(x) = \ln{(x+1)}$ and $h(x) = \sin{x}$, then:

$g'(x) = \frac{1}{x+1}$ and $g''(x) = \frac{-1}{(x+1)^2}$

$h'(x) = \cos{x}$ and $h''(x) = -\sin{x}$

So the Taylor-expansions to second degree about 0 is:

$T_{g(x)}(x) = x - \frac{1}{2}x^2$ and $T_{h(x)}(x) = x$, giving the Taylor-expansion for $f$:

$$T_{f(x)}(x) = 1 - \frac{1}{2}x$$

I don't know if this is true, however it doesn't seem to be very inaccurate if $x=0.01$...

Why polynoimal defined on endomorphism is in $L(V)$

Posted: 23 Jan 2022 12:14 AM PST

Let $L(V)={f:V\to V|}$ $f$ is linear map

$h=a_0+a_1x+..+a_nx^n$

the value of polynomial $h(x)$ on $f\in L(V)$ is defined like this

$h(f)=a_01_V+a_1f+a_2f^2+...+a_nf^n$

where $f^n=f\circ f^{n-1}$

now book says it's clear that $h(f) \in L(V)$.Can you explain why?

We should check that $h(f)$ is linear map from $V\to V$

should check that $h(f+g)=h(f)+h(g)$ and $h(cf)=ch(f)$

$h(f+g)=a_01_V+a_1(f+g)+..+a_n(f+g)^n$

$h(f)=a_01_V+a_1(f)+..+a_n(f)^n$

$h(g)=a_01_V+a_1(g)+..+a_n(g)^n$

from here I don't know how to continue.

EDIT.

$f\circ f(v_1+v_2)=f(f(v_1)+f(v_2))=f(f(v_1))+f(f(v_2))=f^2(v_1)+f^2(v_2)$

Is it true that $\gamma\subset\overline{\Omega}$ implies $int(\gamma)\subset\Omega$ if $\Omega$ is simply connected and bounded?

Posted: 23 Jan 2022 12:47 AM PST

A simply connected domain $\Omega$ in a plane has following property that if A Jordan curve $\gamma$ lies in $\Omega$ ($\gamma\subset\Omega$) then the interior of $\gamma$ is contained in $\Omega$ ($int(\gamma)\subset\Omega$).

Now I do some extension to the proposition above, that if $\gamma\subset\overline{\Omega}$ then $int(\gamma)\subset\Omega$, which means $\gamma $ can touch the boundary of $\Omega$. Is it true or false? I think it should be false, otherwise it would be an important property published in every textbook but actually not. Unfortunately, I cannot find a counter example.

Thanks for your precious answer.

p.s. Thanks to Brian's answer, teaching me the proposition is false. However, actually, I want to know whether the proposition is true when $\Omega$ is bounded. Apologize for my negligence.

If $n=a^2+b^2=c^2+d^2$ then $n=\frac{(ac+bd)(ac-bd)}{(a+d)(a-d)}$

Posted: 23 Jan 2022 12:13 AM PST

If $n=a^2+b^2=c^2+d^2$ then $$n=\frac{(ac+bd)(ac-bd)}{(a+d)(a-d)}$$

or we get that it's enought to show that $$(a^2-d^2)(a^2+b^2+c^2+d^2)=2(a^2c^2-b^2d^2)$$

Any hints?

Dimension of Zariski closure of the image of an algebraic map

Posted: 23 Jan 2022 12:40 AM PST

Let $\phi: \mathbb{R}^m \rightarrow \mathbb{R}^n$ be a polynomial map with rational coefficients. Let $M$ be the Zariski closure of the image of $\phi$ (in the complex space $\mathbb{C}^n$). Then the following holds

$$ \dim(M) = \dim(\mathbb{R}^m) - \dim(\phi^{-1}(\phi(p))) $$ where $p$ is a generic point in $\mathbb{R}^m$ (generic means its coordinates are algebraic independent over the rational field $Q$).

My question is: why is this true? This seems to be well-known as it is given without a reference in a paper I read. Does anyone know a reference of this equality?

Sobolev function in $X = W_0^{1,2}(\Omega) \cap W^{2,2}(\Omega) $

Posted: 23 Jan 2022 12:28 AM PST

Let $\Omega =\left(0,1 \right)$. Given a function $u \in X = W_0^{1,2}(\Omega) \cap W^{2,2}(\Omega) $ I am looking for a function $v \in X$, such that $$v'' =max(0,u'')$$.

Not obvious to me that this is true since $max(0,u'')$ is not necessarily in $X$

Can one always find such a function ? I am a little confused about this, would appreciate any help/hints.

Determine the greatest of the numbers $\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$

Posted: 23 Jan 2022 12:29 AM PST

Determine the greatest of the numbers $$\sqrt2,\sqrt[3]3,\sqrt[4]4,\sqrt[5]5,\sqrt[6]6$$ The least common multiple of $2,3,4,5$ and $6$ is $LCM(2,3,4,5,6)=60$, so $$\sqrt2=\sqrt[60]{2^{30}}\\\sqrt[3]3=\sqrt[60]{3^{20}}\\\sqrt[4]4=\sqrt[60]{4^{15}}=\sqrt[60]{2^{30}}\\\sqrt[5]{5}=\sqrt[60]{5^{12}}\\\sqrt[6]{6}=\sqrt[60]{6^{10}}=\sqrt[60]{2^{10}\cdot3^{10}}$$ Now how do we compare $2^{30},3^{20},4^{15},5^{12}$ and $6^{10}$? I can't come up with another approach.

Given a triangular array of r.v.s so that each column converges in probability to $0$. Can we say that the row-wise averages converge to $0$?

Posted: 23 Jan 2022 12:08 AM PST

Let $X_{n, i}$ is a triangular arrays of r.v.s

$X_{1, 1}$

$X_{2, 1}$, $X_{2, 2}$

$X_{3, 1}$, $X_{3, 2}$, $X_{3, 3}$

...

$X_{n, 1}$, $X_{n, 2}$, ..., $X_{n, n}$

...

so that each column converges in probability to $0$, i.e., for every fixed $i$, $\{X_{n, i}\}\xrightarrow{p} 0$ (convergence with respect to $n$).

Is it correct that the row-wise average $\dfrac{1}{n}\sum\limits_{i=1}^nX_{n, i}\xrightarrow{p} 0$?

Any neighbourhood of a closed decreasing sequence in a compact space must contain one of the sets: why is the space required also to be Hausdorff?

Posted: 23 Jan 2022 12:12 AM PST

$\newcommand{\o}{\mathcal{O}}$Apologies for the perhaps strangely phrased title - the character limit can be a pain.

Let $X$ be a compact Hausdorff space and $\{F_n\subseteq X:n\in\Bbb N\}$ a decreasing sequence of nonempty closed sets in $X$. For any neighbourhood $\o$ of their intersection $\bigcap_{n\in\Bbb N}F_n$, show that there must exist an $N\in\Bbb N$ such that for all $n\ge N$, $F_n\subseteq\o$.

I have done this exercise - but my proof nowhere used the Hausdorff property. I ask: why is it needed?

My solution:

As the $F_n$ have the finite intersection property, and the space is compact, their intersection is nonempty. $\o$ is then nonempty, and wlog may be considered open. Consider then the family of sets $G_n=F_n\cap X\setminus\o$. For all $n$, $G_n$ is the intersection of two closed sets and thus closed. As $F_{n+1}\subseteq F_n,\,G_{n+1}\subseteq G_n$ for all $n$. Assume for the sake of contradiction that $F_n$ is not a subset of $\o$ for any $n$; this is equivalent to the assumption that $G_n$ is nonempty for all $n$.

If none of the $G_n$ are empty, then $\{G_n:n\in\Bbb N\}$ is a decreasing sequence of closed nonempty sets and thus has the finite intersection property, and thus has nonempty intersection due to the compactness of $X$. Then $\bigcap_{n\in\Bbb N}F_n\cap X\setminus\o\neq\varnothing$, which implies the existence of $x\in\bigcap_{n\in\Bbb N}F_n$ such that $x\notin\o$, which contradicts the assertion that $\o$ is a neighbourhood of $\bigcap_{n\in\Bbb N}$. Since $\{G_n\}$ will always have nonempty intersection unless there is an $N\in\Bbb N$ such that $G_N=\varnothing$, this shows that such an $N$ must exist and that $F_N\cap X\setminus\o=\varnothing,\,F_N\subseteq\o$. It follows from the decreasing property that $\forall n\ge N,\,F_n\subseteq\o\quad\blacksquare$

This exercise was left by Royden in the fourth edition of their text "Real Analysis". My proof nowhere requires the Hausdorff assertion (as far as I can tell) so I'm left wondering if Royden made an oversight. Is this the case?

Betting system and upper bounds

Posted: 23 Jan 2022 12:05 AM PST

Define $X_0=\alpha\in(0,1)$ the initial capital and $X_n$ as the remaining capital after each game. A player bets $1-X_n$ if $X_n>1/2$ and $X_n$ if $X_n\leq 1/2$ such that each game is a Bernoulli$(1/2)$. Define $A_n=\{X_n\in(0,1)\}$, the event that the player neither wins everything or reachs ruin. Show by induction that $P(A_n)\leq 2^{-n}$.

If $\alpha< 1/2$ then the player either goes broken with probability $1/2$ or owns $2X_n$ in the next game. If $\alpha >1/2$ then the player owns every available resource with probability $1/2$ or owns $2X_n -1$ in the next game. Should I use total law of probability and try to work with the conditionals upon the last fortune as in

$$P(A_n) = \sum_{k=1}^m P(A_n|B_m)$$ where $\bigcup B_m= \Omega$.

Also if this is the right path, should I split three-ways with a $X_n=1/2$ case where the player reach either $0$ or $1$?

I'm having a little trouble working this problem out.

Quadrilateral $ABCD$ with $AC \cap BD = \{O\}$, $E, F$ diagonal midpoints collinearity question

Posted: 23 Jan 2022 12:32 AM PST

Let $ABCD$ be a convex quadrilateral, with $AC \cap BD = \{O\}$, and $OB > OD$, $OC > OA$. Let $E$ and $F$ be the midpoints of $(AC)$ and $(BD)$, $EF \cap AB = \{J\}$, $EF \cap CD = \{K\}$, $CJ \cap BK = \{L\}$. If $M$ is the midpoint of $(KJ)$, show that $O - M - L$ are collinear.

An attempt:

Figure ABCD

My approach was to prove the affirmation by vectorial calculations. Obviously, $\vec{OM}$ may be expressed by median vector formula depending on $\vec{OJ}$ and $\vec{OK}$. And I want to express $\vec{OL}$ as well as a function of the two vectors, but I'm actually stuck at the part with proving a fromula for $\vec{LJ}$ and $\vec{LK}$. If I proved that $\vec{OL} = a (\vec{OK} + \vec{OJ})$, by collinearity rule, $\vec{OM}$ and $\vec{OL}$ would be parallel vectors, so $O -M - L$ collinear points, the conclusion may be derived.

Find $\text{QR}$ where $\text{PQ} = 39$ and $\text{PR} = 17$ and altitude of $\triangle \text{PQR}$ is $15$?

Posted: 23 Jan 2022 12:44 AM PST

Stumped on the second part of this question for a tutoring student. The problem states:

In $\triangle \text{PQR}$, $\text{PQ} = 39 \text{ in.}$, $\text{PR} = 17 \text{ in.}$, and the altitude $\text{PN} = 15 \text{ in.}$ Find $\text{QR}$. Consider all cases.

The first case, which is more obvious, is that $\text{PN}$ and $\text{NQ}$ satisfy the Pythagorean Theorem because $\triangle \text{QPN}$ and $\triangle \text{PNR}$ are right triangles. Therefore:

$15^2+\text{QN}^2=39^2 \implies \text{QN}=36$

$15^2+\text{NR}^2=17^2 \implies \text{NR}=8$

$\text{QR}=\text{QN}+\text{NR}=36+8=44$

The system allows my student to fill in two possible answers. Of course, the first answer is $44$ in., but what is the case that would give the second answer? At first I thought that we could consider the case where $\angle \text{QPR}=90^{\circ}$ and use the Pythagorean Theorem, but I'm pretty sure that would mean that $\text{PN}\neq 15$. What would the other case be?

(keep in mind this student has not covered trig yet)

Prove that $|PF_{1}|+|PF_{2}|$ is Constant in an Elipse

Posted: 23 Jan 2022 12:47 AM PST

Given an elipse with two focus $F_{1}$ an $F_{2}$, and $A$ is an arbitrary point at the elipse. Stright line $AF_{1}$ has another intersection point $B$ with the elipse, and $AF_{2}$ has another intersection point $C$ with the elipse.

And point $P$ is the intersection of line $BF_{2}$ and $CF_{1}$, now the problem is to prove that $|PF_{1}|+|PF_{2}|$ is constant.

Thanks in advance!

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