Homework Statement
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Given a general triangle ABC, find the geometric locus of points such that the three orthoprojection onto the sides of the triangle are aligned.
Homework Equations
Let's call A', B', and C' the orthoprojection of a given point M onto (AB) , (BC) , and (AC).
M satisfies...
Yes, this is the way I've done it, and I see no content in this kind of exercise. As you say, it is almost without words!
Furthermore, a 13yo who doesn't know or forgot that the median line splits the area of a triangle in two is done without trigonometry. He will never find, unless someone sees...
Homework Statement
On the picture, compare the area of triangle ABC to the area of A'B'C'.
This problem was shown to me by a 13 years old. Trigonometry forbidden. It seems to me that this is the kind of problem you either solve in 2 minutes, or never solve. In both cases, you don't learn...
Lol, this is not a lonely question in a problem set, it is the conclusion of a lengthy problem on sequences, and all the steps needed to answer this question were worked in previous questions. There was almost no work involved here, but honestly, this equality looks too good, and I found it hard...
Yes, we had the analytical answer, as the limit of two converging sequences, but the way the question was put, I suddenly had a doubt and needed to see it work on a computer program. But thanks to @phyzguy 's program I am convinced now :-)
Homework Statement
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I'm helping a 12th grader with his homework, and he is asked to prove the following equality as the conclusion of a problem :
## 1 + \frac{1}{1+ \frac{1}{1+...}} = \sqrt{1+ \sqrt{1+ \sqrt{1+...}}}##
Written like this, the formula is intimidating while it really...
In an orthonormal coordinate system : ##<x,y> = {}^T X Y ##
In a skewed coordinate system ## <x,y > = {}^T X' ({}^T PP ) Y' ##,
where ##X,Y## (resp. ##X',Y'##) are the coordinates of ##x,y## in the orthonormal coordiate system (resp. skewed coordinate system), and ##P## the change of basis...
switch to an orthonormal coordinate system with the appropriate change of basis matrix.
In this new coordinate system, if ## V_1 = P v_1 ##, then you know that ## V_2 = ( B_1, -A_1, 0) ## is orthogonal to ## V_1##, and switch again : ## v_2 = P^{-1} V_2 ##
Ok, it's good you explain, it was needed :-)
So with your notations the horizon points that are to be projected are
##H = \{ M(x,y,z),\ \vec{PM}.\vec{OM} = 0,\ ||\vec{OM}||^2 = r^2,\ z\ge \epsilon \} = \{ M(x,y,z),\ y = \frac{r^2}{ r+h }, z \ge \epsilon, x^2 + z^2 = r^2 (1 - \frac{1}{r+h})...
Lol this is just a copy paste of your post #1!
I think that since your camera can move vertically, if man holding the camera has position ##M_0(x_0,y_0,z_0)## at the surface of the earth, the position of the camera is ##M_c(t) = M_0 + t \begin{pmatrix} x_0 + \alpha \\ y_0 + \beta \\ z_0 + \gamma...
I don't suggest anything at this point. We've just talked about some equations that must satisfy the horizon points (at A) according to the definition you wrote in post #9. The only thing I understand for the moment is that a subset of these points must appear on final screen.
What do we need...
Well if you consider the the sphere
##S|x^2 + y^2 + z^2 + 2\alpha x + 2\beta y + 2\gamma z + \delta = 0##,
center: ##(-\alpha,-\beta,-\gamma)##
radius: ##\sqrt{\alpha ^2 + \beta ^2 + \gamma ^2 -\delta}##
then its tangent planes are the ## T_{(x,y,z)} | xX+yY+zZ + \alpha (x+X) + \beta (y+Y) +...
So at this point we have an equation for this set of points, looking like :
## xx_A + yy_A + zz_A + \alpha (x + x_A) + \beta ( y + y_A) + \gamma (z+z_A) + \delta = 0##
What are we supposed to do next ?
I'm not sure I understand the problem.
Are you looking for the points of the sphere (earth) limited below by a horizontal plane and above by its tangent planes meeting ##A## ? Is it something like this ?
At second thought I think my previous post was unclear.
What I meant is that if ##S## is the set of one to one maps from ##A## to ##B##,
##U## the subset of ##S## verifying your constraint, then ##S## is the disjoint union of ##U## and of its complement in ##S## (say ##V##). Then you know that...
Use contraposition : Let ## f: A\to B ##
Not## ( \forall i=1...5,\ f(i) \neq i \text{ and } f(1) \notin \{0,1\} ) \iff (\exists i = 1...5, f(i) = i \text{ or } f(1)\in \{0,1\} )##
With that kind of reasoning, if ##N## is the total number of function from A to B, and ##M## the number you are...
The trick is to see that there is a surjection that sends exactly six permutations of ##\Omega## to exactly one output ##\omega' \in \Omega'## of the real random experiment where marbles are undistinguishable inside the same color.
The same remark applies between ##A## and ##A'## (event where...
Assume that your marbles are artificially distinguishable and numbered : 1,2 for the two red marbles, 3,4 for the two green marbles, and 5,6 for the two yellow marbles.
Then the set of all possible output for your random experiment can be modeled by ##\Omega = \{ \omega =...
I don't think you are correct. Unless I'm wrong ##D_3 = -1##, but with your answer you get 1.
You don't change the determinant by adding to a column a multiple of another column, and you multiply the determinant by ##-1## by exchanging 2 columns. So after your transformation, you exchange...
Oh Ok, I asked the the OP what was the meaning of this notation but did not reply about that so I assumed it was the number of combinations. So I agree then !
I disagree with these answers.
Let's take (e) :
P(7,4) is the number of sets of 5 person in the pool containing Hal but not Ida (or Ida but not Hal). In each of these sets, people can be ordered in ##5!## different ways. So the answer should be ##2\times (5!)\times P(7,4)## and not ##2\times...
If the notation P(a,b) is the number of combinations of b elements among a, then your answers are incorrect but not stupid.
You forgot to account for a number of configurations in each case. So the correct answers are a multiple of your answers.
EDIT: your answers to e,f,g
You must have worked on this problem for too long :-)
Think about it 10 more minutes
EDIT: or don't think about it the next 10 minutes :-)
EDIT (+10 minutes) : What did we just say ? ##P_n(z) = (z-\alpha_1)...(z-\alpha_n)##. So what is ##P_n(-1)##?
yes, but if you want to convince yourself, you can justify that ## \text{deg}{(A)} = 0##, so that ##A## is a non-zero constant, and with your argument above, ##A = 1##.
Ok, so now you have the answer, what is ##P_n(-1)## ?
Ok so it has ##n## roots without counting multiplicity.
So your polynomial has the form ##P_n(z) = A(z) (z-\alpha_1)...(z-\alpha_n) ##.
Explain why ##A = 1##
Thank you Samy, your answer is very clear ! I think that in order to avoid this double counting, I must impose that the ##k##-blocs ##A_k## contain a random element ##x\in S_{n+1}##.
In which case, each ##A_k## is the union of ##x## and of a ##(k-1)##-block of ##S_{n+1}-\{x\}##, so that ##A_k##...
Homework Statement
Find a recursive relation on the number of partitions ##P_n## for a set ##S_n## of cardinal ##n##. ##P_0 = 1## is given.
Homework Equations
The Attempt at a Solution
A partition of ##S_{n+1}## is given by the choice of a non-empty ##k##-block ##A_k## of ##S_{n+1}## and a...
Oh I see what you mean, I have been sloppy in my proof. In each cases, I assumed ##X## was a solution. With that assumption, I found restrictions for each cases, which are:
##A\subset B##
##B\subset A##
##A\cap B=\emptyset##.
So that outside these restrictions, there aren't any solution. Then...
Would you mind sharing your counterexamples for the 3 different cases? Assuming ##A\subset B## for q1, ## B\subset A## for q2, ##A\cap B =\emptyset ## for q3.
Homework Statement
Let ##A,B \in {\cal P}(E)##. Solve in ##{\cal P}(E)## the following equations:
##X\cup A = B##
##X\cap A = B##
##X - A = B##
Homework Equations
The Attempt at a Solution
We have ##A\cup B = (A\cup X)\cup A = A\cup X = B##. So ##A\subset B## and the solution cannot be...
In general, you have ##(x+y)^n = \sum_{k = 0}^n \binom{n}{k} x^k y^{n-k} ##
Take the derivative with respect to ##x## on each sign of the equation and multiply by ##x##.
You get ##nx(x+y)^{n-1} = \sum_{k = 0}^n k \binom{n}{k} x^k y^{n-k} ##.
Now do it again and set ##x## and ##y## to one...
Oops you're right, I'm just at ## \begin{pmatrix}a\\b\end{pmatrix} \in \mathbb{R} \begin{pmatrix}u.w \\ -u.v \end{pmatrix} ##.
Thanks again, I will try your proof.
Thank you for your answers, but is it possible to avoid lengthy calculations ?
In the OP, I am to the point where ## (a,b) \in \{\pm (u.w, - u.v) \} ##.
How should I decide for the sign ?
Homework Statement
If ##u,v,w\in\mathbb{R}^3##, show that ## u\times(v\times w) = (u.w) v - (u.v) w ##.
Homework Equations
The Attempt at a Solution
Since ## u\times(v\times w)##, ##v## and ##w## are orthogonal to ##v\times w##, these vectors are coplanar. Therefore, there must be reals ##...
The formula is coming from the binomial theorem (https://en.wikipedia.org/wiki/Binomial_theorem).
I am not as confident as you are for the following reason: even though we find the same 'law' for the score of the dealer, an important verification is to make sure that the probability of all...
I agree with your answer but don't follow your reasoning !
Mine is that if you've got ##p_1,p_2,p_3## coins in each box, all these numbers being odd, you can write ## p_i = 2q_i - 1, \ q_i \ge 1 ##. Therefore your problem is equivalent to ## q_1 + q_2 + q_3 = 29 ## which is a classic situation...
Yes, ##C_a^b = {a\choose b} ##.
##C_{k-1}^{t-1}## is the number of ways you can find integers ##k_1,...,k_t > 0## such that ##k_1+...+k_t = k##. Otherwise called ##t##-composition of ##k## (https://en.wikipedia.org/wiki/Composition_%28combinatorics%29).
*******
But now you have to explain...
Oh sorry, I read your post too fast ! So we have
## P(\{ \text{player wins} \} ) = 1- P( \{ \text{dealer wins} \} ) = 1 - \sum_{k = m+1}^{100} P(S_d = k) ##
Do you agree ?
If you call ## S_d^{(t)} := \{ \text{score of the dealer after } t \text{ deals} \} ## and ##D_t := \{ \text{output of...
I don't feel comfortable with probabilities but here is what I think.
Player wins if the dealer scores less than ##m## or more than 100. If you call ##S_d## the score of the dealer, in terms of events, you have :
##
\begin{align}
\{ \text{player wins} \} &= \{ 1\le S_d \le m-1 \} \cup \{ S_d >...