Hammering with barycentric coordinates

by Keith50, Feb 7, 2022, 7:21 AM

Malaysia IMO Training 1 Weekly Test 4 P1 wrote:

Let $\triangle ABC$ be a triangle with orthocenter $H$ and let $M$ be the midpoint of $BC$ . Pick two points $X,Y$ on $AB,AC$ respectively such that $X,H,Y$ are collinear and $XY\perp MH.$ Prove that $HX=HY.$

Solution by hammering
(I actually attempted to use complex bashing during the test, but ended up screwing everything as the algebra are too complicated and not realising this bary bashing solution. :wallbash:

)

This post has been edited 1 time. Last edited by Keith50, Feb 7, 2022, 7:26 AM

geometry

barycentric coordinates

Triangles

orthocenter

midpoint

BIMO 1

Astro Problem of the week#1

by Keith50, May 19, 2021, 8:20 AM

This marks the beginning of Astro Problem of the week! :welcomeani:

Astro Problem of the week#1 wrote:

$\textbf{\color{red}{MYAO 2019:}}$ Hipparchus found that the celestial longitudes increases about $50^{''}$ in a year. This phenomenon is later known as the precession of the Earth's equinox. Based on Hipparchus measurement, calculate the duration of the Earth's equinox precession, in years(s).
$(\textbf{A}) 1 \quad \quad \quad (\textbf{B}) 25740 \quad \quad \quad (\textbf{C}) 25920 \quad \quad \quad (\textbf{D}) 7.2$

Solution

Physics Problem of the week#1

by Keith50, May 18, 2021, 7:32 AM

This marks the beginning of Physics Problem of the week! :jump:

Physics Problem of the week#1 wrote:

Suppose you watch a video backwards and slowed down by a factor $\alpha$ , so that if the video is one hour long when played normally, you watch it from end to beginning and it takes $\alpha$ hours to do so.

The video shows some physical motion. If the motion in the video has acceleration $a$ , when played normally, what if the acceleration $a'$ you would measure when watching the video backwards and in slow motion? Briefly explain your reasoning.

Solution

physics

kinematics

classical mechanics

Math Problem of the week#4

by Keith50, May 15, 2021, 2:14 AM

Math Problem of the week#4 wrote:

$\color{green}{\textbf{JBMO SL 2006:}}$ Determine all numbers $\overline{abcd}$ such that $\overline{abcd}=11(a+b+c+d)^2.$

Solution

This post has been edited 1 time. Last edited by Keith50, May 15, 2021, 2:15 AM

JBMO

number theory

Perfect Squares

Integral of the week#4

by Keith50, May 14, 2021, 11:28 AM

Integral of the week#4 wrote:

Find: $\int \frac{dx}{x+\sqrt{x^2-1}}$

Solution

This post has been edited 2 times. Last edited by Keith50, May 18, 2021, 3:04 AM

calculus

integration

algebra

Kinematics: Getting Started

by Keith50, May 13, 2021, 9:50 AM

$\Large \textbf{\color{green}{Kinematics: Getting Started}}$
Hey guys, this marks my first blog on a physics topic, let's get started with kinematics!
$\large \textbf{\color{blue}{(i) Position, Velocity, and Acceleration}}$
In kinematics, we want to describe the motion of a particle and we can do this by using vectors to specify its position, velocity and acceleration.
At any particular time $t$ , the position of the particle with coordinates $(x,y,z)$ in three-dimension can be represented as a $\textit{position vector}$ $\overset{\rightarrow}{\mathbf{r}}$ : $\overset{\rightarrow}{\mathbf{r}}=x\mathbf{\hat{i}}+y\mathbf{\hat{j}}+z\mathbf{\hat{k}}$ where $\mathbf{\hat{i}},\mathbf{\hat{j}},$ and $\mathbf{\hat{k}}$ are Cartesian unit vectors.
If the particle is located at position $\overset{\rightarrow}{\mathbf{r_1}}$ at time $t_1$ and it moves along its path to position $\overset{\rightarrow}{\mathbf{r_2}}$ at time $t_2$ , we denote its change in position during that interval by the $\textit{displacement vector}$ $\Delta \overset{\rightarrow}{\mathbf{r}}$ : $\Delta \overset{\rightarrow}{\mathbf{r}}=\overset{\rightarrow}{\mathbf{r_2}} -\overset{\rightarrow}{\mathbf{r_1}}.$ We can then define its $\textit{average velocity}$ $\overset{\rightarrow}{\mathbf{v}_{\textrm{av}}}$ in that interval as the displacement divided by the time interval during which the displacement occurs, $\overset{\rightarrow}{\mathbf{v}_{\textrm{av}}}=\frac{\Delta \overset{\rightarrow}{\mathbf{r}}}{\Delta t}.$ Taking the limit as $\Delta t \rightarrow 0$ , we then have the $\textit{instantaneous velocity}$ $\overset{\rightarrow}{\mathbf{v}}$ as $\overset{\rightarrow}{\mathbf{v}}=\lim_{\Delta t \rightarrow 0} \frac{\Delta \overset{\rightarrow}{\mathbf{r}}}{\Delta t}=\frac{d \overset{\rightarrow}{\mathbf{r}}}{dt}.$ The derivative of the position vector is found by taking the derivative of its components: $\frac{d \overset{\rightarrow}{\mathbf{r}}}{dt}=\frac{d}{dt}(x\mathbf{\hat{i}}+y\mathbf{\hat{j}}+z\mathbf{\hat{k}})=\frac{dx}{dt}\mathbf{\hat{i}}+\frac{dy}{dt}\mathbf{\hat{j}}+\frac{dz}{dt}\mathbf{\hat{k}}.$ On the other hand, the velocity vector can also be written in terms of its components $\overset{\rightarrow}{\mathbf{v}}=v_x\mathbf{\hat{i}}+v_y\mathbf{\hat{j}}+v_z\mathbf{\hat{k}}$ where $v_x=\frac{dx}{dt}, \quad v_y=\frac{dy}{dt}, \quad v_z=\frac{dz}{dt}.$ In a similar fashion, we can define the $\textit{average acceleration}$ of the particle in that interval as the change in velocity per time, $\overset{\rightarrow}{\mathbf{a}_{\textrm{av}}}=\frac{\Delta \overset{\rightarrow}{\mathbf{v}}}{\Delta t}.$ The change in velocity means $\overset{\rightarrow}{\mathbf{v}}_{\textrm{final}} -\overset{\rightarrow}{\mathbf{v}}_{\textrm{initial}}$ . By taking the limit as $\Delta t \rightarrow 0$ , we then have the $\textit{instantaneous acceleration}$ $\overset{\rightarrow}{\mathbf{a}}$ as $\overset{\rightarrow}{\mathbf{a}}=\lim_{\Delta t \rightarrow 0} \frac{\Delta \overset{\rightarrow}{\mathbf{v}}}{\Delta t}=\frac{d\overset{\rightarrow}{\mathbf{v}}}{dt}=\frac{d^2\overset{\rightarrow}{\mathbf{r}}}{dt^2}.$ Analoguosly, we can write the instantaneous acceleration vector in terms of its components $\overset{\rightarrow}{\mathbf{a}}=a_x\mathbf{\hat{i}}+a_y\mathbf{\hat{j}}+a_z\mathbf{\hat{k}}$ where $a_x=\frac{dv_x}{dt}=\frac{d^2x}{dt}, \quad a_y=\frac{dv_y}{dt}=\frac{d^2y}{dt}, \quad a_z=\frac{dv_z}{dt}=\frac{d^2z}{dt}.$ $\large \textbf{\color{blue}{(ii) Kinematics Equations for Constant Acceleration}}$
Now, we shall derive several kinematics equations for a particle with constant acceleration. Let's consider a particle moving in one-dimension with a constant acceleration, $a$ , the following equations are true,
1. $\Delta x=\frac{v_{\textrm{initial}}+v_{\textrm{final}}}{2}\Delta t.$ This follows from the graphical interpretation of displacement being the area under a velocity vs. time graph, which is in this case a trapezoid with bases $v_{\textrm{initial}}$ and $v_{\textrm{final}}$ and height $\Delta t$ .
2. $\Delta x=v_{\textrm{initial}}\Delta t+\frac{1}{2}a(\Delta t)^2$ . This again follows from the graphical interpretation of displacement being area under a velocity vs. time graph, but this time, it's a trapezoid with bases $v_{\textrm{initial}}$ and $v_{\textrm{final}}=v_{\textrm{initial}}+at$ , and height $t$ .
3. $\Delta x=v_{\textrm{final}}\Delta t-\frac{1}{2}a(\Delta t)^2$ . This is very similar to the previous case, but now the bases of the trapezoid are $v_{\textrm{final}}-at$ and $v_{\textrm{final}}$ .
4. $v_{\textrm{final}}^2=v_{\textrm{initial}}^2+2a\Delta x$ . This is perhaps the least intuitive of all the kinematics equations, this can be shown true by a simple algebra, as $v_{\textrm{initial}}+v_{\textrm{final}}=2\frac{\Delta x}{\Delta t}$ and $v_{\textrm{initial}}-v_{\textrm{final}}=a\Delta t$ , multiplying both of them yields the result.

Exercise wrote:

For acceleration which is not constant, show that this identity is generalized by the equation $v_{1}^2-v_{0}^2=2\int_{t_0}^{t_1} a \, dx.$ Proof

Now, in general, these kinematics equations holds for a particle moving in $n-$ dimension with constant acceleration, $\overset{\rightarrow}{\mathbf{a}}=\langle a_1,a_2,\ldots a_n\rangle$ as we can decompose its acceleration, velocity and position vectors into their components and work with each component separately. Thus, we still have $\Delta \overset{\rightarrow}{\mathbf{r}}=\frac{\overset{\rightarrow}{\mathbf{v}_1}^2-\overset{\rightarrow}{\mathbf{v_0}}^2}{2}\Delta t$ $\Delta \overset{\rightarrow}{\mathbf{r}}=\overset{\rightarrow}{\mathbf{v}_0}\Delta t+\frac{1}{2}\overset{\rightarrow}{\mathbf{a}}(\Delta t)^2$ $\Delta \overset{\rightarrow}{\mathbf{r}}=\overset{\rightarrow}{\mathbf{v}_1}\Delta t-\frac{1}{2}\overset{\rightarrow}{\mathbf{a}}(\Delta t)^2$ $\overset{\rightarrow}{\mathbf{v}_1}^2=\overset{\rightarrow}{\mathbf{v}_0}^2+2\overset{\rightarrow}{\mathbf{a}} \Delta \overset{\rightarrow}{\mathbf{r}}.$ And also, for acceleration which is not constant, this equation also holds: $\overset{\rightarrow}{\mathbf{v}_1}^2-\overset{\rightarrow}{\mathbf{v}_0}^2=2\int_{t_0}^{t_1} \overset{\rightarrow}{\mathbf{a}} \, d\overset{\rightarrow}{\mathbf{r}}.$ $\large \textbf{\color{blue}{(iii) Free Falling Bodies}}$
A particle which undergoes free fall will experience a acceleration of $-g$ which is the gravitational acceleration, approximately $9.81 \textrm{m/s}^2$ . We can then describe the motion of the particle by using kinematics equations, $a_y=-g,$ $v_y=v_{0y}-gt,$ $y=y_0+v_{0y}t-\frac{1}{2}gt^2.$ $\large \textbf{\color{blue}{(iv) Problems}}$

JEE Mock Test wrote:

Two students simultaneously start from the same place on a circular track and run for $2$ minutes. In this time, one of them completes three and the other four revolutions with uniform speed. Due to thick vegetation in a circular area inside the track, either of the boys can see only one third of the track at a time (a sixth of the track both in front and behind them). How long during their run do they remain visible to each other?
Solution

HRK wrote:

An object, constrained to move along the $x$ axis, travels a distance $d_1$ with constant velocity $v_1$ for a time $t_1$ . It then instantaneously changes its velocity to a constant $v_2$ for a time $t_2$ , travelling a distance $d_2$ .
(a) Show that $\frac{v_1d_1+v_2d_2}{d_1+d_2}\ge \frac{v_1t_1+v_2t_2}{t_1+t_2}.$ (b) Under what condition is this an equality?
Solution

HRK wrote:

A basketball player, about to "dunk" the ball, jumps $76$ cm vertically. How much time does the player spend (a) in the top $15cm$ of this jump and $(b)$ in the bottom $15$ cm?
Solution

This post has been edited 1 time. Last edited by Keith50, May 13, 2021, 11:03 AM

physics

kinematics

classical mechanics

Linear Algebra Notes - Determinants

by Keith50, May 12, 2021, 12:34 PM

$\Large \textbf{\color{blue}{Linear Algebra - Determinants}}$
$\large \textbf{\color{magenta}{(a) 2-order and 3-order Determinants}}$
We know that if we are given a system of two linear equations $\begin{cases} a_{11}x_1+a_{12}x_2=b_1 \\ a_{21}x_1+a_{22}x_2=b_2 \end{cases}$ we can solve it by elimination method and get: $x_1=\frac{b_1a_{22}-a_{12}b_2}{a_{11}a_{22}-a_{12}a_{21}}, x_2=\frac{a_{11}b_2-a_{21}b_1}{a_{11}a_{22}-a_{12}a_{21}}$ notice that the expressions in the numerator and the denominator of $x_1,x_2$ consist of $4$ numbers with $2$ numbers multiplied together respectively and subtracted with the other. In particular, the numbers in the expression $a_{11}a_{22}-a_{12}a_{21}$ are the coefficients in our system of two linear equations, so let's put them in rows and columns of a matrix, $\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ we can then represent $a_{11}a_{22}-a_{12}a_{21}$ as a $2 \times 2$ determinant, $\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}.$ Therefore, we can rewrite $x_1,x_2$ as $x_1=\frac{\begin{vmatrix} b_{1} & a_{12} \\ b_{2} & a_{22} \end{vmatrix}}{\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}}, x_2=\frac{\begin{vmatrix} a_{11} & b_{1} \\ a_{21} & b_{2} \end{vmatrix}}{\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}}.$ Now, we then define the 3x3 determinant as $\begin{vmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}= a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{12}a_{21}a_{33}-a_{11}a_{23}a_{32}.$ $\large \textbf{\color{magenta}{(b) Permutations and Inversions}}$
Let's consider the following example:
Given three numbers $1,2,3$ , how many different arrangements are there for these numbers?
It's easy, there are $3!=6$ of these, $123, 231, 312, 132, 213, 321.$ In general, these numbers like in the example above, $1,2,3$ are called elements. So, given $n$ different elements, there are $n!$ arrangements and these arrangements are also called permutations of the elements.
We then define the inversion number of a permutation. First, an inversion in a permutation is a pair of numbers such that the larger number appears to the left of the smaller one in the permutation. The inversion number of a permutation is the total number of inversions. One way to help calculate the inversion number is to look at each position in the permutation and count how many bigger numbers are to the left, and then add those numbers up.

Exercise wrote:

Calculate the inversion number of the permutation, $9236815$ .
Answer

Lastly, we say that a permutation is odd if the inversion number is odd and a permutation of even if the inversion number is even.
$\large \textbf{\color{magenta}{(c) The Meaning of n-order Determinants}}$
Let's go back to the $3\times 3$ determinant, $\begin{vmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}= a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{12}a_{21}a_{33}-a_{11}a_{23}a_{32}.$ It's easy to see that there are three elements in each term of the above expression and these three elements are from different row and columns. Thus, each term of the expression (neglecting the positive and negative signs) can be rewritten as $a_{1p_1}a_{2p_2}a_{3p_3}$ . Here, the row numbers of each term are assigned $123$ and the column numbers of each term are assigned $p_1p_2p_3$ , there are the permutations of elements $1,2,3$ . There are $6$ permutations which is the same as the number of terms.
By a close inspection, we realise that terms with positive sign have the column numbers being the permutations, $123,231,312$ and terms with negative sign are the permutations $132,213,321$ . Also, by some calculations, we find out that the first three are even permutations and the latter are odd permutations. Therefore, the sign of each term can be represented as $(-1)^t$ , where $t$ is the inversion number of the column numbers of each term.
Hence, the $3-$ order determinant can be rewritten as $\begin{vmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{vmatrix}= \sum (-1)^t a_{1p_1}a_{2p_2}a_{3p_3},$ where $t$ is the inversion number of the permutation $p_1p_2p_3$ .
In general, given a $n-$ order matrix, $\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n} \\ & \cdots & \cdots &\\ a_{n1} & a_{n2} & \cdots & a_{nn}\end{pmatrix}$ we can form products of $n$ elements from different rows and columns and with their respective signs $(-1)^t$ , $(-1)^ta_{1p_1}a_{2p_2}\cdots a_{np_n}$ . The numbers $p_1p_2\cdots p_n$ are permutations of $1,2, \cdots ,n$ and $t$ as the inversion number. Since there are $n!$ permutations, there are total $n!$ terms and their sum $\sum (-1)^t a_{1p_1}a_{2p_2}\cdots a_{np_n}$ is the $n-$ order determinant, denoted by $D=\begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\end{vmatrix}.$

Exercise wrote:

Prove the following $n-$ order determinant $\begin{vmatrix}\lambda_1 & &\\ & \lambda_2 & \\ & &\ddots & \\ & & & \lambda_n \end{vmatrix}=\lambda_1 \lambda_2 \cdots \lambda_n$ $\begin{vmatrix} & & & \lambda_1 \\ & & \lambda_2 \\ & \iddots \\ \lambda_n & \end{vmatrix}=(-1)^{\frac{n(n-1)}{2}}\lambda_1 \lambda_2 \cdots \lambda_n,$ where the unfilled elements are all $0$ .

Exercise wrote:

Prove the following determinant of a lower triangular matrix: $D=\begin{vmatrix} a_{11} \\ a_{21} & a_{22} \\ \vdots & \vdots & \ddots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\end{vmatrix}=a_{11}a_{22}\cdots a_{nn}.$

algebra

linear algebra

determinant

Simultaneous Equations

Math Problem of the week#3

by Keith50, May 9, 2021, 6:53 AM

Math Problem of the week#3 wrote:

$\textbf{\color{blue}{AMC 12 2000/25:}}$ Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.) $[asy]import three; import math; size(180); defaultpen(linewidth(.8pt)); currentprojection=orthographic(2,0.2,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); draw(A--B--E--cycle); draw(A--C--D--cycle); draw(F--C--B--cycle); draw(F--D--E--cycle,dotted+linewidth(0.7));[/asy]$
$\textbf{(A)}\ 210 \qquad \textbf{(B)}\ 560 \qquad \textbf{(C)}\ 840 \qquad \textbf{(D)}\ 1260 \qquad \textbf{(E)}\ 1680$

Solution

Integral of the week#3

by Keith50, May 8, 2021, 2:12 PM

Integral of the week#3 wrote:

$\textbf{\color{blue}{JHMT 2020:}}$ Compute the largest integer less than or equal to $\int_{\pi-1}^{\pi+1} \frac{e^{\cos x}\cos ^3 x+\cot x}{\cot x}\, dx.$