299/ Potw

For a triangle ABC with the sides of a,b,c such that a + c = 2b prove that $\cot(\frac{A}{2}) + \cot(\frac{C}{2}) = 2 \cot(\frac{B}{2})$

Solution:
We have using law of cot for a triangle using standard notation s ,and area of triangle $\triangle$

$\cot(\frac{A}{2}) = \dfrac{s(s-a)}{\triangle}\cdots(1)$

$\cot(\frac{B}{2}) = \dfrac{s(s-b)}{\triangle}\cdots(2)$

$\cot(\frac{C}{2}) = \dfrac{s(s-c)}{\triangle}\cdots(3)$

add (1) and (3) to get

 $\cot(\frac{A}{2}) + \cot(\frac{C}{2}) = \dfrac{s(s-a)}{\triangle} + \dfrac{s(s-c)}{\triangle}$
$= \dfrac{s(s-a)+s(s-c)}{\triangle}$
$= \dfrac{s(s-a+s-c)}{\triangle}$
$= \dfrac{s(2s-(a+c))}{\triangle}$
$= \dfrac{s(2s-2b)}{\triangle}$ (from given conditon)
$= 2\dfrac{s(s-b)}{\triangle}$
$=2\cot(\frac{B}{2})$ (using (2))

proved 

298/ Secondary school/ High school students

Problem:

Given that $ \sum_{k=1}^{35}\sin\,5k =\tan(\frac{a}{b}) $ where angles are measured in degrees and a and b are relatively prime positive integers that satisfy $\frac{a}{b} < 90$ evaluate a + b


Solution

We have by sum of sin kx as below

$$ \sum_{k=1}^{n}\sin\,kx = \frac{\sin\frac{nx}{2}\sin\frac{(n+1)x}{2}}{\sin \frac{x}{2}}$$

put n = 35 and $ x = 5^\circ$


to get $$ \sum_{k=1}^{35}\sin\,5k^\circ = \frac{\sin\frac{175}{2}^\circ\sin\,90^\circ}{\sin \frac{5}{2}^\circ}$$
$$ = \frac{\sin\frac{175}{2}^\circ}{\cos \frac{175}{2}^\circ}=\tan  \frac{175}{2}^\circ$$

from given condition $a=175,b=2,\frac{a}{b} = 87.5 <90$ so a+b = 177

003/POTW 297 Secondary school/ High school students

https://mathhelpboards.com/potw-secondary-school-high-school-students-35/problem-week-297-jan-17th-2018-a-23325.html#post104312

Problem:
if $k_1,k_2,\cdots k_{15}$ are the roots of the equation $x^{15} - 2x^{14} + 3 x^{13} + \cdots + 15x-16= 0$ evaluate $(1+k_1)(1+k_2)\cdots(1+k_{15})$

Because $k_1,k_2,\cdots k_{15}$ are the root

Hence $(x-k_1)(x-k_2)\cdots(x-k_{15}) = x^{15} - 2x^{14} + 3 x^{13} + \cdots + 15x-16$ as RHS is a degree 15 polynomial and coefficient is 1

putting $x= -1 $ we get
$(-1-k_1)(-1-k_2)\cdots(-1-k_{15}) = -1 - 2 -3 - \cdots - 15-16= -136$
or
$(-1)^{15}(1+k_1)(1+k_2),\cdots(-1+k_{15}) = -136$
or $(1+k_1)(1+k_2)\cdots(-1+k_{15}) = 136$

002/ POTW 297 University

http://mathhelpboards.com/potw-university-students-34/problem-week-297-jan-18-2018-a-23332.html#post104342l


Problem 
For which nonnegative integers n and k $(k+1)^n +(k +2)^n +(k +3)^n + (k+4)^n +(k +5)^n$ is divisible by 5.
Solution
Let us define
$f(k,n) = (k+1)^n +(k +2)^n +(k +3)^n + (k+4)^n +(k +5)^n$

Hence $f(k+1,n) - f(k,n  ) = (k+6)^n - (k+1)^n$ and it is divisible by $(k+6) - (k+1)$ or 5. this is independent of n

So $f(k+1,n)$ is divisible by 5 iff $f(k,n)$ is divisible by 5

Further as 5 is prime we have as per Fermat's Little theorem $x^5=x$ for all x .

Hence if $f(k,p)$ is divisible by 5 then $f(k,p+5)$ is divisible by 5

So we need to check for $f(0,0),f(0,1),f(0,2),f(0,3),f(0,4)$ and we get
$f(0,0)=5,f(0,1) = 15, f(0,2) = 55, f(0,3) = 225, f(0,4) = 979$

All except $f(0,4)$ is divisible by 5

So it is divisible by 5 for all k and  n which is not of the form 5p+ 4 for non negative k and p  

001/ Challenge problem http://mathhelpboards.com/challenge-questions-puzzles-28/dice-rolling-running-total-probability-23339.html#post104363

A standard six–sided dice is rolled repeatedly and a running total is kept of
all the numbers rolled. Which of 2,6,1006 is more likely to be one of these
totals? Prove your answer.

We http://mathhelpboards.com/challenge-questions-puzzles-28/dice-rolling-running-total-probability-23339.html#post104363

total number of rolls maximum considered shall be 1006.( Because 1007 counts shall give a minumum sum 1007
let us count number of ways partial sums can be
2 = 2 or 1 + 1 so 2 followed by any number or 1 ,1 followed by any number can give sum 2 so 

$2 *6^{1005} + 1 * 1 * 6^{1004}$ ways
6 csn come as (6), (5,1), (4,2)  and other compinations so number of ways 6 can come
$> 6^{1005} + 2 * 6^{1004}$ taking care if above 3 cases and there are more cases also but not requires

for 1 to come we can have all 1 and for $1006^{th}$ element to be 1 the number of ways is $6^{1005}$ out of which some and lot all can have partial sum 1006.

comparing above 3 we see that  number of ways partial sum > 7 is largest. then comes number of ways partial sum 2 then 1006.
so 6 is most likely to occur followed by 2 followed by 1006.

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