2015/100)n any triangle ABC, prove that

$a^2(\cos^2 B - \cos^2 C) + b^2(\cos^2 C -\cos^2 A) + c^2(\cos^2 A - \cos^2 B) = 0$

proof:

$a^2(\cos^2 B - \cos^2 C) + b^2(\cos^2 C - \cos^2 A) + c^2(\cos^2 A - \cos^2 B)$
= $a^2(\sin^2 C - \sin^2 B) + b^2 (\sin^2 A - \sin^2 C) + c^2( \sin^2 B - \sin^2 A)$

using law of sines we have

let $\dfrac{a}{\sin A} = \dfrac{b}{sin B} = \dfrac{c}{\sin C} = k$ (say)

we get

$a^2(\sin^2 C - \sin^2 B) + b^2 (\sin^2 A- \sin^2 C) + c^2( \sin^2 B - \sin^2 A)$

=$k^2 \sin ^2 A(\sin^2 C - \sin^2 B) + k^2 \sin ^2 B (\sin^2 A -\sin^2 C) + k^2 \sin ^2 C( \sin^2 B - \sin^2 A)$

= $k^2( \sin ^2 A \sin ^2 C – \sin ^2 A \sin ^2 B + \sin ^2 B \sin ^2 A – \sin ^2B \sin^2 C + \sin ^2 C \sin ^2 B – \sin ^2C \sin ^2 A)$

= 0

21015/099) Solve the system of equations

$a(b+c+d+e+f)=184$

$c(a+b+d+e+f)=301$

$e(a+b+c+d+f)=400$

$b(a+c+d+e+f)=225$

$d(a+b+c+e+f)=225$

$f(a+b+c+d+e)=525$

Solution

We note that all the rest are of the form $n(g-n)$
where $g=a+b+c+d+e+f$ and observe that smaller the value, the smaller n and hence b = d

Factoring, we see that
$a(g-a) = 2^2 * 3 * 7$
$b(g-b) = d(g-d) = 3^2 * 5*2$
$e(g-e) = 2^4 * 5^2$
$f(g-f) = 3 * 5^2 * 7$
and
a < b = d < e < f

Because b and d must be greater than a,

so we have the set for a 2,3,4,7

b = 3,5,

d= 3,5

e= 2,5,10

f = 3,5,15 so on

we can quickly find values that satisfy that and the last four equations: they all work out if g=50, which implies that c=7.

This produces the solution
$a=4, b=d=5, c=7, e=8, f=21$

 

2015/098) Prove $(4 \cos\, 20^\circ + 1) \tan 20^\circ = \sqrt3$

we have $tan\, 60^\circ – \tan\, 20^\circ$
= $\dfrac{\sin\, 60^\circ}{\cos\,60^\circ} – \dfrac{\sin\, 20^\circ}{\cos\,20^\circ}$
= $\dfrac{\sin\, 60 ^\circ\cos\, 20^\circ – \cos\, 60^\circ \sin\, 20^\circ}{\cos\, 60^\circ\, \cos\, 20^\circ}$
= $\dfrac{\sin\, 40^\circ}{\cos\, 60^\circ \cos\, 20^\circ}$
=  $2\dfrac{\sin\, 40^\circ}{\cos\, 20^\circ}$
= $2 \dfrac{2 sin\, 20^\circ \cos\, 20^\circ}{\cos\, 20^\circ} = 4 \sin\, 20^\circ$
hence $4\sin \, 20^\circ + \tan\, 20^\circ = \sqrt3$
or $4 \cos\, 20^\circ \tan\, 20^\circ + \tan \,20^\circ = \sqrt3$
or $(4 \cos\, 20^\circ + 1) \tan\, 20^\circ = \sqrt3$  

2015/097) Find the equation of the line through the point (3,5) that cuts off the least area from the first quadrant.

Let the slope of line be m

clearly m is -ve as if m is positive or zero then it shall not cut x and y both in 1st quadrant
so the equation of line is

 $y-5 = m(x-3)$


as it goes via $(3,5)$
 
now x intercept when x = 0 is $y=5-3m$

y intercept when y = 0 is given by $x = \dfrac{3m-5}{m}$    
 so area of the triangle in 1st quadrant is $\dfrac{xy}{2}$
so we need to minimize $xy = \dfrac{(3m-5)^2}{m}$

let m = - p where p > 0

$xy=\dfrac{(5+3p)^2}{p}=(\frac{5}{\sqrt{p}}+3\sqrt{p})^2$
or  $xy=(\frac{5}{\sqrt{p}}-3\sqrt{p})^2+60$

clearly it is lowest when  $\frac{5}{\sqrt{p}}-3\sqrt{p}=0$


 so $p=\frac{5}{3}$

so equation of line is
$y=5-\frac{5}{3}(x-3)$
or $3x+5y=30$

Ths problem appeared as Problem of the week in http://mathhelpboards.com/potw-university-students-34/problem-week-115-june-9th-2014-a-10908.html

where you can find some solution. I answered the question correctly but my solution is not provided there.

2015/096) $ax^2 + bx + c = 0$ has imaginary roots and $a + c \lt b$ then prove that $4a + c \lt 2b$

$f(x) = ax^2 + bx + c$ has imaginary roots so this expression is positive or -ve for all x

$f(-1)= a - b + c < 0$ as$a + c < b$

so

$f(-2) = 4a - 2b +c < 0$ or $4a + c < 2b$

2015/095) If $x^{a}= (x^\frac{b}{2}) (z^\frac{b}{2}) =z^c$ then a,b,c are in?

A) H.P, B) G.P, C) H.P ,D)non of these

we have $x^a = z^c$
so $x =z^\frac{c}{a}\cdots(1)$
further
$x^{a-\frac{b}{2}}=z^{\frac{b}{2}}$
or $x^{2a-b} = z^{b}$
from 1
$z^{(2a-b)*^\frac{c}{a}} = z^{b}$
or
$(2a-b)c= ab$
or
$\dfrac{2}{b} =  \dfrac{1}{a} + \dfrac{1}{c}$
hence they are in HP

2015/094) One roots of a quadratic equation $ax^2+bx+c=0$ is three times the other.

Prove that $3b^2=16ac$

Solution

we are given

$ax^2 + bx+ c = 0 \cdots(1)$

let the roots be t and 3t

so we have $(x-t)(x-3t) = 0$

or $x^2 – 4tx + 3t^2 = 0\cdots(2)$

the roots of(1) and (2) are same so coefficients are proportionate

so

$\dfrac{a}{1} = \dfrac{-b}{4t} = \dfrac{c}{3t^2}$

we need to eliminate t

$4at = -b\cdots(3)$

and $-b = \dfrac{4c}{3t}\cdots(4)$

 

multiplying (3) with (4) we get

$b^2 = \dfrac{16ac}{3}$ or $3b^2 = 16ac$

 

2015/093) Show that $\dfrac{\sec\,A +\tan\, A - 1}{\tan\, A- \sec\, A +1} = \dfrac{\cos\, A}{1-\sin\,A}$

we have
$\dfrac{\sec\,A +\tan\, A - 1}{\tan\, A- \sec\, A +1}$
= $\dfrac{\sec\,A +\tan\, A - (\sec^2A-\tan ^2 A}{\tan\, A- \sec\, A +1}$
=  $\dfrac{\sec\,A +\tan\, A - (\sec\,A+\tan\, A)(\sec\,A-\tan\, A)}{\tan\, A- \sec\, A +1}$
= $\dfrac{(\sec\,A +\tan\, A)(1 - \sec\,A+\tan\, A)}{\tan\, A- \sec\, A +1}$
= $(\sec\,A +\tan\, A)$
= $\dfrac{1}{\cos\, A }+  \dfrac{\sin\, A}{\cos\, A }$
=  $\dfrac{1+ \sin\, A}{\cos\, A }$
=  $\dfrac{(1+ \sin\, A)(1-\sin\, A)}{\cos\, A(1-\sin\, A) }$
= $\dfrac{1-\sin^2A}{\cos\, A(1-\sin\, A) }$
=  $\dfrac{\cos^2A}{\cos\, A(1-\sin\, A) }$
= $\dfrac{\cos\,A}{1-\sin\, A}$

2015/092) Solve $\dfrac{x-a}{b+c} + \dfrac{x-b}{a+c} + \dfrac{x-c}{a+b} = 3$ given a,b,c positive

$\dfrac{x-a}{b+c} + \dfrac{x-b}{a+c} + \dfrac{x-c}{a+b} = 3$
or $(\dfrac{x-a}{b+c}-1) + (\dfrac{x-b}{a+c}-1) + (\dfrac{x-c}{a+b}-1) =0$
or $\dfrac{x-a-b-c}{b+c} + \dfrac{x-b-a-c}{a+c} + \dfrac{x-c-a-b}{a+b} =0$
or  $(x-a-b-c)(\dfrac{1}{b+c} + \dfrac{1}{a+c} + \dfrac{1}{a+b}) =0$
so $x-a-b-c) =0$ or  $\dfrac{1}{b+c} + \dfrac{1}{a+c} + \dfrac{1}{a+b} =0$
as a,b,c are positive so 2nd expression >0 and hence x = a+b+c 

2015/091) Let p be a prime number. Prove that 6[(p-4)!] = 1 (mod p )

we have 6 = 2 * 3
p is prime
now (p-2) mod p = -2
(p-3) mod p = - 3

so 6(p-4)! = - (p-2)(p-3) (p-1)(p-4)! = - (p-1)!

so 6(p-4)! mod p = - 1(p-1)! mod p = -1 * (-1) = 1 mod p

2015/090) Find $\frac{1+a}{1-a}$ if $a = \cos\beta + i\sin\beta$.

$a = \cos\beta + i\sin\beta$
so
 $1+ a = 1+ \cos\beta + i\sin\beta= 2\cos^2\frac{\beta}{2}+2i\cos\frac{\beta}{2}\sin\frac{\beta}{2}= 2 \cos\frac{\beta}{2}(\cos\frac{\beta}{2} +i\sin\frac{\beta}{2})$

further
$1 - a = 1 - \cos\beta -i\sin\beta = ( 2 \sin ^2 \frac{\beta}{2} - 2 i \cos\frac{\beta}{2} \sin\frac{\beta}{2}) = 2 \sin \frac{\beta}{2}(\sin \frac{\beta}{2}- i \cos\frac{\beta}{2})$
= $- 2i \sin \frac{\beta}{2}(\cos\frac{\beta}{2} + i \sin \frac{\beta}{2})$

so $\frac{1+a}{1-a}= \frac{\cos\frac{\beta}{2}}{- i sin \frac{\beta}{2}} = i\cot \frac{\beta}{2}$