Tag: AM-GM Inequality

If you got stuck in this problem start this problem using this hint :

Start with the given hint

(x-p) (x-13) +4 = (x+q)(x+r)

Let’s try to minimize the expression by taking x= -q

so , (-q -p)(-q -13) = -4 , which becomes (q+p) (q+13) = -4

As it is already given p and q are integers we can come up with many cases .

Try to find out the different cases we can have ………………………..

Starting after the last hint :

p+q = 4 and q+13 = -1 ………………………………..(1)

q+p = -4 and q +13 = 1……………..(2)

p+q = 2 and q+13 = -2 and …………………………(3)

p+q = -2 and q+13 = 2 …………………………(4)

For 1st case its simple calculation that we get q = -14 and p = 8

The initial expression becomes (x-p) (x-13) +4 = (x-14) (x-17)

For 2 nd case :

q= -12 and p = 8

so the initial expression becomes : (x-p)(x-13)+4 = (x-9)(x-12)

try the rest of the cases…………

Now let’s talk about 3rd case ,

q= -15 , p = 17 and

hence (x – p) (x – 13) +4 = \(( x – 15)^2\) which is not true to this problem .

For last case , we obtain q = -11, q = 9 so the initial

(x- p) (x-13) +4 = \((x-11)^2\)

So p is 8 and 18 which add upto 8 +18 = 26 -(Answer)

If you really got stuck into this sum we can start from here

\(\frac {cos 100^\circ}{1-4 sin 25^\circ cos 25^\circ cos 50^\circ}\)

= \(\frac {cos 100^\circ}{1-2sin 50^\circ cos 50^\circ}\)

Now let’s check with some basic values in trigonometry

\( Cos 2 A = cos^2 A – sin^2 A \) and

\(2 sin A cos A = sin 2 A\)

Now try the rest of the sum by using these two above mentioned values………………

Let’s continue from the last hint :

\( cos 100^\circ = cos^2 50^\circ – sin^2 50^\circ \)

\( 2 sin 25^\circ cos 25^\circ = sin 50^\circ\)

\(\frac {cos^2 50^\circ – sin^2 50^\circ}{2sin 50^\circ cos 50^\circ}\)

\(\frac {cos^2 50^\circ – sin^2 50^\circ }{(cos 50^\circ – sin 50^\circ)^2}\)

Using \(a^2 – b^2 = (a+b) (a-b)\) formula

\(\frac {cos 50^\circ + sin 50^\circ}{cos 50^\circ – sin 50^\circ}\)

Do the rest of the steps ……………..

Starting from right after the last hint:

\(\frac {cos 50^\circ + sin 50^\circ}{cos 50^\circ – sin 50^\circ}\)

= \(\frac {1+ tan 50^\circ}{1-tan 50^\circ}\)

= \(\frac {tan 45^\circ + tan 50^\circ}{1-tan 45^\circ tan 50 ^\circ}\)

= \( tan 95^\circ\) – Answer

If you got stuck in this sum how to get started you can start by consider a general case where \(40^\circ = A\) , \(60^\circ = B\) and \(80^\circ = C\).

So , A+B+C = \( 180 ^\circ\)

\( A+B = 180^\circ – C\)

(tan (A+B) = tan \(180^\circ – C)\)……………………….(1)

Now try to implement the basic formula and try to do this sum………………

In this we can continue from the last hint:

the formula of tan (A + B) = \(\frac {tan A + tan B}{1- tan A . tan B}\)

From the equation (1) …….

tan (A+B) = tan (180 – C)

\(\frac {tan A + tan B}{1- tan A . tan B} = tan (180^\circ – c)\)

(frac {tan A + tan B}{1- tan A . tan B} = -tan C )

Now just rearrange this expression and you will get the final answer……………..

Here is the final solution:

{tan A + tan B} = -tan C {1- tan A . tan B}

tan A + tan B = – tan C + tan C tan A tan B

tan A + tan B + tan C = tan A tan B tan C

\(\frac {tan A tan B tan C}{tan A + tan B + tan C} = 1\)

Which is the given question. It can be a proof also……………….

If you got stuck in this sum we can start from here:

Let d is the highest common factor of n-11 and 3n +20 which is greater than 1.

So d|(n-11) and d|(3n + 20) .

If we compile this two then d|(3n +20 -3(n-11) when d|53 .

Now one thing is clear that 53 is a prime number and also d >1

so we can consider d = 53.

Now try the rest………………

Now from the previous hint

n-11 = 53 k (let kis the +ve integer)

n = 53 k +11

So for any k 3n +20 is a multiple of 53.

so 3n + 20 = 3(53k +11) +20 = 53(3k+1)

Finish the rest………..

Here is the final solution :

After the last hint :

n = 64 (if k = 1) which is the smallest integer. as hcf of (n – 11,3n +20)>1(answer)

We can start this kind some by using the concept of series and sequence …….

In this problem we can see that the series as

\(\frac {m}{n}\) =\(\frac {1}{10 \times 12}\) +\(\frac {1}{12 \times 14}\) +\( \cdot \cdot\)+

\(\frac {1}{2012 \times 2014} \)

So sum of this series is

\(\frac {m}{n} = \frac {1}{4} \displaystyle\sum _{k = 5}^{1006} \frac {1}{k(k+1)}\)

Now do the rest of the sum ………………

If you are really stuck after the first hint here is the rest of the sum……………

From the above hint we can continue this problem by breaking the formula more we will get :

= \(\frac {1}{4} \displaystyle\sum_{k=5}^{1006} \frac {1}{k} – \frac {1}{k+1}\)

Now replacing by the values:

\(\frac {1}{4} (\frac {1}{5} – \frac {1}{1007}) \)

Please try to do the rest…………………

This is the last hint as well as the final answer….

If we continue after the last hint…

\(\frac {m}{n} = \frac {501}{10070}\)

Since gcd(501,10070) = 1

we can conclude by the values of m= 501 and n = 10070

So the sum is m+n = 10571 (Answer).

We can try this sum from taking

OA = \(\frac {30^2}{20} = 45\)

So the area of \(\triangle {ABC} = \frac {(20+45)\times 30}{2} = 975\)

Try to do the rest of the sum………………………

Now lets try to find height of \(\triangle {CGF}\)

Suppose height of \(\triangle {CGF}\) be ‘h’. Then

\((\frac {h}{30})^2 = \frac {351}{975} = (\frac {3}{5})^2\)

\(\frac {h}{30 – h} = \frac {3}{2}\)

Now we have almost reach the answer . Try to find the area of Rectangle DEFG……

Note that the rectangle DEFG has the same base as \(\triangle {CGF}\). Then its area is

\( 351 \times \frac {2}{3} \times 2 = 468 \) (Answer )

We can start the sum from here :

The region S is the hexagon enclosed by lines

\(x = \pm 10 \) ; \( y = \pm 10 \) and \( x + \frac {1}{2} y = \pm 10\).

So the largest circle contained in S must be tangent to \(x + \frac {1}{2} y = \pm 10\)

Try to find the radius from this ………………..

We can start this hint by using a diagram:

So following the last hint we can understand that its radius is the distance from the origin (0,0) to \( x + \frac {1}{2} y = 10 \)

r = \( \frac {10}{ \sqrt {1+ ({\frac {1}{2}})^2}} = 4\sqrt {5} \).

Now it very easy to find the rest of the solutiion………

So the area of a circle is \(\pi {r}^2\)

= \(\pi {(4\sqrt 5)}^2 \)

= \(80 \pi\) (Answer).

This can be a very good example to show Triangle Inequality

Let ‘ x ‘ be the length of the first side of the given triangle. So the length of the second side will be 3 x and that of the third side be 15 . Now apply triangle inequality and try to find the possible values of the sides.

If we apply Triangle Inequality here then the expression will be like

\(3 x < x + 15 \)

\( 2 x < 15 \)

\( x < \frac {15}{2}\)

x < 7.5

Now do the rest of the problem ………..

I am sure you have already got the answer but let me show the rest of the steps for this sum

If x < 7.5 then

The largest integer satisfying this inequality is 7.

So the largest perimeter is 7 + 3.7 +15 = 7 + 21 + 15 = 43.