**MATHMAA**

Answers from questions 3 onwards of Sample Paper 1 are here .

**3Q)
Prove that in any triangle ABC, if one angle is \(120^{\circ}\), the
triangle formed by the feet of the angle bisectors is a right angled.**

Solution : Let Trianlge ABC, \(\angle A=120^{\circ}\).

Let AD, BE, CF are angle bisectors of A, B, C respectively.

In \(\Delta ABD\), AC is exterior angle bisector(as\(\angle DAC=\angle CAX =60^{\circ}\), BE is internal angle bisector meet at E . E is excentre of \(\Delta ABD\).

So DE is exterior angle bisector of \(\Delta ABD\).

Then \(\angle ADE = \angle EDC \) , \(\angle ADC= 2\angle ADE \).

Similarly we can prove that for the \( \Delta ADC \) F is excenter as AB is exterior angle bisector and CF is interior angle bisector.

Hence DF is exterior angle bisector .

\(\angle ADF =\angle FDB \), \(\angle BDA=2\angle ADF\)

BC is straight angle .

\(\angle BDA + \angle ADC =180^{\circ}\)

\(2\angle ADF + 2\angle ADE =180^{\circ} \).

\( \angle ADF + \angle ADE = 90^{\circ}\)

Hence DEF is right trianlge.

Therefore the triangle formed by the feets of angle bisectors of triangle whose one angle measure is \(120^{\circ}\) is *Right Triangle. *

**4Q) a) Find all the integers which are equal to 11 times the sum of their digits.**

Answers :

We have to find the integers which are equal to 11 times the sum of their digits.

To satisfy the requirement , sum of the digits of n-digit number is less than 9n (all 9).

11 times the sum of the n-digit number is less than 11*9n=99n and for n\(\geq\)4 , number is less than n digits, that is 99*4=396 but no four digit number is less than 396, similarly for n>4 also.

So we have to check for n<4 only.

if n=0 then we get,

One digit integer we get 0 only as 0=11(0) .

n=1 , 99n=99 so check the two digit numbers

Two digit numbers:

Let a and b be the digits in tens and ones place respectively and \(0<a\leq 9 , 0\leq b\leq 9 \).

The number = 10a + b

It is equal to 11 times the sum of a and b

10a + b = 11(a+b)

10a + b = 11a + 11b

a + 10b = 0 , no such digits for a and b exist.

So no solution for two digit numbers.

Let us check for three digit numbers.

100a+10b+c =11(a+b+c)

100a + 10b +c =11a + 11b + 11c

89a - b - 10c =0

89a =10c+b

right side we have 10c + b which is 2 digit number.

89 a is 2 digit number if a=1 only

then 89 =10c+b which is 10*8+9 =10c+b

c=8 b=9

hence we get the number abc as 198 which satisfy the given requiremens.

Therefore we get only two integers 0 and 198 which are equal to 11 times the sum of the digits.

**b) Prove that \(3a^{4}-4a^{3}b+b^{4}\geq 0\)** .

Answers :

\(3a^{4} =2a^{4}+a^{4} \)

\( 3a^{4}-4a^{3}b+b^{4} \)=\(2a^{4}+a^{4}+b^{4} -4a^{3}b\)

\( =(a^{2}-b^{2})^{2} +2a^{2}b^{2}+2a^{4} -4a^{3}b\)

\( = (a^{2}-b^{2})^{2} +2a^{2}(a^{2}-2ab +b^{2})\)

\( =(a^2 -b^{2})^{2} +2a^{2}(a-b)^{2}\)

square of any number is positive and sum of any positive numbers is also positive.

\((a^{2}-b^{2})^{2} \geq 0, 2a^{2} \geq 0, (a-b)^{2} \geq 0 \)

Hence \( =(a^2 -b^{2})^{2} +2a^{2}(a-b)^{2} \geq 0 \)

Therefore \(3a^{4} - 4a^{3}b + b^{4} \geq 0 \).

Hence we proved the Answers for both parts a and b of Q4.

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