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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.

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 $$3a^{4}-4a^{3}b+b^{4}\geq 0$$ .

$$3a^{4} =2a^{4}+a^{4}$$

$$3a^{4}-4a^{3}b+b^{4}$$=$$2a^{4}+a^{4}+b^{4} -4a^{3}b$$

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

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

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

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.