Math Olympiad Contest!-1

Even this does not come under Olypmpiad problems.Please delete this question and post another OLYMPIAD problem.

Attention:The actual problem 16 is not this.

here in the daigram it ab need not be the tangent

the answer lies in the defination of hcf and lcm.we can represent it as a venn daigram.the elements which are perfectly common are taken out which is the hcf and what remains behind are the elements which not perfectly common which is the lcm . when we do the union of the set hcf and lcm we get the original set which the union of the 2 numbers

Case 1:When $a$ and $b$ are co-prime to each other
Since $a$ and $b$ are co-prime,the G.C.D. is $1.$The $L.C.M.$ will be $a\times b$ because $a$ and $b$ should divide the $L.C.M.$So since $a$ and $b$ do not have any common factors,the $L.C.M$ should be the product of $a$ and $b.$Therefore $ab=1\times(a\times b).ab=ab.$Hence proved.
Case 2:When a and b are not co-prime to each other
Let the $G.C.D$ be $k$ of $a$ and $b.$Well,$a\times b$ will contain $k$ since it is the common divisor of $a$ and $b.$So let $a\times b=k\times m.$Since $k$ is already contained in $a$ and $b,$the L.C.M will have to remaining part[that is $m$] for it to divisible by both $a$ and $b.$So,$a\times b=gcd(a,b)\times lcm(a,b)=k\times m=a\times b.$Hence proved.

2,3,4 in any order. Solved by using two new terms, xy+yz+zx and xyz and then using newton's identities

r=(a+b-c)/2. Take general pythagorean triplet, x^2+y^2,x^2-y^2,2xy to get the result.

c is hypotenuse=x^2+y^2

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Make a polynomial with 1/alpha,1/beta,1/gamma as roots.

We have to find A^3+B^3+C^3=3ABC+(A+B+C)((A+B+C)^2-3(AB+BC+CA). from the new polynomial we know each of the following term.

Answer is -42

Wrong reasoning

good try

for a right angled triangle a^2+b^2=c^2 where c is the largest side. the equality holds even for the angles that is 90 =90 now for an acute angled triangle , when we decrease the 90 degree angle the other degrees will be added to the other two angles ,thus the equality breaks and would point towards a and b .

$8(a^4+b^4) \ge 4(a^2+b^2)^2 \ge 4 \left(\dfrac{(a+b)^2}{2} \right)^2=(a+b)^4$. Used Holder and QM-AM inequality

sorry guyes the question is wrong

8(a^4+b^4)greater than or equal to (a+b)^2 if a, b, are positive numbers

Circumradius=abc/4delta=65/8.

OI=SQRT (R (R-2r))=1.008 approximately

Wrong answer!

Wrong

1 hr and 15 minutes have passed

As the problem is too hard to solve and going against rule 3,we are continuing with a new problem.

@Prince Loomba the answer is 288

ANSWER IS 636

Nobody got right. @Ayush Rai give my points to me.

Along the same lines I had a different idea: $p(\omega) = -a_{5}\omega ^2 + a_4 \omega - a_3 +a_2 \omega ^2 - a_1\omega + a_0 = 0$

Therefore,

$(a_2 - a_5)\omega ^2 + (a_4 - a_1)\omega + (a_0 - a_3) = 0$

Hence we get,

$a_2 - a_5 = a_4-a_1=a_0-a_3$

And similarly for $p(\omega ^2)$

Bro 6 hrs passed i suppose

Where's the figure??

5892? Might be a calculating error Am I right?

First get that x-2=b+c/a -->x-1=(a+b+c)/a --->1/(X-1)=a/(a+b+c) Now get eqns for 1/(y-1) ,1/(z-1) Their sum is 1 Thus(x-1)(y-1)+(y-1)(z-1)+(x-1)(z-1)=(x-1)(y-1)(z-1 Now expand and put values to get answer

449

Are x,y,z integers?

x,y,z are integers.

I am feeling too lazy to write the whole solution and even finding the correct answer. I will only share the idea that I had in mind. This question actually boils down to finding coefficient of $m^k$ ($k=x+y+z$) in the expansion of : $(1+m+m^2+m^3+\dots m^60)^3=(1-m^{61})^3(1-m)^{-3}$

I hope from here on its quite simple!

From above condition we can get (60-x)+(60-y)+(60-z)<=30 Then introduce a dummy variable w and change the equation to (60-x)+(60-y)+(60-z)+w=30. Counting these cases is equal to counting the ways of distributing 30 apples in 4 people such that anyone can get any number of apples which are equal to 33C3.

No

I think graph itself is bounded as rectangles

2? a=1 b=2

LaTeX: \begin{aligned} S & = \displaystyle \sum^{infty}_{n=1}\frac{n}{n^4+n^2+1} \\ & = \displaystyle \sum^{infty}_{n=1} \frac{n}{(n^2+1+n)(n^2+1-n)} \\ & = \frac{1}{2} \displaystyle \sum^{infty}_{n=1} \left(\frac{1}{n^2+1-n} - \frac{1}{n^2+1+n}\right) \\ & = \frac{1}{2} \left(1-\frac{1}{3}+\frac{1}{3} - \frac{1}{7} + \frac{1}{7} -\frac{1}{13} + \dots =1/2

n^4+n^2+1=(n^2+n+1)(n^2-n+1) now convert it into telescoping sum to get answer

Firstly if it has roots, one has to be integral as conjugate complex root pairs are formed and , now to prove no integral root is possible, consider y as a root, take it to be 5k,5k+1..,5k+4

Break given equation...

y (y-1)(y+1)(y^2+1)=a, a is not multiple of 5

Then y=5k is eliminated as first term (y)

y=5k+1 and 5k+4 are eliminated by 2nd and 3rd terms respectively.

y=5k+2 and 5k+3 are eliminated by the last term, forming type 5m. Square them and add 1 to get the result.

Thus no y is possible.

So no real root. So cant be broken into factors, as one has to be linear if it can be broken ( the other has imaginary roots)

If or if and only if

Clarification: its $f ((x-y)^{2})$

f(X)=X I dont think any other Is possible Am I right?

First put x=y=0 then we have f(0)=((f(0)))^2 Thus f(0)=1 or f(0)=0 Now put x=y but they are not equal to zero then we have f(0)=(f(x))^2 -2x(f(x))+x^2 write rhs as (f(x)-x))^2 So if f(0)=0 Then f(x)=x Or else f(x)-x=1 f(x)-x=-1 Thus possible values of f(x) are x,x+1,x-1 Now put values of f(x) To see which satisfy and u will get the answer

5/4??

Multiplying both sides by √(x+√x). Then after transposing and squaring, the result follows

iam banned from slack

If you found the circumradius of the triangle then use the property that the centroid of a triangle divides the median ib the ratio two is to one .No need to do such a long calculation.

do u want something?

You are wrong.The points A and D do not coincide..

i think no

Directly use cosine rule and put cos>0 so simple! :)