Olympiad preparation begins ...here

Here are some Olympiad preparatory problems. Please share your views on the problems and spread them by resharing ..

(Solutions are invited)

Note by Kislay Raj
5 years, 1 month ago

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No offense, but #6 is super ugly and bashy... I got

x32p2qx25px+2p3q2qx^3-\frac{2p^2}{q}x^2-5px+\frac{2p^3-q^2}q

Cody Johnson - 5 years, 1 month ago

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To this I say #NoBash. :P

Finn Hulse - 5 years, 1 month ago

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no.6 was designed by the overlord of hell.

Sharvin Jondhale - 5 years, 1 month ago

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hey cody ....yeah u are right ..#6 was a long one.....what do u think of the other problems...?

Kislay Raj - 5 years, 1 month ago

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For #1,

We know that 25!=2a1×3a2×5a3××23a9 25! = 2^{a_1} \times 3^{a_2} \times 5^{a_3} \times \dots \times 23^{a_9} for some natural numbers a1,a2,,a9 a_1,a_2,\ldots, a_9.

Since p p and q q are coprime and pq=25! pq =25! , if mnq m_n|q for some prime mn m_n , then mnanp m_n^{a_n}|p . Therefore, the factors of p p are from the set [2a1,3a2,5a3,,23a9] [2^{a_1},3^{a_2},5^{a_3},\dots, 23^{a_9}] .Since there are two possible choices for each of the 9 factors(i.e either mnanp m_n^{a_n}|p or mnan∤p) m_n^{a_n}\not|p ) there are 29 2^9 possible values of p p and therefore 29 2^9 values of pq \frac{p}{q}

Now we have to find how many of these values satisfy 0<pq<1 0 < \frac{p}{q} < 1 .The first half of the inequality is trivially true for all pairs. Now for every fraction piqi \frac{p_i}{q_i} , there exists a fraction pjqj \frac{p_j}{q_j} such that pi=qj p_i = q_j and qi=pj q_i = p_j . Now, since piqi×pjqj=piqi×qipi=1 \dfrac{p_i}{q_i} \times \dfrac{p_j}{q_j} = \dfrac{p_i}{q_i} \times \dfrac{q_i}{p_i} = 1 and piqi1 \frac{p_i}{q_i} \not = 1 for all i i , it implies that for each pair of fractions piqi \frac{p_i}{q_i} and pjqj \frac{p_j}{q_j} , one fraction is > 1 and the other < 1. Since there are 292=28 \frac{2^9}{2} = 2^8 pairs of fractions, there are 28=256 \boxed{2^8 = 256} fractions which satisfy the given conditions

Siddhartha Srivastava - 5 years, 1 month ago

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wow ...a very good solution .. i think u cracked the toughest part :)

Kislay Raj - 5 years, 1 month ago

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Sol #3 (Recommend to draw with me.) (All these numbers and variables are in degrees.)

Midpoint of AD,BC\overline{AD}, \overline{BC} is F,GF,G respectively. Diagonal of the square meets at HH. Draw AE\overline{AE} bisects BD\overline{BD} at II.

Let AE^H=a,EA^H=b,AE^B=cA\hat{E}H = a, E\hat{A}H = b, A\hat{E}B = c.

We know that AH^F=45A\hat{H}F = 45, BE^G=75B\hat{E}G = 75, EA^B=45bE\hat{A}B = 45 - b, AI^B=c+30A\hat{I}B = c+30

AEH;a+b=45\triangle AEH; a + b = 45

Straight angle at E;a+c+75=180E; a + c + 75 = 180

EAB;45b+c+30+75=180\triangle EAB; 45-b+c+30+75 = 180

Solve three equations we get a=30,b=15,c=75a = 30, b = 15, c = 75.

Since EA^D=45+b=60E\hat{A}D = 45 + b = 60, by symmetry, ED^A=60E\hat{D}A = 60.

Therefore, AED\triangle AED is equilateral triangle. Meow!

If someone have better solution without solving equations, that'd be great! ^__^

Samuraiwarm Tsunayoshi - 5 years, 1 month ago

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There exist a very easy solution to this problem I will add the diagram, can somebody tell me how to link images here? I put this text, but apparently it doesn't work.![alttext](/Users/JordiBosch/Desktop/TDR/DEMO.png"Title")![alt text](/Users/JordiBosch/Desktop/TDR/DEMO.png "Title")

Jordi Bosch - 5 years, 1 month ago

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I haven't done this before, but you should upload this img into website instead. Sorry I don't know the website =="

Samuraiwarm Tsunayoshi - 5 years, 1 month ago

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If you can't do that, you can just tell everyone how to draw that. I know everyone can.

Samuraiwarm Tsunayoshi - 5 years, 1 month ago

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hey Samuraiwarm Tsunayoshi, yor solution is really a good and short one

Kislay Raj - 5 years, 1 month ago

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Isn't this Resonance opt paper?

Lavisha Parab - 5 years, 1 month ago

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(2) is interesting. The answer seems to be 73/64 73/64 . If the polynomial is xn+17xn1+1 x^n + 17x^{n-1} + 1 , the answer only depends on n n mod 6 6 . I have most of a proof but will have to wait until tomorrow to write it up.

Patrick Corn - 5 years, 1 month ago

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sol #4.
we have to find gcd of 20+ n2n^{2} and20+ (n+1)2(n+1)^{2} .
which is the same as gcd(20+ n2n^{2} , 20+ n2n^{2}+2n+1).
which is the same as gcd(20+ n2n^{2} , 2n+1).
let 2n+1=p ==> n=(p-1)/2 we have to find gcd(20+((p1)2(p-1)^{2}/4) , p).
opening p-1 square we have gcd( (81+p2p^{2}-2p)/4 ,p).
for 81+p2p^{2}-2p to share a factor with p, p is power of three.
clearly possible gcds are 1,3,9,27and 81(all powers of three not greater than 81)

Adit Mohan - 5 years, 1 month ago

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sol #5.
let angleAOB=X.
thus COD =X .
BOC=180-X.
AOD=180-X.
let AO=a, BO=b, CO=c, DO=d.
ab sinX=4. (area of AOB). eq1.
cd sinx=9. (area of COD). eq2.
area of triangle BOC= bd sin(180-X)= bd sinX.
similarly area(AOD)=ac sinX.
area(quadrilateral)= 4+9+ bd sinX+ac sinX.
=13+sinX(ac+bd).
(from eq2 sinX=9/cd).
=13+9/cd(ac+bd).
=13+9(b/c+a/d).
from am gm inequality b/c+a/d>=2sqrt(ab/cd)=4/3.
therefore min =13+9(4/3).
=25.

Adit Mohan - 5 years, 1 month ago

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For (4), dn=1,3,9,27,81 d_n = 1,3,9,27,81 are the possibilities. These are attained at n=2,1,4,13,40 n = 2, 1, 4, 13, 40 respectively. To see that these are the only ones, note that if dan d | a_n and dan+1 d | a_{n+1} , then d(4an(2n1)(an+1an)) d | (4a_n - (2n-1)(a_{n+1}-a_n)) , and 4an(2n1)(an+1an) 4a_n - (2n-1)(a_{n+1}-a_n) happens to equal 81 81 .

Patrick Corn - 5 years, 1 month ago

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I think that 1,4 are very well known problems. First has appeared in RMO too

A Former Brilliant Member - 5 years, 1 month ago

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Sorry, I MEANT 1,3

A Former Brilliant Member - 5 years, 1 month ago

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