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

(Solutions are invited)

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

(Solutions are invited)

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TopNewestNo offense, but #6 is super ugly and bashy... I got

\[x^3-\frac{2p^2}{q}x^2-5px+\frac{2p^3-q^2}q\] – Cody Johnson · 3 years ago

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– Finn Hulse · 3 years ago

To this I say #NoBash. :PLog in to reply

– Sharvin Jondhale · 3 years ago

no.6 was designed by the overlord of hell.Log in to reply

– Kislay Raj · 3 years ago

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

For #1,

We know that \( 25! = 2^{a_1} \times 3^{a_2} \times 5^{a_3} \times \dots \times 23^{a_9} \) for some natural numbers \( a_1,a_2,\ldots, a_9\).

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

Now we have to find how many of these values satisfy \( 0 < \frac{p}{q} < 1 \).The first half of the inequality is trivially true for all pairs. Now for every fraction \( \frac{p_i}{q_i} \), there exists a fraction \( \frac{p_j}{q_j}\) such that \( p_i = q_j \) and \( q_i = p_j \). Now, since \( \dfrac{p_i}{q_i} \times \dfrac{p_j}{q_j} = \dfrac{p_i}{q_i} \times \dfrac{q_i}{p_i} = 1 \) and \( \frac{p_i}{q_i} \not = 1 \) for all \( i \), it implies that for each pair of fractions \( \frac{p_i}{q_i} \) and \( \frac{p_j}{q_j} \), one fraction is > 1 and the other < 1. Since there are \( \frac{2^9}{2} = 2^8 \) pairs of fractions, there are \( \boxed{2^8 = 256} \) fractions which satisfy the given conditions – Siddhartha Srivastava · 3 years ago

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– Kislay Raj · 3 years ago

wow ...a very good solution .. i think u cracked the toughest part :)Log in to reply

Isn't this Resonance opt paper? – Lavisha Parab · 3 years ago

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

Midpoint of \(\overline{AD}, \overline{BC}\) is \(F,G\) respectively. Diagonal of the square meets at \(H\). Draw \(\overline{AE}\) bisects \(\overline{BD}\) at \(I\).

Let \(A\hat{E}H = a, E\hat{A}H = b, A\hat{E}B = c\).

We know that \(A\hat{H}F = 45\), \(B\hat{E}G = 75\), \(E\hat{A}B = 45 - b\), \(A\hat{I}B = c+30\)

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

Straight angle at \(E; a + c + 75 = 180\)

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

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

Since \(E\hat{A}D = 45 + b = 60\), by symmetry, \(E\hat{D}A = 60\).

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

If someone have better solution without solving equations, that'd be great! ^__^ – Samuraiwarm Tsunayoshi · 3 years ago

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– Jordi Bosch · 3 years ago

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.\(![alt text](/Users/JordiBosch/Desktop/TDR/DEMO.png "Title")\)Log in to reply

– Samuraiwarm Tsunayoshi · 3 years ago

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

– Samuraiwarm Tsunayoshi · 3 years ago

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

– Kislay Raj · 3 years ago

hey Samuraiwarm Tsunayoshi, yor solution is really a good and short oneLog in to reply

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 · 3 years ago

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sol #4.

we have to find gcd of 20+ \(n^{2}\) and20+ \((n+1)^{2}\) .

which is the same as gcd(20+ \(n^{2}\) , 20+ \(n^{2}\)+2n+1).

which is the same as gcd(20+ \(n^{2}\) , 2n+1).

let 2n+1=p ==> n=(p-1)/2 we have to find gcd(20+(\((p-1)^{2}\)/4) , p).

opening p-1 square we have gcd( (81+\(p^{2}\)-2p)/4 ,p).

for 81+\(p^{2}\)-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 · 3 years ago

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(2) is interesting. The answer seems to be \( 73/64 \). If the polynomial is \( x^n + 17x^{n-1} + 1 \), the answer only depends on \( n \) mod \( 6 \). I have most of a proof but will have to wait until tomorrow to write it up. – Patrick Corn · 3 years ago

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I think that 1,4 are very well known problems. First has appeared in RMO too – Subrata Saha · 3 years ago

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– Subrata Saha · 3 years ago

Sorry, I MEANT 1,3Log in to reply

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

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