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

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`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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

\[x^3-\frac{2p^2}{q}x^2-5px+\frac{2p^3-q^2}q\]

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To this I say #NoBash. :P

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no.6 was designed by the overlord of hell.

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

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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

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

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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! ^__^

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

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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 =="

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

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

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Isn't this Resonance opt paper?

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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.

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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)

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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.

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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 \( 81 \).

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

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Sorry, I MEANT 1,3

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