warm-up problems for national BdMO'14

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photon
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warm-up problems for national BdMO'14

Unread post by photon » Sun Feb 09, 2014 1:12 pm

This thread is for some practice as National Math Olympiad is knocking at door . Others may get benefitted , share and learn things together from here . One thing should be clear that this is not just about to get prize- you try , solve , learn and improve your skill and most important is having fun with solving ! :) I wish everyone give problems here and co-operate . :)

I am a bit sick , so didn't search well looking problems :( . I hope you would give cool probs . Remember to give problem number while posting a problem and giving solution . Here I start ...

1. Determine if the number $\lambda _n = \sqrt {3n^2+2n+2}$ is irrational for all non-negative integers $n$.

2.Find the last 2 digits of $7^{1997}$.

3. Two circles $\omega _1 , \omega _2$ with centers $O_1,O_2$ intersect at $A$ and $B$ . D is an arbitrary point on $\omega_1$ and $DB$ meets $\omega _2 $ at $E$. chord $EA$ extended meets circle $\omega_1$ at $F$, $DF$ and $BA$ extended meet at $G$. $GE$ intersects circle $\omega_2$ at $H$. Prove that $O_1H$ is perpendicular to $GE$ .

Source:
[1] Spain MO 2012 , [2]brilliant , [3] a geometry website
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sowmitra
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Re: warm-up problems for national BdMO'14

Unread post by sowmitra » Sun Feb 09, 2014 2:11 pm

*Solution of No.(2):
We need to calculate $\pmod {100}$.
First of all, $100=2^2\times 5^2$.
$\therefore \varphi(100)=100\times(1-\frac{1}{2})\times(1-\frac{1}{5})=40$.
$\because gcd(7,100)=1$, $\therefore 7^{40}\equiv 1\pmod{100}$.
Now, $1996\equiv36\pmod{40}$.
$\displaystyle \therefore 7^{1996}\equiv 7^{36} \equiv \frac{7^{40}}{7^4}\equiv \frac{1}{49^2}\equiv \frac{1}{2401}\equiv \frac{1}{1}\equiv1 \pmod{100}$.
So, the remainder is $\boxed{01}$.
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sowmitra
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Re: warm-up problems for national BdMO'14

Unread post by sowmitra » Sun Feb 09, 2014 2:27 pm

A much easier solution to No. (2):
$7^{2}\equiv (-1)\pmod{25} \Rightarrow 7^4\equiv 1\pmod{25}$.
Again, $7^2\equiv 1 \pmod4 \Rightarrow 7^4\equiv 1\pmod4$
$\therefore 7^4\equiv 1\pmod{100}$
Now, $1996 \equiv 0 \pmod4$
$\displaystyle \therefore 7^{1996}\equiv 7^0\equiv 1\pmod{100}$
And, the result follows.
Last edited by sowmitra on Sun Feb 09, 2014 3:40 pm, edited 1 time in total.
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Fatin Farhan
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Re: warm-up problems for national BdMO'14

Unread post by Fatin Farhan » Sun Feb 09, 2014 3:31 pm

$$\lambda _n = \sqrt {3n^2+2n+2}$$
$$=>(\lambda _n)^2 = 3n^2+2n+2$$
$$=>3(\lambda _n)^2 = 9n^2+6n+6$$
$$=>3(\lambda _n)^2 = (3n+1)^2+5$$
$$L.H.S \equiv 0,3,4\pmod{8}$$
$$R.H.S\equiv 5,6,9\pmod{8}$$.
L.H.S and R.H.S cannot be same.
$$\therefore \lambda _n \neq \sqrt {3n^2+2n+2} $$
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asif e elahi
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Re: warm-up problems for national BdMO'14

Unread post by asif e elahi » Sun Feb 09, 2014 5:29 pm

Solution of problem $3$:
As $ABHE$ and $ABDF$ cyclic $BHG=180^{\circ}-BHE=BAE=180^{\circ}-FAB=FDB=180^{\circ}-BDG$
or $BHG+BDG=180^{\circ}$
So $BHGD$ cyclic.
By power of point theorem
$GO_{1}^{2}-R^{2}=GB\times GA$
and $EO_{1}^{2}-R^{2}=EB\times ED$ where $R$ is the circumradius of $ABDF$
subtracting this equalities we get
$GO_{1}^{2}-EO_{1}^{2}=GB\times GA-EB\times ED$
=$GH\times GE-EH\times EG$ [As $ABHE$ and $DBHG$ cyclic]
=$GE(GH-EH)=(GH+EH)(GH-EH)$
=$GH^{2}-EH^{2}$
So $GO_{1}^{2}-EO_{1}^{2}=GH^{2}-EH^{2}$
So by perpendicular lemma $O_{1}H\perp GE$ :mrgreen:

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sowmitra
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Re: warm-up problems for national BdMO'14

Unread post by sowmitra » Sun Feb 09, 2014 6:22 pm

I've got some more:
$[4]$(JBMO'07)
Let $ABCD$ be a convex quadrilateral with $\angle DAC = \angle BDC = 36^\circ , \angle CBD = 18^\circ$ and
$\angle BAC = 72^\circ$. The diagonals and intersect at point $P$ . Determine the measure of $\angle APD$ in degrees.

$[5]$(Self-Made)
Show that, there is no positive integer $a$, which satisfies the given inequality,
$$\displaystyle \frac{1+\sqrt{1+8N}}{2}<a<\frac{1+\sqrt{9+8N}}{2}$$
where, $N \in \mathbb{N}$

$[6]$(APMO'08)
$\Gamma$ is the circum-circle of $\triangle ABC$. A circle passing through $A$ and $C$ meets $BC$ and $BA$ at $D$ and $E$ respectively. $AD$ and $CE$ intersect $\Gamma$ again at $G$ and $H$. The tangents of $\Gamma$ at $A$ and $C$ meet $DE$ at $L$ and $M$. Prove that, the intersection of $LH$ and $MG$ lies on $$\Gamma$$.
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photon
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Re: warm-up problems for national BdMO'14

Unread post by photon » Sun Feb 09, 2014 7:10 pm

@Sowmitra , I had solved this using totient theorem ,but $7^4$ made it clean . Though it was $7^{1997}$ , doesn't matter ...
@Fatin , it is not $LHS\not=RHS$ , it is that considering $\lambda_n$ as integer the equation got wrong , so contradiction for integer :) . $\lambda_n \notin \mathbb{N}$ [$\lambda_n$ can't be a fraction]

Problems:

7. Positive integers $a$ and $b$ satisfy the condition $log_2(log_{2^a}(log_{2^b}(2^{1000})))=0$ . Find the sum of all possible values of $a+b$.

8. Find all positive integers $n$ and $k$ such that $(n+1)^n=2n^k+3n+1$ .

Source:
[7] AIME 2013 , [8] Spain MO 2012
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Fatin Farhan
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Re: warm-up problems for national BdMO'14

Unread post by Fatin Farhan » Sun Feb 09, 2014 8:33 pm

$$log_2(log_{2^a}(log_{2^b}(2^{1000})))=0$$
$$=> log_{2^a}(log_{2^b}(2^{1000}))=1$$
$$=> log_{2^b}(2^{1000})=2^a$$
$$=> 2^{1000}=(2^b)^{2^a}$$.
$$a=[1,3]$$. So,
$$(a,b)= (1,500),(2,250), (3,125)$$.
So value of all $$a+b=881$$
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Fatin Farhan
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Re: warm-up problems for national BdMO'14

Unread post by Fatin Farhan » Sun Feb 09, 2014 9:24 pm

8.
$$\binom{n}{n-1}n=3n$$
So, $$n=3$$
now,
$$(3+1)^3=2*3^k+3*3+1$$
$$=> 64-10=2*3^k$$
$$=>2*3^k=54$$
$$k=3$$
$$(n,k)=(3,3)$$
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Fatin Farhan
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Re: warm-up problems for national BdMO'14

Unread post by Fatin Farhan » Mon Feb 10, 2014 5:48 pm

Let $$x=\sqrt{1+8N}$$ and $$y=\sqrt{9+8N}$$
$$y^2-x^2=8$$
So, $$(y+x)=4,8$$ and $$(y-x)=2,1$$
Solving these we get $$(x,y)=(1,3)$$
Now,
$$ \dfrac{1+\sqrt{1+8N}}{2}=\dfrac{1+x}{2}=\dfrac{2}{2}=1$$.
$$ \dfrac{1+\sqrt{9+8N}}{2}=\dfrac{1+y}{2}=\dfrac{4}{2}=2$$.
And there exists no positive integer $$a$$ such that
$$1<a<2$$
or, $$\displaystyle \frac{1+\sqrt{1+8N}}{2}<a<\frac{1+\sqrt{9+8N}}{2}$$
"The box said 'Requires Windows XP or better'. So I installed L$$i$$nux...:p"

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