BdMO National Secondary: Problem Collection(2016)

[Kazi_Zareer]

Bangladesh National Mathematical Olympiad 2016 : Secondary

Problem 1:
(a) Show that $n(n + 1)(n + 2)$ is divisible by $6$.

(b) Show that $1^{2015} + 2^{2015} + 3^{2015} + 4^{2015} + 5^{2015} + 6^{2015}$ is divisible by $7$.

Problem 2:
(a) How many positive integer factors does $6000$ have?

(b) How many positive integer factors of $6000$ are not perfect squares?

Problem 3:
$ \triangle ABC$ is isosceles $AB = AC$. $ P $ is a point inside $ \triangle ABC $ such that $\angle BCP = 30^{\circ} $ and $\angle APB = 150^{\circ}$ and $\angle CAP = 39^{\circ}$. Find $\angle BAP$.

Problem 4:
Consider the set of integers $ \left \{ 1, 2, ......... , 100 \right \} $. Let $ \left \{ x_1, x_2, ......... , x_{100} \right \}$ be some arbitrary arrangement of the integers $ \left \{ 1, 2, ......... , 100 \right \}$, where all of the $x_i$ are different. Find the smallest possible value of the sum,

$S = \left | x_2 - x_1 \right | + \left | x_3 - x_2 \right | + ................+ \left |x_{100} - x_{99} \right | + \left |x_1 - x_{100} \right | $.

Problem 5:
Suppose there are $m$ Martians and $n$ Earthlings at an intergalactic peace conference. To ensure the Martians stay peaceful at the conference, we must make sure that no two Martians sit together, such that between any two Martians there is always at least one Earthling.


(a) Suppose all $m + n$ Martians and Earthlings are seated in a line. How many ways can the Earthlings and Martians be seated in a line?

(b) Suppose now that the $m+n$ Martians and Earthlings are seated around a circular round-table. How many ways can the Earthlings and Martians be seated around the round-table?

Problem 6:
$\triangle ABC$ is an isosceles triangle with $AC = BC$ and $\angle ACB < 60^{\circ}$. $I$ and $O$ are the incenter and circumcenter of $\triangle ABC$. The circumcircle of $\triangle BIO$ intersects $BC$ at $D \neq B$.

(a) Do the lines $AC$ and $DI$ intersect? Give a proof.

(b) What is the angle of intersection between the lines $OD$ and $IB$?

Problem 7:
Aasma is a mathematician and devised an algorithm to find a husband. The strategy is:

$\circ$ Start interviewing a maximum of $1000$ prospective husbands. Assign a ranking $r$ to each person that is a positive integer. No two prospects will have same the rank $r$.
$\circ$ Reject the first $k$ men and let $H$ be highest rank of these $k$ men.
$\circ$ After rejecting the first $k$ men, select the next prospect with a rank greater than $H$ and then stop the search immediately. If no candidate is selected after $999$ interviews, the $1000^{th}$ person is selected.

Aasma wants to find the value of $k$ for which she has the highest probability of choosing the highest ranking prospect among all $1000$ candidates without having to interview all $1000$ prospects.

(a) What is the probability that the highest ranking prospect among all $1000$ prospects is the $(m + 1)^{th}$ prospect?

(b) Assume the highest ranking prospect is the $(m + 1)^{th}$ person to be interviewed. What is the probability that the highest rank candidate among the first $m$ candidates is one of the first $k$ candidates who were rejected?

(c) What is the probability that the prospect with the highest rank is the $(m+1)^{th}$ person and that Aasma will choose the $(m+1)^{th}$ boy using this algorithm?

(d) The total probability that Aasma will choose the highest ranking prospect among the $1000$ prospects is the sum of the probability for each possible value of $m+1$ with $m+1$ ranging between $k+1$ and $1000$. Find the sum. To simplify your answer use the formula
ln $N$ $ \approx \frac{1}{N-1} + \frac{1}{N-2} +........+ \frac{1}{2} +\frac{1}{1} $

(e) Find that value of $k$ that maximizes the probability of choosing the highest ranking prospect without interviewing all $1000$ candidates. You may need to know that the maximum of the function $x$ ln $\frac{A}{x-1}$ is approximately $\frac{A+1}{e}$, where $A$ is a constant and $e$ is Euler's number, $e = 2.718....$.

Problem 8:
$\triangle ABC$ is inscribed in circle $\omega$ with $AB = 5$, $BC = 7$, $AC = 3$. The bisector of $\angle A$ meets side $BC$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $DE$. Circles $\omega$ and $\gamma$ meet at $E$ and at a second point $F$. Then $AF^{2} = \frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

[Kazi_Zareer]

Problem 1(a):

(a)
Here, $n$, $n+1$, $n+2$ are there consecutive numbers. At least one of it will be even and at least one of it will be divisible by 3. So, in results, $6$ | $n(n+1)(n+2)$.
Easy Solution:
$\binom{n+2}{3}$ = $\frac{n(n+1)(n+2)}{3!}$ = $\frac{n(n+1)(n+2)}{6}.$

Problem 1(b):

(b)

$1^{2015} + 6^{2015} \equiv 0 $ (mod $7$)
$2^{2015} + 5^{2015} \equiv 0$ (mod $7$)
$3^{2015} + 4^{2015} \equiv 0 $(mod $7$)
From this we can write, $1^{2015} + 2^{2015} + 3^{2015} + 4^{2015} + 5^{2015} + 6^{2015 } \equiv 0 $ (mod $7$)

[Kazi_Zareer]

Problem 2 (a):

(a)
Factoring, $600 = 2^4 \times 3 \times 5^3$
$\therefore$ $600$ has $(4+1) (1+1) (3+1) = 40$ positive integers factor.

Problem 2 (b):

(b)
Positive integer factors of $6000$ are perfect squares are $6$ in numbers.
$\therefore$ $600$ has $(40 - 6) = 34$ positive integer factors which are not perfect squares.

[Kazi_Zareer]

Problem 4:
viewtopic.php?f=13&t=3656

[Kazi_Zareer]

Problem 6(a):

(a)
NO, the lines $AC$ and $DI$ doesn't intersect because they are parallel.

Claim 1: $I, O, C$ are collinear.
Proof: $AC = BC$, So, $\angle BAC = \angle ABC$.
O is the circumcenter of $\triangle$, $\angle OAB = \angle OBA$.
So, $\angle OAC = \angle OBC$.
Again, $\triangle OAC $ $\cong $ $\triangle OBC$. [By ASA theorem]
So, $d(O,BC)=d(O,CA)$
We know, $I$ is the incenter of $\triangle ABC$,so the difference between $IC$ & $AC$ and $IC$ and $BC$ are same.

Thus, $O$ lies on the line $IC$.

Now by doing some angle chasing, see that,
$\angle IDB = \angle IOB = \angle OCB + \angle CBO = 2\angle OCB = \angle ACB$
So, this proves that, $AC$ and $DI$ are parallel.

[Kazi_Zareer]

Problem 6(b):

(b)
$DO \cap BI $ $= E$.
In $\triangle EIO,$ $\angle IOE + \angle EOI + \angle EIO = 180^{\circ}$. and also, $\angle EIO = \angle ICB+\angle CBI$
$\angle IEO = 180^{\circ} - ( \angle EOI + \angle EIO)$ $\dots (1)$
$AC = BC$ and $BIOD$ is a cyclic quadrilateral.
$\angle EOI + \angle EIO = \angle DBI + \angle ODB = \angle CBI + (\angle ICB+\angle CBI) = \angle ICB + 2\angle CBI = \angle ICB+\angle ABC = 90^{\circ}$
Thus, $OD$ and $IB$ are perpendicular.

[Kazi_Zareer]

Problem 8:
viewtopic.php?f=13&t=3855

[dshasan]

$\text{PROBLEM 5 (a)}$

Answer : $(n-1)C(m-2)$

[dshasan]

Another solution to Problem 6(a)

It can be easily proved that $C, O, I$ are collinear. Let $\angle DBO = \angle a, \angle IDO = \angle b, \angle IDB = \angle c, \angle BOD = \angle d$. $O$ is the circumcenter of $\bigtriangleup ABC$.

So, $\angle OCB = \angle OBC = \angle CID = \angle a$. So, $\angle a + \angle a = \angle c$.

Now, $\angle a + \angle b + \angle c + \angle d = 180$. And,

$\angle ACD + \angle IDC = 2\angle a + \angle b + \angle a + \angle d = \angle c + \angle b + \angle c + \angle d = 180$.

So , $AC$ and $DI$ are parallel.

[Tasnood]

5(a)

When there are $m$ Martians and $n$ Earthlings. If we consider the Earthlings set, there are $(n+1)$ places for the Martians. So, the place can be chosen in ${n+1 \choose m}$. But the Earthlings can change their place in $n!$ ways where, the Martians can in $m$ ways.
So, total combination =$m!n!{n+1 \choose m}$