# Pakistan IMO First Training Camp Test (Juniors)

Hello fellow Brilliantians!

As most of you know, the IMO is one of the biggest and most famous of all international Maths Olympiads. Following are the problems which appeared in the test at the end of the First Training Camp of Pakistan for the IMO 2017. Enjoy!:

$\textbf{Time: 4 hours}$

$\textbf{Number of Problems: 4}$

$\underline{\textbf{Problem 1}}$

Let $\Delta ABC$ be a triangle. Let $D,E,F$ be the feet of the perpendiculars from $A$ to $BC$,$B$ to $CA$ and $C$ to $AB$ respectively. Let $P,Q,R$ and $S$ be the feet of the perpendiculars from $D$ to $AB,BE,CF$ and $CA$ respectively. Prove that $P,Q,R,S$ are collinear.

$\underline{\textbf{Problem 2}}$

Find the number of ordered pairs of sets $(A,B)$ such that $A\cup B=\{1,2,3,4,5,\cdots,n\}$.Compute the answer for $n=5$.

Note:

$(A,B)$ and $(C,D)$ are the same pair if $A=C$ and $B=D$.

$A$ or $B$ can be an empty set (the pairs in which either one of $A,B$ is an empty set are to be counted as well)

$\underline{\textbf{Problem 3}}$

Let $a,b$ be two positive integers (not necessarily distinct) and let $p$ be a prime such that $lcm(a,a+p)=lcm(b,b+p)$.Prove that $a=b$

Note:

$lcm(x,y)$ denotes the Least Common Multiple of the numbers $x$ and $y$.

$\underline{\textbf{Problem 4}}$

Find all polynomials $P(x)$ with real coefficients such that the polynomial $Q(x)=(x+1)P(x-1)-(x-1)P(x)$ is constant.

Note by Abdur Rehman Zahid
2 years, 11 months ago

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Problem 4

I think my solution is correct. Please check if it is or not.

By observation, if $P(x)=c$, then $Q(x)$ is also a constant function. If $P$ isn't constant, we can do as follows:

We can assume $P$ is monic since if it wasn't, we can divide out by the leading coefficient to still have a constant on the RHS. Let $P(x)=(x-r_1)(x-r_2) \ldots (x-r_n)$. By Vieta's Formula,

\begin{aligned} P(x) &= x^n - (r_1+r_2+\ldots+r_n) x^{n-1} + \ldots\\ \implies (x-1)P(x) &= x^{n+1} - (r_1+r_2+\ldots+r_n+1) x^n+\ldots\\ P(x-1) &= (x-r_1-1)(x-r_2-1)\ldots (x-r_n-1)\\ &=x^n - (r_1+r_2+\ldots + n) + \ldots\\ \implies (x+1) P(x-1) &= x^{n+1} - (r_1 + r_2 + \ldots + r_n + n - 1) x^n + \ldots \end{aligned}

The coefficients for the $x^n$ in the expression for $(x+1)P(x)$ and $(x-1)P(x)$ must cancel each other out, so they must be equal. Thus, $r_1+r_2+\ldots+r_n+1 = r_1 + r_2 + \ldots + r_n + n - 1$, so $n=2$. Thus, $P(x)=x^2 - (r_1 + r_2) x + (r_1 r_2)$. We now have (from above)

\begin{aligned} (x-1) P(x) &= x^3 - (r_1 + r_2 + 1) x^2 + (r_1 + r_2 + r_1 r_2) x - r_1 r_2\\ P(x-1) &= x^2 - (r_1 + r_2 + 2) x + (r_1 r_2 + r_1 + r_2 + 1)\\ \implies (x+1) P(x-1) &= x^3 - (r_1 + r_2 + 1) x^2 + (r_1 r_2 - 1) x + (r_1 r_2 + r_1 + r_2 + 1) \end{aligned}

The coefficient for $x$ in the expression for $(x+1)P(x)$ and $(x-1)P(x)$ must cancel each other out, so they must be equal. Thus, $r_1 + r_2 + r_1 r_2 = r_1 r_2 - 1$, so $r_1 + r_2 = -1$. Substituting back into $P(x)$, we get $P(x)=x^2+x+k$.

Checking, we have $P(x-1)=x^2-x+k$, so $(x+1)P(x-1)=x^3+(k-1)x+k$ whereas $(x-1)P(x)=x^3 - (k-1)x - k$. Subtracting these two, we get $Q(x)=2k$, which is a constant. Therefore, our solution has been verified.

Therefore, all solutions that satisfy are $P(x)=c$ and $P(x)=A(x^2+x+k)$, where $c$, $A$ and $k$ are constants.

- 2 years, 11 months ago

My solution is exactly the same in the sense that it computes the coefficent of $x^n$ in $Q (x)$ but I used the binomial expansion after using a generic polynomial

$P (x)= \sum_{i=0}^n a_i x^i$

Expanding and comparing coefficents kills it

- 2 years, 11 months ago

Problem 3

I found this problem pretty trivial.

First, we will prove the $\text{gcd}(a, a+p)$ is either $1$ or $p$. Let $k$ be the $\text{gcd}$. We have

\begin{aligned} k &\mid a\\ k &\mid a+p\\ \implies k &\mid a+p-a\\ \implies k &\mid p \end{aligned}

Thus, the $\text{gcd}$ is either $1$ or $p$. The same goes for $b$ and $b+p$. We now have three cases:

Case 1: $\text{gcd} (a, a+p) = \text{gcd} (b, b+p) = 1$

Thus, their $\text{LCM}$ for each is equal to their product, so $a(a+p)=b(b+p)$, which simplifies as $a^2 - b^2 = p(b-a)$ so $-(a+b)=p$ or $b-a=0$. The former cannot be true since $a, b, p$ are all positive integers. Therefore, $b-a=0$ so $a=b$.

Case 2: $\text{gcd} (a, a+p) = \text{gcd} (b, b+p) = p$

Thus, their $\text{LCM}$ for each is equal to their product divided by $p$, so $\dfrac {a(a+p)}{p} = \dfrac {b(b+p)}{p}$, which simplifies to the statement in the previous case. Thus, $a=b$.

Case 3: $\text{gcd} (a, a+p) = 1$, $\text{gcd} (b, b+p) = p$

Thus, each of their $\text{lcm}$'s are $a(a+p)$ and $\dfrac {b(b+p)}{p}$. Therefore,

$ap(a+p) = b(b+p)$ However, we know that $p \mid b$ and $p \mid b+p$. Thus, $p^2 \mid b(b+p)$, which means $p^2 \mid ap(a+p)$, so $p \mid a(a+p)$. However, this means either $p \mid a$ or $p \mid a+p$, which is contradictory since either statement results in the other being also true, so their $\text{gcd}=p\neq 1$. Thus, we have a contradiction, so this case cannot occur.

Therefore, over all the cases $a=b$.

- 2 years, 11 months ago

This is pretty trivial...

- 2 years, 11 months ago

Exactly! But I wanted to case bash one of the problems.

- 2 years, 11 months ago

Problem 2

This was a fairly simple problem.

Let $S$ be the set $\{1, 2, \ldots n\}$. Let $A=\{a_1, \ldots, a_k\}$ with $k \leq n$. Then, $B$ must contain at least $\{S/A\}$ as well as any element chosen from $A$. This gives us $2^k$ choices since each element in $A$ may or may not be in $B$. The number of ways to choose $k$ elements in $A$ for this to satisfy is $n \choose k$. Thus, we must sum $n \choose k$ $\times 2^k$ from $k=0$ to $n$.

\begin{aligned} \displaystyle \sum_{k=0}^n \dbinom{n}{k} 2^k &= \displaystyle \sum_{k=0}^n \dbinom{n}{k} 2^k 1^{n-k}\\ &= (2+1)^n\\ &= 3^n \end{aligned}

Thus, the general formula is $3^n$, so if $n=5$, $3^n=243$.

- 2 years, 11 months ago

Here is another solution I found:

Note that we have exactly 3 choices for each element, It goes in A,B or A and B. Since the choices for each element is exclusive we have that the number of ways is exactly $3^n$

- 2 years, 11 months ago

Yep, that works as well, and is fairly quick. I saw my way first but this is an even quicker solution.

- 2 years, 11 months ago

Problem 1 is from India's RMO 2006. Check the solution here

- 2 years, 11 months ago

The question isn't too hard. Angle chasing is all it takes.

- 2 years, 11 months ago

I found at least 2 solutions for this:

A nicer approach, Consider the triangle $BFH$. PQR is its simson line. Similarly QRS is collinear and we are done!

- 2 years, 11 months ago

Yeah, I considered using Simson's line but I saw the angle chasing solution straight off.

- 2 years, 11 months ago

I probably should have seen the Simson's line solution first, but I saw this instead.

- 2 years, 11 months ago

Problem 1

This was fairly trivial, with a good diagram.

Firstly, since $DP$ is perpendicular to $AB$ and $CF$ was perpendicular to $AB$, $DP$ and $CF$ are parallel. Similarly, $DS$ and $BE$ are parallel. Also, since $APDS$ and $BFEC$ are cyclic, we have $\angle APS = \angle ADS = \angle ECB = \angle EFA$, Thus, $PS$ and $EF$ are parallel.

Note that $QPBD$ is cyclic, as well as $EFBC$. Since they both share angle $ABC$, both $P$ and $F$ are on $AB$, and $D$ and $C$ are on $BC$, it follows that $EF$ is parallel to $PQ$. From the previous result, we get that $P,Q,S$ are collinear and parallel to $EF$. Using a similar argument for $R$, we get $P,Q,R,S$ are all collinear parallel to $EF$.

Alternate solution (Kudos to Sualeh Asif)

This is the slick solution.

Since $HFBD$ is cyclic, where $H$ is the orthocentre, and $DP, DQ, DR$ are perpendiculars to $BF$, $BH$ and $HF$ respectively, $PQR$ is the Simson Line of $HBF$ ($D$ is the point on the circumcircle). Thus, $P,Q,R$ are collinear. Similarly, $Q,R,S$ are collinear so we are done.

- 2 years, 11 months ago

sigma upper limit 2001 and lower limit k=1 (x+k)^2=y^2 has no integer solutions prove this was a problem at the imo training camp at Romania Please help me with the solution

- 2 years, 10 months ago

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