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!:


Time: 4 hours\textbf{Time: 4 hours}

Number of Problems: 4\textbf{Number of Problems: 4}


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

Let ΔABC\Delta ABC be a triangle. Let D,E,FD,E,F be the feet of the perpendiculars from AA to BCBC,BB to CACA and CC to ABAB respectively. Let P,Q,RP,Q,R and SS be the feet of the perpendiculars from DD to AB,BE,CFAB,BE,CF and CACA respectively. Prove that P,Q,R,SP,Q,R,S are collinear.


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

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

Note:

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

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


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

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

Note:

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


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

Find all polynomials P(x)P(x) with real coefficients such that the polynomial Q(x)=(x+1)P(x1)(x1)P(x)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)=cP(x)=c, then Q(x)Q(x) is also a constant function. If PP isn't constant, we can do as follows:

We can assume PP 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)=(xr1)(xr2)(xrn)P(x)=(x-r_1)(x-r_2) \ldots (x-r_n). By Vieta's Formula,

P(x)=xn(r1+r2++rn)xn1+    (x1)P(x)=xn+1(r1+r2++rn+1)xn+P(x1)=(xr11)(xr21)(xrn1)=xn(r1+r2++n)+    (x+1)P(x1)=xn+1(r1+r2++rn+n1)xn+\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 xnx^n in the expression for (x+1)P(x)(x+1)P(x) and (x1)P(x)(x-1)P(x) must cancel each other out, so they must be equal. Thus, r1+r2++rn+1=r1+r2++rn+n1r_1+r_2+\ldots+r_n+1 = r_1 + r_2 + \ldots + r_n + n - 1, so n=2n=2. Thus, P(x)=x2(r1+r2)x+(r1r2)P(x)=x^2 - (r_1 + r_2) x + (r_1 r_2). We now have (from above)

(x1)P(x)=x3(r1+r2+1)x2+(r1+r2+r1r2)xr1r2P(x1)=x2(r1+r2+2)x+(r1r2+r1+r2+1)    (x+1)P(x1)=x3(r1+r2+1)x2+(r1r21)x+(r1r2+r1+r2+1)\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 xx in the expression for (x+1)P(x)(x+1)P(x) and (x1)P(x)(x-1)P(x) must cancel each other out, so they must be equal. Thus, r1+r2+r1r2=r1r21r_1 + r_2 + r_1 r_2 = r_1 r_2 - 1, so r1+r2=1r_1 + r_2 = -1. Substituting back into P(x)P(x), we get P(x)=x2+x+kP(x)=x^2+x+k.

Checking, we have P(x1)=x2x+kP(x-1)=x^2-x+k, so (x+1)P(x1)=x3+(k1)x+k(x+1)P(x-1)=x^3+(k-1)x+k whereas (x1)P(x)=x3(k1)xk(x-1)P(x)=x^3 - (k-1)x - k. Subtracting these two, we get Q(x)=2kQ(x)=2k, which is a constant. Therefore, our solution has been verified.

Therefore, all solutions that satisfy are P(x)=cP(x)=c and P(x)=A(x2+x+k)P(x)=A(x^2+x+k), where cc, AA and kk are constants.

Sharky Kesa - 2 years, 11 months ago

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

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

Expanding and comparing coefficents kills it

Sualeh Asif - 2 years, 11 months ago

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Problem 3

I found this problem pretty trivial.

First, we will prove the gcd(a,a+p)\text{gcd}(a, a+p) is either 11 or pp. Let kk be the gcd\text{gcd}. We have

kaka+p    ka+pa    kp\begin{aligned} k &\mid a\\ k &\mid a+p\\ \implies k &\mid a+p-a\\ \implies k &\mid p \end{aligned}

Thus, the gcd\text{gcd} is either 11 or pp. The same goes for bb and b+pb+p. We now have three cases:

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

Thus, their LCM\text{LCM} for each is equal to their product, so a(a+p)=b(b+p)a(a+p)=b(b+p), which simplifies as a2b2=p(ba)a^2 - b^2 = p(b-a) so (a+b)=p-(a+b)=p or ba=0b-a=0. The former cannot be true since a,b,pa, b, p are all positive integers. Therefore, ba=0b-a=0 so a=ba=b.

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

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

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

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

ap(a+p)=b(b+p)ap(a+p) = b(b+p) However, we know that pbp \mid b and pb+pp \mid b+p. Thus, p2b(b+p)p^2 \mid b(b+p), which means p2ap(a+p)p^2 \mid ap(a+p), so pa(a+p)p \mid a(a+p). However, this means either pap \mid a or pa+pp \mid a+p, which is contradictory since either statement results in the other being also true, so their gcd=p1\text{gcd}=p\neq 1. Thus, we have a contradiction, so this case cannot occur.

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

Sharky Kesa - 2 years, 11 months ago

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This is pretty trivial...

Sualeh Asif - 2 years, 11 months ago

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

Sharky Kesa - 2 years, 11 months ago

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Problem 2

This was a fairly simple problem.

Let SS be the set {1,2,n}\{1, 2, \ldots n\}. Let A={a1,,ak}A=\{a_1, \ldots, a_k\} with knk \leq n. Then, BB must contain at least {S/A}\{S/A\} as well as any element chosen from AA. This gives us 2k2^k choices since each element in AA may or may not be in BB. The number of ways to choose kk elements in AA for this to satisfy is (nk)n \choose k. Thus, we must sum (nk)n \choose k ×2k\times 2^k from k=0k=0 to nn.

k=0n(nk)2k=k=0n(nk)2k1nk=(2+1)n=3n\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 3n3^n, so if n=5n=5, 3n=2433^n=243.

Sharky Kesa - 2 years, 11 months ago

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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 3n3^n

Sualeh Asif - 2 years, 11 months ago

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

Sharky Kesa - 2 years, 11 months ago

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

Brilliant Member - 2 years, 11 months ago

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

Sharky Kesa - 2 years, 11 months ago

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I found at least 2 solutions for this:

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

Sualeh Asif - 2 years, 11 months ago

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

Sharky Kesa - 2 years, 11 months ago

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

Sharky Kesa - 2 years, 11 months ago

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Problem 1

This was fairly trivial, with a good diagram.

Firstly, since DPDP is perpendicular to ABAB and CFCF was perpendicular to ABAB, DPDP and CFCF are parallel. Similarly, DSDS and BEBE are parallel. Also, since APDSAPDS and BFECBFEC are cyclic, we have APS=ADS=ECB=EFA\angle APS = \angle ADS = \angle ECB = \angle EFA, Thus, PSPS and EFEF are parallel.

Note that QPBDQPBD is cyclic, as well as EFBCEFBC. Since they both share angle ABCABC, both PP and FF are on ABAB, and DD and CC are on BCBC, it follows that EFEF is parallel to PQPQ. From the previous result, we get that P,Q,SP,Q,S are collinear and parallel to EFEF. Using a similar argument for RR, we get P,Q,R,SP,Q,R,S are all collinear parallel to EFEF.


Alternate solution (Kudos to Sualeh Asif)

This is the slick solution.

Since HFBDHFBD is cyclic, where HH is the orthocentre, and DP,DQ,DRDP, DQ, DR are perpendiculars to BFBF, BHBH and HFHF respectively, PQRPQR is the Simson Line of HBFHBF (DD is the point on the circumcircle). Thus, P,Q,RP,Q,R are collinear. Similarly, Q,R,SQ,R,S are collinear so we are done.

Sharky Kesa - 2 years, 11 months ago

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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

RK Rohit - 2 years, 10 months ago

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