RMO 2023 Orange Mock One

22 October 20235 minutesviews

This article contains spoilers for the mock test.

I recommend attempting the problems before proceeding to read the article. You can download the problems and the answer key using the buttons below.

This is the first RMO Mock test from MOMC Season 2. All the credit and authorship of the problems belongs to their respective sources.

I will be discussing the solutions of the problems along with how I motivated them. So if you are planning to attempt the mock, beware of the spoilers ahead.

Problem 1 (Canada 1993) Show that the number xx is rational if and only if three distinct terms that form a geometric progression can be chosen from the sequence

x, x+1, x+2, x+3,.x, ~ x+1, ~ x+2,~ x+3,\ldots .

Solution: One direction of the problem is much easier than the other. If there exist three distinct terms x+a,x+b,x+cx+a,x+b,x+c such that (x+a)(x+c)=(x+b)2(x+a)(x+c) = (x+b)^2 then xx can be shown to be equal to b2aca+c2b\frac{b^2-ac}{a+c-2b}. Therefore xx is rational.

Let x=pqx = \frac{p}{q} where p,qp,q are integers and q0q \neq 0. For the converse, consider the following triple:

(pq,pq+p,pq+p(q+2))\left (\frac{p}{q}, \frac{p}{q}+p, \frac{p}{q} + p(q+2) \right )

It is easy to verify that this triple works. The triple can be found by considering a general triple (x,x+a,x+b)(x,x+a,x+b) and then working out the divisibility constraints.

However, we are not done yet! We need to ensure that each term of the triple is pq\geq \frac{p}{q}. Thus our proof fails if x<0x < 0. In fact, it fails for x=0x = 0 as well, as the terms of the triple won't be distinct. For x=0x = 0 we can just pick (1,2,4)(1,2,4). For negative xx the trick is to see that we can just consider a large positive integer mm such that x+m>0x+m > 0. Then letting x+m=pqx+m = \frac{p}{q} and picking our tuple as before works.

Problem 2 (239 2014 S6) Given positive real numbers a1,a2,,ana_1,a_2,\dots,a_n such that a12+2a23++nann+1<1a_1^2+2a_2^3+\dots+na_n^{n+1} < 1 . Prove that 2a1+3a22++(n+1)ann<32a_1+3a_2^2+\dots+(n+1)a_{n}^n < 3 .

Solution: The difficult observation is that the infinite sum:

(12)2+2(12)3+3(12)4+\left( \frac{1}{2} \right)^2 + 2\left( \frac{1}{2} \right)^3 + 3 \left( \frac{1}{2} \right)^4 + \cdots

is equal to 11. This suggests that all aia_i equal to 12\frac{1}{2} is a pseudo equality case. Therefore, we are motivated to apply AM-GM Inequality to akk+1a_k^{k+1} and 12k+1\frac{1}{2^{k+1}}:

kakk+1+12k+1k+1akk2\frac{ka_k^{k+1} + \frac{1}{2^{k+1}}}{k+1} \geq \frac{a_k^k}{2}

(Had we taken akka_k^k and 12k\frac{1}{2^{k}}, the exponents would not be nice after AM-GM). This implies that (k+1)akk2kakk+1+12k(k+1) a_k^k \le 2ka_k^{k+1} + \frac{1}{2^k}. Summing this inequality over 1kn1 \leq k \leq n gives:

2a1+3a22++(n+1)ann2(a12+2a23++nann+1)+(12+14++12n)<2+1=32a_1+3a_2^2+\dots+(n+1)a_{n}^n \leq 2(a_1^2+2a_2^3+\dots+na_n^{n+1}) + \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) < 2 + 1 =3

Problem 3 (HMMT 2023 Feb Teams) Let ABCDA B C D be a convex quadrilateral such that ABC=BCD=θ\angle A B C=\angle B C D=\theta for some angle θ<90\theta < 90^{\circ}. Point XX lies inside the quadrilateral such that XAD=XDA=90θ\angle X A D=\angle X D A=90^{\circ}-\theta. Prove that BX=XCB X=X C.

Solution: The official source provides 4 amazing solutions to the problem here. My solution was the same as the first one.

Problem 4 (All Russian Olympiad 2000) Let 1<x1<x2,<xn<1-1 < x_1 < x_2 , \cdots < x_n < 1 and x113+x213++xn13=x1+x2++xnx_1^{13} + x_2^{13} + \cdots + x_n^{13} = x_1 + x_2 + \cdots + x_n. Prove that if y1<y2<<yny_1 < y_2 < \cdots < y_n, then

x113y1++xn13yn<x1y1+x2y2++xnyn.x_1^{13}y_1 + \cdots + x_n^{13}y_n < x_1y_1 + x_2y_2 + \cdots + x_ny_n.

Solution: This is a non standard inequality as all regular methods fail. The key observation is that S=i=1nyi(xi13xi)S = \sum_{i =1}^n y_i(x^{13}_i-x_i)is invariant when we shift the {yi}i=1n\{y_i\}_{i=1}^n sequence by any constant. Further observe that x121x^{12} - 1 is always <0 < 0 if 1<x<1-1 < x < 1. It would be easy to show S<0S < 0 if each of the individual summands are negative. For that we need that yixiy_ix_i is positive. But we can ensure that by shifting the sequence yiy_i such that yiy_i and xix_i have the same sign for all 1in1 \leq i \leq n. Thus we are done!

Problem 5 (HMMT 2023 Feb Teams) Let ABCA B C be a triangle. Point DD lies on segment BCB C such that BAD=DAC\angle B A D=\angle D A C. Point XX lies on the opposite side of line BCB C as AA and satisfies XB=XDX B=X D and BXD=ACB\angle B X D=\angle A C B. Analogously, point YY lies on the opposite side of line BCB C as AA and satisfies YC=YDY C=Y D and CYD=ABC\angle C Y D=\angle A B C. Prove that lines XYX Y and ADA D are perpendicular.

Solution: The official source provides 33 beautiful proofs to this superb problem here

Problem 6 (239 2017 J1) Let {a1,a2,,an}\{a_1,a_2, \cdots, a_n\} be a permutation of {1,2,3,n}\{1,2,3 \cdots,n \}. Prove that the sum

1(a1)(a1+a2)(a1+a2+a3)(a1+a2++an)\sum \frac{1}{(a_1)(a_1+a_2)(a_1+a_2+a_3)\dots(a_1+a_2+\dots+a_n)}

taken over all permutations equals 1n!\frac{1}{n!}.

Solution: We show the result by induction on nn. The base case n=1n=1 is clear. We would, in fact, show the general result for a permutation of any set of real numbers (The reason for this would be soon clear). For any 1in+11 \leq i \leq n+1, consider all the permutations in which aia_i appears last. By using the induction hypothesis, the corresponding sum would be equal to 1(a1+a2++an+1)jin+1aj\frac{1}{(a_1 + a_2 + \cdots + a_{n+1})\prod\limits_{j \neq i}^{n+1}a_j}. Note that we used the general form of the result here. Hence the total sum is:

1a1+a2++an+1(i=1n+11jin+1aj)=1a1a2a3an+1\frac{1}{a_1 + a_2 + \cdots + a_{n+1}}\left (\sum_{i=1}^{n+1} \frac{1}{\prod\limits_{j \neq i}^{n+1}a_j} \right) = \frac{1}{a_1a_2a_3\cdots a_{n+1}}

which completes the induction hypothesis.

Side note: I had not anticipated the huge workload that is associated with the college admissions process. So I was not able to put out any mocks in the past two months. Nonetheless, I plan to post 1-2 more mocks in the coming days. Have a look out for those!