RMO 2023 Orange Mock Two

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

Sticking to my usual, 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 (Simon Marais 2017 A2) Let a1,a2,a3,a_1,a_2,a_3,\ldots be the sequence of real numbers defined by a1=1a_1=1 and

am=1a12+a22++am12for m2.a_m=\frac1{a_1^2+a_2^2+\ldots+a_{m-1}^2}\qquad\text{for }m\ge2.

Determine whether there exists a positive integer NN such that


Solution: Suppose for the sake of contradiction that the sum of the sequence was bounded. Thus there doesn't exist a constant cc such that aica_i \geq c for all ii. Hence ai0a_i \rightarrow 0 as ii \rightarrow \infty. Now we could relate ai\sum a_i with ai2\sum a_i^2 if 0<am10 < a_m \leq 1 or am1a_m \geq 1 for all mm. The former luckily holds as a12+a22++am12a12=1    am1a_1^2 + a_2^2 + \dots + a_{m-1}^2 \geq a_1^2 = 1 \implies a_m \leq 1. Hence:

a1+a2++aNa12+a22++aN2=1aN+1a_1 + a_2 + \dots + a_N \geq a_1^2 + a_2^2 + \dots + a_N^2 = \frac{1}{a_{N+1}} \rightarrow \infty

A contradiction!

Problem 2 (China Northern Mathematical Olympiad 2020 P4) In ABC\triangle ABC, BAC=60\angle BAC = 60^{\circ}, point DD lies on side BCBC, O1O_1 and O2O_2 are the centers of the circumcircles of ABD\triangle ABD and ACD\triangle ACD, respectively. Lines BO1BO_1 and CO2CO_2 intersect at point PP. If II is the incenter of ABC\triangle ABC and HH is the orthocenter of PBC\triangle PBC, then prove that the four points B,C,I,HB,C,I,H are on the same circle.

Solution: This is a straightforward angle chasing problem. It suffices to show that BHC=BIC\angle BHC = \angle BIC. It is well-known that BIC=90+12BAC\angle BIC = 90 + \frac{1}{2} \angle BAC which is 120120 in our case. It is also well-known that BHC=180BPC\angle BHC = 180 - \angle BPC. So BIC=120    BPC=60    O1BC+O2CB=120\angle BIC = 120 \iff \angle BPC = 60 \iff \angle O_1BC + \angle O_2CB = 120. Now O1BD=180BO1D2=1802BAD2=90BAD\angle O_1BD = \frac{180-\angle BO_1D}{2} = \frac{180 - 2\angle BAD}{2} = 90 - \angle BAD. Likewise O2CB=90CAD\angle O_2CB = 90 - \angle CAD. Adding we get the desired result. For more insightful solutions you can check out the AoPS thread.

Problem 3: (China Northern Math Olympiad 2017 P4) Let QQ be a set of permutations of 1,2,...,1001,2,...,100 such that for all 1a,b1001\leq a,b \leq 100, aa can be found to the left of bb and adjacent to bb in at most one permutation in QQ. Find the largest possible number of elements in QQ.

Solution: Let MM be the maximum number of elements in QQ. For each permutation, there exist exactly 9999 consecutive pairs of numbers. So, there are 99M99M total pairs. Each consecutive pair of numbers appears exactly once. Hence 99M9910099M \leq 99\cdot 100 as there are 9910099 \cdot 100 total possible pairs of consecutive integers in a permutation. Therefore M100M \leq 100. It remains to show that a construction exists.

Consider a 100×100100 \times 100 matrix where each row represents a permutation of 1,2,1001,2, \dots 100. Initially, all elements of the first row are 11, all elements of the second row are 22 and so on. We cycle upwards, the kthk^{\text{th}} column other than the first one, by tkt_k units relative to the k1thk-1^{\text{th}} column. It suffices to find distinct positive integers tkt_k for 2k1002 \leq k \leq 100 such that the sum of any contiguous sum of tkt_k isn't divisible by 100100. The following sequence works:

1,98,3,96,5,94,,97,2,991,98,3,96,5,94, \cdots , 97,2,99

Problem 4 (Austrian Math Olympiad 2006 P4) The function f:RRf: \mathbb{R} \rightarrow \mathbb{R} is defined as f(x)=x2+{x}f(x) = \lfloor x^2 \rfloor + \{ x \}. Call a real number regional if it doesn't lie in the range of ff. Show that there exists an infinite arithmetic progression of distinct regional positive rational numbers, which all have denominator 33 when written in lowest form.

Solution: If 33 is the denominator of a rational number, then it can be written as m+13m+\frac{1}{3} or m+23m + \frac{2}{3}. For simplicity, consider the former one. Then f(x)=m+13    {x}=13f(x) = m + \frac{1}{3} \implies \{x\} = \frac{1}{3}. Hence x=+13x = \ell + \frac{1}{3} for some integer \ell. By using the definition of f(x)f(x) and taking cases on \ell modulo 33, we deduce that 9m+1,9m+49m+1, 9m+4 and 9m+79m+7 are perfect squares. Picking m6(mod8)    9m+17,9m+42,9m+75(mod8)m \equiv 6 \pmod 8 \implies 9m+1 \equiv 7, 9m+4 \equiv 2, 9m+7 \equiv 5 \pmod 8. None of 2,5,72,5,7 are quadratic residues modulo 88, as required. Hence the arithmetic sequence 6+13,14+13,22+13,6 + \frac{1}{3}, 14 + \frac{1}{3}, 22 + \frac{1}{3}, \dots works.

Problem 5 (China Northern Math Olympiad) Find all integers nn such that there exists a concave pentagon which can be dissected into nn congruent triangles.

Solution: The solution to this difficult problem can be found here.

Problem 6 (Hong Kong National Olympiad 2013 P1) Let a,b,ca,b,c be positive real numbers such that ab+bc+ca=1ab+bc+ca=1. Prove that


Solution: Check out the solution of this inequality here.

The final mock would be posted on 26th October. Stay in the loop for that! :)