RMO 2023 Orange Mock Three

26 October 20234 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 third and final RMO 2023 Mock test from MOMC Season 2. All the credit and authorship of the problems belongs to their respective sources.

As early-action college deadlines are nearing, I won't be able to get a proper look at the RMO 2023 problems. But I wish you all the best of luck!

Following my typical approach, 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 (Dutch BxMO TST 2018 P2) Let ABC\triangle ABC be a triangle of which the side lengths are positive integers which are pairwise coprime. The tangent in AA to the circumcircle intersects line BCBC in DD. Prove that BDBD is not an integer.

Solution: This problem entails some trigonometric calculation. The key is to have faith that spamming sine rule and cosine rule will result in something manageable at the end. Without loss of generality let AB<ACAB < AC. By using the tangency, see that DAB=ACB\angle DAB = \angle ACB. Now Sine rule in ABD\triangle ABD and ABC\triangle ABC together implies

BDsinC=ABsin(BC)    BD=ABsinCsin(BC)=ABsinCsinBcosCsinCcosB=ABsinBsinCcosCcosB\frac{BD}{\sin C} = \frac{AB}{\sin (B-C)} \implies BD = \frac{AB \sin C}{\sin (B-C)} = \frac{AB \sin C}{\sin B \cos C - \sin C \cos B} = \frac{AB}{\frac{ \sin B }{\sin C} \cos C - \cos B}

Now writing sinBsinC=ACAB\frac{\sin B}{\sin C} = \frac{AC}{AB} (by Sine rule), applying cosine rule to evaluate cosC\cos C and cosB\cos B and simplifying the expression yields that BD=BCAB2AC2AB2=ac2b2c2BD = \frac{BC \cdot AB^2}{AC^2-AB^2} = \frac{a c^2}{b^2-c^2}. Suppose for the sake of contradiction that BDBD is an integer. Then b2c2ac2b^2-c^2 \mid ac^2. If pp is a prime such that pp divides b2c2b^2-c^2 then if pp divides c2c^2 then pp divides b2b^2 which is a contradiction to the fact b,cb,c are coprime. Hence b2c2ab^2-c^2 \mid a, so b2c2a<b+c    bc<1b^2-c^2 \leq a < b +c \implies b -c < 1, impossible. Here we have used the triangle inequality.

Problem 2 (Japan) Prove that there doesn't exist any positive integer nn such that 2n2+1,3n2+12n^2+1,3n^2+1 and 6n2+16n^2+1 are perfect squares.

Solution: The key idea is to multiply different perfect squares so that their resulting difference is 11:

(3n2+1)(6n2+1)(3n)2(2n2+1)=1(3n^2+1)(6n^2+1) - (3n)^2(2n^2+1) = 1

They only two consecutive perfect squares are 00 and 11, which would imply n=0n=0, a contradiction!

Problem 3 (Swiss TST 2023 P7) Find all monic polynomials P(x)=x2023+a2022x2022++a1x+a0P(x)=x^{2023}+a_{2022}x^{2022}+\ldots+a_1x+a_0 with real coefficients such that a2022=0a_{2022}=0, P(1)=1P(1)=1 and all roots of PP are real and less than 11.

Solution: Upon inspection, x2023x^{2023} seems to be the only trivial solution. Let the roots be y1,y2,y2023y_1,y_2, \cdots y_{2023}. We are given that yi=0\sum y_i = 0, (1yi)=1\prod (1-y_i) = 1 and yi<1y_i < 1. Therefore 1yi>01-y_i > 0. Applying AM-GM over the product gives:

2023yi2023((1yi))12023=1\frac{2023-\sum y_i}{2023} \geq \left ( \prod (1-y_i) \right)^{\frac{1}{2023}} = 1

As yi=0\sum y_i = 0, equality holds and so all the roots are equal. Therefore (1y1)2023=1    y1=0    P(x)=x2023(1-y_1)^{2023} = 1 \implies y_1 = 0 \implies P(x) = x^{2023}. We are done!

Problem 4 (Singapore TST 2004) Let 0<a,b,c<10 < a, b, c < 1 with ab+bc+ca=1ab + bc + ca = 1. Prove that

a1a2+b1b2+c1c2332.\frac{a}{1-a^2} + \frac{b}{1-b^2} + \frac{c}{1-c^2} \geq \frac {3 \sqrt{3}}{2}.

Determine when equality holds.

Solution: For this problem, there exist a lot of solutions which use Jensen's Inequality. I would advise against using it unless you can write the proof of Jensen during the exam. Here, we provide a different solution. As a2+b2+c2ab+bc+ca=1a^2 + b^2 + c^2 \geq ab + bc + ca = 1, we can wish that maybe a1a233a22\frac{a}{1-a^2} \geq \frac{3\sqrt{3}a^2}{2} will hold. This is fortunately the case as:

a1a233a22    a(1a2)233\frac{a}{1-a^2} \geq \frac{3\sqrt{3}a^2}{2} \iff a(1-a^2) \leq \frac{2}{3\sqrt{3}}

We can differentiate to find that this is true and equality holds at a=b=c=13a=b=c = \frac{1}{\sqrt{3}}.

Problem 5 (Benelux 2019 P3) Two circles Γ1\Gamma_1 and Γ2\Gamma_2 intersect at points AA and ZZ (with AZA\neq Z). Let BB be the centre of Γ1\Gamma_1 and let CC be the centre of Γ2\Gamma_2. The exterior angle bisector of BAC\angle{BAC} intersects Γ1\Gamma_1 again at XX and Γ2\Gamma_2 again at YY. Prove that the interior angle bisector of BZC\angle{BZC} passes through the circumcenter of XYZ\triangle{XYZ}.

Solution: Check out the solution of this problem here.

Problem 6 (2007 Peru Iberoamerican TST P2) Find all positive integers mm and nn that satisfy the equality:


Solution: The hard step is to factorise n5+n4+1n^5+n^4+1 into (n3n+1)(n2+n+1)(n^3-n+1)(n^2+n+1). By using the euclidean algorithm we can deduce that gcd(n3n+1,n2+n+1)=gcd(n2,7){1,7}\gcd(n^3-n+1, n^2+n+1) = \gcd(n-2,7) \in \{1,7\}. We conclude that n=2n=2 and m=2m=2 is the only solution.