NYC RMO STUDY GUIDE EXAM REVIEW
QUESTIONSWITH CORRECT SOLUTIONS
UPDATED 2025/2026 SYLLABUS
ALREADY GRADED!!100%
GUARANTEED PASS!!!<<BRAND NEW
VERSION>>
The NYC RMO is a highly competitive exam that serves as the first major stepping stone
towards the USAMO (United States of America Mathematical Olympiad) and the IMO
(International Mathematical Olympiad). The problems require deep conceptual
understanding, creativity, and flawless proof-writing skills, not just computational speed.
Key Characteristics of the Exam:
Proof-Based: All answers must be rigorous mathematical proofs.
Thematic Focus: Problems primarily come from Algebra, Geometry, Number Theory,
and Combinatorics.
Depth over Breadth: A few, very challenging problems (typically 4-6) must be solved
within a 3-hour window.
How to Use This Guide: Work through each topic systematically. Understand the
underlying principles in the solutions, not just the final answer. The goal is to internalize
the problem-solving strategies.
1. Algebra
Algebra problems test skills in manipulating equations, inequalities, sequences, series,
and functional equations.
, Core Concepts: Polynomials, Inequalities (AM-GM, Cauchy-Schwarz, Jensen's),
Sequences & Series, Functional Equations.
Practice Questions:
1. Find all real numbers xx such that x+3−4x−1+x+8−6x−1=1x+3−4x−1+x+8−6x−1
=1.
o ANSWER ✓ We observe that both expressions under the square roots are perfect
squares.
Let t=x−1t=x−1, so x−1=t2x−1=t2 and x=t2+1x=t2+1, with t≥0t≥0.
Substitute into the first radical:
x+3−4x−1=(t2+1)+3−4t=t2−4t+4=(t−2)2x+3−4x−1
=(t2+1)+3−4t=t2−4t+4=(t−2)2.
Similarly, the second radical:
x+8−6x−1=(t2+1)+8−6t=t2−6t+9=(t−3)2x+8−6x−1
=(t2+1)+8−6t=t2−6t+9=(t−3)2.
The equation becomes ∣t−2∣+∣t−3∣=1∣t−2∣+∣t−3∣=1.
This is an absolute value equation. We consider cases based on the value of tt:
o Case 1: t<2t<2. Then ∣t−2∣=2−t∣t−2∣=2−t, ∣t−3∣=3−t∣t−3∣=3−t. The equation
is (2−t)+(3−t)=5−2t=1(2−t)+(3−t)=5−2t=1 ⇒ t=2t=2. But t=2t=2 is not in the
interval t<2t<2. No solution.
o Case 2: 2≤t≤32≤t≤3. Then ∣t−2∣=t−2∣t−2∣=t−2, ∣t−3∣=3−t∣t−3∣=3−t. The equation
is (t−2)+(3−t)=1(t−2)+(3−t)=1 ⇒ 1=11=1. This is true for all tt in this interval.
o Case 3: t>3t>3. Then ∣t−2∣=t−2∣t−2∣=t−2, ∣t−3∣=t−3∣t−3∣=t−3. The equation
is (t−2)+(t−3)=2t−5=1(t−2)+(t−3)=2t−5=1 ⇒ t=3t=3. But t=3t=3 is not in the
interval t>3t>3. No solution.
Therefore, the equation holds for 2≤t≤32≤t≤3.
Recalling t=x−1t=x−1, we have 2≤x−1≤32≤x−1
≤3 ⇒ 4≤x−1≤94≤x−1≤9 ⇒ 5≤x≤105≤x≤10.
The solution set is [5,10][5,10].
2. Let a,b,ca,b,c be positive real numbers such that abc=1abc=1. Prove
that ab+bc+ca≥a+b+cba+cb+ac≥a+b+c.
o ANSWER ✓ By the AM-GM inequality, we know that for positive reals, the arithmetic
mean is at least the geometric mean.
Consider the three terms ab,ab,bcba,ba,cb. Their sum is 2ab+bc2ba+cb.
By AM-GM: 2ab+bc≥3ab⋅ab⋅bc3=3a2bc32ba+cb≥33ba⋅ba⋅cb=33bca2.
Since abc=1abc=1, we have bc=1/abc=1/a, so a2bc=a3bca2=a3. Thus, 2ab+bc≥3a2ba
+cb≥3a.
QUESTIONSWITH CORRECT SOLUTIONS
UPDATED 2025/2026 SYLLABUS
ALREADY GRADED!!100%
GUARANTEED PASS!!!<<BRAND NEW
VERSION>>
The NYC RMO is a highly competitive exam that serves as the first major stepping stone
towards the USAMO (United States of America Mathematical Olympiad) and the IMO
(International Mathematical Olympiad). The problems require deep conceptual
understanding, creativity, and flawless proof-writing skills, not just computational speed.
Key Characteristics of the Exam:
Proof-Based: All answers must be rigorous mathematical proofs.
Thematic Focus: Problems primarily come from Algebra, Geometry, Number Theory,
and Combinatorics.
Depth over Breadth: A few, very challenging problems (typically 4-6) must be solved
within a 3-hour window.
How to Use This Guide: Work through each topic systematically. Understand the
underlying principles in the solutions, not just the final answer. The goal is to internalize
the problem-solving strategies.
1. Algebra
Algebra problems test skills in manipulating equations, inequalities, sequences, series,
and functional equations.
, Core Concepts: Polynomials, Inequalities (AM-GM, Cauchy-Schwarz, Jensen's),
Sequences & Series, Functional Equations.
Practice Questions:
1. Find all real numbers xx such that x+3−4x−1+x+8−6x−1=1x+3−4x−1+x+8−6x−1
=1.
o ANSWER ✓ We observe that both expressions under the square roots are perfect
squares.
Let t=x−1t=x−1, so x−1=t2x−1=t2 and x=t2+1x=t2+1, with t≥0t≥0.
Substitute into the first radical:
x+3−4x−1=(t2+1)+3−4t=t2−4t+4=(t−2)2x+3−4x−1
=(t2+1)+3−4t=t2−4t+4=(t−2)2.
Similarly, the second radical:
x+8−6x−1=(t2+1)+8−6t=t2−6t+9=(t−3)2x+8−6x−1
=(t2+1)+8−6t=t2−6t+9=(t−3)2.
The equation becomes ∣t−2∣+∣t−3∣=1∣t−2∣+∣t−3∣=1.
This is an absolute value equation. We consider cases based on the value of tt:
o Case 1: t<2t<2. Then ∣t−2∣=2−t∣t−2∣=2−t, ∣t−3∣=3−t∣t−3∣=3−t. The equation
is (2−t)+(3−t)=5−2t=1(2−t)+(3−t)=5−2t=1 ⇒ t=2t=2. But t=2t=2 is not in the
interval t<2t<2. No solution.
o Case 2: 2≤t≤32≤t≤3. Then ∣t−2∣=t−2∣t−2∣=t−2, ∣t−3∣=3−t∣t−3∣=3−t. The equation
is (t−2)+(3−t)=1(t−2)+(3−t)=1 ⇒ 1=11=1. This is true for all tt in this interval.
o Case 3: t>3t>3. Then ∣t−2∣=t−2∣t−2∣=t−2, ∣t−3∣=t−3∣t−3∣=t−3. The equation
is (t−2)+(t−3)=2t−5=1(t−2)+(t−3)=2t−5=1 ⇒ t=3t=3. But t=3t=3 is not in the
interval t>3t>3. No solution.
Therefore, the equation holds for 2≤t≤32≤t≤3.
Recalling t=x−1t=x−1, we have 2≤x−1≤32≤x−1
≤3 ⇒ 4≤x−1≤94≤x−1≤9 ⇒ 5≤x≤105≤x≤10.
The solution set is [5,10][5,10].
2. Let a,b,ca,b,c be positive real numbers such that abc=1abc=1. Prove
that ab+bc+ca≥a+b+cba+cb+ac≥a+b+c.
o ANSWER ✓ By the AM-GM inequality, we know that for positive reals, the arithmetic
mean is at least the geometric mean.
Consider the three terms ab,ab,bcba,ba,cb. Their sum is 2ab+bc2ba+cb.
By AM-GM: 2ab+bc≥3ab⋅ab⋅bc3=3a2bc32ba+cb≥33ba⋅ba⋅cb=33bca2.
Since abc=1abc=1, we have bc=1/abc=1/a, so a2bc=a3bca2=a3. Thus, 2ab+bc≥3a2ba
+cb≥3a.
