Kenneth ChengMathematical modeling

Competition · Competition archive

Macau School Mathematics Competition Archive, 2019–2026

A source-verified archive of 214 Junior and Senior competition solutions, with strategy notes, knowledge maps, diagrams, and three expandable examples.

This teaching archive covers eight competition cycles and both secondary divisions. Use the official DSEDJ paper archive for the authoritative questions, then consult the independent solution companion for strategies and complete arguments.

Cover of the Macau School Mathematics Competition solution companion

2019–2025/2026 · Junior and Senior Divisions

Move from an archived paper to an inspectable solution strategy.

This archive is not a pile of examination PDFs. Every entry is organised as an official-paper locator, strategy, Olympiad knowledge map, complete argument, and final answer. The DSEDJ facsimile remains authoritative.

Download the 248-page companionOpen the official DSEDJ archive
8
competition cycles
16
Junior and Senior papers
214
worked solutions
46
vector diagrams

Coverage matrix

Each row is one competition cycle; every solution is keyed to its official division and question number.

CycleJuniorSeniorTotal
2019131528
2020191635
2021131023
2022161228
2022/2023161228
2023/2024161127
2024/2025161026
2025/202610919
Total11698214

Which mathematical domains appear?

A problem can carry several tags, so domain counts overlap and do not sum to 214.

Algebra and equations103
Number theory64
Geometry and trigonometry48
Inequalities and optimisation44
Combinatorics and games36
Sequences, functions, and analysis26
Mixed or foundational3

Three practical ways to use the archive

  1. 01

    One-problem practice

    Open the official question, spend 10–20 minutes finding a foothold, then compare your route with the strategy and proof.

  2. 02

    Topic training

    Work through algebra, number theory, geometry, combinatorics, or inequalities and record the transformation that actually unlocked each problem.

  3. 03

    Full mock paper

    Use the official time limit for one division, then assess the answer, central idea, completeness, and communication separately.

Expandable examples

Do not stop at the answer: find the structure controlling the problem.

The three examples show estimation, finite-state recurrence, and a complete geometry proof. Every disclosure is keyboard accessible.

2025/2026 · Junior Q2 · Floors and estimation

When does a difference of radicals fall below one?

N=⌊16−6⌋+⌊26−16⌋+⋯+⌊2026−2016⌋N = \left\lfloor\sqrt{16}-\sqrt{6}\right\rfloor + \left\lfloor\sqrt{26}-\sqrt{16}\right\rfloor + \cdots + \left\lfloor\sqrt{2026}-\sqrt{2016}\right\rfloor
Strategy hint

Rationalise term k as 10/(√(10k+6)+√(10k−4)). The denominator increases, so only the first threshold cases need exact checking.

10k+6−10k−4=1010k+6+10k−4\sqrt{10k+6}-\sqrt{10k-4}=\frac{10}{\sqrt{10k+6}+\sqrt{10k-4}}
Open the worked solution

Write the sum for k=1,…,202. For k=1,2, the rationalised value lies strictly in (1,2), so each floor is 1. For k≥3, the denominator is at least √36+√26>10; every remaining term lies in (0,1) and has floor 0. Hence N=1+1=2.

1<1010k+6+10k−4<2 (k=1,2),0<1010k+6+10k−4<1 (k≥3),N=21<\frac{10}{\sqrt{10k+6}+\sqrt{10k-4}}<2\ (k=1,2),\qquad 0<\frac{10}{\sqrt{10k+6}+\sqrt{10k-4}}<1\ (k\geq3),\qquad N=2

2025/2026 · Senior Q3 · Finite-state recurrence

How many length-nine strings avoid AA and BBB?

Classify every valid string by whether it ends in A, one trailing B, or BB.

Strategy hint
un+1=vn+wn,vn+1=un,wn+1=vnu_{n+1}=v_n+w_n,\qquad v_{n+1}=u_n,\qquad w_{n+1}=v_n
Open the worked solution

The initial state is (u₁,v₁,w₁)=(1,1,0). Iterating to n=9 gives (9,7,5), so the count is 9+7+5=21. The official choices do not contain 21; the companion records that mismatch and keeps the facsimile authoritative instead of rewriting the choices.

(u1,v1,w1)=(1,1,0),(u9,v9,w9)=(9,7,5),u9+v9+w9=21(u_1,v_1,w_1)=(1,1,0),\qquad (u_9,v_9,w_9)=(9,7,5),\qquad u_9+v_9+w_9=21

2025/2026 · Senior Q9 · Excircle and collinearity

How does a midpoint condition force three points to be collinear?

Triangle with circumcircle and A-excircleLabels A, B, C, incenter I, circumcenter O, tangency points K, L, M, and midpoint N of KL.ABCKLNMOI
Strategy hint

First use coordinates along the A-angle bisector. Equal tangents give AK=AL=s, and N∈ω yields a side-length identity. Then put BC on the x-axis and evaluate the signed-area determinant of I, M, O.

AK=AL=s,N∈ω,[I,M,O]=0AK=AL=s,\qquad N\in\omega,\qquad [I,M,O]=0
Open the worked solution

Let a=BC, b=CA, c=AB, s=(a+b+c)/2, and θ=A/2. Put A at the origin with the angle bisector as the x-axis. Then K,L=(s cosθ,±s sinθ), so N=(s cosθ,0). Writing the circumcircle as x²+y²−λx−μy=0 and substituting B,C gives λ=(b+c)/(2cosθ). Since N is the other x-axis intersection, s cosθ=λ, hence 2s cos²θ=b+c. The cosine law reduces this to F(a,b,c)=0.

s=a+b+c2,θ=A2,K,L=(scos⁡θ,±ssin⁡θ),N=(scos⁡θ,0),2scos⁡2θ=b+cs=\frac{a+b+c}{2},\quad \theta=\frac{A}{2},\quad K,L=(s\cos\theta,\pm s\sin\theta),\quad N=(s\cos\theta,0),\quad 2s\cos^2\theta=b+c
F=a3+a2b+a2c−ab2−2abc−ac2−b3+b2c+bc2−c3F=a^3+a^2b+a^2c-ab^2-2abc-ac^2-b^3+b^2c+bc^2-c^3

Now put B=(0,0), C=(a,0), A=(p,q), where p=(a²+c²−b²)/(2a). The excircle contact gives M=(s−c,0); the incenter and circumcenter are I=(a(p+c)/(2s), aq/(2s)) and O=(a/2,(c²−ap)/(2q)). Their signed-area determinant simplifies to Δ=(b−c)F/(8aq)=0. Therefore I, M, O are collinear.

p=a2+c2−b22a,M=(s−c,0),I=(a(p+c)2s,aq2s),O=(a2,c2−ap2q),Δ=(b−c)F8aq=0p=\frac{a^2+c^2-b^2}{2a},\quad M=(s-c,0),\quad I=\left(\frac{a(p+c)}{2s},\frac{aq}{2s}\right),\quad O=\left(\frac{a}{2},\frac{c^2-ap}{2q}\right),\quad \Delta=\frac{(b-c)F}{8aq}=0

Source: 2025/2026 School Mathematics Competition. The DSEDJ facsimile is authoritative.

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