§ 6.1 · The Cartesian Plane
Coordinates
Core Concept: Four quadrants, two axes, one point at a time
The two perpendicular number lines — the x-axis (horizontal) and y-axis (vertical) — split the plane into four quadrants. Every point is written as (x, y): x first (horizontal), then y (vertical).
-
Quadrant I (top right): x > 0, y > 0 →
(+, +)
-
Quadrant II (top left): x < 0, y > 0 →
(−, +)
-
Quadrant III (bottom left): x < 0, y < 0 →
(−, −)
-
Quadrant IV (bottom right): x > 0, y < 0 →
(+, −)
The origin is (0, 0). A point on the x-axis has y = 0; a point on the y-axis has x = 0.
In the graph shown, which of the following statements is true about the coordinates of the point P(x, y)? (P is to the right of the y-axis and below the x-axis.)
- A. Both x and y are positive
- B. x positive, y negative
- C. x negative, y positive
- D. Both negative
- E. x = 0, y negative
Concept
Sign of the coordinates is determined by the quadrant alone. Right of y-axis ⇒ x > 0. Below x-axis ⇒ y < 0.
Steps
- The diagram places P in the fourth quadrant (bottom-right region).
- Quadrant IV has the sign pattern
(+, −).
- So x is positive and y is negative.
Pitfall
Don't try to read off exact numerical values — the problem only asks about signs. The diagram is schematic, not to scale.
Answer
B — x positive, y negative
In the diagram shown, what are the coordinates of point P? (P is plotted at 2 units right and 4 units up.)
- A. (4, 0)
- B. (2, 2)
- C. (2, 0)
- D. (4, 4)
- E. (2, 4)
Steps
- Count rightward from the origin: x = 2.
- Count upward: y = 4.
- Coordinates:
(2, 4).
Pitfall
Always write x before y. (2, 4) ≠ (4, 2).
A dot starts at (20, 19). The first move is 1 unit horizontally or vertically; each subsequent move is one unit longer than the previous one and perpendicular to it. After 10 moves the dot stops. Which final location is not possible?
- A. (27, 33)
- B. (30, 40)
- C. (21, 21)
- D. (42, 44)
- E. (37, 37)
Strategy
Moves alternate between horizontal and vertical. Lengths are 1, 2, 3, …, 10. Split them by parity of the move number: odd-indexed moves (lengths 1, 3, 5, 7, 9) go in one direction; even-indexed moves (lengths 2, 4, 6, 8, 10) go in the other. Each step gets an independent sign (+ or −).
Steps
- Let
a = x − 20 and b = y − 19 be the total horizontal and vertical displacements.
-
Case: first move is horizontal. Then
a = ±1 ± 3 ± 5 ± 7 ± 9 and b = ±2 ± 4 ± 6 ± 8 ± 10.
-
Case: first move is vertical. Then
a = ±2 ± 4 ± 6 ± 8 ± 10 and b = ±1 ± 3 ± 5 ± 7 ± 9.
- Test each option by computing (a, b) and checking whether the needed signed sums are possible:
- A. (27, 33) → a = 7 (odd), b = 14 (even) → first move horizontal. Achievable: a = 1+3+5+7−9 = 7 ✓, b = 2+4+6−8+10 = 14 ✓.
-
B. (30, 40) → a = 10 (even), b = 21 (odd) → first move vertical. Need a from {±2,±4,±6,±8,±10} summing to 10, and b from {±1,±3,±5,±7,±9} summing to 21. For b = 21: total of odds is 25, so flip a subset summing to (25−21)/2 = 2. But no subset of {1,3,5,7,9} sums to 2. Impossible.
- C. (21, 21) → a = 1, b = 2. Both achievable.
- D. (42, 44) → a = 22, b = 25. Both achievable.
- E. (37, 37) → a = 17, b = 18. Both achievable.
Insight
Parity of the displacement tells you which direction the first move went. Then check whether each target sum is reachable using signed subsets.
The points (2, 1), (4, 1), and (2, 5) are three vertices of a rectangle. What are the coordinates of the fourth vertex?
- A. (5, 2)
- B. (4, 4)
- C. (1, 5)
- D. (4, 5)
- E. (2, 4)
Strategy
A rectangle has opposite sides parallel to the axes (here). The fourth vertex shares its x-coordinate with one neighbour and its y-coordinate with the other.
Steps
- (2, 1) and (4, 1) share y = 1: one horizontal side.
- (2, 1) and (2, 5) share x = 2: one vertical side.
- The fourth vertex sits opposite to (2, 1). Its x-coord matches (4, 1), so x = 4. Its y-coord matches (2, 5), so y = 5.
- Fourth vertex:
(4, 5).
§ 6.2 · Cuboids & Cubes
3D Figures: Surface Area & Volume
Core Concept: Two key formulas for boxes
A cuboid (rectangular prism) has 6 rectangular faces, 12 edges, and 8 vertices. Opposite faces are identical. A cube is a special cuboid where all edges are equal.
Cuboid (length l, width w, height h):
Surface Area: SA = 2(lw + lh + wh)
Volume: V = l × w × h
Cube (edge a):
Surface Area: SA = 6a²
Volume: V = a³
Two-step habit: always identify the three pair-of-faces first (top-bottom, front-back, left-right), compute one of each, then double the sum.
The length of each edge of a cube is 1 cm. The surface area of the cube, in cm², is:
Steps
Apply SA = 6a²: SA = 6 × 1² = 6 cm².
Variation
Edge 2 cm → SA = 6 × 4 = 24 cm². Edge 3 cm → SA = 54 cm². Surface area scales with the square of the edge length.
What is the surface area of a 1 cm by 2 cm by 2 cm rectangular prism?
- A. 10 cm²
- B. 20 cm²
- C. 12 cm²
- D. 24 cm²
- E. 16 cm²
Steps
- Pair 1 (top & bottom): each is 2 × 2 = 4 cm². Together: 8 cm².
- Pair 2 (front & back): each is 1 × 2 = 2 cm². Together: 4 cm².
- Pair 3 (left & right): each is 1 × 2 = 2 cm². Together: 4 cm².
- Total:
8 + 4 + 4 = 16 cm².
Check
Using the formula directly: SA = 2(1·2 + 1·2 + 2·2) = 2(2 + 2 + 4) = 16. ✓
A rectangular prism has integer edge lengths and volume V. All six faces are painted, then the prism is cut into 1×1×1 cubes. Exactly 50 of these cubes have no paint. What is the mean of all possible values of V?
- A. 224
- B. 310
- C. 396
- D. 288
- E. 348
Strategy
The unpainted cubes form an inner prism with one layer peeled off every face. If the outer prism is a × b × c, the inner prism is (a−2) × (b−2) × (c−2) — and that product is 50.
Steps
- Find all ways to write 50 as a product of three positive integers (the inner dimensions).
50 = 1·1·50 = 1·2·25 = 1·5·10 = 2·5·5 — four cases.
- Add 2 to each inner dimension to get the outer prism. Compute V for each:
- Inner 1×1×50 → outer 3×3×52 → V =
468
- Inner 1×2×25 → outer 3×4×27 → V =
324
- Inner 1×5×10 → outer 3×7×12 → V =
252
- Inner 2×5×5 → outer 4×7×7 → V =
196
- Mean:
(468 + 324 + 252 + 196) / 4 = 1240 / 4 = 310.
Pitfall
Don't forget that each dimension must be at least 3 in the outer prism (otherwise that direction has no interior). This is automatic if the inner dimensions are at least 1.
A closed rectangular prism with height 8 cm stands on a 2 cm × 5 cm face. It contains water to a depth of 6 cm. When the prism is tipped to stand on its largest face, what is the new depth of the water?
- A. 0.75 cm
- B. 1 cm
- C. 1.25 cm
- D. 1.5 cm
- E. 1.75 cm
Strategy
Volume of water is conserved. Compute the volume now, then divide by the new base area to find the new depth.
Steps
- Initial water volume: base 2 × 5 = 10 cm², depth 6 cm →
V = 10 × 6 = 60 cm³.
- Faces of the prism: 2×5 = 10, 2×8 = 16, 5×8 = 40. Largest face is 5 × 8.
- Standing on the 5×8 face, the new base area is
40 cm². Let the new depth be d.
-
40 · d = 60 ⇒ d = 60/40 = 1.5 cm.
Insight
Larger base ⇒ shallower water (volume is fixed). The depth is inversely proportional to the base area.
§ 6.3 · Numbers That Read the Same Both Ways
Palindrome Numbers
Core Concept: Counting palindromes by digit structure
A palindrome reads the same forwards and backwards: 121, 3443, 12321, 9779.
-
3-digit palindromes have the form
aba: 9 choices for a (1–9), 10 for b (0–9) → 90 total.
-
4-digit palindromes:
abba → 9 × 10 = 90.
-
5-digit palindromes:
abcba → 9 × 10 × 10 = 900.
The pattern: a palindrome with n digits has roughly 10^⌈n/2⌉ × 9/10 choices — pick the first half freely (first digit non-zero), then mirror it.
A palindrome is a positive integer that reads the same forwards or backwards. How many palindromes are there between 100 and 1000?
- A. 10
- B. 90
- C. 100
- D. 900
- E. 1000
Strategy
"Between 100 and 1000" means 3-digit palindromes aba.
Steps
- Choose a (the first/last digit): a ∈ {1, 2, …, 9} → 9 choices.
- Choose b (the middle digit): b ∈ {0, 1, …, 9} → 10 choices.
- Total:
9 × 10 = 90 palindromes.
Pitfall
Don't forget a ≠ 0, otherwise you'd be looking at "0b0" — a 1-digit number, not a 3-digit one.
A palindrome is a positive integer whose digits read the same forwards or backwards. 13931 is a palindrome. What is the sum of the digits of the next palindrome greater than 13931?
Strategy
A 5-digit palindrome has the form abcba. The next one after 13931 must have a, b at least as large as 1, 3 (or larger when needed).
Steps
- Try keeping a = 1, b = 3. Then the palindrome is
13c31. We need 13c31 > 13931, so c must satisfy 13c31 > 13931. But c is a single digit (0–9), so the largest "13c31" is 13931 itself — no room.
- Bump b up to 4: palindromes
14c41. The smallest is 14041 (c = 0). Check: 14041 > 13931 ✓. So the next palindrome is 14041.
- Sum of digits:
1 + 4 + 0 + 4 + 1 = 10.
Pitfall
"Increase the middle digit" only works if there's room. When the middle of 13931 is already 9, you must bump up the next layer (b: 3 → 4) and reset the middle to 0.
How many 5-digit palindromes (greater than 10000 and less than 100000) are multiples of 18?
- A. 41
- B. 42
- C. 43
- D. 44
- E. 45
Concept
A number is a multiple of 18 ⇔ it's a multiple of both 2 and 9. Multiple of 2: last digit even. Multiple of 9: sum of digits divisible by 9.
Steps
- Write the palindrome as
abcba. Divisible by 2 ⇒ last digit a is even ⇒ a ∈ {2, 4, 6, 8}.
- Divisible by 9 ⇒
2a + 2b + c is divisible by 9.
-
Case a = 2:
4 + 2b + c ≡ 0 (mod 9), with 0 ≤ 2b + c ≤ 27. So 2b + c ∈ {5, 14, 23}. Counting (b, c) solutions: 3 + 5 + 3 = 11.
-
Case a = 4:
2b + c ∈ {1, 10, 19} → 1 + 5 + 5 = 11.
-
Case a = 6:
2b + c ∈ {6, 15, 24} → 4 + 5 + 2 = 11.
-
Case a = 8:
2b + c ∈ {2, 11, 20} → 2 + 5 + 4 = 11.
- Total:
11 × 4 = 44.
Lesson
For divisibility by 9, only the digit sum matters. For palindromes abcba, the digit sum is 2a + 2b + c — even pairs telescope nicely.
§ 6.4 · Patterns in Triangular Arrays
Number Tables
Core Concept: Find where row n starts, then count
For a triangular table where row n has n numbers (1, then 2, 3, …), the count of numbers before row n is 1 + 2 + … + (n−1) = n(n−1)/2. So row n starts at position n(n−1)/2 + 1.
Sum of an arithmetic block: sum = (count) × (first + last) / 2.
The positive integers are arranged into a triangular table — row 1: 1; row 2: 2, 3; row 3: 4, 5, 6; row 4: 7, 8, 9, 10; and so on. What is the sum of all numbers in the 12th row?
- A. 540
- B. 864
- C. 870
- D. 1020
Strategy
Find the first and last number in row 12, then use the arithmetic-series sum.
Steps
- Numbers before row 12:
1 + 2 + … + 11 = 11 · 12 / 2 = 66.
- Row 12 starts at
66 + 1 = 67.
- Row 12 has 12 numbers, so it ends at
67 + 11 = 78.
- Sum:
12 × (67 + 78) / 2 = 12 × 145 / 2 = 6 × 145 = 870.
Lesson
Row n sum = n × (first + last) / 2 where first = n(n−1)/2 + 1, last = n(n+1)/2. For n = 12: first = 67, last = 78, sum = 870.
All even numbers starting from 2 are placed in a triangular table: row 1: 2; row 2: 4, 6; row 3: 8, 10, 12; row 4: 14, 16, 18, 20; … What is the last number in row 20?
Strategy
Count how many even numbers appear up to the end of row 20.
Steps
- Total even numbers used through row 20:
1 + 2 + … + 20 = 20 · 21 / 2 = 210.
- The k-th even number is
2k. The 210th even number is 2 × 210 = 420.
- So row 20 ends at 420.
Check
Row 20 has 20 numbers; previous rows used 190 numbers; so row 20 contains the 191st through 210th even numbers — that is, 382, 384, …, 420. ✓
In a triangle table, the d-th diagonal line begins with d and each subsequent number on that diagonal is d larger than the previous one. (So diagonal 1 is 1, 2, 3, …; diagonal 2 is 2, 4, 6, …; etc.) The n-th term of the d-th diagonal sits in horizontal row n + d − 1. In which horizontal row does 2016 first appear?
- A. 90
- B. 94
- C. 88
- D. 91
- E. 89
Strategy
2016 appears on diagonal d at the n-th position when n · d = 2016. The row number is n + d − 1. To minimize the row, minimize n + d subject to nd = 2016 — by AM–GM, this happens when n and d are close to √2016 ≈ 44.9.
Steps
- Factor 2016 =
2⁵ · 3² · 7. List factor pairs (d, n) with d ≤ n closest to √2016:
- (d, n) = (42, 48): n + d = 90 → row = 89
- (d, n) = (36, 56): n + d = 92 → row = 91
- (d, n) = (32, 63): n + d = 95 → row = 94
- (d, n) = (28, 72): n + d = 100 → row = 99
- The smallest row is
89, achieved by d = 42, n = 48 (or d = 48, n = 42).
Pitfall
"First appears" means smallest row number. The factor pair closest to √2016 gives the answer — but make sure both factors are valid divisors.
§ 6.5 · Mixed Practice
Homework
Points P(15, 55), Q(26, 55), and R(26, 35) are three vertices of rectangle PQRS. What is the area of this rectangle?
- A. 360
- B. 800
- C. 220
- D. 580
- E. 330
Steps
- PQ is horizontal (both y = 55): length =
26 − 15 = 11.
- QR is vertical (both x = 26): length =
55 − 35 = 20.
- Area =
11 × 20 = 220.
In the diagram, the point (3, −4) is labelled — among five points P, Q, R, S, T arranged on a grid.
Steps
- x = 3 means right of the y-axis (Quadrant IV since y is negative).
- y = −4 means 4 units below the x-axis.
- From the grid, the point at (3, −4) is labelled S.
What is the volume of the cuboid below? (Dimensions: 8 cm × 2 cm × 3 cm.)
- A. 46
- B. 92
- C. 48
- D. 96
- E. 102
Steps
Volume = l × w × h = 8 × 2 × 3 = 48 cm³.
Four identical cubes form a rectangular solid as shown (a 2 × 2 × 1 arrangement of cubes — four cubes in an L/block shape). The surface area of each cube is 24 cm². What is the surface area of the whole solid?
- A. 80 cm²
- B. 64 cm²
- C. 40 cm²
- D. 32 cm²
- E. 24 cm²
Strategy
Find the edge length of each cube first, then compute the surface area of the combined solid using its outer dimensions.
Steps
- Each cube:
6a² = 24 ⇒ a² = 4 ⇒ a = 2 cm.
- Four cubes in a 2 × 2 × 1 arrangement form a cuboid with outer dimensions
4 × 4 × 2 cm.
- SA =
2(4·4 + 4·2 + 4·2) = 2(16 + 8 + 8) = 2 × 32 = 64 cm².
Pitfall
When several cubes are joined, hidden faces (where two cubes touch) are not part of the outer surface. Always re-compute using the outer bounding-box and the formula, not by adding the cubes' individual surfaces.
A tank with dimensions 50 cm × 10 cm × 18 cm will be filled to the brim after 5 litres of water are added. What is the current volume of water in the tank, in cm³? (Note: 1 L = 1000 cm³.)
- A. 2840
- B. 4000
- C. 3500
- D. 3840
- E. 4200
Steps
- Tank capacity:
50 × 10 × 18 = 9000 cm³.
- Volume needed to fill: 5 L =
5000 cm³.
- Current water:
9000 − 5000 = 4000 cm³.
Four vertices of a quadrilateral are located at (7, 6), (−5, 1), (−2, −3), and (10, 2). What is the area of the quadrilateral, in square units?
- A. 60
- B. 63
- C. 67
- D. 70
- E. 72
Strategy
Use the shoelace formula on the vertices in order. For vertices (x₁, y₁), (x₂, y₂), (x₃, y₃), (x₄, y₄):
Area = ½ |x₁(y₂ − y₄) + x₂(y₃ − y₁) + x₃(y₄ − y₂) + x₄(y₁ − y₃)|
Steps
- Label:
A = (7, 6), B = (−5, 1), C = (−2, −3), D = (10, 2).
- Shoelace:
½ |7(1 − 2) + (−5)(−3 − 6) + (−2)(2 − 1) + 10(6 − (−3))|.
- Compute:
½ |7(−1) + (−5)(−9) + (−2)(1) + 10(9)| = ½ |−7 + 45 − 2 + 90| = ½ × 126.
- Area =
63.
Alternative
You can also enclose the quadrilateral in a bounding rectangle (here, from (−5, −3) to (10, 6) → 15 × 9 = 135) and subtract the four right triangles around the perimeter. Same answer, more bookkeeping.
The digits 1–9 are each used exactly once to write three one-digit integers and three two-digit integers. The one-digit integers are the length, width, and height of a rectangular prism. The two-digit integers are the areas of the prism's faces. What is the surface area of the rectangular prism?
- A. 176
- B. 184
- C. 186
- D. 198
- E. 212
Strategy
Let the dimensions be l, w, h (one-digit). The face areas are lw, lh, wh (two-digit). Together, six digits are used across the three two-digit areas, and three digits are used as dimensions — all nine digits 1–9 appear exactly once.
Steps
- None of l, w, h can be 5, because 5 times any non-5 digit ends in 0 or 5 — but 0 isn't available, and using digit 5 twice (as factor and as a 5 in the area) isn't allowed.
- So 5 must appear inside one of the two-digit face areas. It must be the tens digit (since being the units digit would force one factor to be 5, ruled out above).
- Two-digit products with tens digit 5 (from non-5 single digits):
54 = 6 × 9 and 56 = 7 × 8.
- Case
56 = 7 × 8: digits 5, 6, 7, 8 used. Remaining digits {1, 2, 3, 4, 9} — find the third dimension so the other two areas use exactly the remaining four digits (and don't repeat 5, 6, 7, 8). Systematic check rules out all five candidates.
- Case
54 = 6 × 9: digits 5, 4, 6, 9 used. Try third dimension = 3 → areas are 3 × 6 = 18 and 3 × 9 = 27. Digits used in areas: 5, 4, 1, 8, 2, 7. Dimensions: 3, 6, 9. All nine digits 1–9 appear exactly once ✓.
- So dimensions are 3, 6, 9. SA =
2(18 + 27 + 54) = 2 × 99 = 198.
Lesson
"Each digit used exactly once" is a strong constraint. Always check it carefully against your candidate.
Palindrome years (like 1991 and 2002) read the same forwards and backwards. 1991 and 2002 are 11 years apart. How many years apart are 2002 and the first palindrome year after 2002?
Steps
- A 4-digit palindrome has the form
abba. After 2002, the next palindrome of form 2bb2 would need bb > 00: smallest is 2112.
- Check there's nothing in between:
2002, 2112, 2222, …. So the next palindrome year is 2112.
- Difference:
2112 − 2002 = 110 years.
A palindrome is a number whose digits read the same forwards as backwards (for example, 2002). In how many years from 2013 will the next year be a palindrome?
- A. 11
- B. 33
- C. 99
- D. 111
- E. 131
Steps
- The next 4-digit palindrome after 2013 has form
2bb2 with 2bb2 ≥ 2013. The smallest is 2112 (since 2002 < 2013 < 2112).
- Difference:
2112 − 2013 = 99 years.
An 8 × 8 × n rectangular prism is built from 1 × 1 × 1 cubes. Let A be the surface area of the prism and B be the combined surface area of all the 1 × 1 × 1 cubes. For how many values of n is B/A an integer? (Compute the sum of all such n.)
- A. 86
- B. 90
- C. 70
- D. 78
- E. 96
Strategy
Express A and B in terms of n, simplify B/A, and look for n that makes it an integer.
Steps
- Surface area of the prism: A has two 8×8 faces (area 64 each) and four 8×n faces (area 8n each). So
A = 128 + 32n.
- Total cubes:
8 × 8 × n = 64n. Each unit cube has SA = 6, so B = 6 · 64n = 384n.
- Ratio:
B/A = 384n / (128 + 32n) = 12n / (4 + n) after dividing numerator and denominator by 32.
- Write
12n / (n + 4) = 12 − 48/(n + 4). For this to be a positive integer, n + 4 must be a positive divisor of 48.
- Divisors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Corresponding n: −3, −2, −1, 0, 2, 4, 8, 12, 20, 44. Positive n: {2, 4, 8, 12, 20, 44}.
- Sum:
2 + 4 + 8 + 12 + 20 + 44 = 90.
Lesson
The trick of rewriting kn/(n+m) as k − (km)/(n+m) converts a "when is this an integer" question into a divisor-counting question.