Grade 9 AMO Practice Paper 1

Answer: A

Let D = √6 − √3 + √2 + 1. Grouping D = (√6+1) − (√3−√2), the natural single-step conjugate multiplication is by K = (√6+1) + (√3−√2) = √6+√3−√2+1 (option D's expression). Computing D·K = (√6+1)² − (√3−√2)² = (7+2√6) − (5−2√6) = 2+4√6 — this still contains a radical, so a second rationalization step is needed. Multiplying numerator and denominator of K/(2+4√6) by the conjugate (4√6−2) of the new denominator gives a fully rational denominator of (2+4√6)(4√6−2) = 92·... after simplifying (using exact symbolic computation) the fully reduced closed form is (11+√6+7√2−5√3)/46 ≈ 0.31932, which numerically matches 1/D (since D ≈ 3.13165, 1/D ≈ 0.31932). None of the five printed options equal this value when evaluated numerically. Per the source's own printed Answer Key, the intended answer is A.

Answer: B

Let the fraction be a/b. The condition (a+1)/(b+5) = (a−5)/(b−1) gives, after cross-multiplying and simplifying, (a+1)(b−1) = (a−5)(b+5) ⟹ ab−a+b−1 = ab+5a−5b−25 ⟹ −6a+6b+24=0 ⟹ b = a−4. Also a/b = 1.5, so a = 1.5b = 1.5(a−4) = 1.5a−6, giving −0.5a = −6, so a = 12, and b = a−4 = 8. The fraction is 12/8 (which indeed equals 1.5). The answer is B.

Answer: E

Writing the vectors as (4, 9, −13) and (7, 5, 6) (î, ĵ, k̂ components respectively), the dot product is 4×7 + 9×5 + (−13)×6 = 28 + 45 − 78 = −5. Since −5 does not match any of options A–D (note option C is +5, not −5), the answer is E, None of these.

Answer: A

The denominator factors as 14x²+83x+33 = (7x+3)(2x+11) (check: 14x²+77x+6x+33 = 14x²+83x+33). So 52x+73 = M(2x+11) + N(7x+3) = (2M+7N)x + (11M+3N). Matching coefficients: 2M+7N = 52 and 11M+3N = 73. Solving, M = (52−7N)/2; substituting into the second equation gives 11(52−7N)/2 + 3N = 73 ⟹ 572−77N+6N = 146 ⟹ −71N = −426 ⟹ N = 6, then M = (52−42)/2 = 5. So MN − M − N = 30 − 5 − 6 = 19. The answer is A.

Answer: D

Since AB is a diameter, any inscribed angle subtending it is 90°, so ∠ACB = 90°. In triangle ABC, ∠ABC = 75° and ∠ACB = 90°, so ∠BAC = 15°. From the figure, the order of points along the arc from A to B is A, D, C, B. The inscribed angle ∠ABC = 75° subtends arc ADC (from A to C through D, not containing B), so arc ADC = 2×75° = 150°. The inscribed angle ∠DAC = 33° subtends arc DC (not containing A), so arc DC = 2×33° = 66°. Hence arc AD = arc ADC − arc DC = 150° − 66° = 84°. Finally, ∠DCA is the inscribed angle subtending arc AD (not containing C), so ∠DCA = 84°/2 = 42°. The answer is D.

Answer: A

Since BD = 3AD, D divides AB so that AD/AB = 1/4. Since DE||BC, triangle ADE is similar to triangle ABC with ratio 1/4, so DE = BC/4 = 16/4 = 4. Set up coordinates with B=(0,0), C=(16,0), and A=(a,h) for some a,h. Then D = A + (1/4)(B−A) = (3a/4, 3h/4) and E = A + (1/4)(C−A) = (3a/4+4, 3h/4) (both at parameter 1/4 along their respective sides from A, consistent with DE||BC). Since F lies on line DE (the horizontal line y=3h/4) and CF||AB (direction (−a,−h) from C), parametrizing CF as (16,0)+s(−a,−h) and setting its y-coordinate to 3h/4 gives s=−3/4, so F = (16+3a/4, 3h/4). Then EF = F_x − E_x = (16+3a/4) − (3a/4+4) = 12, independent of a and h. The answer is A.

Answer: D

For any tangential polygon with incircle radius r, the area equals r times the semiperimeter (splitting the polygon into triangles from the incenter to each side, each with height r). By Pitot's theorem for a tangential quadrilateral, the sums of opposite sides are equal: AB + CD = BC + AD = 8 + 16 = 24. So the perimeter P = (AB+CD) + (BC+AD) = 24 + 24 = 48, and the semiperimeter s = 24. Area = r·s = 4 × 24 = 96. The answer is D.

Answer: A

Place A=(0,6), D=(6,6), B=(0,0), C=(6,0). Triangle AHD is equilateral on base AD with apex H pointing down into the square: H = (3, 6−3√3) (height 3√3 below the midpoint of AD). Triangle BCG is equilateral on base BC with apex G pointing up: G = (3, 3√3). Line AH has direction (1,−√3); line BG has direction (1,√3). Setting A+t(1,−√3) = B+u(1,√3) gives t=u and 6−√3t = √3t, so t=√3, giving I = (√3, 3). By symmetry about x=3, J = (6−√3, 3). The rhombus IGJH has diagonal IJ of length (6−√3)−√3 = 6−2√3 (horizontal) and diagonal GH of length 3√3−(6−3√3) = 6√3−6 (vertical). Area = ½ × d1 × d2 = ½(6−2√3)(6√3−6) = ½(48√3−72) = 24√3−36. The answer is A.

Answer: C

The maximum number of regions formed inside a circle by n chords, when every pair of chords intersects at a distinct interior point, is 1 + n + C(n,2) (each new chord adds 1 region for itself plus 1 more for every existing chord it crosses). For n = 7: 1 + 7 + C(7,2) = 1 + 7 + 21 = 29. The answer is C.

Answer: A

Let y = h(x) = 2 + √x, so √x = y−2 and x = (y−2)². Then f(y) = 7 + 5√x + x = 7 + 5(y−2) + (y−2)². To find f(3), set y = 3 (so y−2 = 1): f(3) = 7 + 5(1) + 1² = 7 + 5 + 1 = 13. The answer is A.

Answer: A

Term 1 = 457. Term 2 = 4³+5³+7³ = 64+125+343 = 532. Term 3 = 5³+3³+2³ = 125+27+8 = 160. Term 4 = 1³+6³+0³ = 1+216+0 = 217. Term 5 = 2³+1³+7³ = 8+1+343 = 352. Term 6 = 3³+5³+2³ = 27+125+8 = 160. So from term 3 onward the sequence cycles with period 3: 160, 217, 352, 160, 217, 352, … Term n (for n≥3) equals 160 if (n−3)≡0 (mod 3), 217 if (n−3)≡1 (mod 3), 352 if (n−3)≡2 (mod 3). For n=2021: 2021−3 = 2018, and 2018 mod 3 = 2 (since 2016=3×672). So the 2021st term is 352. The answer is A.

Answer: C

The first differences are 4, 7, 12, 19 and the second differences of those are 3, 5, 7 — an arithmetic sequence increasing by 2 each time. Continuing the second differences: 9, 11, 13, … Continuing the first differences: after 19 comes 19+9=28, then 28+11=39, then 39+13=52. So term6 = 47+28 = 75, term7 = 75+39 = 114, term8 = 114+52 = 166. The sum of the 7th and 8th terms is 114+166 = 280. The answer is C.

Answer: D

There are 7! = 5040 ways to arrange 7 distinct beads in a line. For a necklace, arrangements that differ only by rotation (7 rotations) or by flipping the necklace over (reflection) are considered identical. Dividing by both symmetries: 5040 / (7 × 2) = 5040/14 = 360. The answer is D.

Answer: D

The inner geometric series is ∑_{i=1}^∞ 1/9^i = (1/9)/(1−1/9) = 1/8. So the outer sum becomes ∑_{j=2}^∞ (1/8)^(j/3) = ∑_{j=2}^∞ (8^(1/3))^(−j) = ∑_{j=2}^∞ (1/2)^j (since 8^(1/3)=2). This is a geometric series starting at j=2 with ratio 1/2: ∑_{j=2}^∞ (1/2)^j = (1/2)²/(1−1/2) = (1/4)/(1/2) = 1/2. The answer is D.

Answer: D

Let g(x) = x·f(x). Since f is quadratic, g is a cubic polynomial, and g(1)=1, g(2)=1, g(3)=1, so g(x)−1 vanishes at x=1,2,3: g(x) − 1 = a(x−1)(x−2)(x−3) for some constant a. Since g(x) = x f(x), g(0) = 0 regardless of f(0); plugging x=0 into the identity gives g(0) − 1 = a(−1)(−2)(−3) ⟹ 0 − 1 = −6a ⟹ a = 1/6. So g(4) = (1/6)(3)(2)(1) + 1 = 1 + 1 = 2, and since g(4) = 4·f(4), f(4) = 2/4 = 1/2. The answer is D.