[Advanced Algebra] 15 Challenge Problems: Direct & Inverse Variation (Contest & SAT Prep)
Advanced Challenge Series | AMC 8 & SAT Math Prep
[Advanced Algebra] 15 Challenge Problems: Direct & Inverse Variation
Conquer contest-level functional thinking: composite coordinate geometry, extreme parameter bounding, and hyperbola area invariance.
Includes 15 Master-Level Problems with detailed algebraic derivations and strategic insights.
Includes 15 Master-Level Problems with detailed algebraic derivations and strategic insights.
CHALLENGE 01
Three-Variable Composite Variation Chain
The variable $z$ varies directly as the square of $x$ and inversely as the square root of $y$. When $x = 3$ and $y = 16$, the value of $z$ is $18$. Determine the value of $z$ when $x = 4$ and $y = 64$.
π‘ View Detailed Solution & Derivation
Answer: $16$
• Setup model: $z = k \cdot \frac{x^2}{\sqrt{y}}$.
• Substitute $(x, y, z) = (3, 16, 18)$: $18 = k \cdot \frac{3^2}{\sqrt{16}} = k \cdot \frac{9}{4} \implies k = 8$.
• Formula: $z = \frac{8x^2}{\sqrt{y}}$.
• For $x = 4, y = 64$: $z = \frac{8 \times 4^2}{\sqrt{64}} = \frac{8 \times 16}{8} = 16$.
• Setup model: $z = k \cdot \frac{x^2}{\sqrt{y}}$.
• Substitute $(x, y, z) = (3, 16, 18)$: $18 = k \cdot \frac{3^2}{\sqrt{16}} = k \cdot \frac{9}{4} \implies k = 8$.
• Formula: $z = \frac{8x^2}{\sqrt{y}}$.
• For $x = 4, y = 64$: $z = \frac{8 \times 4^2}{\sqrt{64}} = \frac{8 \times 16}{8} = 16$.
πΏ Strategy: Always isolate the universal variation constant $k$ first before evaluating secondary states.
CHALLENGE 02
Percentage Sensitivity & Scaling Analysis
If $y$ varies inversely as $x$, and $x$ increases by $25\%$, find the percentage change (increase or decrease) in $y$.
π‘ View Detailed Solution & Derivation
Answer: Decreases by $20\%$
• $xy = k$ is constant.
• When $x$ becomes $x' = 1.25x = \frac{5}{4}x$, the new value $y'$ must satisfy $x' y' = k$.
• $\left(\frac{5}{4}x\right) y' = xy \implies y' = \frac{4}{5}y = 0.80y$.
• The change is $y' - y = -0.20y$, which represents a decrease of $20\%$.
• $xy = k$ is constant.
• When $x$ becomes $x' = 1.25x = \frac{5}{4}x$, the new value $y'$ must satisfy $x' y' = k$.
• $\left(\frac{5}{4}x\right) y' = xy \implies y' = \frac{4}{5}y = 0.80y$.
• The change is $y' - y = -0.20y$, which represents a decrease of $20\%$.
πΏ Strategy: In inverse variation, scaling factors are reciprocals: multiplying by $\frac{5}{4}$ forces the conjugate quantity to scale by $\frac{4}{5}$.
CHALLENGE 03
Linear-Inverse System Intersection & Origin Symmetry
The line $y = ax$ and the hyperbola $y = \frac{k}{x}$ intersect at points $A$ and $B$. If the coordinates of $A$ are $(2, 6)$, find the coordinates of $B$, the value of $k$, and the distance between $A$ and $B$.
π‘ View Detailed Solution & Derivation
Answer: $B(-2, -6)$, $k = 12$, and $AB = 4\sqrt{10}$
• Point $A(2, 6)$ on $y = \frac{k}{x} \implies k = 2 \times 6 = 12$.
• Since both $y = ax$ and $y = \frac{12}{x}$ are symmetric with respect to the origin $(0,0)$, point $B$ is $(-2, -6)$.
• $\overline{AB} = \sqrt{(2 - (-2))^2 + (6 - (-6))^2} = \sqrt{4^2 + 12^2} = \sqrt{16 + 144} = \sqrt{160} = 4\sqrt{10}$.
• Point $A(2, 6)$ on $y = \frac{k}{x} \implies k = 2 \times 6 = 12$.
• Since both $y = ax$ and $y = \frac{12}{x}$ are symmetric with respect to the origin $(0,0)$, point $B$ is $(-2, -6)$.
• $\overline{AB} = \sqrt{(2 - (-2))^2 + (6 - (-6))^2} = \sqrt{4^2 + 12^2} = \sqrt{16 + 144} = \sqrt{160} = 4\sqrt{10}$.
πΏ Strategy: Odd symmetry guarantees that the midpoint of the chord connecting the intersections is always the origin $(0,0)$.
CHALLENGE 04
Quadrant Trajectory & Non-Zero Coefficient Deduction
If the graph of the linear equation $ax + by = 0$ passes through Quadrants II and IV, determine which quadrants contain the graph of the hyperbola $y = \frac{ab}{x}$.
π‘ View Detailed Solution & Derivation
Answer: Quadrants I and III
• Rearrange to slope-intercept form: $by = -ax \implies y = -\frac{a}{b}x$.
• Passing through Quadrants II and IV means the slope is negative: $-\frac{a}{b} < 0 \implies \frac{a}{b} > 0$.
• Since $\frac{a}{b} > 0$, $a$ and $b$ share the exact same sign $\implies ab > 0$.
• Because the numerator $ab > 0$, $y = \frac{ab}{x}$ must lie in Quadrants I and III.
• Rearrange to slope-intercept form: $by = -ax \implies y = -\frac{a}{b}x$.
• Passing through Quadrants II and IV means the slope is negative: $-\frac{a}{b} < 0 \implies \frac{a}{b} > 0$.
• Since $\frac{a}{b} > 0$, $a$ and $b$ share the exact same sign $\implies ab > 0$.
• Because the numerator $ab > 0$, $y = \frac{ab}{x}$ must lie in Quadrants I and III.
πΏ Strategy: The ratio $\frac{a}{b}$ and the product $ab$ always share identical algebraic signs for non-zero real numbers.
CHALLENGE 05
Rotational Bounding of Polygon Intersections
A triangle has vertices at $A(2, 6)$, $B(5, 2)$, and $C(2, 2)$. Find the complete range of values of $m$ such that the line $y = mx$ intersects the triangle $\triangle ABC$.
π‘ View Detailed Solution & Derivation
Answer: $\frac{2}{5} \le m \le 3$
• Slope from origin to vertex $A(2, 6)$: $m_A = \frac{6}{2} = 3$.
• Slope from origin to vertex $B(5, 2)$: $m_B = \frac{2}{5}$.
• Slope from origin to vertex $C(2, 2)$: $m_C = \frac{2}{2} = 1$.
• The extreme angular bounds occur at vertices $B$ and $A$. Therefore, $\frac{2}{5} \le m \le 3$.
• Slope from origin to vertex $A(2, 6)$: $m_A = \frac{6}{2} = 3$.
• Slope from origin to vertex $B(5, 2)$: $m_B = \frac{2}{5}$.
• Slope from origin to vertex $C(2, 2)$: $m_C = \frac{2}{2} = 1$.
• The extreme angular bounds occur at vertices $B$ and $A$. Therefore, $\frac{2}{5} \le m \le 3$.
πΏ Strategy: Slopes through the origin correspond directly to the tangent of inclination angles; check all polygon vertices to determine the extremal envelope.
CHALLENGE 06
Extremal Parameter Bounding for Hyperbolas
For the line segment joining $P(2, 8)$ and $Q(6, 2)$, determine the range of values of $k$ such that the curve $y = \frac{k}{x}$ intersects the segment $\overline{PQ}$.
π‘ View Detailed Solution & Derivation
Answer: $\frac{49}{4} \le k \le 16$ (or $12.25 \le k \le 16$)
• Equation of line $\overline{PQ}$: slope $m = \frac{2 - 8}{6 - 2} = -\frac{3}{2}$. Line: $y - 2 = -\frac{3}{2}(x - 6) \implies y = -\frac{3}{2}x + 11$.
• Endpoints: $k(P) = 2 \times 8 = 16$; $k(Q) = 6 \times 2 = 12$.
• The hyperbola is tangent to the line when $-\frac{3}{2}x + 11 = \frac{k}{x} \implies 3x^2 - 22x + 2k = 0$.
• Discriminant $\Delta = (-22)^2 - 4(3)(2k) = 0 \implies 484 - 24k = 0 \implies k = \frac{484}{24} = \frac{121}{6} \approx 20.17$?
• Note geometry: between $x=2$ and $x=6$, the product $xy = x(11 - 1.5x) = -1.5x^2 + 11x$. Vertex at $x = \frac{11}{3} \approx 3.67$, giving $xy_{\max} = \frac{121}{6}$.
• Thus, minimum on the segment occurs at endpoint $Q$: $k_{\min} = 12$; maximum occurs at tangency vertex: $k_{\max} = \frac{121}{6}$. Range: $12 \le k \le \frac{121}{6}$.
• Equation of line $\overline{PQ}$: slope $m = \frac{2 - 8}{6 - 2} = -\frac{3}{2}$. Line: $y - 2 = -\frac{3}{2}(x - 6) \implies y = -\frac{3}{2}x + 11$.
• Endpoints: $k(P) = 2 \times 8 = 16$; $k(Q) = 6 \times 2 = 12$.
• The hyperbola is tangent to the line when $-\frac{3}{2}x + 11 = \frac{k}{x} \implies 3x^2 - 22x + 2k = 0$.
• Discriminant $\Delta = (-22)^2 - 4(3)(2k) = 0 \implies 484 - 24k = 0 \implies k = \frac{484}{24} = \frac{121}{6} \approx 20.17$?
• Note geometry: between $x=2$ and $x=6$, the product $xy = x(11 - 1.5x) = -1.5x^2 + 11x$. Vertex at $x = \frac{11}{3} \approx 3.67$, giving $xy_{\max} = \frac{121}{6}$.
• Thus, minimum on the segment occurs at endpoint $Q$: $k_{\min} = 12$; maximum occurs at tangency vertex: $k_{\max} = \frac{121}{6}$. Range: $12 \le k \le \frac{121}{6}$.
πΏ Strategy: The product $xy$ along a straight line segment forms a downward quadratic function; check both endpoints and the parabolic apex.
CHALLENGE 07
Concentric Hyperbola Area Invariance Law
In Quadrant I, two curves $C_1: y = \frac{12}{x}$ and $C_2: y = \frac{32}{x}$ are given. From a point $A(p, q)$ on $C_2$, lines parallel to the $x$-axis and $y$-axis intersect $C_1$ at points $B$ and $C$, respectively. Prove that the area of rectangle $ABDC$ is invariant and calculate its exact value.
π‘ View Detailed Solution & Derivation
Answer: $\frac{25}{2}$ (or $12.5$)
• $A = \left(p, \frac{32}{p}\right)$.
• Horizontal line $y = \frac{32}{p}$ meets $C_1$: $\frac{32}{p} = \frac{12}{x_B} \implies x_B = \frac{12}{32}p = \frac{3}{8}p$.
• Width $\overline{AB} = p - \frac{3}{8}p = \frac{5}{8}p$.
• Vertical line $x = p$ meets $C_1$: $y_C = \frac{12}{p}$.
• Height $\overline{AC} = \frac{32}{p} - \frac{12}{p} = \frac{20}{p}$.
• $\text{Area} = \overline{AB} \times \overline{AC} = \left(\frac{5}{8}p\right) \times \left(\frac{20}{p}\right) = \frac{100}{8} = \frac{25}{2}$. The variable $p$ vanishes!
• $A = \left(p, \frac{32}{p}\right)$.
• Horizontal line $y = \frac{32}{p}$ meets $C_1$: $\frac{32}{p} = \frac{12}{x_B} \implies x_B = \frac{12}{32}p = \frac{3}{8}p$.
• Width $\overline{AB} = p - \frac{3}{8}p = \frac{5}{8}p$.
• Vertical line $x = p$ meets $C_1$: $y_C = \frac{12}{p}$.
• Height $\overline{AC} = \frac{32}{p} - \frac{12}{p} = \frac{20}{p}$.
• $\text{Area} = \overline{AB} \times \overline{AC} = \left(\frac{5}{8}p\right) \times \left(\frac{20}{p}\right) = \frac{100}{8} = \frac{25}{2}$. The variable $p$ vanishes!
πΏ Strategy: General formula for concentric hyperbolas: $\text{Area} = \frac{(k_2 - k_1)^2}{k_2} = \frac{(32 - 12)^2}{32} = \frac{400}{32} = 12.5$.
CHALLENGE 08
Triangle Area Bounded by Reciprocal Curves
A point $P(a, b)$ lies on the curve $y = \frac{18}{x}$ in Quadrant I. The tangent-like projection right triangle has vertices at the origin $O(0,0)$, $A(a, 0)$, and $B(0, b)$. Find the area of triangle $\triangle OAB$, and state whether it depends on the choice of point $P$.
π‘ View Detailed Solution & Derivation
Answer: $9$, completely independent of $P$
• Coordinates: $O(0,0), A(a, 0), B(0, b)$.
• Base $= a$, Height $= b$.
• $\text{Area of } \triangle OAB = \frac{1}{2}ab$.
• Since $P(a, b)$ lies on $y = \frac{18}{x}$, $ab = 18$.
• $\text{Area} = \frac{1}{2} \times 18 = 9$.
• Coordinates: $O(0,0), A(a, 0), B(0, b)$.
• Base $= a$, Height $= b$.
• $\text{Area of } \triangle OAB = \frac{1}{2}ab$.
• Since $P(a, b)$ lies on $y = \frac{18}{x}$, $ab = 18$.
• $\text{Area} = \frac{1}{2} \times 18 = 9$.
πΏ Strategy: The coordinate projection triangle under any inverse variation curve always has area $\frac{1}{2}|k|$.
CHALLENGE 09
Horizontal Chord Decomposition Across Dual Functions
A horizontal line $y = 6$ intersects the line $y = \frac{3}{2}x$ at point $A$, and intersects the hyperbola $y = \frac{k}{x}$ ($k > 0$) at point $B$. If the length of segment $\overline{AB}$ is $5$ and point $B$ lies to the right of point $A$, find the value of $k$.
π‘ View Detailed Solution & Derivation
Answer: $k = 54$
• Find $x_A$: $6 = \frac{3}{2}x_A \implies x_A = 4$. Thus $A(4, 6)$.
• Segment length $\overline{AB} = x_B - x_A = 5 \implies x_B = 4 + 5 = 9$.
• Since $B(9, 6)$ lies on $y = \frac{k}{x}$: $k = 9 \times 6 = 54$.
• Find $x_A$: $6 = \frac{3}{2}x_A \implies x_A = 4$. Thus $A(4, 6)$.
• Segment length $\overline{AB} = x_B - x_A = 5 \implies x_B = 4 + 5 = 9$.
• Since $B(9, 6)$ lies on $y = \frac{k}{x}$: $k = 9 \times 6 = 54$.
πΏ Strategy: On horizontal lines ($y = c$), translate distance directly to $\Delta x = x_{\text{right}} - x_{\text{left}}$.
CHALLENGE 10
Lattice Points on Hyperbolic Integer Domains
Find the number of points $(x, y)$ on the curve $y = \frac{144}{x}$ such that both $x$ and $y$ are integers, and determine how many of these points lie in Quadrant I with $x \le y$.
π‘ View Detailed Solution & Derivation
Answer: $30$ total integer points; $8$ points in Quadrant I with $x \le y$
• Prime factorization: $144 = 2^4 \times 3^2$.
• Total positive divisors: $(4 + 1)(2 + 1) = 5 \times 3 = 15$.
• Total integer points (positive + negative): $15 \times 2 = 30$.
• For Quadrant I ($x > 0, y > 0$): Exactly one point has $x = y$ ($12 \times 12 = 144$).
• Remaining $14$ points split equally: $7$ points with $x < y$, $7$ with $x > y$.
• Points with $x \le y$: $7 + 1 = 8$.
• Prime factorization: $144 = 2^4 \times 3^2$.
• Total positive divisors: $(4 + 1)(2 + 1) = 5 \times 3 = 15$.
• Total integer points (positive + negative): $15 \times 2 = 30$.
• For Quadrant I ($x > 0, y > 0$): Exactly one point has $x = y$ ($12 \times 12 = 144$).
• Remaining $14$ points split equally: $7$ points with $x < y$, $7$ with $x > y$.
• Points with $x \le y$: $7 + 1 = 8$.
πΏ Strategy: Perfect squares contribute an odd number of divisors; account for the symmetric diagonal point $( \sqrt{k}, \sqrt{k} )$ separately.
CHALLENGE 11
Multi-Branch Gear Train Velocity Ratios
A drivetrain contains three gears: $A$ ($30$ teeth), $B$ ($45$ teeth), and $C$ ($x$ teeth) meshed sequentially. When gear $A$ makes $12$ revolutions, the combined sum of revolutions made by gears $B$ and $C$ is $17$. Find the number of teeth on gear $C$.
π‘ View Detailed Solution & Derivation
Answer: $40$ teeth
• Total tooth engagements: $N = 30 \times 12 = 360$ teeth.
• Revolutions of gear $B$: $R_B = \frac{360}{45} = 8$ revolutions.
• Revolutions of gear $C$: $R_C = 17 - 8 = 9$ revolutions.
• Tooth count of gear $C$: $x \times 9 = 360 \implies x = \frac{360}{9} = 40$ teeth.
• Total tooth engagements: $N = 30 \times 12 = 360$ teeth.
• Revolutions of gear $B$: $R_B = \frac{360}{45} = 8$ revolutions.
• Revolutions of gear $C$: $R_C = 17 - 8 = 9$ revolutions.
• Tooth count of gear $C$: $x \times 9 = 360 \implies x = \frac{360}{9} = 40$ teeth.
πΏ Strategy: For any continuous gear sequence, the total number of meshed tooth contacts is completely conserved across all gears.
CHALLENGE 12
Fluid Dynamics & Inflow-Outflow Inverse Variation
A reservoir is drained through $n$ identical pipes at a constant rate in $T$ hours, satisfying the model $nT = 48$. If $2$ pipes are clogged and unavailable, the remaining pipes take $4$ additional hours to completely empty the reservoir. Find the initial number of pipes $n$.
π‘ View Detailed Solution & Derivation
Answer: $6$ pipes
• Original relationship: $T = \frac{48}{n}$.
• New condition: $(n - 2)(T + 4) = 48$.
• Expand: $nT + 4n - 2T - 8 = 48 \implies 48 + 4n - 2\left(\frac{48}{n}\right) - 8 = 48$.
• $4n - \frac{96}{n} - 8 = 0 \implies n^2 - 2n - 24 = 0 \implies (n - 6)(n + 4) = 0$.
• Since $n > 0$, $n = 6$.
• Original relationship: $T = \frac{48}{n}$.
• New condition: $(n - 2)(T + 4) = 48$.
• Expand: $nT + 4n - 2T - 8 = 48 \implies 48 + 4n - 2\left(\frac{48}{n}\right) - 8 = 48$.
• $4n - \frac{96}{n} - 8 = 0 \implies n^2 - 2n - 24 = 0 \implies (n - 6)(n + 4) = 0$.
• Since $n > 0$, $n = 6$.
πΏ Strategy: Convert work-rate conservation into product invariants: $\text{Rate} \times \text{Time} = \text{Constant}$.
CHALLENGE 13
Perpendicular Bisectors and Inverse Variation Projections
The line $y = x$ intersects $y = \frac{16}{x}$ at $A$ in Quadrant I. A line passing through $A$ perpendicular to $y = x$ intersects the $x$-axis at $P$ and the $y$-axis at $Q$. Calculate the area of triangle $\triangle OPQ$..png)
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π‘ View Detailed Solution & Derivation
Answer: $32$
• Intersection $A$: $x = \frac{16}{x} \implies x^2 = 16 \implies x = 4$. So $A(4, 4)$.
• Perpendicular line slope $m = -1$. Line through $(4, 4)$: $y - 4 = -(x - 4) \implies y = -x + 8$.
• Intercepts: $P(8, 0)$ and $Q(0, 8)$.
• $\text{Area of } \triangle OPQ = \frac{1}{2} \times 8 \times 8 = 32$.
• Intersection $A$: $x = \frac{16}{x} \implies x^2 = 16 \implies x = 4$. So $A(4, 4)$.
• Perpendicular line slope $m = -1$. Line through $(4, 4)$: $y - 4 = -(x - 4) \implies y = -x + 8$.
• Intercepts: $P(8, 0)$ and $Q(0, 8)$.
• $\text{Area of } \triangle OPQ = \frac{1}{2} \times 8 \times 8 = 32$.
πΏ Strategy: At the symmetric vertex $A(\sqrt{k}, \sqrt{k})$, the normal line forms an isosceles right triangle with area $2k$.
CHALLENGE 14
Asymmetric Rational Functions & Translation Links
The curve $y = \frac{k}{x - 2} + 3$ passes through the origin $(0,0)$. Determine the constant $k$, identify the vertical and horizontal asymptotes, and find the coordinates of the other intersection with the line $y = 3x$.
π‘ View Detailed Solution & Derivation
Answer: $k = 6$, Asymptotes: $x = 2, y = 3$, and Intersection: $(3, 9)$
• Passes through $(0,0)$: $0 = \frac{k}{0 - 2} + 3 \implies \frac{k}{2} = 3 \implies k = 6$.
• Equation: $y = \frac{6}{x - 2} + 3$. Asymptotes: $x = 2$ and $y = 3$.
• Intersect with $y = 3x$: $3x = \frac{6}{x - 2} + 3 \implies 3(x - 1) = \frac{6}{x - 2}$.
• $(x - 1)(x - 2) = 2 \implies x^2 - 3x + 2 = 2 \implies x(x - 3) = 0$.
• The origin is $x = 0$; the other point is $x = 3 \implies y = 3(3) = 9 \implies (3, 9)$.
• Passes through $(0,0)$: $0 = \frac{k}{0 - 2} + 3 \implies \frac{k}{2} = 3 \implies k = 6$.
• Equation: $y = \frac{6}{x - 2} + 3$. Asymptotes: $x = 2$ and $y = 3$.
• Intersect with $y = 3x$: $3x = \frac{6}{x - 2} + 3 \implies 3(x - 1) = \frac{6}{x - 2}$.
• $(x - 1)(x - 2) = 2 \implies x^2 - 3x + 2 = 2 \implies x(x - 3) = 0$.
• The origin is $x = 0$; the other point is $x = 3 \implies y = 3(3) = 9 \implies (3, 9)$.
πΏ Strategy: Shifted hyperbola equations $y = \frac{k}{x - h} + v$ have center of symmetry at $(h, v)$.
CHALLENGE 15
AMC-Level Coordinate Optimization & Area Minimization
A line with negative slope passes through the point $P(2, 8)$ and intersects the positive $x$-axis at $A$ and the positive $y$-axis at $B$. Find the minimum possible area of triangle $\triangle OAB$, and state the coordinates of $A$ and $B$ when this minimum is achieved.
π‘ View Detailed Solution & Derivation
Answer: Minimum Area $= 32$, achieved when $A(4, 0)$ and $B(0, 16)$
• Intercept form: $\frac{x}{a} + \frac{y}{b} = 1$ with $a > 0, b > 0$.
• Since $P(2, 8)$ lies on the line: $\frac{2}{a} + \frac{8}{b} = 1$.
• By AM-GM Inequality: $1 = \frac{2}{a} + \frac{8}{b} \ge 2\sqrt{\frac{16}{ab}} = \frac{8}{\sqrt{ab}}$.
• $\sqrt{ab} \ge 8 \implies ab \ge 64$.
• $\text{Area} = \frac{1}{2}ab \ge \frac{1}{2} \times 64 = 32$.
• Equality holds when $\frac{2}{a} = \frac{8}{b} = \frac{1}{2} \implies a = 4, b = 16$.
• Intercept form: $\frac{x}{a} + \frac{y}{b} = 1$ with $a > 0, b > 0$.
• Since $P(2, 8)$ lies on the line: $\frac{2}{a} + \frac{8}{b} = 1$.
• By AM-GM Inequality: $1 = \frac{2}{a} + \frac{8}{b} \ge 2\sqrt{\frac{16}{ab}} = \frac{8}{\sqrt{ab}}$.
• $\sqrt{ab} \ge 8 \implies ab \ge 64$.
• $\text{Area} = \frac{1}{2}ab \ge \frac{1}{2} \times 64 = 32$.
• Equality holds when $\frac{2}{a} = \frac{8}{b} = \frac{1}{2} \implies a = 4, b = 16$.
πΏ Strategy: When a line passes through a fixed point $(p, q)$, the minimum enclosing triangle area is always $2pq$, with $P$ acting as the exact midpoint of hypotenuse $\overline{AB}$.
πΏ
Yul's Insight: The Deep Geometric Symmetries Behind Proportions
These 15 challenge problems demonstrate that variation is not merely an introductory algebra topic—it is the direct gateway to analytic coordinate geometry, optimization, and contest-level reasoning.
Notice the repeated themes: origin symmetry ($180^\circ$ rotation) eliminates quadratic equations; area invariance ($xy = k$) cancels unknown parameters; and the AM-GM inequality turns coordinate geometry into pure algebraic elegance.
Do not settle for mechanically plugging numbers into formulas. When you train your intuition to see the geometric symmetries behind each algebraic equation, competitive math transforms into an invigorating intellectual journey.
SELF-TEST WORKSHEET
[Practice Test] 15 Challenge Problems: Direct & Inverse Variation
π‘ Instructions: Clicking [Print Practice Test] formats this page into a clean, 2-column A4 exam paper with all solutions hidden. Set a timer for 50 minutes and test your mastery!
Subject: Algebra 1 / Pre-Calculus Fusion
Name: ____________________
Score: _________ / 100
01. [Composite Variation]
The variable $z$ varies directly as $x^2$ and inversely as $\sqrt{y}$. When $x = 3$ and $y = 16$, $z = 18$. Find $z$ when $x = 4$ and $y = 64$.
02. [Percentage Sensitivity]
If $y$ varies inversely as $x$, and $x$ increases by $25\%$, find the percentage change in $y$.
03. [Linear-Inverse Symmetry]
The line $y = ax$ and the hyperbola $y = \frac{k}{x}$ intersect at points $A(2, 6)$ and $B$. Find the coordinates of $B$, the constant $k$, and the distance $AB$.
04. [Quadrant Deduction]
If the graph of $ax + by = 0$ passes through Quadrants II and IV, in which quadrants does the hyperbola $y = \frac{ab}{x}$ lie?
05. [Rotational Bounding]
A triangle has vertices $A(2, 6)$, $B(5, 2)$, and $C(2, 2)$. Find the complete range of slopes $m$ such that the line $y = mx$ intersects $\triangle ABC$.
06. [Hyperbola Tangency Bounds]
For the segment joining $P(2, 8)$ and $Q(6, 2)$, determine the range of values of $k$ such that $y = \frac{k}{x}$ intersects $\overline{PQ}$.
07. [Concentric Hyperbola Area]
From a point $A$ on $y = \frac{32}{x}$, horizontal and vertical lines meet $y = \frac{12}{x}$ at $B$ and $C$, respectively. Calculate the invariant area of rectangle $ABDC$.
08. [Projection Triangle Area]
Point $P(a, b)$ lies on $y = \frac{18}{x}$ in Quadrant I. Find the area of the projection triangle with vertices $O(0,0)$, $A(a, 0)$, and $B(0, b)$.
09. [Horizontal Chord Span]
The line $y = 6$ intersects $y = \frac{3}{2}x$ at $A$, and $y = \frac{k}{x}$ at $B$. If $\overline{AB} = 5$ and point $B$ lies to the right of $A$, find $k$.
10. [Lattice Points Counting]
Find the total number of integer coordinate points on $y = \frac{144}{x}$, and find how many of these lie in Quadrant I with $x \le y$.
11. [Gear Train Conservation]
Three gears $A$ ($30$ teeth), $B$ ($45$ teeth), and $C$ ($x$ teeth) mesh in sequence. When $A$ makes $12$ revolutions, gears $B$ and $C$ make a combined total of $17$ revolutions. Find $x$.
12. [Fluid Rate Modeling]
A reservoir is drained through $n$ identical pipes in $T$ hours such that $nT = 48$. If $2$ pipes are unavailable, drainage takes $4$ extra hours. Find $n$.
13. [Perpendicular Bisector Triangle]
The line $y = x$ intersects $y = \frac{16}{x}$ at $A$ in Quadrant I. The line perpendicular to $y = x$ at $A$ meets the $x$-axis at $P$ and $y$-axis at $Q$. Find the area of $\triangle OPQ$.
14. [Shifted Rational Function]
The curve $y = \frac{k}{x - 2} + 3$ passes through $(0,0)$. Find $k$, state its asymptotes, and find its other intersection point with $y = 3x$.
15. [AMC-Level Area Optimization]
A line with negative slope passes through $P(2, 8)$ and meets the axes at $A(a, 0)$ and $B(0, b)$ ($a, b > 0$). Find the minimum possible area of $\triangle OAB$.
π [Quick Answer Key] Practice Test Solution Matrix (Click to Expand)
| # | Answer | # | Answer | # | Answer |
|---|---|---|---|---|---|
| 01 | $16$ | 06 | $12 \le k \le \frac{121}{6}$ | 11 | $40$ teeth |
| 02 | Decreases by $20\%$ | 07 | $\frac{25}{2}$ ($12.5$) | 12 | $6$ pipes |
| 03 | $B(-2, -6), k=12, AB=4\sqrt{10}$ | 08 | $9$ | 13 | $32$ |
| 04 | Quadrants I and III | 09 | $k = 54$ | 14 | $k=6, x=2, y=3, (3, 9)$ |
| 05 | $\frac{2}{5} \le m \le 3$ | 10 | $30$ total; $8$ points | 15 | $\text{Area} = 32$ |
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