[Transformations: Practice Lab ①] Translations of Points & Curves: 15 Advanced Killer Problems & Self-Diagnosis
In our foundational essay, [Math Essay: The Hidden Relativity of f(x - a)], we resolved the classic paradox: "Why do coordinates add: $(x + a, y + b)$, while equations subtract: $f(x - a, y - b) = 0$?" We established that an active forward shift of a coordinate point requires looking backward to satisfy past constraints ($X = x + a \iff x = X - a$).
However, competitive exams and standardized assessments (such as AP Precalculus, SAT Math Module 2, IB Math AA HL, and AMC 10/12) test translation mechanics at a much higher level. They demand that you evaluate shortest distances between simultaneously moving curves parameterized by time $t$, track rigid motions using representative anchor points (centers and vertices), resolve piecewise absolute-value phase boundaries and collision counts, and compute Cavalieri-invariance integrals without calculus.
In this Practice Lab, we dissect 15 Advanced Killer Problems and their 1:1 Parallel Self-Diagnosis Variants. Sketch the trajectories on your scratchpad first, factor out leading coefficients rigorously, and open each toggle to compare your reasoning with Yul's Pro-Tips.
1. Track Geometric Anchor Points, Not Entire Curves: For circles, track center $(a, b)$; for parabolas, track vertex $(p, q)$; for lines, track fixed intercepts. Translating points eliminates 90% of algebraic errors.
2. The Factorization Iron Rule: When the coefficient of $x$ is not $1$, always factor it out completely into $k(x - a)$ before reading horizontal displacement.
3. Relative Motion Framing: When two objects move simultaneously along distinct vectors, fix one object at the origin and combine the velocity vectors onto the other to reduce 2D kinematics to a 1D quadratic distance function.
Theme 1. Simultaneous Point & Curve Translations (01 ~ 04)
[Problem 01] Invariance Condition for a Line Mapped onto Itself Under Translation
A translation $T: (x, y) \to (x + 2k - 1, y - 3k + 2)$ maps the line $l: 3x + 2y - 5 = 0$ onto line $l'$. Determine all values of the constant $k$ such that $l$ and $l'$ are completely identical.
π Solution & Step-by-Step Breakdown
Answer: No such real constant $k$ exists (Geometric Inconsistency)
[Solution]
1) For a line to map onto itself under a translation, the translation displacement vector must be parallel to the direction of the line.
Equivalently, the translation vector $(\Delta x, \Delta y)$ must be perpendicular to the line's normal vector $(3, 2)$.
2) Set the dot product with the normal vector to zero:
$$3(\Delta x) + 2(\Delta y) = 0 \implies 3(2k - 1) + 2(-3k + 2) = 0$$
$$6k - 3 - 6k + 4 = 1 \ne 0$$
3) Because the constant evaluates to $1 \ne 0$ identically for all $k \in \mathbb{R}$, the line shifts by a constant non-zero normal distance ($d = \dfrac{1}{\sqrt{13}}$). Thus, $l$ and $l'$ can never coincide.
A line $ax + by + c = 0$ remains invariant under translation $(\Delta x, \Delta y)$ if and only if $a\Delta x + b\Delta y = 0$. The translation vector must glide along the line's slope without producing any perpendicular offset.
π― [Self-Diagnosis Variant 01] (Line Self-Coincidence Parameter)
Translation $(x, y) \to (x + k + 1, y + 2k - 4)$ maps $2x - y + 3 = 0$ onto itself. Find the value of constant $k$.
View Variant Solution
Breakdown: $2(k + 1) - (2k - 4) = 2k + 2 - 2k + 4 = 6 \ne 0$. Invariance is impossible.
[Problem 02] Parabola Vertex Translation and Image of Curve Points
The parabola $y = x^2 - 4x + 7$ is translated by $a$ units along the $x$-axis and $b$ units along the $y$-axis so that its vertex moves to the origin $(0, 0)$. Find the coordinates of the image point $P'$ of point $P(3, 4)$ on the original parabola.
π Solution & Step-by-Step Breakdown
Answer: $(1, 1)$
[Solution]
1) Complete the square for the original parabola:
$$y = (x - 2)^2 + 3 \implies \text{Vertex: } (2, 3)$$
2) Moving the vertex $(2, 3)$ to $(0, 0)$ requires shifting by $-2$ horizontally and $-3$ vertically ($a = -2, b = -3$).
The transformation rule is: $(x, y) \to (x - 2, y - 3)$.
3) Apply this translation to point $P(3, 4)$:
$$P' = (3 - 2, 4 - 3) = \mathbf{(1, 1)}$$
Do not expand full equations with $(x - a)$ and $(y - b)$. Extract the vertex $(h, k)$ first, compute the translation vector directly from vertex displacement, and apply it to any given point.
π― [Self-Diagnosis Variant 02] (Envelope of Vertices Constrained to a Line)
Parabola $y = x^2$ is translated such that its vertex moves along $y = 2x - 1$. Find the boundary curve of the region never traversed by these parabolas.
View Variant Solution
Breakdown: Family is $y = (x - t)^2 + 2t - 1 \implies t^2 - 2(x - 1)t + x^2 - y - 1 = 0$. Discriminant condition $D/4 \ge 0$ yields boundary line $\mathbf{y = 2x - 2}$.
[Problem 03] Translated Circle Tangent to a Line with Center on an Axis
Circle $C: (x - 1)^2 + (y + 2)^2 = 9$ is translated by $m$ horizontally and $n$ vertically into circle $C'$. If the center of $C'$ lies on the $x$-axis and $C'$ is tangent to line $3x - 4y + 5 = 0$, find the positive constant $m$.
π Solution & Step-by-Step Breakdown
Answer: $m = \dfrac{7}{3}$
[Solution]
1) The original circle has center $(1, -2)$ and radius $r = 3$. Radius is invariant under translation.
2) The new center is $(1 + m, -2 + n)$. Because it lies on the $x$-axis, its $y$-coordinate is zero:
$$-2 + n = 0 \implies n = 2 \implies \text{New Center: } (1 + m, 0)$$
3) Apply the tangency condition ($d = r$) to line $3x - 4y + 5 = 0$:
$$d = \dfrac{|3(1 + m) - 4(0) + 5|}{\sqrt{3^2 + (-4)^2}} = \dfrac{|3m + 8|}{5} = 3$$
$$|3m + 8| = 15 \implies 3m + 8 = 15 \implies 3m = 7 \implies \mathbf{m = \dfrac{7}{3}}$$
Rigid translations preserve radius ($r$ is an invariant). Never substitute into the general circle equation; simply shift the center point and invoke the point-to-line distance formula $d = r$.
π― [Self-Diagnosis Variant 03] (Simultaneous Circle & Line Shift Tangency)
Circle $x^2 + y^2 = 4$ and line $x - y + 1 = 0$ are simultaneously shifted by $(a, 2a)$. Find the sum of all constants $a$ for which they are tangent.
View Variant Solution
Breakdown: Shifted center $(a, 2a)$ and shifted line $x - y + a + 1 = 0$. Equate distance $d = r$ to isolate parameter $a$.
[Problem 04] Factoring Composite Arguments in Function Translations
The equation $f(x, y) = 0$ is translated by $+2$ along the $x$-axis and $-3$ along the $y$-axis. Analyze the horizontal and vertical shift amounts represented in $f(3 - x, 2y + 1) = 0$.
π Solution & Step-by-Step Breakdown
Answer: Horizontal shift $+3$ with reflection; Vertical shift $-\dfrac{1}{2}$ with stretch
[Solution]
1) The pure translation of $f(x, y) = 0$ by $(+2, -3)$ is $f(x - 2, y + 3) = 0$.
2) To read the transformations inside $f(3 - x, 2y + 1) = 0$, factor out leading coefficients:
$$3 - x = -(x - 3), \quad 2y + 1 = 2\left(y + \dfrac{1}{2}\right)$$
3) This represents a reflection across the vertical axis followed by a horizontal translation of $+3$, and a vertical dilation by factor $2$ followed by a translation of $-\dfrac{1}{2}$.
Never read $3 - x$ as a translation of $-3$. Always write $-(x - 3)$. The subtraction sign inside the factored parentheses proves the horizontal translation is $+3$.
π― [Self-Diagnosis Variant 04] (Radical Function Starting Point Shift)
The curve $y = \sqrt{4 - 2x} + 3$ is obtained by shifting $y = \sqrt{-2x}$ by $(m, n)$. Find $m + n$.
View Variant Solution
Breakdown: $y = \sqrt{-2(x - 2)} + 3 \implies m = 2, n = 3$. Total: $2 + 3 = \mathbf{5}$.
Theme 2. Area Invariance & Cavalieri’s Principle (05 ~ 08)
[Problem 05] Area Bounded by Translated Parabolas and Two Horizontal Lines
Find the area enclosed by the parabola $C_1: y = x^2 - 2x + 3$, its translated image $C_2: y = x^2 - 8x + 18$, and the two horizontal lines $y = 3$ and $y = 6$.
π Solution & Step-by-Step Breakdown
Answer: $9$
[Solution - Calculus-Free Geometric Invariance]
1) Compare the vertices of both parabolas:
$C_1: y = (x - 1)^2 + 2 \implies \text{Vertex: } (1, 2)$
$C_2: y = (x - 4)^2 + 2 \implies \text{Vertex: } (4, 2)$
2) Both curves share the identical quadratic curvature (coefficient $1$) and identical vertical elevation ($y = 2$). $C_2$ is simply $C_1$ translated horizontally by $\Delta x = 4 - 1 = 3$.
3) Any horizontal slice between $y = 3$ and $y = 6$ yields a constant cross-sectional width of $\text{Base} = 3$.
4) By Cavalieri's Principle (shear invariance), the bounded area is equivalent to a parallelogram with base $3$ and vertical height $h = 6 - 3 = 3$:
$$\text{Area} = \text{Base} \times \text{Height} = 3 \times 3 = \mathbf{9}$$
Do not integrate. When two congruent curves are related purely by a horizontal shift $\Delta x$, the horizontal width across all slices is invariant. The enclosed area is simply $\text{Shift Distance} \times \text{Vertical Height}$.
π― [Self-Diagnosis Variant 05] (Area Between Translated Rational Curves)
Find the area bounded by $y = \dfrac{4}{x}$, $y = \dfrac{4}{x - 5} + 2$, and the lines $y = 3$ and $y = 5$.
View Variant Solution
Breakdown: Horizontal shift is $5$, vertical height is $5 - 3 = 2$. Area $= 5 \times 2 = \mathbf{10}$.
[Problem 06] Overlapping Circular Lens Area Under Center Translation
Circle $C_1: x^2 + y^2 = 16$ is translated rightward by $4$ units into $C_2$. Find the area of the overlapping region shared by $C_1$ and $C_2$.
π Solution & Step-by-Step Breakdown
Answer: $\dfrac{32\pi}{3} - 8\sqrt{3}$
[Solution]
1) Centers are $O_1(0, 0)$ and $O_2(4, 0)$, both with radius $R = 4$. Center distance equals radius: $d = R = 4$.
2) The common chord is the perpendicular bisector line $x = 2$.
3) In $C_1$, the central angle subtending half the chord satisfies $\cos(\theta/2) = \dfrac{2}{4} = \dfrac{1}{2} \implies \theta = 120^\circ = \dfrac{2\pi}{3}$.
4) Circular segment area: $\text{Sector} - \text{Triangle} = \dfrac{1}{2}(16)\left(\dfrac{2\pi}{3}\right) - \dfrac{1}{2}(16)\sin(120^\circ) = \dfrac{16\pi}{3} - 4\sqrt{3}$.
5) Overlap consists of two symmetric segments: $2 \times \left(\dfrac{16\pi}{3} - 4\sqrt{3}\right) = \mathbf{\dfrac{32\pi}{3} - 8\sqrt{3}}$.
Whenever center separation equals radius ($d = R$), the radii and common chord form two joined equilateral triangles, locking the central angle to exactly $120^\circ$.
π― [Self-Diagnosis Variant 06] (Orthogonal Overlapping Lens Area)
Circle $x^2 + y^2 = 8$ is shifted horizontally by $4$. Find the area of the overlapping lens.
View Variant Solution
Breakdown: $R = 2\sqrt{2}, d = 4 \implies d^2 = R^2 + R^2$. Central angle is $90^\circ$. Area $= 2(\text{Sector} - \text{Triangle}) = \mathbf{4\pi - 8}$.
[Problem 07] Overlapping Area Extremum Under Right Triangle Translation
Right triangle $OAB$ with vertices $O(0, 0), A(6, 0), B(0, 6)$ is translated by vector $(t, t)$ for $0 < t < 3$ to become $O'A'B'$. Determine the maximum overlapping area $S(t)$.
π Solution & Step-by-Step Breakdown
Answer: $9$
[Solution]
1) Original hypotenuse is $x + y = 6$. The translated origin is $O'(t, t)$ with legs along $x = t$ and $y = t$.
2) The common region is bounded inside $OAB$ where $x \ge t$ and $y \ge t$.
3) This intersection forms a right isosceles triangle with leg length $6 - 2t$.
4) Setting up the area function and maximizing under the symmetric configuration yields maximum area $S = \mathbf{9}$.
Express the boundaries of the moving polygon as linear constraints parameterized by $t$, converting polygon overlap into a single quadratic optimization problem.
π― [Self-Diagnosis Variant 07] (Square Translation Overlap Parameter)
A square of side length $4$ is shifted along $(t, 2t)$. Find positive $t$ such that the overlapping area is half the original area ($8$).
View Variant Solution
Breakdown: $(4 - t)(4 - 2t) = 8 \implies 2t^2 - 12t + 8 = 0$. Positive root gives $\mathbf{4 - 2\sqrt{2}}$.
[Problem 08] Parallelogram Diagonal Translation Vector
Four points $A(1, 2), B(5, 3), C(7, 7), D(a, b)$ form the consecutive vertices of parallelogram $ABCD$. Find the coordinates of image $D'$ when $D$ is shifted under the translation mapping $A$ onto $C$.
π Solution & Step-by-Step Breakdown
Answer: $(9, 11)$
[Solution]
1) Diagonals of a parallelogram bisect each other: $A + C = B + D$.
$$(1, 2) + (7, 7) = (5, 3) + (a, b) \implies (8, 9) = (5 + a, 3 + b) \implies D(3, 6)$$
2) Vector mapping $A(1, 2)$ to $C(7, 7)$:
$$\vec{v} = (7 - 1, 7 - 2) = (+6, +5)$$
3) Apply vector $\vec{v}$ to point $D(3, 6)$:
$$D' = (3 + 6, 6 + 5) = \mathbf{(9, 11)}$$
Parallelograms embody translation: side $\overline{AB}$ is translated into $\overline{DC}$. Use the midpoint identity $A + C = B + D$ to find missing vertices instantaneously.
π― [Self-Diagnosis Variant 08] (Rhombus Diagonal Intersection Shift)
The diagonal intersection of a rhombus is $(2, 3)$. If shifted by $(-4, +1)$, find the sum of the $x$-coordinates of all four new vertices.
View Variant Solution
Breakdown: Center of mass is the diagonal intersection. New center $x = 2 - 4 = -2$. Sum is $4 \times (-2) = \mathbf{-8}$.
Theme 3. Simultaneous Translation Kinematics (09 ~ 11)
[Problem 09] Minimum Distance Between Two Moving Particles Over Time
Particle $P(2, 0)$ moves along the positive $x$-axis at speed $1$, and particle $Q(0, 1)$ moves along the positive $y$-axis at speed $2$. Find the minimum distance between $P$ and $Q$ for time $t \ge 0$.
π Solution & Step-by-Step Breakdown
Answer: $\sqrt{5}$ (Boundary Minimum at $t = 0$)
[Solution]
1) Position vectors at time $t$: $P(t) = (2 + t, 0)$ and $Q(t) = (0, 1 + 2t)$.
2) Distance squared function:
$$L(t)^2 = (t + 2)^2 + (2t + 1)^2 = 5t^2 + 8t + 5$$
3) Complete the square: $L(t)^2 = 5\left(t + \dfrac{4}{5}\right)^2 + \dfrac{9}{5}$.
4) The vertex occurs at $t = -\dfrac{4}{5}$, which lies outside the domain $t \ge 0$.
Because $L(t)^2$ is strictly increasing for $t \ge 0$, the minimum occurs at the boundary $t = 0$:
$$L(0) = \sqrt{2^2 + 1^2} = \mathbf{\sqrt{5}}$$
Always verify whether the axis of symmetry falls within the physical domain ($t \ge 0$). When the axis is negative, the minimum is locked at the initial boundary $t = 0$.
π― [Self-Diagnosis Variant 09] (Moving Particle to Line Distance Minimum)
Find the minimum distance between point $P(t, t + 1)$ and line $x - 2y + 4 = 0$ for $t \in \mathbb{R}$.
View Variant Solution
Breakdown: Point $P$ lies on the line $y = x + 1$, which intersects $x - 2y + 4 = 0$. The minimum distance is $\mathbf{0}$.
[Problem 10] Relative Translation and Collision Time Window for Two Circles
Circle $C_1: x^2 + y^2 = 4$ translates horizontally at $+1$ unit/sec, while $C_2: (x - 8)^2 + (y - 6)^2 = 9$ translates vertically at $-2$ units/sec. Find the time interval during which the two circles intersect.
π Solution & Step-by-Step Breakdown
Answer: $3 \le t \le 5$ seconds
[Solution]
1) Radii are $r_1 = 2$ and $r_2 = 3$. Intersection requires: $|r_1 - r_2| \le d(t) \le r_1 + r_2 \implies 1 \le d(t) \le 5$.
2) Center coordinates at time $t$: $O_1(t) = (t, 0)$ and $O_2(t) = (8, 6 - 2t)$.
3) Center distance squared:
$$d(t)^2 = (8 - t)^2 + (6 - 2t)^2 = 5t^2 - 40t + 100$$
4) Solve the external tangency threshold $d(t)^2 \le 25$:
$$5t^2 - 40t + 100 \le 25 \implies t^2 - 8t + 15 \le 0 \implies (t - 3)(t - 5) \le 0 \implies \mathbf{3 \le t \le 5}$$
When both objects move, write the square of their center separation $d(t)^2$ as a quadratic in $t$, and bound it between $(r_1 - r_2)^2$ and $(r_1 + r_2)^2$.
π― [Self-Diagnosis Variant 10] (Moving Circle and Stationary Line Collision)
Circle $x^2 + y^2 = 5$ translates along velocity vector $(1, 1)$. Find the duration for which it intersects line $2x - y + 5 = 0$.
View Variant Solution
Breakdown: $d = \dfrac{|t + 5|}{\sqrt{5}} \le \sqrt{5} \implies -10 \le t \le 0$. Length of interval is $\mathbf{10}$.
[Problem 11] Minimum Distance Between a Curve and a Line via Parallel Shift
Find the minimum distance between any point $P$ on parabola $y = x^2$ and any point $Q$ on line $y = 2x - 4$.
π Solution & Step-by-Step Breakdown
Answer: $\dfrac{3\sqrt{5}}{5}$
[Solution]
1) Translate line $y = 2x - 4$ toward the parabola until it becomes tangent: $y = 2x + k$.
2) Set up intersection: $x^2 = 2x + k \implies x^2 - 2x - k = 0$.
Tangency condition: $D/4 = (-1)^2 - (-k) = 0 \implies k = -1$. The tangent line is $2x - y - 1 = 0$.
3) Distance between parallel lines $2x - y - 4 = 0$ and $2x - y - 1 = 0$:
$$d = \dfrac{|-4 - (-1)|}{\sqrt{2^2 + (-1)^2}} = \dfrac{3}{\sqrt{5}} = \mathbf{\dfrac{3\sqrt{5}}{5}}$$
The shortest distance from a curve to a line is attained at the point where the tangent is parallel to the line. Compute the parallel tangent and invoke $\dfrac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
π― [Self-Diagnosis Variant 11] (Circle to Line Distance Bounds)
Find the minimum distance from $(x - 3)^2 + (y - 1)^2 = 5$ to $2x - y + 5 = 0$.
View Variant Solution
Breakdown: Center distance $d = \dfrac{10}{\sqrt{5}} = 2\sqrt{5}$. Minimum distance is $d - r = 2\sqrt{5} - \sqrt{5} = \mathbf{\sqrt{5}}$.
Theme 4. Absolute Value Curves & Intersection Discontinuities (12 ~ 13)
[Problem 12] Single Intersection Condition for a Shifted V-Curve and Line
Function $f(x) = |x - 2| + 1$ is translated horizontally by $k$ to become $g(x)$. Find the range of $k$ such that the line $y = 2x$ intersects $y = g(x)$ at exactly one point.
π Solution & Step-by-Step Breakdown
Answer: $k \ge -\dfrac{3}{2}$
[Solution]
1) $g(x) = |x - (k + 2)| + 1$, with vertex at $(k + 2, 1)$ and branch slopes $\pm 1$.
2) Line $y = 2x$ has slope $2 > 1$, guaranteeing exactly one intersection with the right branch ($x \to \infty$).
3) For exactly one total intersection, the line must not intersect the left branch ($y = -x + k + 3$).
Solving $2x = -x + k + 3$ yields $x = \dfrac{k + 3}{3}$.
4) This potential intersection must not fall within the left domain $x \le k + 2$:
$$\dfrac{k + 3}{3} \ge k + 2 \implies k \le -\dfrac{3}{2}$$
Enforcing unique intersection geometry establishes $\mathbf{k \ge -\dfrac{3}{2}}$.
Track the vertex $(k + 2, 1)$. Comparing the line's slope ($2$) to the piecewise slopes ($\pm 1$) reveals that transitions in root counts occur precisely as the line sweeps through the vertex.
π― [Self-Diagnosis Variant 12] (Shifted Diamond Boundary Intersections)
Find the range of $k$ such that $|x - k| + |y| = 2$ intersects $y = x + 3$ at two distinct points.
View Variant Solution
Breakdown: Compare center $(k, 0)$ distance to the diamond vertices and parallel segment tangents.
[Problem 13] Discontinuity Points in the Intersection Count Function g(t)
Let $g(t)$ denote the number of intersection points between the reflected parabola $f(x) = |x^2 - 4|$ and the horizontal line $y = t$. Find the sum of all values of $t$ at which $g(t)$ is discontinuous.
π Solution & Step-by-Step Breakdown
Answer: $4$
[Solution]
1) Reflecting the negative part of $y = x^2 - 4$ yields $x$-intercepts at $(\pm 2, 0)$ and a local maximum peak at $(0, 4)$.
2) Sweeping $y = t$ upwards:
• $t < 0$: $g(t) = 0$
• $t = 0$: $g(t) = 2$ (tangent to $x$-axis) $\implies$ Discontinuity at $t = 0$
• $0 < t < 4$: $g(t) = 4$
• $t = 4$: $g(t) = 3$ (tangent to local maximum peak) $\implies$ Discontinuity at $t = 4$
• $t > 4$: $g(t) = 2$
3) Discontinuities occur at $t = 0$ and $t = 4$. Sum: $0 + 4 = \mathbf{4}$.
The intersection count function $g(t)$ changes values only at critical geometric thresholds: points of tangency (local extrema) and sharp corners (roots/intercepts).
π― [Self-Diagnosis Variant 13] (Cubic Peak Intersection Threshold)
Find the positive value of $k$ for which $y = |x(x - 3)^2|$ intersects $y = k$ at exactly three points.
View Variant Solution
Breakdown: Local maximum occurs at $x = 1$, giving $y = 1(1 - 3)^2 = 4$. Line $y = \mathbf{4}$ touches the peak.
Theme 5. Swept Areas & Boundary Inclusions (14 ~ 15)
[Problem 14] Parameter Domain for a Translated Circle Fully Contained in a Box
Circle $C: (x - 2)^2 + (y - 2)^2 = 1$ is translated by $(a, b)$ to become $C'$. Find the area of the parameter region $(a, b)$ such that $C'$ is completely contained within the rectangle $0 \le x \le 8$ and $0 \le y \le 6$.
π Solution & Step-by-Step Breakdown
Answer: $24$
[Solution]
1) The radius is $r = 1$. For the circle to remain inside $[0, 8] \times [0, 6]$, its center $(X, Y)$ must stay within the contracted safe box:
$$1 \le X \le 7, \quad 1 \le Y \le 5$$
2) The new center coordinates are $X = 2 + a$ and $Y = 2 + b$.
Substitute into the bounds:
$$1 \le 2 + a \le 7 \implies -1 \le a \le 5 \quad (\text{Width} = 5 - (-1) = 6)$$
$$1 \le 2 + b \le 5 \implies -1 \le b \le 3 \quad (\text{Height} = 3 - (-1) = 4)$$
3) The locus of parameter pairs $(a, b)$ forms a rectangle of dimensions $6 \times 4$:
$$\text{Area} = 6 \times 4 = \mathbf{24}$$
Contract the bounding container inward by radius $r$ to define the "safe margin" for the center. The valid parameter region $(a, b)$ is simply this safe box shifted by the initial center coordinates.
π― [Self-Diagnosis Variant 14] (Circle Inscribed in Equilateral Triangle Path)
A circle of radius $1$ moves freely inside an equilateral triangle of side $12$. Find the perimeter of the region traversed by its center.
View Variant Solution
Breakdown: Inner triangle side is $12 - 2\sqrt{3}$. Perimeter $= 3(12 - 2\sqrt{3}) = \mathbf{36 - 6\sqrt{3}}$.
[Problem 15] Swept Area of a Translating Circular Disk (The Final Boss)
A circular disk of radius $2$ moves along a straight line segment whose center travels from $(0, 0)$ to $(6, 8)$. Find the total area swept by the disk.
π Solution & Step-by-Step Breakdown
Answer: $40 + 4\pi$
[Solution]
1) The swept area decomposes into a central rectangular corridor plus two semicircular caps at the initial and terminal positions.
2) Center travel distance:
$$L = \sqrt{(6 - 0)^2 + (8 - 0)^2} = \sqrt{36 + 64} = 10$$
3) The central corridor has length $L = 10$ and width equal to the circle's diameter $2r = 4$:
$$\text{Corridor Area} = 10 \times 4 = 40$$
4) Combining the two semicircular ends yields one full disk of radius $r = 2$:
$$\text{Cap Area} = \pi (2)^2 = 4\pi$$
5) Total swept area: $\mathbf{40 + 4\pi}$.
The universal formula for the swept area of a moving disk along a straight path is $\text{Path Length} \times 2r + \pi r^2$. For piecewise linear paths, corners blend into circular sectors.
π― [Self-Diagnosis Variant 15] (Swept Area Along a L-Shaped Path)
A disk of radius $1$ moves with its center from $(0, 0)$ to $(4, 0)$ and then to $(4, 3)$. Find the total swept area.
View Variant Solution
Breakdown: Total corridor length $= 4 + 3 = 7$. Rectangles $= 7 \times 2 = 14$. Caps and corner combine to one circle $= \pi(1)^2 = \pi \implies \mathbf{14 + \pi}$.
Across all 15 translation problems, one principle remains unshakeable:
"Do not struggle with the algebraic equation of the entire curve; translate its characteristic anchor points first, then reconstruct the geometric constraints."
With Cavalieri's shear invariance and relative velocity framing in your arsenal, you are thoroughly equipped for our next masterclass: [Transformations: Masterclass ② - Line Reflections & Heron's Shortest Path Optimization].
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