The Ultimate Guide to Shortest Distance in Geometry: Linear Sum vs. Squared Sum (2D to 3D Optimization)

 

📐 Introduction & Overview

Optimization problems involving distances in coordinate geometry and 3D spatial figures are among the most frequently tested concepts in secondary mathematics, IB Math (AA HL/SL), AP Precalculus, and competitive math exams (AMC/AIME).

A common pitfall among students is confusing the sum of direct linear lengths ($AP + PB$) with the sum of squared distances ($AP^2 + PB^2$). The former is fundamentally a geometric path problem solved via reflections and triangle inequalities, while the latter is an algebraic optimization problem governed by Pappus’s Median Theorem and Centroids.

This master guide covers the rigorous geometric principles, essential algebraic proofs, deep-dive theoretical insights, and worked examples across 2D coordinate planes and 3D spatial surfaces.

PART 1. Sum of Segment Lengths ($AP + PB$) — Reflections & Straight-Line Paths

Given two fixed points $A, B$ and a point $P$ moving along a boundary line $l$, the objective is to minimize the total perimeter path $AP + PB$.

🔍 Core Principle & Case Breakdown:

• Opposite Half-Planes: If $A$ and $B$ lie on opposite sides of $l$, the shortest path is simply the straight line segment connecting them: $\min(AP + PB) = \overline{AB}$.
• Same Half-Plane (Reflection Principle): Reflect point $A$ across the line $l$ to obtain $A'$. Because $AP = A'P$ by perpendicular bisector symmetry: $$AP + PB = A'P + PB \ge \overline{A'B}$$ The minimum distance is the direct segment $\overline{A'B}$, achieved where line $A'B$ intersects $l$.
• Optical Insight (Angle of Incidence = Angle of Reflection): At the optimal point $P$, the incoming angle $\alpha$ and outgoing angle $\beta$ with respect to the line $l$ satisfy $\alpha = \beta$. This mirrors Heron’s Shortest Path Principle for light reflection.

💡 Deep Dive: Reflection Across an Arbitrary Slanted Line ($y = mx + n$)

When the reflecting boundary is not an axis but an arbitrary linear function $y = mx + n$, find the reflection $A'(a, b)$ of point $A(x_1, y_1)$ using the Two Fundamental Conditions:

1. Perpendicular Condition (Slope Product $= -1$): $$\left(\frac{b - y_1}{a - x_1}\right) \times m = -1$$
2. Midpoint Condition (Midpoint lies on the line): $$M\left(\frac{x_1 + a}{2}, \; \frac{y_1 + b}{2}\right) \implies \left(\frac{y_1 + b}{2}\right) = m\left(\frac{x_1 + a}{2}\right) + n$$

📝 Example 1: Reflection across a Slanted Boundary Line

Let $A(1, 6)$ and $B(7, 4)$ be two fixed points. Find the minimum value of $AP + PB$ where $P$ is a point on the line $l: y = x - 1$, and find the coordinates of $P$.

[Solution]

1. Step 1: Check Position: $f(x, y) = x - y - 1$. For $A(1, 6)$, $1 - 6 - 1 = -6 < 0$. For $B(7, 4)$, $7 - 4 - 1 = 2 > 0$. Since they have opposite signs, points $A$ and $B$ already lie on opposite sides of the line $l$.
2. Step 2: Calculate Shortest Distance: Since they are on opposite sides, no reflection is required! The minimum path is simply the straight line segment $AB$: $$\min(AP + PB) = \overline{AB} = \sqrt{(7 - 1)^2 + (4 - 6)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$$
3. Step 3: Find Intersection $P$: The equation of line $AB$ passing through $(1, 6)$ and $(7, 4)$ has slope $m = \frac{4-6}{7-1} = -\frac{1}{3}$. $$y - 6 = -\frac{1}{3}(x - 1) \implies x + 3y = 19$$ Solving the system $\begin{cases} x + 3y = 19 \\ x - y = 1 \end{cases}$ yields $4y = 18 \implies y = \frac{9}{2}, \; x = \frac{11}{2}$.

Answer: Minimum distance = $2\sqrt{10}$,   Point $P = \left(\frac{11}{2}, \frac{9}{2}\right)$

PART 2. Sum of Squared Distances ($AP^2 + BP^2$) — Pappus's Theorem & Quadratic Minima

When distances are squared, reflections cannot be applied because the triangle inequality does not hold for squares ($\sqrt{a^2+b^2} \neq a+b$). Instead, we apply Pappus’s Median Theorem (Apollonius' Theorem) or complete the square algebraically.

🔍 Geometric Insight (Pappus’s Theorem): Let $M$ be the midpoint of segment $AB$. For any point $P$: $$AP^2 + BP^2 = 2(AM^2 + PM^2)$$ Since $A$ and $B$ are fixed, $AM$ is a constant. Thus, minimizing $AP^2 + BP^2$ is strictly equivalent to minimizing the distance $PM$.

• If $P$ is unrestricted in the plane $\implies P = M$ (the midpoint).
• If $P$ is constrained to a line $l \implies P$ is the orthogonal projection (foot of perpendicular) dropped from $M$ onto $l$.
• Concentric Circles Interpretation: The level curves $AP^2 + BP^2 = k$ form concentric circles centered at $M$. The smallest $k$ touching line $l$ occurs when the circle is tangent to $l$.

📝 Example 2: Minimizing Sum of Squares via Median Theorem

Given $A(-1, 5)$ and $B(3, 1)$, find the minimum value of $AP^2 + BP^2$ where $P$ moves along the line $l: 2x - y + 4 = 0$.

[Solution]

1. Midpoint $M$: $M\left(\frac{-1 + 3}{2}, \frac{5 + 1}{2}\right) = (1, 3)$.
2. Fixed Length $AM^2$: $AM^2 = (1 - (-1))^2 + (3 - 5)^2 = 4 + 4 = 8$.
3. Minimum Distance $PM_{\min}$ (Perpendicular distance from $M$ to $l$): $$d = \frac{|2(1) - (3) + 4|}{\sqrt{2^2 + (-1)^2}} = \frac{3}{\sqrt{5}} \implies PM^2 = d^2 = \frac{9}{5}$$
4. Apply Pappus's Theorem: $$AP^2 + BP^2 = 2\left(AM^2 + PM^2\right) = 2\left(8 + \frac{9}{5}\right) = 2 \times \frac{49}{5} = \frac{98}{5}$$

Answer: Minimum value = $\frac{98}{5}$

PART 3. Three Segments: Linear Sum vs. Squared Sum

1) Linear Sum ($AP + BP + CP$) — The Fermat-Torricelli Point

For $\triangle ABC$, the point $P$ minimizing $AP + BP + CP$ is the Fermat Point.

• When all angles $< 120^\circ$: Point $P$ satisfies $\angle APB = \angle BPC = \angle CPA = 120^\circ$.
• When one angle $\ge 120^\circ$: Point $P$ collapses to the obtuse vertex itself (e.g., if $\angle A \ge 120^\circ$, minimum is $AB + AC$ at $P = A$).

📐 Rigorous Proof via $60^\circ$ Rotation:
Rotate $\triangle ABP$ by $60^\circ$ around vertex $B$ to form $\triangle C'BP'$. Because $\triangle BPP'$ is equilateral, $BP = P'P$ and $AP = C'P'$. Therefore: $$AP + BP + CP = C'P' + P'P + PC \ge \overline{C'C}$$ The minimum occurs when the four points $C', P', P, C$ lie on a single straight line, which forces $\angle BPC = 120^\circ$ and $\angle APB = 120^\circ$.

2) Sum of Squares ($AP^2 + BP^2 + CP^2$) — The Centroid Theorem

For any triangle $\triangle ABC$, the point $P$ that minimizes $AP^2 + BP^2 + CP^2$ is uniquely the Centroid $G$. $$G = \left(\frac{x_1 + x_2 + x_3}{3}, \; \frac{y_1 + y_2 + y_3}{3}\right)$$ Algebraically, by vector identity: $AP^2 + BP^2 + CP^2 = AG^2 + BG^2 + CG^2 + 3PG^2$, which is minimized when $PG = 0 \implies P = G$.

📝 Example 3: Centroid Minimization for 3 Points

Find the coordinates of $P(x, y)$ that minimizes $AP^2 + BP^2 + CP^2$ for vertices $A(2, 6)$, $B(-4, 1)$, and $C(5, 2)$, and find the minimum value.

[Solution]

1. Centroid Coordinates $G$: $$P = G = \left(\frac{2 + (-4) + 5}{3}, \; \frac{6 + 1 + 2}{3}\right) = (1, 3)$$
2. Sum of Squared Distances to $G$: $$AG^2 = (1 - 2)^2 + (3 - 6)^2 = 1 + 9 = 10$$ $$BG^2 = (1 - (-4))^2 + (3 - 1)^2 = 25 + 4 = 29$$ $$CG^2 = (1 - 5)^2 + (3 - 2)^2 = 16 + 1 = 17$$ $$\text{Minimum Sum} = 10 + 29 + 17 = 56$$

Answer: $P(1, 3)$,   Minimum sum = $56$

PART 4. 3D Surface Shortest Paths — The "Unfolding" Principle

Geodesics on developable surfaces (shapes with Gaussian curvature $K=0$ like prisms, pyramids, and cones) are solved by unfolding the adjacent faces into a single 2D net and connecting the endpoints with a straight line.

1) Rectangular Cuboid Paths ($a \le b \le c$)

Across three possible net unfoldings, the shortest distance between opposite diagonal vertices is achieved by pairing the two smaller dimensions together: $$\text{Shortest Path} = \sqrt{(a + b)^2 + c^2} \quad (\text{where } c \text{ is the longest edge})$$

2) Regular Pyramid: Centroid-to-Centroid Path

For adjacent lateral faces sharing an edge in regular pyramid $O-ABCD$, the distance between centroids $G_1$ and $G_2$ relates to the base midpoints $M_1, M_2$ by a $2:3$ similarity ratio: $$\overline{G_1G_2} = \frac{2}{3} \times \overline{M_1M_2}$$

3) Conical Surfaces: Full Turn vs. Half Turn vs. Frustum

• Full Loop on a Cone: Net sector angle $\theta = 360^\circ \times \frac{r}{l}$. The shortest path starting and returning to base point $A$ is the chord $\overline{AA'} = 2l \sin\left(\frac{\theta}{2}\right)$.
• Half-Turn (Opposite Generatrix): If traveling to the diametrically opposite side, the sector angle is halved ($\frac{\theta}{2}$).
• Cone Frustum (Virtual Apex Reconstruction): For a frustum with radii $R, r$ and slant height $L$, reconstruct the total virtual cone slant height $l_2 = \frac{R}{R-r}L$ to calculate the central angle $\theta = 360^\circ \times \frac{R-r}{L}$.

📝 Example 4: Cone Surface Shortest Path (Half-Loop)

A cone has a base radius of $r = 3$ and a slant height of $l = 12$. An ant starts at point $A$ on the base perimeter and climbs along the lateral surface to point $B$ located on the opposite generatrix at a distance of $8$ from the apex $O$. Find the minimum path length.

[Solution]

1. Total Sector Angle $\theta$: $$\theta = 360^\circ \times \frac{r}{l} = 360^\circ \times \frac{3}{12} = 90^\circ$$
2. Half-Turn Sector Angle: Because $B$ is on the diametrically opposite generatrix, the angle in the unfolded net is: $$\angle AOB = \frac{\theta}{2} = \frac{90^\circ}{2} = 45^\circ$$
3. Law of Cosines on $\triangle AOB$ ($OA = 12, OB = 8, \angle AOB = 45^\circ$): $$AB^2 = 12^2 + 8^2 - 2(12)(8)\cos(45^\circ) = 144 + 64 - 192\left(\frac{\sqrt{2}}{2}\right) = 208 - 96\sqrt{2}$$ $$AB = \sqrt{208 - 96\sqrt{2}} = 4\sqrt{13 - 6\sqrt{2}}$$

Answer: Shortest path = $4\sqrt{13 - 6\sqrt{2}}$

📌 Master Comparison Matrix: Optimization in Geometry

Category	Target Objective	Mathematical Tool	Optimal Condition / Closed Form
2D Linear Sum	$AP + PB$ on line $l$	Reflection & Triangle Inequality	Intersection of $l$ with straight segment $A'B$
2D Squared Sum	$AP^2 + PB^2$ on line $l$	Pappus’s Median Theorem	Foot of perpendicular from midpoint $M$ to $l$
3-Point Linear	$AP + BP + CP$	$60^\circ$ Rotation Construction	Fermat Point ($\angle APB = \angle BPC = \angle CPA = 120^\circ$)
3-Point Squared	$\sum AP_i^2$	Completing the Square	Centroid $G = \frac{1}{3}\sum (x_i, y_i)$
3D Cuboid	Diagonal corner path	2D Net Unfolding	$\sqrt{(a+b)^2 + c^2}$ ($c$: maximum dimension)
3D Conical Net	Lateral loop around cone	Sector Angle + Chord Length	$\theta = 360^\circ(\frac{r}{l}) \implies 2l\sin(\frac{\theta}{2})$

🌿 Yulcho’s Pedagogical Insight

"The beauty of geometric optimization lies in recognizing the exact mathematical structure beneath the problem statement. When dealing with pure distance sums, your instinct must immediately look for straight lines—whether by reflecting across boundaries in 2D or unfolding curved manifolds into flat nets in 3D. Conversely, the moment you see squared distances, shift your perspective entirely from Euclidean distance to quadratic variance, where midpoints, perpendicular projections, and centroids govern the equilibrium. Memorizing answers will fail under slight question variations; mastering the underlying geometric invariance will grant you total clarity on every test."