---
title: "Affine Spaces"
book: "University Mathematics — Year 2"
subject: math
language: en
chapter: 17
exercises: 12
source: https://one-course.com/books/math/4/en/chapter/17-affine-spaces
---

# Chapter 17 — Affine Spaces

Vector spaces have a privileged point — the origin — that geometry does not want. An *[affine space](#def-b2-affine-def)* is a vector space that has forgotten its origin: points and vectors become different species, related by translation. This short chapter builds the dictionary (points, [barycenters](#def-b2-affine-barycenter), [affine](#def-b2-affine-subspace) subspaces and maps), the [affine](#def-b2-affine-subspace) view of convexity, and the classification tools used in the geometry chapters ahead.

## 17.1 Points and vectors

**Definition 17.1.**

An *affine space* directed by a real vector space $E$ is a nonempty set $\mathcal{E}$ with a map $(A, B) \mapsto \vect{AB} \in E$ satisfying

$$
\vect{AB} + \vect{BC} = \vect{AC}
\quad \text{(Chasles)},
\qquad
\text{for each } A,\ B \mapsto \vect{AB} \text{ is a bijection }
\mathcal{E} \to E .
$$

One writes $B = A + u$ for the unique point with $\vect{AB} = u$. The *dimension* of $\mathcal{E}$ is $\dim E$. Every vector space is an [affine](#def-b2-affine-subspace) space over itself ($\vect{AB} = B - A$); every choice of origin $O \in \mathcal{E}$ identifies $\mathcal{E}$ with $E$ via $M \mapsto \vect{OM}$.

**Example 17.2 (An affine space with no natural origin).**

The solution plane $\mathcal E = \{(x, y, z) \in \R^3 : x + y
+ z = 1\}$ is not a vector subspace ($0 \notin \mathcal E$), but it is an [affine space](#def-b2-affine-def) directed by $E = \{x + y + z =
0\}$: for $A, B \in \mathcal E$ the difference $\vect{AB} = B
- A$ lands in $E$ (the sums cancel), Chasles is inherited from $\R^3$, and $B \mapsto \vect{AB}$ is bijective onto $E$. No point of $\mathcal E$ is distinguished — any choice of “origin” $O \in \mathcal E$ works equally well, and all the identifications $M \mapsto \vect{OM}$ differ by translations. This is the typical situation: solution sets of inhomogeneous linear problems (linear systems, linear differential equations in [Chapter 16](https://one-course.com/books/math/4/en/chapter/16-differential-equations#ch-b2-diffeq)) are [affine](#def-b2-affine-subspace), never linear, and the slogan “particular solution plus kernel” is exactly the statement $\mathcal F = A + F$ of the next definition.

**Definition 17.3 (Barycenter).**

Let $(A_i, \lambda_i)_{i \leq k}$ be weighted points with $\sum\lambda_i \neq 0$. The *barycenter* $G
= \operatorname{bar}\bigl((A_i, \lambda_i)\bigr)$ is the unique point with

$$
\sum_i \lambda_i\, \vect{GA_i} = 0
\qquad\text{equivalently}\qquad
\vect{OG} = \frac{1}{\sum\lambda_i}\sum_i \lambda_i\,\vect{OA_i}
\quad (\text{any } O).
$$

Barycenters are *associative* (subgroups of points may be replaced by their partial barycenter with the summed weight) and invariant under rescaling of all weights.

**Proof of existence and the formulas.** Fix $O$ and write $s = \sum_i\lambda_i \neq 0$. By Chasles,

$$
\sum_i\lambda_i\,\vect{GA_i} = 0
\iff \sum_i\lambda_i\bigl(\vect{GO} + \vect{OA_i}\bigr) = 0
\iff s\,\vect{OG} = \sum_i\lambda_i\,\vect{OA_i},
$$

which determines $G = O + \frac1s\sum_i\lambda_i\vect{OA_i}$ uniquely. *Independence of $O$:* for another origin $O'$,

$$
\frac1s\sum_i\lambda_i\,\vect{O'A_i}
= \frac1s\sum_i\lambda_i\bigl(\vect{O'O} + \vect{OA_i}\bigr)
= \vect{O'O} + \frac1s\sum_i\lambda_i\,\vect{OA_i}
= \vect{O'G} :
$$

the same point $G$. *Associativity:* split the index set as $I \sqcup J$ with $s_I = \sum_{i\in I}\lambda_i \neq 0$, and let $G_I$ be the [barycenter](#def-b2-affine-barycenter) of $(A_i, \lambda_i)_{i\in
I}$, so that $\sum_{i\in I}\lambda_i\vect{OA_i} =
s_I\,\vect{OG_I}$. Then

$$
s\,\vect{OG}
= \sum_{i\in I}\lambda_i\vect{OA_i} +
\sum_{j\in J}\lambda_j\vect{OA_j}
= s_I\,\vect{OG_I} + \sum_{j\in J}\lambda_j\vect{OA_j} :
$$

$G$ is the [barycenter](#def-b2-affine-barycenter) of $(G_I, s_I)$ together with $(A_j,
\lambda_j)_{j\in J}$, as claimed. *Rescaling:* replacing each $\lambda_i$ by $t\lambda_i$ ($t \neq 0$) multiplies $s$ and the weighted sum by $t$, leaving $\vect{OG}$ unchanged. ∎

**Remark 17.4 (Common pitfalls).**

Two traps surround the definition. First, if the weights sum to *zero*, there is no [barycenter](#def-b2-affine-barycenter): the map $O \mapsto
\sum\lambda_i\vect{OA_i}$ is then independent of $O$ and defines a *vector*, not a point — for instance $(A, -1;\ B, 1)$ encodes $\vect{AB}$. Keeping track of which of the two objects a computation produces is half of barycentric hygiene. Second, weights are only meaningful up to a common nonzero factor; formulas like “the coordinates of $G$ are $\lambda_1, \dots, \lambda_k$” presuppose a normalization (usually $\sum\lambda_i = 1$), and forgetting to normalize is the standard source of wrong ratios on a figure.

**Definition 17.5 (Affine subspaces; affine maps).**

An *affine subspace* is a set $\mathcal{F} = A + F = \{A + u :
u \in F\}$ with $F$ a vector subspace (its *direction*); equivalently, a nonempty set stable under [barycenters](#def-b2-affine-barycenter). Affine subspaces of $\R^n$ are exactly the solution sets of linear systems $MX = B$ (Year 1: particular solution plus kernel). A map $f \colon
\mathcal{E} \to \mathcal{E}'$ is *affine* when it preserves [barycenters](#def-b2-affine-barycenter) — equivalently when

$$
f(A + u) = f(A) + \varphi(u)
$$

for a (unique) linear map $\varphi = \vec f$, the *linear part*. Affine maps of $\R^n$: $X \mapsto MX + C$. Compositions are affine with composed linear parts; $f$ is bijective iff $\vec f$ is.

**Proof of the equivalence for maps.** If $f(A + u) = f(A) + \varphi(u)$: for a [barycenter](#def-b2-affine-barycenter) $G$ of $(A_i,
\lambda_i)$, expanding every point from $A$, $f(G) = f(A) +
\varphi(\vect{AG})$ and $\varphi(\vect{AG}) =
\frac{\sum\lambda_i\varphi(\vect{AA_i})}{\sum\lambda_i}$: $f(G)$ is the [barycenter](#def-b2-affine-barycenter) of the images. Conversely, fix $A$ and define $\varphi(u) = \vect{f(A)\,f(A + u)}$. *Homogeneity:* $A + tu =
\operatorname{bar}\bigl(A, 1-t;\ A + u, t\bigr)$ for every real $t$, so preservation of [barycenters](#def-b2-affine-barycenter) (with arbitrary real weights, as hypothesized) gives $\varphi(tu) = t\,\varphi(u)$ directly. *Additivity:* $A + u + v = \operatorname{bar}\bigl(A + 2u,
\tfrac12;\ A + 2v, \tfrac12\bigr)$, so $\varphi(u + v) =
\tfrac12\varphi(2u) + \tfrac12\varphi(2v) = \varphi(u) +
\varphi(v)$, using homogeneity. Hence $\varphi$ is linear. ∎

**Remark 17.6.**

The proof used [barycenters](#def-b2-affine-barycenter) with *arbitrary real* weights: the homogeneity step takes $t$ outside $\intcc01$. If a map is only assumed to preserve [barycenters](#def-b2-affine-barycenter) with nonnegative weights — equivalently, midpoints and segments — linearity of the vector map no longer comes for free: one only gets $\Q$-linearity, and a [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) hypothesis is needed to conclude, exactly as in [Exercise 17.5](#exo-b2-affine-5). Distinguishing “preserves all [barycenters](#def-b2-affine-barycenter)” from “preserves [convex](#def-b2-affine-convex) combinations” is a small but real subtlety of the [affine](#def-b2-affine-subspace) vocabulary.

**Example 17.7 (Classical barycenter geometry).**

The centroid of a triangle $ABC$ is the [barycenter](#def-b2-affine-barycenter) $G =
\operatorname{bar}(A,1; B,1; C,1)$. Associativity with the midpoint $A' = \operatorname{bar}(B, 1; C, 1)$ shows

$$
G = \operatorname{bar}(A, 1;\ A', 2) :
$$

$G$ lies on the median $AA'$ at two-thirds of it — and likewise for the other two medians: the three medians are concurrent, in one line of barycentric calculus.

**Example 17.8 (The bimedians of a quadrilateral).**

Let $ABCD$ be any quadrilateral (planar or not!) and consider its *bimedians*: the segments joining the midpoints of opposite sides, $M_{AB}M_{CD}$ and $M_{BC}M_{DA}$. Introduce the [barycenter](#def-b2-affine-barycenter) $G$ of $(A,1; B,1; C,1; D,1)$ and group the weights two ways:

$$
G = \operatorname{bar}\bigl(M_{AB}, 2;\ M_{CD}, 2\bigr)
= \operatorname{bar}\bigl(M_{BC}, 2;\ M_{DA}, 2\bigr) :
$$

$G$ is the midpoint of *both* bimedians — so the two bimedians always bisect each other, and the quadrilateral of the four midpoints is a parallelogram (its diagonals are the bimedians). No case analysis, no coordinates, and the argument survives unchanged for a skew quadrilateral in $\R^3$, where a picture-based proof would already be delicate: associativity does not care about dimension.

**Example 17.9 (Classifying an affine map, start to finish).**

Let $f(x, y) = (2x - 1,\ 3y - 4)$ on $\R^2$. Its linear part is $\varphi = \operatorname{diag}(2, 3)$, whose [spectrum](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\{2, 3\}$ avoids $1$: by the fixed-point criterion proved below ([Proposition 17.17](#prop-b2-affine-fixedpoint)), $f$ has exactly one fixed point, found by solving

$$
x = 2x - 1, \qquad y = 3y - 4
\qquad\Longrightarrow\qquad \Omega = (1, 2).
$$

Recentering at $\Omega$ (set $x = 1 + u$, $y = 2 + v$):

$$
f(1 + u,\ 2 + v) = (1 + 2u,\ 2 + 3v) :
$$

in the frame at $\Omega$, $f$ *is* its linear part, an anisotropic dilation stretching by $2$ horizontally and $3$ vertically from the center $(1, 2)$. The general lesson: an [affine map](#def-b2-affine-subspace) is “linear map plus location data”, and the location data collapses to one well-chosen origin whenever $1$ is not an [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen). Conversely, translating the origin badly *creates* the constant terms: [affine](#def-b2-affine-subspace) geometry is the art of choosing where to put $0$.

**Remark 17.10 (Method: concurrency and alignment by barycenters).**

[Example 17.7](#ex-b2-affine-median) is an instance of a general recipe. To prove that three cevians of a triangle are concurrent, exhibit a *single* weighted system $(A,
\alpha; B, \beta; C, \gamma)$ and use associativity three ways: grouping $(B, C)$ shows the [barycenter](#def-b2-affine-barycenter) lies on the cevian from $A$, grouping $(C, A)$ on the cevian from $B$, grouping $(A, B)$ on the third. For the medians, the system $(A, 1; B, 1; C, 1)$ does all the work; for cevians cutting the sides in prescribed ratios, the weights are read off the ratios. To prove three points *aligned*, write one as a [barycenter](#def-b2-affine-barycenter) of the other two ([Exercise 17.2](#exo-b2-affine-2)), or use the [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) criterion of [Exercise 17.11](#exo-b2-affine-11). Both recipes replace geometric ingenuity by weight bookkeeping — this is precisely what barycentric calculus is for.

## 17.2 Convexity, affinely

**Definition 17.11.**

A subset $C$ of an [affine space](#def-b2-affine-def) is *convex* when it contains every [barycenter](#def-b2-affine-barycenter) with *nonnegative* weights of its points — equivalently, every segment $\intcc{A}{B} = \{\operatorname{bar}(A,
1-t; B, t) : t \in \intcc{0}{1}\}$ between its points. The *convex hull* $\operatorname{conv}(S)$ is the set of all nonnegative-weight [barycenters](#def-b2-affine-barycenter) of points of $S$ — the smallest convex set containing $S$.

**Example 17.12 (Epigraphs are convex sets).**

The region $C = \{(x, y) : y \geq x^2\}$ above the parabola is [convex](#def-b2-affine-convex): for $(x_1, y_1), (x_2, y_2) \in C$ and $t \in
\intcc01$, the convexity inequality of the square function gives

$$
\bigl((1-t)x_1 + tx_2\bigr)^2 \leq (1-t)x_1^2 + tx_2^2
\leq (1-t)y_1 + ty_2 ,
$$

so the [barycenter](#def-b2-affine-barycenter) stays above the parabola. The computation is general: $\{y \geq f(x)\}$ is [convex](#def-b2-affine-convex) exactly when $f$ is a convex function — [convex](#def-b2-affine-convex) *sets* and [convex](#def-b2-affine-convex) *functions* ([Chapter 8](https://one-course.com/books/math/4/en/chapter/8-functions-of-a-real-variable#ch-b2-realfun)) are two faces of one notion, epigraphs being the dictionary. This is the geometric reason support lines exist for convex functions, the fact that will prove Jensen’s inequality in [Chapter 22](https://one-course.com/books/math/4/en/chapter/22-discrete-random-variables#ch-b2-randomvar).

**Example 17.13 (Redundant generators of a convex hull).**

Let $S = \{(0,0), (2,0), (2,2), (0,2), (1,1)\}$. The fifth point is the [barycenter](#def-b2-affine-barycenter)

$$
(1,1) = \operatorname{bar}\bigl((0,0), \tfrac12;\ (2,2),
\tfrac12\bigr),
$$

so it already lies in the hull of the other four: $\operatorname{conv}(S)$ is the square with the four corners as vertices. In general, a point of $S$ that is a nonnegative-weight [barycenter](#def-b2-affine-barycenter) of the *other* points of $S$ can be deleted without changing the hull; the points that can never be deleted (here the four corners) are the *extreme points* of the hull. Determining them is a pure [barycenter](#def-b2-affine-barycenter) computation: $(2,0)$, say, cannot be written as $\operatorname{bar}$ of the remaining points with nonnegative weights, because the first coordinate would force all weight onto points with $x = 2$, and the second coordinate then fails. Convexity questions reduce, again and again, to solving small weighted systems.

**Theorem 17.14 (Carathéodory).**

In an [affine space](#def-b2-affine-def) of dimension $n$, every point of $\operatorname{conv}(S)$ is a [barycenter](#def-b2-affine-barycenter) of at most $n + 1$ points of $S$.

**Proof.** Let $G = \operatorname{bar}(A_0, \lambda_0; \dots; A_k, \lambda_k)$ with $\lambda_i > 0$, $\sum\lambda_i = 1$, and $k + 1 > n + 1$ points. The $k$ vectors $\vect{A_0A_i}$ ($i \geq 1$) are linked ($k > n$): $\sum_{i\geq1}\mu_i \vect{A_0A_i} = 0$ nontrivially; setting $\mu_0 = -\sum_{i\geq1}\mu_i$, we get weights $(\mu_i)$ with $\sum\mu_i = 0$, $\sum \mu_i\,\vect{OA_i} = 0$ (any $O$), not all zero. Then for every real $t$ the weights $\lambda_i - t\mu_i$ still sum to $1$ and, since $\sum_i\mu_i\vect{OA_i} = 0$,

$$
\sum_i(\lambda_i - t\mu_i)\,\vect{OA_i}
= \sum_i\lambda_i\,\vect{OA_i} :
$$

they produce the *same* point $G$. Now slide $t$ from $0$: some $\mu_i$ is positive (they sum to zero and are not all zero), so

$$
t^* = \min\Bigl\{\frac{\lambda_i}{\mu_i} : \mu_i > 0\Bigr\}
$$

is well defined and positive. At $t = t^*$: for indices with $\mu_i > 0$, $\lambda_i - t^*\mu_i \geq 0$ by minimality, with equality at a minimizing index; for indices with $\mu_i
\leq 0$, $\lambda_i - t^*\mu_i \geq \lambda_i > 0$. All weights remain nonnegative and at least one has died: $G$ is rewritten as a [barycenter](#def-b2-affine-barycenter) of fewer points. Iterate while more than $n + 1$ points remain. ∎

**Example 17.15.**

In the plane ($n = 2$): every point of the [convex hull](#def-b2-affine-convex) of a finite set lies in a triangle with vertices in the set — the geometric content of Carathéodory, used in optimization and probability (mixtures) alike.

**Example 17.16 (Running Carathéodory’s algorithm).**

Write the center of the square of [Example 17.13](#ex-b2-affine-squarehull) with its four corners $A_1 =
(0,0)$, $A_2 = (2,0)$, $A_3 = (2,2)$, $A_4 = (0,2)$:

$$
(1,1) = \operatorname{bar}\bigl(A_1, \tfrac14;\ A_2,
\tfrac14;\ A_3, \tfrac14;\ A_4, \tfrac14\bigr),
$$

four points in dimension $2$ — one too many. The proof’s recipe asks for weights $(\mu_i)$ with $\sum\mu_i = 0$ and $\sum\mu_i\vect{OA_i} = 0$: here $\mu = (1, -1, 1, -1)$ works (the two diagonals share their midpoint). Sliding $\lambda_i
\mapsto \lambda_i - t\mu_i$ keeps the [barycenter](#def-b2-affine-barycenter) fixed for every $t$; the extremal admissible value $t = \frac14$ makes the weights $(0, \tfrac12, 0, \tfrac12)$, killing $A_1$ and $A_3$ simultaneously:

$$
(1,1) = \operatorname{bar}\bigl(A_2, \tfrac12;\ A_4,
\tfrac12\bigr),
$$

a representation by two points — even better than the three that the theorem guarantees, because the center happens to lie on a segment between generators. The algorithm is entirely mechanical: find a dependence, slide until a weight dies, repeat.

## 17.3 Affine classification tools

**Proposition 17.17 (Fixed points of affine maps).**

Let $f$ be an [affine](#def-b2-affine-subspace) endomorphism of a finite-dimensional [affine space](#def-b2-affine-def) with linear part $\varphi$. If $1 \notin
\operatorname{Sp}(\varphi)$, then $f$ has exactly one fixed point $\Omega$, and in the vectorialization at $\Omega$, $f$ *is* its linear part. (Translations, with $\varphi = \mathrm{id}$ and no fixed point, are the basic obstruction.)

**Proof.** Fix $O$ and write $f(O + x) = f(O) + \varphi(x)$. The point $O + x$ is fixed iff $O + x = f(O) + \varphi(x)$, i.e.

$$
(\mathrm{id} - \varphi)(x) = \vect{O f(O)} .
$$

In finite dimension, $\mathrm{id} - \varphi$ is invertible iff $0$ is not an [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of $\mathrm{id} - \varphi$, iff $1 \notin \operatorname{Sp}\varphi$ — and in that case the displayed equation has exactly one solution $x^*$, giving the unique fixed point $\Omega = O + x^*$. Recentering: for any vector $u$,

$$
f(\Omega + u) = f(\Omega) + \varphi(u) = \Omega + \varphi(u),
$$

so in the frame with origin $\Omega$ the map reads $u \mapsto
\varphi(u)$: purely linear. When $1 \in
\operatorname{Sp}\varphi$, either no fixed point exists (the displayed equation may be unsolvable, as for a translation) or a whole [affine](#def-b2-affine-subspace) subspace of them does (add any [eigenvector](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $1$ to a solution): uniqueness is exactly the spectral condition. ∎

**Example 17.18 (Plane isometries, completed).**

An [affine](#def-b2-affine-subspace) isometry of the Euclidean plane has linear part in $O(2)$: a rotation $R_\theta$ or a reflection (Year 1 volume). If $\theta \neq 0$: $1 \notin \operatorname{Sp} R_\theta$, so the map is a *rotation about a unique center* ([Proposition 17.17](#prop-b2-affine-fixedpoint)). If the linear part is a reflection: either a reflection in an axis (fixed points exist) or a *glide reflection* (reflection composed with a translation along the axis, no fixed point). With translations, this is the [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) classification of plane isometries.

**Remark 17.19 (The plane isometries, at a glance).**

Collecting the cases: identity; translations ($\vec f =
\mathrm{id}$, no fixed point unless trivial); rotations (linear part $R_\theta$, $\theta \neq 0$: one center); reflections (linear part a reflection, a line of fixed points); glide reflections (same linear part, no fixed point). Four families plus the identity, each recognized by two data only: the linear part and the fixed-point set — the pattern of [Proposition 17.17](#prop-b2-affine-fixedpoint) made exhaustive.

**Example 17.20 (A glide reflection, caught in the act).**

Let $f(x, y) = (y + 1,\ x + 1)$. The linear part $(x, y)
\mapsto (y, x)$ is the reflection in the diagonal $y = x$, so $1 \in \operatorname{Sp}\vec f$ and [Proposition 17.17](#prop-b2-affine-fixedpoint) is silent. Fixed points would need $x = y + 1$ and $y = x + 1$ simultaneously: impossible — none exist, so $f$ is not a reflection. Squaring settles the classification:

$$
f\bigl(f(x, y)\bigr) = f(y + 1,\ x + 1) = (x + 2,\ y + 2),
$$

the translation by $(2, 2)$: $f$ is the *glide reflection* with axis the line $y = x$ (shifted appropriately: the midpoint of $M$ and $f(M)$ always lies on $y = x + {}$constant, here $y = x$, as one checks on $M = (0,
0) \mapsto (1,1)$) and glide vector $(1, 1)$, half of $f
\circ f$. Compare with [Exercise 17.6](#exo-b2-affine-6), where the same linear part but a different constant produced an honest reflection: with [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $1$ present, the constant term decides everything.

**Example 17.21 (Affine recursions are affine dynamics).**

The classical recursion $u_{n+1} = au_n + b$ ($a \neq 1$) iterates the [affine map](#def-b2-affine-subspace) $f(x) = ax + b$ of the line, whose linear part $a$ avoids the [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $1$: there is a unique fixed point $\omega = \frac{b}{1-a}$, and recentering there (the one-dimensional case of the proposition above) turns $f$ into multiplication by $a$:

$$
u_{n+1} - \omega = a\,(u_n - \omega)
\qquad\Longrightarrow\qquad
u_n = \omega + a^n(u_0 - \omega) .
$$

For $u_{n+1} = \frac{u_n}2 + 3$: $\omega = 6$ and $u_n = 6 +
(u_0 - 6)2^{-n} \to 6$. The recipe taught for such recursions in [Chapter 7](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#ch-b2-series) — “subtract the fixed point” — is exactly the vectorialization of an [affine map](#def-b2-affine-subspace) at its fixed point; convergence for $\abs a < 1$ is the contraction phenomenon that [Chapter 4](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#ch-b2-metric) turned into the Banach fixed-point theorem. One idea, three chapters.

**Example 17.22 (Finding the center of a rotation).**

Let $f(x, y) = (-y + 2,\ x)$. The linear part is $\varphi(x, y) = (-y, x)$: the rotation of angle $\frac\pi2$, whose [spectrum](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\{\iu, -\iu\}$ avoids $1$. By [Proposition 17.17](#prop-b2-affine-fixedpoint) there is exactly one fixed point: $x = -y + 2$ and $y = x$ give $x = 1$, $y = 1$, so $\Omega = (1, 1)$, and $f$ *is* the rotation of center $(1, 1)$ and angle $\frac\pi2$. The general lesson: when $1
\notin \operatorname{Sp}\vec f$, classifying $f$ costs one linear system — the geometry is entirely in the linear part, the arithmetic entirely in locating the center.

**Remark 17.23 (Where affine language is used next).**

[Barycenters](#def-b2-affine-barycenter) and [affine maps](#def-b2-affine-subspace) are the grammar of the geometry chapters ahead: tangent lines and planes are [affine](#def-b2-affine-subspace) objects (Chapters [18](https://one-course.com/books/math/4/en/chapter/18-curves#ch-b2-curves) and [19](https://one-course.com/books/math/4/en/chapter/19-surfaces#ch-b2-surfaces)), an [affine](#def-b2-affine-subspace) change of variables multiplies areas and volumes by $\abs{\det \vec f}$ ([Chapter 20](https://one-course.com/books/math/4/en/chapter/20-line-integrals-and-multiple-integrals#ch-b2-multint)), and expectation is a [barycenter](#def-b2-affine-barycenter) with weights given by a probability law, which is why convexity governs Jensen’s inequality ([Chapter 22](https://one-course.com/books/math/4/en/chapter/22-discrete-random-variables#ch-b2-randomvar)). In the Year 3 volume the same convexity vocabulary carries the study of $L^p$ [norms](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm) and of integral inequalities.

**Remark 17.24 (Perspectives within this volume).**

Two threads leave this chapter. The *[affine](#def-b2-affine-subspace)* thread: tangent lines ([Chapter 18](https://one-course.com/books/math/4/en/chapter/18-curves#ch-b2-curves)) and tangent planes ([Chapter 19](https://one-course.com/books/math/4/en/chapter/19-surfaces#ch-b2-surfaces)) are [affine](#def-b2-affine-subspace) subspaces attached to nonlinear objects, and the classification of quadrics in the surfaces chapter runs on this chapter’s center equation $A\Omega = -b$. The *[convex](#def-b2-affine-convex)* thread is longer: convexity of half-planes and disks powers the Helly theory of the weekend problem; convexity of functions gives Jensen’s inequality ([Chapter 22](https://one-course.com/books/math/4/en/chapter/22-discrete-random-variables#ch-b2-randomvar)); and the final theorem of the book — the extinction criterion for branching processes ([Chapter 23](https://one-course.com/books/math/4/en/chapter/23-probability-generating-functions#ch-b2-genfun)) — is decided by the position of a convex curve relative to the diagonal, a picture that belongs to this chapter as much as to probability. [Barycenters](#def-b2-affine-barycenter) return there too: an expectation is a [barycenter](#def-b2-affine-barycenter) with probability weights.

## 17.4 Exercises

**Exercise 17.1 ★.**

In $\R^3$, are the following [affine](#def-b2-affine-subspace) subspaces? Give directions and dimensions. $\{x + y + z = 1\}$; $\;\{x + y + z = 1,\ x - z =
3\}$; $\;\{x^2 + y^2 = 1\}$; the solution set of $MX = B$ for a given compatible system.

**Solution of Exercise 17.1.**

$\{x + y + z = 1\}$: [affine](#def-b2-affine-subspace) plane, direction the vector plane $\{x
+ y + z = 0\}$, dimension $2$. Adding $x - z = 3$: an [affine](#def-b2-affine-subspace) line (two independent equations), direction $\{x + y + z = 0,\ x = z\}
= \operatorname{Vect}\bigl((1, -2, 1)\bigr)$, dimension $1$. $\{x^2
+ y^2 = 1\}$: a cylinder — not stable under [barycenters](#def-b2-affine-barycenter) (the midpoint of $(1,0,0)$ and $(-1,0,0)$ is the origin, off the cylinder): not [affine](#def-b2-affine-subspace). A compatible system $MX = B$: [affine](#def-b2-affine-subspace) subspace $X_0 + \ker M$ of dimension $\dim\ker M$, as recalled in [Definition 17.5](#def-b2-affine-subspace).

**Exercise 17.2 ★.**

Prove that three distinct points $A, B, C$ of an [affine space](#def-b2-affine-def) are aligned iff $C$ is a [barycenter](#def-b2-affine-barycenter) of $A$ and $B$, iff the vectors $\vect{AB}, \vect{AC}$ are linked. Deduce Menelaus-style weight bookkeeping: if $C = \operatorname{bar}(A, 1 - t; B, t)$, locate $C$ for $t = \frac12$, $t = 2$, $t = -1$.

**Solution of Exercise 17.2.**

$C = \operatorname{bar}(A, 1-t; B, t)$ means $\vect{AC} =
t\,\vect{AB}$: existence of such $t$ is exactly linkage of $\vect{AC}$ with $\vect{AB} \neq 0$, i.e. alignment. Positions: $t
= \frac12$: midpoint; $t = 2$: beyond $B$, at $B$’s distance from it ($\vect{AC} = 2\vect{AB}$); $t = -1$: the reflection of $B$ through $A$.

**Exercise 17.3 ★.**

Let $f$ be the [affine map](#def-b2-affine-subspace) of $\R^2$ given by $f(X) = MX + C$ with $M = \frac12\begin{pmatrix} 1 & 1\\ 1 & 1\end{pmatrix}$ and $C =
(1, 0)^{\mathsf T}$. Determine the image of $f$, its fixed points (if any), and $f \circ f$.

**Solution of Exercise 17.3.**

$M$ is the projection matrix onto $\operatorname{Vect}(1,1)$ along $(1,-1)$ (check $M^2 = M$). Image of $f$: $\{MX + C\} = C +
\operatorname{im} M$: the [affine](#def-b2-affine-subspace) line through $(1,0)$ directed by $(1,1)$. Fixed points: $X = MX + C$, i.e. $(I - M)X = C$; but $C =
(1, 0)^{\mathsf T}$ and $\operatorname{im}(I - M) =
\operatorname{Vect}(1,-1)$; is $(1,0)$ in it? $(1, 0) =
\alpha(1,-1)$ forces $\alpha = 1$ and $0 = -1$: no. No fixed points. And

$$
f(f(X)) = M(MX + C) + C = MX + MC + C = f(X) + MC,
\qquad MC = \tfrac12(1,1)^{\mathsf T} :
$$

$f\circ f$ is $f$ followed by a translation along the image line — $f$ is a “glide projection”: projection onto the line composed with a slide.

**Exercise 17.4 ★★.**

(Associativity in action) In a triangle $ABC$, let $I, J, K$ divide $BC$, $CA$, $AB$ in ratios $\vect{BI} = \frac13\vect{BC}$, $\vect{CJ} = \frac13\vect{CA}$, $\vect{AK} = \frac13\vect{AB}$. Express $I, J, K$ as [barycenters](#def-b2-affine-barycenter) and compute the [barycenter](#def-b2-affine-barycenter) of $(I,1;J,1;K,1)$: what do you find, and why was it predictable?

**Solution of Exercise 17.4.**

$I = \operatorname{bar}(B, 2; C, 1)$ (since $\vect{BI} =
\frac13\vect{BC}$ places $I$ closer to $B$: weights $2$ on $B$, $1$ on $C$ — check: $\vect{BI} = \frac{1}{3}\vect{BC}$). Similarly $J = \operatorname{bar}(C, 2; A, 1)$, $K =
\operatorname{bar}(A, 2; B, 1)$. Summing the three weighted systems, the [barycenter](#def-b2-affine-barycenter) of $(I, 1; J, 1; K, 1)$ (each of total weight $3$, so replace $I$ by its system, etc.) is

$$
\operatorname{bar}\bigl(A, 1 + 2;\ B, 2 + 1;\ C, 1 + 2\bigr)
= \operatorname{bar}(A, 1; B, 1; C, 1) = G ,
$$

the centroid of $ABC$: the triangle $IJK$ has the same centroid — predictable, because the construction treats $A, B, C$ cyclically and the centroid is the unique fixed point of the [cyclic](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#def-b2-structures-generated) symmetry of weights.

**Exercise 17.5 ★★.**

Prove that a map $f \colon \R^n \to \R^n$ preserving *midpoints* ($f\bigl(\frac{A+B}{2}\bigr) = \frac{f(A) + f(B)}{2}$) and *[continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)* is [affine](#def-b2-affine-subspace). *(Show the vector map $u \mapsto
f(O + u) - f(O)$ is additive via midpoints, then $\Q$-homogeneous, then $\R$-homogeneous by [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) — the same density strategy used for Cauchy’s functional equation in the Year 1 volume; re-derive the needed steps here.)*

**Solution of Exercise 17.5.**

Set $g(u) = f(O + u) - f(O)$ (working in $\R^n$ vectorialized at $O$), $g(0) = 0$.

*Additivity:* $\frac{(O + u) + (O + v)}{2} = O + \frac{u +
v}{2}$, so midpoint preservation gives $g\bigl(\frac{u+v}{2}\bigr)
= \frac{g(u) + g(v)}{2}$; with $v = 0$: $g(u/2) = g(u)/2$; combining, $g(u + v) = 2g\bigl(\frac{u+v}{2}\bigr) = g(u) + g(v)$.

*$\Q$-homogeneity:* additivity gives $g(nu) = ng(u)$ ($n \in
\N$, induction), then $g(-u) = -g(u)$ (add), then $g(\frac pq u) =
\frac pq g(u)$ (apply $q$, use injectivity of scaling).

*$\R$-homogeneity:* for $t \in \R$, take rationals $t_n \to
t$: $g(t_nu) = t_ng(u)$, and [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) of $g$ (inherited from $f$) passes to the limit: $g(tu) = tg(u)$. Hence $g$ is linear and $f =
f(O) + g$: [affine](#def-b2-affine-subspace).

**Exercise 17.6 ★★.**

Classify the [affine map](#def-b2-affine-subspace) $f(x, y) = (y + 1,\; x - 1)$ of the Euclidean plane: linear part, fixed points, geometric nature (reflection? glide?). Compute $f \circ f$ and conclude.

**Solution of Exercise 17.6.**

Linear part $\varphi(x,y) = (y, x)$: the reflection in the diagonal $y = x$ (orthogonal, [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) $-1$). Fixed points: $(x, y) = (y
+ 1, x - 1)$ amounts to the single equation $y = x - 1$ (the two components are equivalent): every point of the line $y = x - 1$ is fixed. So $f$ fixes that line [pointwise](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def): $f$ is the *reflection* in that axis (an isometry with a line of fixed points and linear part a reflection). Consistently, $f \circ
f(x,y) = f(y+1, x-1) = (x - 1 + 1, y + 1 - 1) = (x, y)$: an involution, as a reflection must be.

**Exercise 17.7 ★★★.**

(Radon) Let $A_1, \dots, A_{n+2}$ be points of an [affine space](#def-b2-affine-def) of dimension $n$. Prove that they can be split into two disjoint groups whose [convex hulls](#def-b2-affine-convex) intersect. *(As in Carathéodory’s proof, find weights $\mu_i$, not all zero, with $\sum\mu_i = 0$ and $\sum\mu_i\vect{OA_i} = 0$; separate positive and negative weights and normalize both sides.)*

**Solution of Exercise 17.7.**

The $n + 1$ vectors $\vect{A_1A_i}$ ($i \geq 2$) are linked in dimension $n$: there are $\mu_i$, not all zero, with $\sum_{i\geq2}
\mu_i\vect{A_1A_i} = 0$; set $\mu_1 = -\sum_{i \geq 2}\mu_i$, so $\sum_{i}\mu_i = 0$ and $\sum_i \mu_i\,\vect{OA_i} = 0$ for every $O$, with not all $\mu_i$ zero. Split indices: $P = \{i : \mu_i >
0\}$, $N = \{i : \mu_i < 0\}$, both nonempty (the $\mu_i$ sum to zero and are not all zero). With $s = \sum_{i\in P}\mu_i =
-\sum_{i \in N}\mu_i > 0$:

$$
\operatorname{bar}\bigl(A_i, \tfrac{\mu_i}{s}\bigr)_{i \in P}
= \operatorname{bar}\bigl(A_i, \tfrac{-\mu_i}{s}\bigr)_{i \in N},
$$

(both sides equal the point $X$ with $\vect{OX} = \frac1s\sum_{i\in
P}\mu_i\vect{OA_i}$, by the relation): a common point of the two [convex hulls](#def-b2-affine-convex), with disjoint index groups.

**Exercise 17.8 ★★★.**

Let $f$ be an [affine](#def-b2-affine-subspace) endomorphism of $\R^n$ with $f \circ f = f$. Prove that $f$ is the [affine](#def-b2-affine-subspace) projection onto the [affine](#def-b2-affine-subspace) subspace $\operatorname{Fix}(f) = \operatorname{im} f$ along the direction $\ker\vec f$, and that conversely all such projections are idempotent. *(Show first that $\operatorname{im} f$ consists of fixed points.)*

**Solution of Exercise 17.8.**

*Image = fixed points:* for $Y = f(X)$, $f(Y) = f(f(X)) =
f(X) = Y$: every image point is fixed; conversely fixed points are images. So $\mathcal{F} = \operatorname{im} f =
\operatorname{Fix}(f)$ is nonempty, and it is an [affine](#def-b2-affine-subspace) subspace (image of an [affine map](#def-b2-affine-subspace)), with direction $\operatorname{im}\vec f$.

*Projection structure:* $\vec f$ is idempotent ($\vec{f\circ
f} = \vec f^{\,2} = \vec f$), so $E = \operatorname{im}\vec f
\oplus \ker \vec f$ ([Example 3.18](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#ex-b2-reduction-projections)). For any point $X$, consider the vector $\vect{f(X)\,X}$; applying $\vec f$:

$$
\vec f\bigl(\vect{f(X)\,X}\bigr) = \vect{f(f(X))\,f(X)} = 0
\qquad (f \circ f = f),
$$

so $\vect{f(X)\,X} \in \ker\vec f$. Hence $X = f(X) +
\vect{f(X)X}$ displays $X$ as a point of $\mathcal{F}$ translated by a vector of $\ker\vec f$: $f$ is exactly the projection onto $\mathcal{F}$ along $\ker\vec f$. Conversely such projections clearly satisfy $f \circ f = f$.

**Exercise 17.9 ★.**

Let $G = \operatorname{bar}(A, 1;\ B, 2;\ C, 3)$ in a triangle $ABC$. Using associativity, show that the line $AG$ meets $BC$ at $M = \operatorname{bar}(B, 2;\ C, 3)$, and locate $G$ on the segment $\intcc AM$; locate likewise the intersection of $BG$ with $CA$.

**Solution of Exercise 17.9.**

Let $M = \operatorname{bar}(B, 2;\ C, 3)$, of total weight $5$. Associativity gives $G = \operatorname{bar}(A, 1;\ M, 5)$, so $\vect{AG} = \frac56\,\vect{AM}$: $G$ lies on the segment $\intcc AM$ at five-sixths of it from $A$. Since $A \notin
(BC)$, the line $(AG) = (AM)$ meets $(BC)$ at the single point $M$, with $\vect{BM} = \frac35\,\vect{BC}$. Likewise, with $N =
\operatorname{bar}(C, 3;\ A, 1)$ (total weight $4$, $\vect{CN}
= \frac14\,\vect{CA}$), associativity gives $G =
\operatorname{bar}(B, 2;\ N, 4)$: the line $(BG)$ meets $(CA)$ at $N$, and $\vect{BG} = \frac46\,\vect{BN} =
\frac23\,\vect{BN}$.

**Exercise 17.10 ★★.**

For $\lambda \neq 0$, the *homothety* $h_{\Omega,
\lambda}$ is the [affine map](#def-b2-affine-subspace) fixing $\Omega$ with linear part $\lambda\,\mathrm{id}$. Prove that the composition $h_{\Omega', \mu} \circ h_{\Omega, \lambda}$ is a homothety of ratio $\lambda\mu$ when $\lambda\mu \neq 1$, and a translation when $\lambda\mu = 1$; in the case $\lambda = \mu = -1$ (two point reflections), compute the translation vector.

**Solution of Exercise 17.10.**

Vectorialize at an origin $O$ and write points as vectors: $h_{\Omega, \lambda}(x) = \omega + \lambda(x - \omega)$ with $\omega = \vect{O\Omega}$. The composition $g = h_{\Omega',
\mu} \circ h_{\Omega, \lambda}$ is [affine](#def-b2-affine-subspace) with linear part $\mu\lambda\,\mathrm{id}$. If $\lambda\mu \neq 1$: $1 \notin
\operatorname{Sp}(\lambda\mu\,\mathrm{id})$, so [Proposition 17.17](#prop-b2-affine-fixedpoint) yields a unique fixed point $\Omega''$ and, vectorialized there, $g =
\lambda\mu\,\mathrm{id}$: the homothety $h_{\Omega'',
\lambda\mu}$. If $\lambda\mu = 1$ the linear part is the identity, so $g$ is a translation; expanding,

$$
g(x) = \omega' + \mu\bigl(\omega + \lambda(x - \omega) -
\omega'\bigr) = x + (1 - \mu)\,\omega' + \mu(1 -
\lambda)\,\omega .
$$

For $\lambda = \mu = -1$ (point reflections) the vector is $2\omega' - 2\omega = 2\,\vect{\Omega\Omega'}$: the composition of the point reflections in $\Omega$ then $\Omega'$ is the translation by $2\,\vect{\Omega\Omega'}$.

**Exercise 17.11 ★★.**

(Menelaus) In a triangle $ABC$, let $A' \in (BC)$, $B' \in
(CA)$, $C' \in (AB)$, all distinct from the vertices, and define $\alpha, \beta, \gamma$ by $\vect{A'B} =
\alpha\,\vect{A'C}$, $\vect{B'C} = \beta\,\vect{B'A}$, $\vect{C'A} = \gamma\,\vect{C'B}$. Prove that $A', B', C'$ are aligned if and only if $\alpha\beta\gamma = 1$. *(Write each point as a [barycenter](#def-b2-affine-barycenter) of two vertices; show that three points are aligned iff their barycentric coordinate rows with respect to $(A, B, C)$ form a singular $3 \times 3$ matrix.)*

**Solution of Exercise 17.11.**

$\vect{A'B} = \alpha\,\vect{A'C}$ says exactly $1\cdot
\vect{A'B} - \alpha\,\vect{A'C} = 0$, i.e. $A' =
\operatorname{bar}(B, 1;\ C, -\alpha)$ (total weight $1 -
\alpha \neq 0$ since $B \neq C$); likewise $B' =
\operatorname{bar}(C, 1;\ A, -\beta)$ and $C' =
\operatorname{bar}(A, 1;\ B, -\gamma)$.

*The alignment criterion.* Give each point $P$ its normalized barycentric row $p = (p_A, p_B, p_C)$, $p_A + p_B +
p_C = 1$, with respect to $(A, B, C)$. If $\sum_i c_i p_i = 0$ with $(c_1, c_2, c_3) \neq 0$ for three points $P_1, P_2,
P_3$, then summing the entries gives $\sum c_i = 0$, and $\sum_i c_i \vect{OP_i} = \sum_j \bigl(\sum_i
c_ip_{ij}\bigr)\vect{OV_j} = 0$: the $P_i$ are affinely dependent, i.e. aligned. Conversely an [affine](#def-b2-affine-subspace) dependence $(t_i)$ gives $w = \sum t_ip_i$ with entries summing to $0$ and $\sum_j w_j\vect{OV_j} = 0$; expanding from $A$, $w_B
\vect{AB} + w_C\vect{AC} = 0$, so $w = 0$ by [affine](#def-b2-affine-subspace) independence of $(A, B, C)$: the rows are linearly dependent. So alignment amounts to a vanishing $3 \times 3$ [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det), and scaling rows by the nonzero factors $1 - \alpha$, $1 -
\beta$, $1 - \gamma$ changes nothing:

$$
\det\begin{pmatrix} 0 & 1 & -\alpha\\ -\beta & 0 & 1\\ 1 &
-\gamma & 0\end{pmatrix}
= 1 - \alpha\beta\gamma .
$$

Hence $A', B', C'$ are aligned iff $\alpha\beta\gamma = 1$: Menelaus’ theorem.

**Exercise 17.12 ★★★.**

Prove that the [convex hull](#def-b2-affine-convex) of a [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) subset $K$ of $\R^n$ is [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact). *(By [Theorem 17.14](#thm-b2-affine-caratheodory), $\operatorname{conv}(K)$ is the image of a [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) set under a [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) map.)* Show by an example in $\R^2$ that the [convex hull](#def-b2-affine-convex) of a *closed* set need not be closed.

**Solution of Exercise 17.12.**

Let $\Delta = \{\lambda \in \R^{n+1} : \lambda_i \geq 0,\
\sum\lambda_i = 1\}$: closed and bounded in $\R^{n+1}$, hence [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact), and $K^{n+1}$ is [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) as a finite product. The map

$$
\Phi \colon \Delta \times K^{n+1} \to \R^n, \qquad
\Phi(\lambda, x_0, \dots, x_n) = \sum_{i=0}^n \lambda_i x_i
$$

is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity), and [Theorem 17.14](#thm-b2-affine-caratheodory) says precisely that $\operatorname{conv}(K) = \Phi(\Delta \times
K^{n+1})$: a [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) image of a [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) set ([Theorem 4.16](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-compactprops)), hence [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact).

For a closed set: take $S = (\R \times \{0\}) \cup \{(0,
1)\}$, closed in $\R^2$. A [convex](#def-b2-affine-convex) combination putting weight $t$ on $(0,1)$ and $1 - t$ on axis points has second coordinate $t$, so

$$
\operatorname{conv}(S) = \bigl(\R \times \intco01\bigr) \cup
\{(0,1)\}
$$

(for $0 \leq t < 1$, $(x, t) = t\,(0,1) + (1-t)\,\bigl(\tfrac
x{1-t}, 0\bigr)$). The point $(1, 1) = \lim_{t \to 1}(1, t)$ is adherent but not in the hull: not closed.

## 17.5 Problem: from Radon to Helly, centerpoints and Jung’s theorem

![The two Radon types for four points of the plane in general position: one point inside the triangle of the others (partition \A_4\ \A_1, A_2, A_3\), or convex position, where the Radon point (orange) is the intersection of the two diagonals.](https://one-course.com/images/onecourse/chapters/math-4/b2-affine/fig-3fd6518615fc.svg)

![The two Radon types for four points of the plane in general position: one point inside the triangle of the others (partition \A_4\ \A_1, A_2, A_3\), or convex position, where the Radon point (orange) is the intersection of the two diagonals.](https://one-course.com/images/onecourse/chapters/math-4/b2-affine/fig-f12d89fd9901.svg)

*The two Radon types for four points of the plane in general position: one point inside the triangle of the others (partition $\{A_4\} \mid \{A_1, A_2, A_3\}$), or [convex](#def-b2-affine-convex) position, where the Radon point (orange) is the intersection of the two diagonals.*

**Problem 17.1.**

Weekend problem — Helly’s theorem and two of its dividends

Radon’s lemma ([Exercise 17.7](#exo-b2-affine-7)) says that $n + 2$ points of an $n$-dimensional [affine space](#def-b2-affine-def) always split into two groups with intersecting [convex hulls](#def-b2-affine-convex). This problem turns that one linear-algebra fact into a chain of theorems of combinatorial geometry: Helly’s intersection theorem, the centerpoint theorem (a two-dimensional median), and Jung’s covering theorem. Throughout, the plane is $\R^2$ with its usual Euclidean structure, and $\det$ is the [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) in the canonical basis.

**Part I — Barycentric coordinates.** Points $A_0, \dots, A_k$ are *affinely independent* when the vectors $\vect{A_0A_1}, \dots, \vect{A_0A_k}$ are linearly independent.

1. Show that [affine](#def-b2-affine-subspace) independence does not depend on the choice of the base point $A_0$ , and that it is equivalent to: whenever two families of weights, each summing to $1$ , define the same [barycenter](#def-b2-affine-barycenter) of $(A_0,  \dots, A_k)$ , the weights coincide.
2. Let $(A, B, C)$ be affinely independent in the plane. Show that every point $M$ admits a unique triple $(\alpha, \beta, \gamma)$ with $\alpha + \beta +  \gamma = 1$ and $M = \operatorname{bar}(A, \alpha; B,  \beta; C, \gamma)$ — its *barycentric coordinates* .
3. Prove the [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) formulas $$\alpha = \frac{\det(\vect{MB},  \vect{MC})}{\det(\vect{AB}, \vect{AC})},  \qquad  \beta = \frac{\det(\vect{MC},  \vect{MA})}{\det(\vect{AB}, \vect{AC})},  \qquad  \gamma = \frac{\det(\vect{MA},  \vect{MB})}{\det(\vect{AB}, \vect{AC})} :$$ barycentric coordinates are ratios of signed areas.
4. The lines $BC$ , $CA$ , $AB$ are the coordinate lines $\{\alpha = 0\}$ , $\{\beta = 0\}$ , $\{\gamma = 0\}$ . Show that $M$ lies in the closed triangle $\operatorname{conv}\{A, B, C\}$ iff $\alpha, \beta,  \gamma \geq 0$ , and that the three lines cut the plane into exactly seven regions, classified by the signs of $(\alpha, \beta, \gamma)$ (the sign pattern $(-, -,  -)$ being impossible).
5. Let $u \colon \R^2 \to \R$ be an [affine map](#def-b2-affine-subspace) (an *[affine](#def-b2-affine-subspace) form* ). Show $u(M) = \alpha\,u(A) +  \beta\,u(B) + \gamma\,u(C)$ , that the level sets of a nonconstant [affine](#def-b2-affine-subspace) form are lines, that every line arises this way, and that the closed half-planes $\{u  \geq c\}$ are [convex](#def-b2-affine-convex) .

**Part II — Radon partitions, refined.** A family of $n + 2$ points of $\R^n$ is in *general position* when every $n + 1$ of them are affinely independent. An *[affine](#def-b2-affine-subspace) dependence* of $(A_1, \dots, A_{n+2})$ is a family $(\mu_i)$ with $\sum_i \mu_i = 0$ and $\sum_i
\mu_i\,\vect{OA_i} = 0$ for one (hence every) origin $O$.

6. Compute a nonzero [affine](#def-b2-affine-subspace) dependence of the four points $A_1 = (0,0)$ , $A_2 = (3,0)$ , $A_3 = (0,3)$ , $A_4 = (1,1)$ ; give the Radon partition and the Radon point.
7. Show that for points in general position the vector space of [affine](#def-b2-affine-subspace) dependences has dimension exactly $1$ , and that a nonzero dependence has *no* vanishing coefficient.
8. Deduce that the Radon partition of $n + 2$ points in general position is unique (up to swapping the two blocks), each block being the set of indices where $\mu_i$ has one fixed sign.
9. For four points of the plane in general position, show the dichotomy: either the partition has type $(1, 3)$ — one point interior to the triangle of the other three — or type $(2, 2)$ : the four points are in [convex](#def-b2-affine-convex) position and the segments joining the two pairs (the diagonals) intersect, at the Radon point.
10. Carry out question 6 for the unit square $(0,0)$ , $(1,0)$ , $(1,1)$ , $(0,1)$ : dependence, partition, Radon point.

**Part III — Helly’s theorem in the plane.**

11. Let $C_1, C_2, C_3, C_4$ be [convex](#def-b2-affine-convex) subsets of $\R^2$ , any three of which have a common point. Pick $x_i \in  \bigcap_{j \neq i} C_j$ and apply Radon’s lemma to $x_1, \dots, x_4$ : show that the Radon point belongs to all four sets. *(For each $k$, the block not containing $x_k$ consists of points of $C_k$.)*
12. (Helly) Let $C_1, \dots, C_m$ ( $m \geq 3$ ) be [convex](#def-b2-affine-convex) subsets of $\R^2$ , any three of which intersect. Prove $\bigcap_{i=1}^m C_i \neq \emptyset$ , by induction on $m$ : replace $C_{m-1}$ and $C_m$ by $C_{m-1} \cap C_m$ and check the hypothesis for the new family using question 11.
13. Three counterexamples, one per hypothesis: (a) the three closed edges of a triangle pairwise intersect but have no common point ( $3$ cannot be lowered to $2$ ); (b) the four sets $S_i = \{x_1, \dots, x_4\}  \setminus \{x_i\}$ , for four points in general position, satisfy the triple-intersection hypothesis but not the conclusion (convexity matters); (c) the closed half-planes $H_k = \intco{k}{+\infty} \times  \R$ , $k \in \N$ , pairwise and triple-wise intersect but $\bigcap_k H_k = \emptyset$ (infinite families need [compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) ).
14. ( [Compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) Helly) Let $(K_i)_{i \in I}$ be an arbitrary family of [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) [convex](#def-b2-affine-convex) subsets of $\R^2$ , any three of which intersect. Using question 12 and the Borel–Lebesgue property ( [Theorem 4.20](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-borellebesgue) ), show $\bigcap_{i \in I} K_i \neq \emptyset$ .
15. (First dividend) Let $S$ be a finite set of points of the plane and $r > 0$ . Show: if every three points of $S$ lie in some closed disk of radius $r$ , then $S$ lies in one closed disk of radius $r$ . *(Apply Helly to the disks $\overline D(p, r)$, $p \in S$.)*

**Part IV — The centerpoint theorem.** A *centerpoint* of a finite set $S$ of $n$ points of the plane is a point $c$ (not necessarily in $S$) such that every closed half-plane containing $c$ contains at least $n/3$ points of $S$.

16. (Dimension $1$ ) For reals $x_1 \leq \dots \leq x_n$ , show that the median $c = x_{\lceil n/2 \rceil}$ satisfies: every closed half-line containing $c$ contains at least $n/2$ of the $x_i$ .
17. (Counting lemma) If $A, B, C$ are subsets of $S$ with $\abs A, \abs B, \abs C > \tfrac{2n}3$ , show $A \cap  B \cap C \neq \emptyset$ .
18. Let $m = \floor{2n/3} + 1$ and let $\mathcal F$ be the (finite) family of the [convex hulls](#def-b2-affine-convex) $\operatorname{conv}(T)$ , $T \subseteq S$ , $\abs T =  m$ . Show that any three members of $\mathcal F$ have a common point, and deduce from Helly a point $c$ common to all of them.
19. Prove that this $c$ is a centerpoint of $S$ : the *centerpoint theorem* . *(If a closed half-plane through $c$ contained fewer than $n/3$ points, its [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) complement would contain a set $T$ of $m$ points, and $\operatorname{conv}(T)$ would avoid $c$.)*
20. Sharpness: let $n = 3k$ and place $k$ points in each of three disks of small radius $\varepsilon$ centered at the vertices of a large triangle. Show that for *every* point $c$ of the plane some closed half-plane containing $c$ contains at most $n/3$ points of $S$ , so the constant $1/3$ cannot be improved. *(Among the three directions from $c$ to the disk centers, two make an angle at most $2\pi/3$.)*

**Part V — Jung’s theorem and synthesis.**

21. (Triangle lemma) Let $P, Q, R$ be three points with pairwise distances $\leq 1$ . Show they lie in a closed disk of radius $1/\sqrt3$ . *(If some angle is $\geq \pi/2$, take the disk on the longest side as diameter, using the median formula $\norm{RM}^2 = \tfrac12\norm{RP}^2 +  \tfrac12\norm{RQ}^2 - \tfrac14\norm{PQ}^2$; if the triangle is acute, bound the circumradius $a/(2\sin  \widehat A)$ using its largest angle, which lies in $\intco{\pi/3}{\pi/2}$.)*
22. (Jung) Deduce: every [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) subset of the plane of diameter $\leq 1$ is contained in a closed disk of radius $1/\sqrt3$ .
23. Sharpness: for the equilateral triangle $A_1A_2A_3$ of side $1$ with centroid $G$ , prove the Leibniz identity $\sum_i \norm{\vect{OA_i}}^2 =  3\norm{\vect{OG}}^2 + \sum_i \norm{\vect{GA_i}}^2$ for every point $O$ , and conclude that any disk containing the three vertices has radius $\geq  1/\sqrt3$ , with equality only for the circumdisk.
24. (Helly in $\R^n$ ) State and prove Helly’s theorem in $\R^n$ : if finitely many [convex sets](#def-b2-affine-convex) are such that any $n + 1$ of them intersect, then all of them intersect. *(Radon’s lemma [Exercise 17.7](#exo-b2-affine-7) handles $n + 2$ sets; then induct as in question 12.)*
25. Synthesis. Assemble the chain $$\text{affine dependence} \Rightarrow \text{Radon}  \Rightarrow \text{Helly} \Rightarrow  \text{centerpoint and Jung},$$ indicating in one sentence each: where linear algebra enters, where the signs of the weights enter, where convexity enters, and which single step used the dimension of the plane. What do the constants $3$ (in Helly), $1/3$ (centerpoint) and $1/\sqrt3$ (Jung) become in $\R^n$? (State without proof.)

**Solution of Problem 17.1.**

**1.** Rebase at $A_j$: for $i \neq j$, $\vect{A_jA_i} =
\vect{A_0A_i} - \vect{A_0A_j}$. If $\sum_{i \neq j}
c_i\vect{A_jA_i} = 0$, expanding gives $\sum_{i \notin \{0,
j\}} c_i\,\vect{A_0A_i} - \bigl(\sum_{i\neq j}
c_i\bigr)\vect{A_0A_j} = 0$; independence of the $\vect{A_0A_i}$ forces $c_i = 0$ for $i \notin \{0, j\}$, then $c_0 = 0$: independence at $A_j$. For the equivalence: two weight families $(\lambda_i)$, $(\lambda_i')$ summing to $1$ with the same [barycenter](#def-b2-affine-barycenter) give, with $\nu = \lambda -
\lambda'$: $\sum\nu_i = 0$ and (origin $A_0$) $\sum_{i \geq
1}\nu_i\,\vect{A_0A_i} = 0$, so $\nu = 0$ under independence. Conversely a nontrivial relation $\sum_{i\geq1}\mu_i
\vect{A_0A_i} = 0$, completed by $\mu_0 = -\sum_{i\geq1}
\mu_i$, lets one add $t(\mu_i)$ to any weight family without moving the [barycenter](#def-b2-affine-barycenter): non-uniqueness.

**2.** $(\vect{AB}, \vect{AC})$ is a basis of $\R^2$: write $\vect{AM} = \beta\,\vect{AB} + \gamma\,\vect{AC}$ (unique) and set $\alpha = 1 - \beta - \gamma$; the [barycenter](#def-b2-affine-barycenter) condition at origin $A$ reads exactly $\vect{AM} =
\beta\,\vect{AB} + \gamma\,\vect{AC}$. Uniqueness is question 1.

**3.** From $\alpha\vect{MA} + \beta\vect{MB} +
\gamma\vect{MC} = 0$ and Chasles, $\vect{MA} =
-\beta\,\vect{AB} - \gamma\,\vect{AC}$. With $D =
\det(\vect{AB}, \vect{AC})$:

$$
\begin{align*}
\det(\vect{MB}, \vect{MC})
&= \det(\vect{MA} + \vect{AB},\ \vect{MA} + \vect{AC})\\
&= \det(\vect{MA}, \vect{AC}) + \det(\vect{AB}, \vect{MA}) +
D = -\beta D - \gamma D + D = \alpha D,
\end{align*}
$$

using bilinearity and $\det(\vect{MA}, \vect{AC}) = -\beta D$, $\det(\vect{AB}, \vect{MA}) = -\gamma D$. The other two formulas follow by the same computation with the roles permuted cyclically.

**4.** By definition $\operatorname{conv}\{A, B, C\}$ is the set of [barycenters](#def-b2-affine-barycenter) with nonnegative weights; normalizing the weights to sum $1$ and invoking uniqueness (question 2), $M \in \operatorname{conv}\{A,B,C\}$ iff $\alpha, \beta,
\gamma \geq 0$. Each coordinate is an [affine](#def-b2-affine-subspace) function of $M$ (question 3: a $2\times2$ [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) with one column [affine](#def-b2-affine-subspace) in $M$), so each [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) sign condition defines an [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) half-plane. The pattern $(-,-,-)$ contradicts $\alpha + \beta
+ \gamma = 1$; each of the remaining seven patterns is realized: scale a sign-respecting triple with at least one $+$ entry so that the (positive) sum is $1$ — e.g. $(-1, 1, 1)$, $(3, -1, -1)$, $(\frac13, \frac13, \frac13)$, and permutations.

**5.** An [affine map](#def-b2-affine-subspace) preserves [barycenters](#def-b2-affine-barycenter) ([Definition 17.5](#def-b2-affine-subspace)), so $u(M) = \alpha u(A) +
\beta u(B) + \gamma u(C)$. Writing $u(x, y) = ax + by + c$ with $(a, b) \neq (0,0)$: $\{u = c'\}$ is a line, and every line $ax + by = c'$ is such a level set. If $u(M), u(N) \geq
c$ and $t \in \intcc01$, then $u\bigl(\operatorname{bar}(M,
1-t; N, t)\bigr) = (1-t)u(M) + tu(N) \geq c$: half-planes are [convex](#def-b2-affine-convex).

**6.** The conditions $\sum\mu_i = 0$, $3\mu_2 + \mu_4 =
0$, $3\mu_3 + \mu_4 = 0$ give (taking $\mu_4 = 3$) the dependence $(\mu_1, \mu_2, \mu_3, \mu_4) = (-1, -1, -1, 3)$. Signs split as $\{A_4\} \mid \{A_1, A_2, A_3\}$, and normalizing each side by $3$:

$$
A_4 = \operatorname{bar}\bigl(A_1, \tfrac13;\ A_2, \tfrac13;\
A_3, \tfrac13\bigr) = (1,1),
$$

the centroid of the triangle: the Radon point is $A_4$ itself, which indeed lies inside the triangle $A_1A_2A_3$.

**7.** The linear map $\Phi \colon \R^{n+2} \to \R
\times \R^n$, $\mu \mapsto (\sum\mu_i,\
\sum\mu_i\vect{OA_i})$, has rank $\leq n + 1$, so $\dim\ker
\Phi \geq 1$. If two independent dependences $\mu, \mu'$ existed, a suitable combination $\nu = \mu'_{n+2}\mu -
\mu_{n+2}\mu'$ (or $\mu$ itself if both last coefficients vanish) would be a *nonzero* dependence with $\nu_{n+2}
= 0$; restricting to $A_1, \dots, A_{n+1}$ and rebasing at $A_1$, some $\nu_i \neq 0$ with $i \geq 2$ (a single nonzero weight cannot sum to zero), giving a nontrivial relation $\sum_{i\geq2}\nu_i\vect{A_1A_i} = 0$: the $n+1$ points would be affinely dependent, against general position. So $\dim\ker
\Phi = 1$. The same restriction argument shows a nonzero dependence has no vanishing coefficient.

**8.** Let $\mu \neq 0$ be a dependence, $P = \{i :
\mu_i > 0\}$ and $N = \{i : \mu_i < 0\}$: both nonempty ($\sum\mu_i = 0$, $\mu \neq 0$) and exhaustive (no zero coefficient). Radon’s construction ([Exercise 17.7](#exo-b2-affine-7)) produces the common hull point from exactly this partition. Since the dependence is unique up to a nonzero scalar (question 7), the unordered pair $\{P, N\}$ — hence the Radon partition — is unique.

**9.** Blocks are nonempty, so the type is $(1,3)$ or $(2,2)$. Type $(1,3)$, block $\{j\}$: the Radon point lies in $\operatorname{conv}\{A_j\} = \{A_j\}$, so $A_j \in
\operatorname{conv}$ of the other three; it cannot lie on an edge (three of the points would be aligned, against general position), so $A_j$ is interior to the triangle. Type $(2,2)$, blocks $\{i,j\} \mid \{k,l\}$: the Radon point $z$ lies on $\intcc{A_i}{A_j} \cap \intcc{A_k}{A_l}$, and $z$ is not an endpoint (that would align three points): the two segments cross at an interior point. Moreover no point lies in the hull of the others: such a containment $A_l =
\operatorname{bar}(A_i, \lambda_i)_{i \neq l}$ with $\lambda_i \geq 0$ is an [affine](#def-b2-affine-subspace) dependence with sign pattern $(+,+,+,-)$, which by uniqueness (question 8) would make the partition $(1,3)$. So in the $(2,2)$ case the four points are in [convex](#def-b2-affine-convex) position and the crossing segments are the diagonals.

**10.** The equations $\mu_2 + \mu_3 = 0$, $\mu_3 +
\mu_4 = 0$, $\sum\mu_i = 0$ give the dependence $(1, -1, 1,
-1)$: partition $\{(0,0), (1,1)\} \mid \{(1,0), (0,1)\}$, and

$$
\operatorname{bar}\bigl((0,0), \tfrac12;\ (1,1),
\tfrac12\bigr) = \bigl(\tfrac12, \tfrac12\bigr) =
\operatorname{bar}\bigl((1,0), \tfrac12;\ (0,1),
\tfrac12\bigr) :
$$

the Radon point is the center of the square, where the two diagonals cross — type $(2,2)$, as the picture predicts.

**11.** Radon applied to $x_1, \dots, x_4$ gives blocks $I \mid J$ and a point $z \in \operatorname{conv}\{x_i : i
\in I\} \cap \operatorname{conv}\{x_j : j \in J\}$. Fix $k
\in \{1, \dots, 4\}$, say $k \in I$. Every $j \in J$ satisfies $j \neq k$, so $x_j \in C_k$ by the choice $x_j \in
\bigcap_{l \neq j}C_l$; since $C_k$ is [convex](#def-b2-affine-convex), $z \in
\operatorname{conv}\{x_j : j \in J\} \subseteq C_k$. As $k$ was arbitrary, $z \in C_1 \cap C_2 \cap C_3 \cap C_4$.

**12.** Induction on $m$. For $m = 3$ the hypothesis is the conclusion; $m = 4$ is question 11. Let $m \geq 4$, assume the statement for $m$ sets, and take $C_1, \dots,
C_{m+1}$ with the triple-intersection property. Set $C_m' =
C_m \cap C_{m+1}$, [convex](#def-b2-affine-convex). The family $C_1, \dots, C_{m-1},
C_m'$ has $m$ members; a triple avoiding $C_m'$ intersects by hypothesis, and a triple $\{C_i, C_j, C_m'\}$ has intersection $C_i \cap C_j \cap C_m \cap C_{m+1}$, nonempty by question 11 applied to $C_i, C_j, C_m, C_{m+1}$ (any three of these meet, by hypothesis). The induction hypothesis now yields a common point of the new family, i.e. of all $m+1$ sets.

**13.** (a) The closed edges $\intcc PQ$, $\intcc QR$, $\intcc RP$ of a nondegenerate triangle: any two share a vertex, but a common point of all three would lie in $\intcc
PQ \cap \intcc RP = \{P\}$ and in $\intcc QR$, which excludes $P$. (b) Any three of the sets $S_i = \{x_1, \dots, x_4\}
\setminus \{x_i\}$ omit three of the four points, leaving exactly one common point; the total intersection omits every point. The $S_i$ are finite, not [convex](#def-b2-affine-convex): convexity is essential. (c) Finitely many $H_k = \intco{k}{+\infty} \times
\R$ intersect in $\intco{k_{\max}}{+\infty} \times \R \neq
\emptyset$, yet no point has $x \geq k$ for all $k \in \N$: for infinite families, [compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) is essential.

**14.** Suppose $\bigcap_{i \in I}K_i = \emptyset$ and fix $i_0$. Every $x \in K_{i_0}$ misses some $K_i$, so $K_{i_0} \subseteq \bigcup_{i \in I}(\R^2 \setminus K_i)$, a cover by [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) sets ($K_i$ is [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact), hence closed). By Borel–Lebesgue ([Theorem 4.20](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-borellebesgue)) finitely many suffice: $K_{i_0} \cap K_{i_1} \cap \dots \cap K_{i_N} =
\emptyset$. But any three members of this finite family of [convex sets](#def-b2-affine-convex) intersect, so question 12 makes the intersection nonempty: contradiction.

**15.** Set $D_p = \overline D(p, r)$ for $p \in S$: [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) [convex sets](#def-b2-affine-convex). For $p, q, s \in S$, the hypothesis gives a closed disk $\overline D(z, r)$ containing $p, q, s$; then $\norm{\vect{zp}}, \norm{\vect{zq}}, \norm{\vect{zs}}
\leq r$, i.e. $z \in D_p \cap D_q \cap D_s$. By Helly (question 12; the family is finite) there is $c \in
\bigcap_{p \in S}D_p$: every $p \in S$ satisfies $\norm{\vect{cp}} \leq r$, so $S \subseteq \overline D(c,
r)$.

**16.** Let $c = x_{\lceil n/2\rceil}$. A closed half-line containing $c$ is $\intoc{-\infty}{t}$ with $t \geq
c$ or $\intco{t}{+\infty}$ with $t \leq c$. The first contains $x_1, \dots, x_{\lceil n/2\rceil}$: at least $\lceil n/2\rceil \geq n/2$ points. The second contains $x_{\lceil n/2\rceil}, \dots, x_n$: exactly $n - \lceil
n/2\rceil + 1 = \floor{n/2} + 1 > n/2$ points.

**17.** $\abs{A \cap B} = \abs A + \abs B - \abs{A \cup
B} \geq \abs A + \abs B - n > \tfrac{4n}3 - n = \tfrac n3$, then

$$
\abs{A \cap B \cap C} \geq \abs{A \cap B} + \abs C - n >
\tfrac n3 + \tfrac{2n}3 - n = 0 .
$$

**18.** Note $m > 2n/3$. For $T_1, T_2, T_3 \subseteq S$ of cardinality $m$, question 17 provides a point $x \in T_1
\cap T_2 \cap T_3$; then $x \in \operatorname{conv}(T_i)$ for each $i$: any three members of $\mathcal F$ meet. The family is finite (finitely many subsets of $S$) and consists of [convex sets](#def-b2-affine-convex), so Helly (question 12) gives $c \in
\bigcap_{\abs T = m}\operatorname{conv}(T)$.

**19.** Suppose some closed half-plane $H \ni c$ contains fewer than $n/3$ points of $S$. Its complement $U$ is an [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) half-plane, [convex](#def-b2-affine-convex), with $\abs{S \cap U} > 2n/3$, hence $\abs{S \cap U} \geq \floor{2n/3} + 1 = m$; choose $T
\subseteq S \cap U$ with $\abs T = m$. Then $\operatorname{conv}(T) \subseteq U$ by convexity of $U$, so $c \in \operatorname{conv}(T) \subseteq U$: contradiction with $c \in H$. Hence every closed half-plane containing $c$ contains at least $n/3$ points: $c$ is a centerpoint.

**20.** Take the triangle equilateral of side $L$ and $\varepsilon = L/100$. Let $c$ be any point; we exhibit a closed half-plane containing $c$ and at most $k$ points.

*Case 1: $c$ is within $L/10$ of a vertex, say $B$.* Directions from $c$ to $A$ and to $C$ deviate from the directions $B \to A$, $B \to C$ by at most $\arcsin\bigl(\tfrac{L/10}{9L/10}\bigr) = \arcsin\tfrac19$, so they make an angle $\leq \tfrac\pi3 + 2\arcsin\tfrac19 <
\tfrac{2\pi}3$. *Case 2: $c$ is at distance $> L/10$ from all vertices.* If $c$ is in the triangle, the three angular gaps between the directions $u_A, u_B, u_C$ from $c$ to the vertices sum to $2\pi$, so some gap is $\leq 2\pi/3$; if $c$ is outside, the three directions lie in an [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) half-plane of directions and two of them make an angle $<
\pi/2$. In every case two directions, say toward $X$ and $Y$, make an angle $\leq 2\pi/3$; let $w$ be their unit bisector, so $\langle u_X, w\rangle, \langle u_Y, w\rangle \geq
\cos\tfrac\pi3 = \tfrac12$. For any point $b$ of the disk around $X$:

$$
\langle \vect{cb}, w\rangle \geq
\tfrac12\norm{\vect{cX}} - \varepsilon > 0,
$$

since $\norm{\vect{cX}} \geq L/10 > 2\varepsilon$ (and likewise for $Y$): the [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) half-plane $\{\langle \vect{cx},
w\rangle > 0\}$ swallows both clusters. Its closed complement contains $c$ and at most the $k$ points of the third cluster. So no point of the plane beats $n/3$: with question 19, the centerpoint constant is exactly $1/3$.

**21.** Order the angles; the largest, $\theta$, satisfies $\theta \geq \pi/3$ (the three sum to $\pi$). If $\theta \geq \pi/2$, say at $R$, let $M$ be the midpoint of the opposite side $\intcc PQ$. The median formula ($\vect{RM}
= \tfrac12(\vect{RP} + \vect{RQ})$, expand and eliminate $\langle\vect{RP}, \vect{RQ}\rangle$ with the law of cosines) gives

$$
\norm{\vect{RM}}^2 = \tfrac12\norm{\vect{RP}}^2 +
\tfrac12\norm{\vect{RQ}}^2 - \tfrac14\norm{\vect{PQ}}^2 \leq
\tfrac12\norm{\vect{PQ}}^2 - \tfrac14\norm{\vect{PQ}}^2 =
\tfrac14\norm{\vect{PQ}}^2,
$$

using $\norm{\vect{PQ}}^2 = \norm{\vect{RP}}^2 +
\norm{\vect{RQ}}^2 - 2\langle\vect{RP}, \vect{RQ}\rangle \geq
\norm{\vect{RP}}^2 + \norm{\vect{RQ}}^2$ (the inner product is $\leq 0$). So the disk of diameter $\intcc PQ$, of radius $\leq \tfrac12 < \tfrac1{\sqrt3}$, contains all three points (degenerate aligned triples fall under $\theta = \pi$). If $\theta < \pi/2$ the triangle is acute; by the law of sines the circumradius is $R_c = a/(2\sin\theta)$ with $a \leq 1$ the side opposite $\theta$, and $\theta \in
\intco{\pi/3}{\pi/2}$ gives $\sin\theta \geq \sqrt3/2$, so $R_c \leq 1/\sqrt3$: the circumdisk does the job.

**22.** For $p \in S$ let $K_p = \overline D(p,
1/\sqrt3)$: [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) [convex](#def-b2-affine-convex). Any three points $p, q, s$ of $S$ are pairwise at distance $\leq 1$, so question 21 gives a disk of radius $1/\sqrt3$ containing them: its center lies in $K_p \cap K_q \cap K_s$. By [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) Helly (question 14, arbitrary families allowed) there is $c \in \bigcap_{p\in
S}K_p$: every $p \in S$ is within $1/\sqrt3$ of $c$, i.e. $S
\subseteq \overline D(c, 1/\sqrt3)$. This is Jung’s theorem in the plane.

**23.** With $G$ the centroid, $\sum_i\vect{GA_i} = 0$, so

$$
\sum_i\norm{\vect{OA_i}}^2 = \sum_i\norm{\vect{OG} +
\vect{GA_i}}^2 = 3\norm{\vect{OG}}^2 + 2\Bigl\langle
\vect{OG}, \sum_i\vect{GA_i}\Bigr\rangle +
\sum_i\norm{\vect{GA_i}}^2,
$$

and the middle term vanishes: the Leibniz identity. For the equilateral triangle of side $1$, $\norm{\vect{GA_i}} =
1/\sqrt3$ (two thirds of the height $\sqrt3/2$), so $\sum_i\norm{\vect{GA_i}}^2 = 1$. If $\overline D(O, r)$ contains the vertices, then $3r^2 \geq
\sum_i\norm{\vect{OA_i}}^2 = 3\norm{\vect{OG}}^2 + 1 \geq 1$: $r \geq 1/\sqrt3$, with equality forcing $O = G$ and all three distances equal to $r$ — the circumdisk. Jung’s constant $1/\sqrt3$ is sharp.

**24.** *Helly in $\R^n$: if $C_1, \dots, C_m$ ($m \geq n + 1$) are [convex](#def-b2-affine-convex) subsets of $\R^n$ and any $n + 1$ of them intersect, then all of them do.* Base case $m = n +
2$: pick $x_i \in \bigcap_{j\neq i}C_j$; Radon’s lemma ([Exercise 17.7](#exo-b2-affine-7)) splits $x_1, \dots, x_{n+2}$ into blocks $I \mid J$ with a common hull point $z$, and for each $k$, the block not containing $k$ consists of points of $C_k$, so $z \in C_k$ by convexity, exactly as in question 11. Induction step for $m \geq n + 2$: replace $C_m,
C_{m+1}$ by $C_m \cap C_{m+1}$; an $(n+1)$-tuple of the new family containing the intersected member amounts to $n + 2$ of the old sets, handled by the base case, and the other tuples are covered by hypothesis. Conclude by the induction hypothesis.

**25.** Linear algebra enters once: $n + 2$ vectors in the $(n+1)$-dimensional space of pairs (total weight, weighted position) must be dependent — that is the [affine](#def-b2-affine-subspace) dependence. The signs of its coefficients split the points into the two Radon blocks and turn one linear relation into an equality of two nonnegative [barycenters](#def-b2-affine-barycenter). Convexity is used exactly twice: in Helly’s step (the hull of points of $C_k$ stays in $C_k$) and in the applications (half-planes and disks are [convex](#def-b2-affine-convex)). The dimension of the plane entered only through the number $4 = 2 + 2$ of points fed to Radon, i.e. the “$3 = 2 + 1$” in Helly’s hypothesis; everything else was dimension-free, as question 24 confirms. In $\R^n$ the constants become: Helly number $n + 1$; centerpoint constant $\frac1{n+1}$ (every finite set has a point every closed half-space through which contains a fraction $\geq
\frac1{n+1}$ of it); Jung radius $\sqrt{\frac{n}{2(n+1)}}$ for sets of diameter $1$ — equal to $1/\sqrt3$ when $n = 2$.