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curriculum-project-hub/render/examples/smoke-parts.typ
T
sjfhsjfh 6f46aa708a fix(render): per-level heading numbering via numbly; config-driven, offline (WU-C)
Fixes the numbering bug (single pattern "一、" reused across levels → level-2
rendered "二、一、…"). display now takes a `config` dict; heading numbering uses
numbly per level — framework default ("{1:一}、","{1:1}.{2:1}","{1:1}.{2:1}.{3:1}")
→ 一、 / 1.1 / 1.1.1 — overridable via config.numbering.heading from the
engineering file (ADR-0009's file-resident override path, the layer whose absence
caused the bug). Verified: grep "二、一" = 0 across all smoke PDFs.

Adds @preview/numbly:0.1.0 (pure typst, zero transitive deps), VENDORED in-repo
at render/vendor/typst-packages/preview/numbly/0.1.0/ for network-free CI;
resolve via --package-cache-path render/vendor/typst-packages. parts contract +
student/teacher field visibility unchanged.

Co-Authored-By: Claude Opus 4.8 (1M context) <noreply@anthropic.com>
2026-06-22 08:42:15 +08:00

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// Shared hand-written sample lesson exercising all 4 kinds.
// Content field values are plain content blocks — exactly what the Rust driver
// would hand us via `include`.
#let info = (
title: "向量与几何 · 示例讲义",
author: ("张老师", "李老师"),
)
#let parts = (
// segment — with NESTED headings so per-level numbering is visible:
// level-1 `=` should render `一、`, level-2 `==` should render `1.1`
// (NOT the old buggy `二、一、`).
(
kind: "segment",
textbook: [
= 平面向量的数量积
本节研究平面向量的基本运算。设 $arrow(a)$$arrow(b)$ 为平面内两个向量,
其数量积定义为 $arrow(a) dot arrow(b) = |arrow(a)| |arrow(b)| cos theta$
其中 $theta$ 为两向量的夹角。
== 坐标表示
在直角坐标系下,若 $arrow(a) = (x_1, y_1)$$arrow(b) = (x_2, y_2)$,则
$arrow(a) dot arrow(b) = x_1 x_2 + y_1 y_2$
== 几何意义
数量积等于一个向量的模与另一向量在其方向上投影之积。
],
),
// example WITH source
(
kind: "example",
source: "2024 高考甲卷",
problem: [
已知 $arrow(a) = (1, 2)$$arrow(b) = (3, -1)$,求 $arrow(a) dot arrow(b)$
],
solution: [
由坐标公式,$arrow(a) dot arrow(b) = 1 times 3 + 2 times (-1) = 3 - 2 = 1$
],
),
// example WITHOUT source
(
kind: "example",
problem: [
求向量 $arrow(a) = (3, 4)$ 的模长 $|arrow(a)|$
],
solution: [
$|arrow(a)| = sqrt(3^2 + 4^2) = sqrt(25) = 5$
],
),
// lemma WITH proof
(
kind: "lemma",
stmt: [
对任意向量 $arrow(a)$$arrow(b)$,有 $|arrow(a) dot arrow(b)| <= |arrow(a)| |arrow(b)|$
],
proof: [
由数量积定义 $arrow(a) dot arrow(b) = |arrow(a)| |arrow(b)| cos theta$
$|cos theta| <= 1$,故 $|arrow(a) dot arrow(b)| = |arrow(a)| |arrow(b)| |cos theta| <= |arrow(a)| |arrow(b)|$
$qed$
],
),
// lemma WITHOUT proof (proof key omitted entirely)
(
kind: "lemma",
stmt: [
两个非零向量垂直当且仅当其数量积为零,即 $arrow(a) perp arrow(b) <==> arrow(a) dot arrow(b) = 0$
],
),
// sop
(
kind: "sop",
sop: [
求两向量夹角的标准步骤:
+ 计算数量积 $arrow(a) dot arrow(b)$
+ 分别计算模长 $|arrow(a)|$$|arrow(b)|$
+ 代入 $cos theta = (arrow(a) dot arrow(b)) / (|arrow(a)| |arrow(b)|)$ 求出 $theta$
],
),
)