Discrete Geometry / Polytopes, 2010 course
Federico Ardila · Free author course
Read lectures 27–28 for mixed-volume context; reproduce this rectangle expansion yourself.
Open the sourceMixed area: a geometric inequality you can prove
Rectangles reveal the mechanism before abstract geometry arrives.
Prerequisites: 10 · Inner products and projection, 19 · Log-concavity: why a sequence forms a hill
Understand
Objectives
- Expand the rectangle widths and heights.
- Keep the binomial factor two in the middle term.
- Prove the square gap and identify the equality condition.
Start with something you can see
In the area lab set K to width 2, height 1, and L to width 1, height 3. Before moving a slider, predict the three terms of area(K+tL). Both rectangles are anchored at the origin; adding their points adds widths and heights.
Give the idea a precise name
A set is convex if it contains the segment joining any two of its points. The Minkowski sum K+tL consists of x+ty with x∈K and y∈L. For rectangles of widths and heights (a,b) and (c,d), let V₀=ab, V₁=(ad+bc)/2, V₂=cd. The factor two is the binomial coefficient in dimension two.
area(K+tL)=V₀+2V₁t+V₂t²; V₁²−V₀V₂=(ad−bc)²/4≥0
Here t≥0; a,b,c,d are positive side lengths, so V₀ and V₂ are areas. Expansion gives (a+tc)(b+td)=ab+(ad+bc)t+cdt². Subtract abcd from (ad+bc)²/4: the numerator is (ad−bc)². Equality means ad=bc, the same width-to-height ratio, not merely the same area.
Work one small world
For K=2×1 and L=1×3, width and height of K+tL are 2+t and 1+3t. The coefficients beside the area lab are:
| Power | Expansion | Normalized value |
|---|---|---|
| 1 | 2 | V₀=2 |
| t | 7t | 2V₁=7; V₁=3.5 |
| t² | 3t² | V₂=3 |
| gap | (6−1)²/4 | V₁²−V₀V₂=6.25 |
Counterexample to treating raw coefficients as mixed areas: the raw middle coefficient 7 is not V₁. Here V₁=3.5 and 2V₁=7; at t=1 the area is 2+7+3=12. Equal areas alone do not give equality: rectangles 2×1 and 1×2 both have area 2, but their gap is 2.25.
THE BRIDGEThe polytope lesson also uses convexity, but a convex hull of discrete selections is not a Minkowski sum of rectangles.
WHERE THIS IDEA STOPSThis direct rectangle calculation illustrates a general geometric inequality; it does not prove the general mixed-volume theorem.
Experiment
Set the four labeled sliders to 2,1,1,3. Compare the lab’s V values with the table, then change one side and predict whether the square gap increases.
What feature of two rectangles survives in the single expression ad−bc?
Check & explain
For rectangles 3×2 and 1×4, what is the normalized middle mixed area?
For positive rectangles, when is the gap zero?
- When ad=bc
- Whenever ab=cd
- Whenever t=1
Teach back
Derive the polynomial and the gap for 3×2 and 1×4 without copying the displayed example. What would a raw coefficient of 14 mean?
- Define pointwise Minkowski addition.
- Separate 2V₁ from V₁.
- Use ad=bc to interpret equality.
Read deeper
Lecture 9: Newton Polytopes and Mixed Volume
MIT OpenCourseWare · Course notes · MIT OCW
Optional lecture 9 on Newton polytopes and mixed volume; it is not a proof of this mixed-area inequality.
Open the source