Kvant Math Problem 697
Let the square have side length $1$.
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 9m51s
Source on kvant.digital
Problem
Let us call the plumpness of a rectangle the ratio of its shorter side to its longer side (the plumpness of a square is 1). Prove that no matter how a square is dissected into rectangles, the sum of their plumpnesses is at least 1.
S. V. Fomin
Exploration
Let the square have side length $1$. For a rectangle with sides $a\le b$, its plumpness is $a/b$. Since the area is $ab$, we have
$$\frac{a}{b}=\frac{a^2}{ab}.$$
Thus the plumpness equals $a^2$ divided by the area.
Suppose a rectangle in the dissection has horizontal width $w$ and vertical height $h$. Then
$$\frac{\min(w,h)}{\max(w,h)} = \frac{\min(w,h)^2}{wh}.$$
Multiplying by the area $wh$ gives exactly $\min(w,h)^2$.
Hence if the rectangles are $R_i$ and $p_i$ denotes the plumpness of $R_i$,
$$p_i,\operatorname{area}(R_i)=s_i^2,$$
where $s_i$ is the shorter side of $R_i$.
Summing,
$$\sum p_i = \sum \frac{s_i^2}{\operatorname{area}(R_i)}.$$
This does not immediately yield a lower bound.
A natural inequality is Cauchy:
$$\Bigl(\sum p_i\Bigr)\Bigl(\sum \operatorname{area}(R_i)\Bigr) \ge \Bigl(\sum s_i\Bigr)^2.$$
Since the areas sum to $1$, it would suffice to prove
$$\sum s_i\ge 1.$$
The problem is now reduced to showing that the sum of the shorter sides of all rectangles in a dissection of the unit square is at least $1$.
Why should this be true? Consider a horizontal line $y=t$. Every rectangle intersecting this line contributes at least its shorter side to the total sum. If a rectangle intersects the line, its vertical side is at least the length of the intersection in the vertical direction needed to reach the line, but that seems unhelpful.
A better idea is to integrate. For a rectangle $R$, let $h_R$ be its height and $w_R$ its width. The set of horizontal lines meeting $R$ has measure $h_R$. Since the shorter side $s_R=\min(h_R,w_R)\le h_R$,
$$\frac{s_R}{h_R}\le 1.$$
Define the contribution of $R$ to a horizontal line $y=t$ as $s_R/h_R$ whenever the line meets $R$. Integrating over all $t$ gives exactly $s_R$.
For a fixed $t$, the line cuts the square into intervals belonging to rectangles meeting the line. These intervals have lengths equal to the widths of those rectangles. Since $s_R\le w_R$, we have
$$\frac{s_R}{h_R}\le \frac{w_R}{h_R}.$$
This still does not sum to something useful.
Try instead
$$\frac{s_R}{h_R} = \min!\left(1,\frac{w_R}{h_R}\right).$$
For the rectangles meeting the line, if $w_R\le h_R$ then this equals $w_R/h_R\le w_R$; if $w_R\ge h_R$ it equals $1\le w_R$ because the square side is $1$ and any width is positive. Hence
$$\frac{s_R}{h_R}\le w_R.$$
Summing over rectangles crossed by the line,
$$\sum_{R\ni t}\frac{s_R}{h_R} \le \sum_{R\ni t} w_R.$$
But the intervals cut out on the horizontal line partition the whole side of the square, so
$$\sum_{R\ni t} w_R=1.$$
Integrating over $t\in[0,1]$ yields
$$\sum s_R = \int_0^1 \sum_{R\ni t}\frac{s_R}{h_R},dt \le 1.$$
This gives the wrong direction. The inequality is reversed from what is needed.
So the previous route cannot work.
Try a vertical line instead. For a rectangle, define contribution $s_R/w_R$ on each vertical line crossing it. Integrating gives $s_R$. For a fixed vertical line,
$$\frac{s_R}{w_R} = \min!\left(1,\frac{h_R}{w_R}\right) \le h_R.$$
Again we get $\sum s_R\le1$.
This suggests the true statement is not $\sum s_R\ge1$.
Check examples. If the square is cut into two vertical rectangles of widths $1/2$, each has $s=1/2$, so $\sum s=1$. If cut into four equal squares, $\sum s=2$. Thus $\sum s\ge1$ might still be true.
The integral argument should instead produce a lower bound. For a horizontal line $y=t$, rectangles meeting it have widths summing to $1$. Since
$$w_R\le \max(w_R,h_R),$$
we have
$$\frac{w_R}{\max(w_R,h_R)} = \frac{s_R}{h_R}.$$
Therefore
$$w_R\le h_R\cdot \frac{s_R}{h_R}=s_R$$
when $w_R\le h_R$, and
$$w_R=w_R\cdot1=w_R\cdot\frac{s_R}{h_R}$$
when $w_R\ge h_R$ is false. So this line fails.
The crucial observation is:
$$w_R \le \max(w_R,h_R),$$
hence
$$\frac{w_R}{\max(w_R,h_R)} \le 1.$$
But
$$\frac{w_R}{\max(w_R,h_R)} = \begin{cases} 1,& w_R\ge h_R,\[2mm] \dfrac{w_R}{h_R},& w_R\le h_R, \end{cases} = \dfrac{s_R}{h_R}.$$
Thus for every horizontal line,
$$\sum_{R\ni t}\frac{s_R}{h_R} = \sum_{R\ni t}\frac{w_R}{\max(w_R,h_R)} \ge \sum_{R\ni t} w_R = 1?$$
No, each term is at most $w_R$, not at least.
A different approach is needed.
Consider Cauchy in the form
$$\sum p_i = \sum \frac{s_i^2}{A_i} \ge \frac{(\sum s_i)^2}{\sum A_i} = (\sum s_i)^2.$$
Thus proving $\sum s_i\ge1$ would finish the problem.
To prove $\sum s_i\ge1$, take a vertical line $x=t$. Let $E(t)$ be the set of rectangles intersected by this line. Their heights sum to $1$ because they partition the line segment inside the square.
For every intersected rectangle $R$,
$$s_R=\min(w_R,h_R)\ge h_R,\mathbf 1_{{w_R\ge h_R}}.$$
Integrating the indicator over $t$, the measure of $t$ for which the line intersects $R$ equals $w_R$. Hence
$$\int_0^1 \sum_{R\in E(t)} \mathbf 1_{{w_R\ge h_R}},h_R,dt = \sum_{w_R\ge h_R} w_Rh_R.$$
Not enough.
A stronger statement is
$$s_R \ge h_R\cdot \frac{s_R}{w_R}.$$
Indeed $s_R/w_R\le1$. Integrating over vertical lines,
$$\sum s_R = \int_0^1 \sum_{R\in E(t)} h_R\frac{s_R}{w_R},dt.$$
For fixed $t$,
$$\sum_{R\in E(t)} h_R=1.$$
Also $s_R/w_R=\min(1,h_R/w_R)$.
Since $\min(1,x)$ is concave,
$$\sum h_R\min!\left(1,\frac{h_R}{w_R}\right) \ge \min!\left(1,\sum \frac{h_R^2}{w_R}\right).$$
This seems messy.
A more direct application of Cauchy on the vertical slice:
$$\sum_{R\in E(t)} \frac{h_R^2}{w_R} \ge \frac{(\sum h_R)^2}{\sum w_R}.$$
But for all intersected rectangles, $w_R>0$ and $\sum h_R=1$. Since each $w_R\le1$,
$$\sum w_R \le #E(t),$$
not useful.
Wait: for a rectangle,
$$\frac{s_R}{w_R} = \min!\left(1,\frac{h_R}{w_R}\right) \ge \frac{h_R}{w_R+h_R}.$$
Indeed for $x\ge0$, $\min(1,x)\ge x/(1+x)$.
Therefore
$$\sum s_R = \int_0^1 \sum_{R\in E(t)} h_R\frac{s_R}{w_R},dt \ge \int_0^1 \sum_{R\in E(t)} \frac{h_R^2}{w_R+h_R},dt.$$
For fixed $t$,
$$\sum \frac{h_R^2}{w_R+h_R} \ge \frac{(\sum h_R)^2}{\sum (w_R+h_R)} = \frac1{\sum (w_R+h_R)}.$$
Need this to be at least $1$, impossible.
Need a cleaner idea.
A known trick: assign to each rectangle weight equal to plumpness. Since plumpness is area divided by square of longer side:
$$p_R=\frac{A_R}{L_R^2}.$$
Hence
$$\sum p_R = \sum \frac{A_R}{L_R^2}.$$
Consider any vertical line. Rectangles crossing it have heights $\le L_R$. Therefore
$$\frac{A_R}{L_R^2} = \frac{h_Rw_R}{L_R^2} \ge \frac{h_R}{L_R}\cdot \frac{w_R}{L_R}.$$
Integrating over lines may yield a covering argument. The classical proof is likely Cauchy plus $\sum s_i\ge1$, and indeed that statement is true. The correct proof of $\sum s_i\ge1$ comes from projecting each rectangle's shorter side onto a side of the square. If the shorter side is horizontal, project its width onto the bottom side; total projected lengths from such rectangles cover at most 1. Likewise for vertical. Need lower bound, not upper.
Use area: color rectangles according to orientation. Let $H$ be rectangles with width $\le h$, so $s=w$. Then $\sum_{H} s$ equals total widths. Since their projections on the bottom side cover at most 1, not enough.
A standard argument: Let $S_H=\sum_{H} w$, $S_V=\sum_{V} h$. Every vertical line intersects $H$-rectangles in total height at most $S_H$? Integrating gives area(H). Since area(H)≤S_H. Similarly area(V)≤S_V. Total area is 1, so
$$1=\text{area}(H)+\text{area}(V)\le S_H+S_V=\sum s_i.$$
This works.
Problem Understanding
A square is partitioned into finitely many rectangles. For each rectangle, its plumpness is the ratio of its shorter side to its longer side. The task is to prove that the sum of the plumpnesses of all rectangles in the dissection is always at least $1$.
This is a Type B problem, a pure proof.
The core difficulty is to convert the geometric information about the dissection into a global inequality. The useful quantity turns out to be the sum of the shorter sides of the rectangles. After obtaining a lower bound for that sum, a Cauchy inequality yields the desired estimate for the sum of plumpnesses.
Proof Architecture
Let $s_i$ denote the shorter side, $L_i$ the longer side, and $A_i=s_iL_i$ the area of the $i$-th rectangle.
Lemma 1: $\sum_i s_i\ge1$. Split the rectangles into those whose shorter side is horizontal and those whose shorter side is vertical, estimate the total area of each class by the sum of its shorter sides, and use that the total area of the square equals $1$.
Lemma 2: $\sum_i \dfrac{s_i^2}{A_i}\ge\dfrac{(\sum_i s_i)^2}{\sum_i A_i}$. This is the Engel form of the Cauchy inequality.
Lemma 3: Since $A_i=s_iL_i$, the plumpness of the $i$-th rectangle equals $\dfrac{s_i^2}{A_i}$.
The hardest part is Lemma 1, where one must obtain the lower bound $\sum s_i\ge1$ from the geometry of the dissection.
Solution
Normalize the square so that its side length is $1$. Let the rectangles of the dissection be $R_1,\dots,R_n$.
For the rectangle $R_i$, let $s_i$ be its shorter side, $L_i$ its longer side, and $A_i=s_iL_i$ its area. Its plumpness is
$$p_i=\frac{s_i}{L_i}=\frac{s_i^2}{A_i}.$$
We first prove that
$$\sum_{i=1}^n s_i\ge1.$$
Divide the rectangles into two classes.
The first class consists of rectangles whose shorter side is horizontal. For such a rectangle, its width equals its shorter side. Let the sum of these widths be $S_H$.
The second class consists of rectangles whose shorter side is vertical. For such a rectangle, its height equals its shorter side. Let the sum of these heights be $S_V$.
Then
$$\sum_{i=1}^n s_i=S_H+S_V.$$
Let $A_H$ be the total area of the rectangles in the first class. Every such rectangle has height at most $1$, because it lies inside the unit square. Hence for each rectangle in the first class,
$$\text{area}\le \text{width}.$$
Summing over that class gives
$$A_H\le S_H.$$
Similarly, if $A_V$ denotes the total area of the second class, every rectangle in that class has width at most $1$, so
$$\text{area}\le \text{height},$$
and therefore
$$A_V\le S_V.$$
Since the rectangles form a dissection of the unit square,
$$A_H+A_V=1.$$
Combining the inequalities,
$$1=A_H+A_V\le S_H+S_V=\sum_{i=1}^n s_i.$$
Thus
$$\sum_{i=1}^n s_i\ge1.$$
Now apply the Engel form of the Cauchy inequality:
$$\sum_{i=1}^n \frac{s_i^2}{A_i} \ge \frac{\left(\sum_{i=1}^n s_i\right)^2} {\sum_{i=1}^n A_i}.$$
The total area of the rectangles is the area of the unit square, so
$$\sum_{i=1}^n A_i=1.$$
Using $\sum s_i\ge1$, we obtain
$$\sum_{i=1}^n \frac{s_i^2}{A_i} \ge \left(\sum_{i=1}^n s_i\right)^2 \ge1.$$
Since $\dfrac{s_i^2}{A_i}=p_i$ for every rectangle,
$$\sum_{i=1}^n p_i\ge1.$$
This completes the proof.
∎
Verification of Key Steps
The most delicate step is the proof that $\sum s_i\ge1$.
For rectangles whose shorter side is horizontal, the width equals the shorter side. Because the square has side length $1$, every such rectangle has height at most $1$. Hence
$$\text{area}=\text{width}\cdot\text{height}\le\text{width}.$$
Summing yields $A_H\le S_H$. The analogous argument gives $A_V\le S_V$. Since $A_H+A_V=1$, the conclusion follows. A careless argument might replace the inequality $\text{height}\le1$ by an unstated assumption; that bound is the essential geometric input.
The second delicate step is the Cauchy application. Writing
$$\sum_{i=1}^n \frac{s_i^2}{A_i} = \sum_{i=1}^n \frac{(s_i)^2}{A_i},$$
the Engel inequality gives
$$\sum \frac{s_i^2}{A_i} \ge \frac{(\sum s_i)^2}{\sum A_i}.$$
The denominator is exactly $1$ because the rectangles partition the unit square. Forgetting this normalization would leave an incorrect factor.
The identity
$$p_i=\frac{s_i}{L_i}=\frac{s_i^2}{s_iL_i}=\frac{s_i^2}{A_i}$$
is the bridge between geometry and the Cauchy inequality. Any error here would change the quantity being estimated.
Alternative Approaches
A different proof starts from the representation
$$p_i=\frac{A_i}{L_i^2},$$
where $L_i$ is the longer side of the rectangle. One can interpret $A_i/L_i$ as the shorter side and derive a lower bound for the sum of these quantities by integrating suitable cross-sections of the square. Afterward, Cauchy's inequality again converts that lower bound into the desired estimate for $\sum p_i$.
The presented argument is preferable because it isolates a simple geometric fact, namely $\sum s_i\ge1$, and then finishes with a single standard inequality. The geometry and the algebra remain completely separate, which makes every step transparent.