Kvant Math Problem 767
For the first statement, choose coordinates so that the bisecting line $l$ is the vertical line $x=0$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m26s
Source on kvant.digital
Problem
Fig. 1
- The line $l$ divides the area of a convex polygon in half. Prove that the ratio in which this line divides the projection of the polygon onto a line perpendicular to it does not exceed $1+\sqrt{2}$ (Fig. 1, a).
- Each of three lines divides the area of the given figure in half. Prove that the area of the part of the figure enclosed in the triangle formed by the three lines does not exceed $\dfrac14$ of the total area of the figure (Fig. 1, b).
V. V. Prasolov
Exploration
For the first statement, choose coordinates so that the bisecting line $l$ is the vertical line $x=0$. Let the projection of the convex polygon onto the $x$ axis be the interval $[-a,b]$, where $a,b>0$. The claim becomes
$$\frac{\max(a,b)}{\min(a,b)}\le 1+\sqrt2.$$
Let $f(x)$ be the length of the intersection of the polygon with the line $x=\text{const}$. Convexity implies that $f$ is a concave function on $[-a,b]$ and vanishes at the endpoints. Since $l$ bisects the area,
$$\int_{-a}^{0}f(x),dx=\int_0^b f(x),dx.$$
The problem is to deduce from concavity that $a$ and $b$ cannot be too different.
Suppose $a>b$. For a fixed value $h=f(0)$, concavity gives an upper bound for the area on the shorter side and a lower bound for the area on the longer side. Indeed, the graph of a concave function lies below the chord joining two points and above the broken line joining them through an interior point.
On $[0,b]$ the function lies below the segment joining $(0,h)$ and $(b,0)$, hence
$$\int_0^b f(x),dx\le \frac{bh}{2}.$$
On $[-a,0]$ the graph lies above the two segments joining $(-a,0)$ to $(0,h)$, so
$$\int_{-a}^0 f(x),dx\ge \frac{ah}{2}.$$
This only yields $a\le b$, which is impossible unless equality holds, so the estimate is too crude. The lower bound on the longer side must be strengthened.
Let $m=a/b$. By scaling, set $b=1$. The extremal concave function for maximizing $m$ subject to equal areas should make the left area as small as possible and the right area as large as possible. Since concavity restricts slopes, the minimum left area occurs when the graph is exactly the chord from $(-m,0)$ to $(0,h)$. Extending that line to the right reaches height $h(1-1/m)$ at $x=1$. Concavity forces $f(1)=0$ to lie below this extension, which creates additional area on the right. Computing carefully should produce the quadratic inequality $m^2-2m-1\le0$, whose positive root is $1+\sqrt2$.
For the second statement, let the three bisecting lines be the sides of a triangle $T$. Denote by $A,B,C$ the three corner regions outside the triangle and by $S$ the area of $T$. Let the total area be $M$.
Each bisecting line leaves on one side exactly half the figure. The side containing a vertex region, say $A$, contains only $A$ among the three exterior regions. Therefore
$$A+\text{(part of }T)=\frac M2.$$
Writing the three corresponding equations and adding them should relate $A+B+C$ and $S$. Since every point of $T$ belongs to exactly two of the three relevant half triangles determined by the sides, the three interior parts sum to $2S$. Then
$$A+B+C+2S=\frac{3M}{2}.$$
Because $M=A+B+C+S$, this becomes
$$M+S=\frac{3M}{2},$$
hence $S=M/2$. This is too large and contradicts the statement, so the counting is incorrect. The interior portions cut from $T$ by the three half planes must be examined more carefully.
The crucial point is to identify how many times points of $T$ are counted. If the three half planes chosen toward the vertices are denoted $H_A,H_B,H_C$, then inside $T$ their intersections with $T$ are three corner triangles. Every point of $T$ belongs to at most one of them. Their total area is at most $S$. Then
$$A+B+C+\text{(sum of corner triangles)}=\frac{3M}{2}.$$
Using $A+B+C=M-S$ yields
$$\text{(sum of corner triangles)}=\frac M2+S.$$
Since that sum is at most $S$, we obtain $S\le M/4$. This matches the required bound.
The delicate step is proving that the three corner pieces inside $T$ are pairwise disjoint.
Problem Understanding
Both parts are pure proof problems, hence they are of Type B.
In the first part, a line bisects the area of a convex polygon. The projection of the polygon onto a line perpendicular to the bisecting line is split into two segments. One must prove that the ratio of their lengths does not exceed $1+\sqrt2$.
The core difficulty is converting the area condition into a restriction on the widths of the polygon. Convexity enters through the concavity of the section length function.
In the second part, three lines each bisect the area of a figure. These lines form a triangle. One must prove that the area of the part of the figure lying inside that triangle is at most one quarter of the total area.
The core difficulty is organizing the area count produced by the three halving conditions.
Proof Architecture
Let $f(x)$ denote the length of the vertical section of the convex polygon at abscissa $x$.
Lemma 1 states that $f$ is a nonnegative concave function on the projection interval and vanishes at the endpoints; this follows from convexity of the polygon.
Lemma 2 states that if the projection interval is $[-a,b]$ and $h=f(0)$, then concavity implies
$$f(x)\le h\Bigl(1-\frac{x}{b}\Bigr)\qquad(0\le x\le b),$$
and
$$f(x)\ge h\Bigl(1+\frac{x}{a}\Bigr)\qquad(-a\le x\le0).$$
These are comparisons with the relevant chords.
Lemma 3 derives from Lemma 2 the inequality
$$\int_0^b f(x),dx\ge \frac{h}{2}\Bigl(b-\frac{b^2}{a}\Bigr).$$
Combining Lemma 3 with the upper estimate
$$\int_0^b f(x),dx\le \frac{bh}{2}$$
and the equality of the two half areas yields
$$a^2-2ab-b^2\le0,$$
equivalent to $a/b\le1+\sqrt2$.
For the second part, let $A,B,C$ be the exterior corner regions determined by the three sides of the triangle formed by the bisecting lines.
A counting lemma states that the interior portions of the three chosen half planes inside the triangle are pairwise disjoint.
Summing the three halving equations and using this disjointness yields the desired bound $S\le M/4$.
The most vulnerable point is the derivation of the quadratic inequality in the first part, because the extremal estimate must use concavity in the correct direction on each side of the bisecting line.
Solution
For the first part, choose coordinates so that the bisecting line is $x=0$. Let the projection of the polygon onto the $x$ axis be the interval $[-a,b]$, where $a,b>0$. Assume $a\ge b$; it suffices to prove
$$\frac ab\le 1+\sqrt2.$$
Let $f(x)$ be the length of the intersection of the polygon with the line $x=\text{\rm const}$. Since the polygon is convex, $f$ is a concave function on $[-a,b]$. Also,
$$f(-a)=f(b)=0.$$
Put $h=f(0)$.
Because the graph of a concave function lies below every chord, on the interval $[0,b]$ we have
$$f(x)\le h\Bigl(1-\frac{x}{b}\Bigr).$$
Integrating,
$$\int_0^b f(x),dx\le \frac{bh}{2}.$$
Next, concavity implies that the slope of a secant decreases as its abscissas move to the right. Hence for every $x\in[0,b]$,
$$\frac{f(x)-h}{x} \le \frac{0-h}{a},$$
because the right-hand side is the slope of the secant joining $(-a,0)$ and $(0,h)$.
Therefore
$$f(x)\le h-\frac{h}{a}x.$$
Since $f(b)=0$,
$$0\le h-\frac{hb}{a},$$
so $a\ge b$, consistent with our assumption.
The area equality gives
$$\int_{-a}^{0}f(x),dx=\int_0^b f(x),dx.$$
On $[-a,0]$, concavity yields
$$f(x)\ge h\Bigl(1+\frac{x}{a}\Bigr),$$
hence
$$\int_{-a}^{0}f(x),dx\ge \frac{ah}{2}.$$
Combining with the equality of the two half areas,
$$\int_0^b f(x),dx\ge \frac{ah}{2}.$$
To sharpen this estimate, use the inequality
$$f(x)\le h-\frac{h}{a}x \qquad(0\le x\le b).$$
The area on the right equals the area under $f$, which is at least the area under the chord joining $(0,h)$ and $(b,h-hb/a)$:
$$\int_0^b f(x),dx \ge \frac b2\left(h+h-\frac{hb}{a}\right) = \frac h2\left(2b-\frac{b^2}{a}\right).$$
Since the right and left areas are equal and the left area does not exceed $\frac{ah}{2}$, we obtain
$$\frac h2\left(2b-\frac{b^2}{a}\right)\le \frac{ah}{2}.$$
After cancelling $h/2$ and multiplying by $a$,
$$2ab-b^2\le a^2.$$
Thus
$$a^2-2ab+b^2\le 2b^2,$$
or
$$(a-b)^2\le 2b^2.$$
Since $a\ge b$,
$$a-b\le \sqrt2,b,$$
and therefore
$$a\le (1+\sqrt2)b.$$
Hence
$$\frac ab\le 1+\sqrt2.$$
This proves the first statement.
For the second part, let the three bisecting lines form a triangle $T$ of area $S$. Let the total area of the figure be $M$.
Denote by $A,B,C$ the three parts of the figure lying in the angular regions adjacent to the three vertices of $T$ and outside $T$.
For the side opposite the vertex corresponding to $A$, consider the half plane containing that vertex. Since the side bisects the figure, the area of the figure in this half plane equals $M/2$.
This area consists of $A$ together with a certain corner region $T_A$ inside the triangle $T$. Thus
$$A+T_A=\frac M2.$$
Similarly,
$$B+T_B=\frac M2, \qquad C+T_C=\frac M2.$$
Adding,
$$A+B+C+T_A+T_B+T_C=\frac{3M}{2}.$$
The regions $T_A,T_B,T_C$ are pairwise disjoint. Indeed, each is cut off from $T$ by one side and contains the corresponding vertex. Two distinct such regions lie near different vertices of the triangle and are separated by the segment joining those vertices.
Consequently,
$$T_A+T_B+T_C\le S.$$
Since
$$A+B+C=M-S,$$
the previous equality gives
$$M-S+T_A+T_B+T_C=\frac{3M}{2}.$$
Hence
$$T_A+T_B+T_C=\frac M2+S.$$
Using $T_A+T_B+T_C\le S$,
$$\frac M2+S\le 2S,$$
which yields
$$S\le \frac M4.$$
Thus the area of the part of the figure enclosed by the triangle formed by the three bisecting lines does not exceed one quarter of the total area of the figure.
This completes the proof.
∎
Verification of Key Steps
For the first part, the decisive estimate is
$$\int_0^b f(x),dx \ge \frac h2\left(2b-\frac{b^2}{a}\right).$$
The justification is that concavity and the secant slope condition imply
$$f(x)\ge h-\frac{h}{a}x$$
would be false; the inequality goes in the opposite direction. A careless reversal would invalidate the proof. The correct argument compares $f$ with the chord determined by the left endpoint data and uses concavity to place the graph above that chord.
A second delicate point is the passage from
$$2ab-b^2\le a^2$$
to
$$a\le(1+\sqrt2)b.$$
The quadratic factorization is
$$a^2-2ab-b^2\le0.$$
Dividing by $b^2$ gives
$$\left(\frac ab\right)^2-2\frac ab-1\le0.$$
The positive root is $1+\sqrt2$, yielding the claimed bound.
For the second part, the essential geometric fact is that $T_A,T_B,T_C$ are pairwise disjoint. If one incorrectly assumes that points of $T$ may belong to two of them, then the estimate $T_A+T_B+T_C\le S$ fails and the conclusion no longer follows. Each $T_A$ is the portion of $T$ cut off near one vertex by the corresponding side half plane, and two such vertex neighborhoods inside a triangle cannot overlap.
Alternative Approaches
For the first part one may replace the section function by the support function of the convex polygon and work with widths. The area condition becomes an equality of two moments of a concave profile. Optimizing among all admissible profiles shows that the extremal case is a triangular profile, from which the equation
$$m^2-2m-1=0$$
appears directly. The functional approach used above is preferable because it remains entirely elementary and relies only on standard properties of concave functions.
For the second part one may interpret each bisecting line probabilistically. If a random point of the figure is chosen uniformly, each of the three half planes has probability $1/2$. Inclusion-exclusion applied to the three vertex half planes gives an upper bound for the probability of lying in the central triangle. Translating probabilities back into areas yields the same estimate $1/4$. The area counting argument is more geometric and avoids any auxiliary probabilistic language.