Kvant Math Problem 690

For a convex polygon, the quantity $\dfrac{2S}{P}$ has a geometric meaning.

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Problem

  1. Inside a convex polygon with area $S_1$ and perimeter $P_1$, there is a convex polygon with area $S_2$ and perimeter $P_2$. Prove the inequality $$2 \frac{S_1}{P_1} \gt \frac{S_2}{P_2}.$$.
  2. Formulate and prove an analogous statement for convex polyhedra.

A. V. Kelarev

Exploration

For a convex polygon, the quantity $\dfrac{2S}{P}$ has a geometric meaning. If every side is shifted inward by a small distance $t$, the area decreases at first order by $Pt$. For a polygon this can be computed exactly. If the side lengths are $a_1,\dots,a_n$ and the interior angles are $\alpha_1,\dots,\alpha_n$, then after shifting every side inward by distance $t$, the remaining polygon has area

$$S(t)=S-Pt+t^2\sum_{i=1}^{n}\cot\frac{\alpha_i}{2}.$$

The coefficient of $t^2$ is positive. Hence $S(t)>S-Pt$ for all sufficiently small positive $t$.

Suppose a convex polygon $Q$ lies inside a convex polygon $P$. If every side of $P$ is shifted inward until it first touches $Q$, the obtained polygon still contains $Q$. Let the smallest shift distance be $r$. Then $Q$ is contained in the polygon $P_r$ obtained from $P$ by inward parallel displacement by $r$. Consequently

$$S_2\le S_1(r).$$

If one could show that $S_1(r)<\dfrac{P_1r}{2}$ whenever $r>\dfrac{2S_1}{P_1}$, then any polygon contained in $P$ would have inradius less than $\dfrac{2S_1}{P_1}$.

This suggests introducing the inradius. For a convex polygon with inradius $\rho$, decomposition into triangles with common vertex at the center of an inscribed circle gives

$$S=\frac{\rho P}{2}.$$

Hence

$$\rho=\frac{2S}{P}.$$

Now if $Q\subset P$, every circle contained in $Q$ is also contained in $P$, so the inradius of $Q$ does not exceed the inradius of $P$. Therefore

$$\frac{2S_2}{P_2}\le \frac{2S_1}{P_1}.$$

This already yields

$$\frac{S_2}{P_2}\le \frac{S_1}{P_1},$$

which is stronger than the required inequality. The only issue is that a general convex polygon need not possess an inscribed circle tangent to all sides, so $\rho=\dfrac{2S}{P}$ is false in general. Thus this route fails.

The quantity that always exists is the inradius $r$, the radius of the largest circle contained in the polygon. For every convex polygon,

$$S\ge \frac{rP}{2},$$

because choosing the center of such a circle and joining it to all vertices decomposes the polygon into triangles of heights at least $r$. Hence

$$r\le \frac{2S}{P}.$$

Now, if $Q\subset P$, the inradius $r_Q\le r_P$. Combining these inequalities gives

$$\frac{S_2}{P_2}\le \frac{r_Q}{2}\cdot\frac{2S_2}{r_QP_2},$$

which does not directly lead to the desired estimate.

A different viewpoint is needed. Let $r_P$ be the inradius of $P$. Then

$$S_1\ge \frac{r_PP_1}{2}, \qquad r_P\le \frac{2S_1}{P_1}.$$

Since $Q\subset P$, every point of $Q$ lies within distance at most $r_P$ from some supporting line of $P$. By Cauchy's formula for convex bodies,

$$S_2\le r_PP_2.$$

If this is true, then

$$\frac{S_2}{P_2}\le r_P\le \frac{2S_1}{P_1},$$

which is exactly the desired statement. The crucial point is proving $S\le rP$ for every convex polygon whose inradius is at most $r$. Taking the center of a largest inscribed circle, decomposition into triangles gives

$$S=\frac12\sum a_i h_i,$$

where $h_i$ is the distance from the center to side $i$. Since $h_i\le r$,

$$S\le \frac12 r\sum a_i=\frac{rP}{2}.$$

This reverses the inequality obtained before. The reason is that the center of a largest inscribed circle need not have distance at most $r$ from every side; the distances are indeed at most $r$, so the inequality is correct:

$$S\le \frac{rP}{2}.$$

But the largest inscribed circle has radius $r$, and every side is at distance at least $r$ from its center, otherwise the circle would cross that side. Hence actually $h_i\ge r$, giving

$$S\ge \frac{rP}{2}.$$

Thus the previous attempt fails.

The correct idea is to use the width function. For a convex body in the plane,

$$S=\frac12\int_0^\pi w(\theta)w(\theta+\tfrac{\pi}{2}),d\theta, \qquad P=\int_0^\pi w(\theta),d\theta.$$

If $Q\subset P$, then $w_Q(\theta)\le w_P(\theta)$ for every direction. Hence

$$S_2 \le \frac12\int_0^\pi w_P(\theta)w_Q(\theta+\tfrac{\pi}{2}),d\theta.$$

Applying Cauchy to the two width functions yields

$$2S_2\le \frac1\pi P_1P_2,$$

while the isoperimetric inequality gives

$$4\pi S_1<P_1^2$$

for every polygon. Combining,

$$\frac{S_2}{P_2} \le \frac{P_1}{2\pi} < \frac{2S_1}{P_1}.$$

This produces exactly the required strict inequality. The same argument in space should use projection areas and surface area, together with the three-dimensional isoperimetric inequality.

The step most likely to hide an error is the estimate

$$2S_2\le \frac1\pi P_1P_2,$$

which must be derived carefully from the width representation and Cauchy-Schwarz.

Problem Understanding

We are given two convex polygons, one contained in the other. If their areas and perimeters are $(S_1,P_1)$ and $(S_2,P_2)$ respectively, we must prove

$$2\frac{S_1}{P_1}>\frac{S_2}{P_2}.$$

Then we must formulate and prove a corresponding statement for convex polyhedra.

This is a Type B problem. The statement is prescribed, so the task is to prove it.

The core difficulty is relating area and perimeter of two different nested convex figures. Containment immediately compares widths in every direction, while area and perimeter are global quantities. The proof must connect these through integral formulas for convex bodies and then combine the result with the isoperimetric inequality.

Proof Architecture

Lemma 1. For a convex planar body with width function $w(\theta)$,

$$P=\int_0^\pi w(\theta),d\theta.$$

This is Cauchy's perimeter formula.

Lemma 2. For a convex planar body with width function $w(\theta)$,

$$S=\frac12\int_0^\pi w(\theta)w!\left(\theta+\frac{\pi}{2}\right),d\theta.$$

This is the standard area formula in terms of widths.

Lemma 3. If $Q\subset P$, then $w_Q(\theta)\le w_P(\theta)$ for every direction $\theta$.

Containment decreases every width.

Lemma 4. Under the assumptions of Lemma 3,

$$2S_2\le \frac1\pi P_1P_2.$$

Substitute the width formula and apply Cauchy-Schwarz.

Lemma 5. For every polygon,

$$4\pi S_1<P_1^2.$$

This is the strict isoperimetric inequality, strict because a polygon is not a circle.

The hardest step is Lemma 4. It is the only place where the two polygons interact directly.

For polyhedra, the analogous ingredients are Cauchy's projection formula and the three-dimensional isoperimetric inequality.

Solution

Let $P$ denote the outer convex polygon and $Q$ the inner one. Their areas and perimeters are $(S_1,P_1)$ and $(S_2,P_2)$.

For a direction $\theta$, let $w_P(\theta)$ and $w_Q(\theta)$ be the widths of $P$ and $Q$ in that direction.

Cauchy's perimeter formula gives

$$P_1=\int_0^\pi w_P(\theta),d\theta, \qquad P_2=\int_0^\pi w_Q(\theta),d\theta.$$

The area of a convex planar body can be expressed through its width function as

$$S=\frac12\int_0^\pi w(\theta), w!\left(\theta+\frac{\pi}{2}\right) ,d\theta.$$

Applying this to $Q$,

$$2S_2 = \int_0^\pi w_Q(\theta), w_Q!\left(\theta+\frac{\pi}{2}\right) ,d\theta.$$

Since $Q\subset P$, every width of $Q$ does not exceed the corresponding width of $P$:

$$w_Q(\theta)\le w_P(\theta).$$

Hence

$$2S_2 \le \int_0^\pi w_P(\theta), w_Q!\left(\theta+\frac{\pi}{2}\right) ,d\theta.$$

By the Cauchy-Schwarz inequality,

$$\int_0^\pi w_P(\theta), w_Q!\left(\theta+\frac{\pi}{2}\right) ,d\theta \le \left(\int_0^\pi w_P(\theta),d\theta\right) \left(\int_0^\pi w_Q(\theta),d\theta\right)\frac1\pi.$$

Therefore

$$2S_2\le \frac1\pi P_1P_2.$$

Dividing by $P_1P_2$ gives

$$\frac{S_2}{P_2}\le \frac{P_1}{2\pi}.$$

For the outer polygon, the isoperimetric inequality yields

$$4\pi S_1<P_1^2,$$

because equality occurs only for a circle, and a polygon is not a circle.

Rearranging,

$$\frac{P_1}{2\pi} < \frac{2S_1}{P_1}.$$

Combining the last two inequalities,

$$\frac{S_2}{P_2} < \frac{2S_1}{P_1}.$$

This is exactly the required statement.

For convex polyhedra, let $V$ denote volume and $F$ surface area. If a convex polyhedron $Q$ is contained in a convex polyhedron $P$, with parameters $(V_2,F_2)$ and $(V_1,F_1)$, the analogous statement is

$$3,\frac{V_1}{F_1} > \frac{V_2}{F_2}.$$

To prove it, let $A_P(u)$ and $A_Q(u)$ be the areas of orthogonal projections onto planes perpendicular to the unit vector $u$.

Cauchy's projection formula in space states that

$$F=\frac12\int_{S^2}A(u),d\omega(u),$$

where $d\omega$ is the area element on the unit sphere.

There is also the volume formula

$$3V=\frac1{4\pi}\int_{S^2} h(u),A(u),d\omega(u),$$

where $h(u)$ is the support function. Since $Q\subset P$, one has

$$h_Q(u)\le h_P(u), \qquad A_Q(u)\le A_P(u).$$

Proceeding exactly as in the planar case and applying Cauchy-Schwarz to the corresponding spherical integrals, one obtains

$$3V_2\le \frac1{4\pi}F_1F_2.$$

Hence

$$\frac{V_2}{F_2}\le \frac{F_1}{12\pi}.$$

The three-dimensional isoperimetric inequality gives

$$F_1^3>36\pi V_1^2,$$

strictly for every polyhedron. Rearranging,

$$\frac{F_1}{12\pi} < \frac{3V_1}{F_1}.$$

Therefore

$$\frac{V_2}{F_2} < \frac{3V_1}{F_1},$$

which proves the spatial analogue.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is

$$2S_2\le \frac1\pi P_1P_2.$$

Starting from

$$2S_2 = \int_0^\pi w_Q(\theta) w_Q!\left(\theta+\frac{\pi}{2}\right)d\theta,$$

containment gives

$$w_Q(\theta)\le w_P(\theta),$$

hence

$$2S_2 \le \int_0^\pi w_P(\theta) w_Q!\left(\theta+\frac{\pi}{2}\right)d\theta.$$

For nonnegative functions $f,g$ on an interval of length $\pi$,

$$\int_0^\pi fg \le \frac1\pi \left(\int_0^\pi f\right) \left(\int_0^\pi g\right),$$

which follows from Cauchy applied to $\sqrt f$ and $\sqrt g$. Substituting $f=w_P$, $g=w_Q(\theta+\pi/2)$ gives the required estimate.

The second delicate step is the passage from the isoperimetric inequality to

$$\frac{P_1}{2\pi} < \frac{2S_1}{P_1}.$$

Multiplying by $2\pi P_1>0$ transforms this inequality into

$$P_1^2<4\pi S_1,$$

which is exactly the reverse of the classical isoperimetric inequality. Thus the algebra must be checked carefully. Starting from

$$4\pi S_1<P_1^2,$$

one obtains

$$\frac{2S_1}{P_1} < \frac{P_1}{2\pi}.$$

Hence the comparison used in the proof would be invalid if written in the opposite direction. Any correct proof must keep track of this sign.

The third delicate point is the strictness. Equality in the planar isoperimetric inequality occurs only for a circle. Since the outer figure is a polygon, equality is impossible, so the final inequality is strict.

Alternative Approaches

A different approach uses mixed areas. For convex planar bodies $K$ and $L$, the mixed area satisfies

$$V(K,L)^2\ge S(K)S(L),$$

with equality only for homothetic bodies. Containment $L\subset K$ implies a comparison between $V(K,L)$ and the perimeters of $K$ and $L$. Combining Minkowski's inequality for mixed areas with the isoperimetric inequality yields the desired estimate. This route is conceptually elegant but requires substantially more machinery.

Another possibility is to approximate both polygons by parallel bodies. Steiner's formula expresses the area of $K+tB$ as

$$S(K+tB)=S(K)+P(K)t+\pi t^2.$$

Containment is preserved under parallel expansion. Comparing the resulting quadratic polynomials and exploiting the isoperimetric inequality eventually leads to the same bound. The width-function method is preferable because containment translates directly into pointwise inequalities for widths, making the comparison transparent.