Kvant Math Problem 1089

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Problem

In the convex quadrilateral $ABCD$ with area $S$, the diagonals intersect at the point $O$. Let $K$, $L$, $M$, $N$ be the centers of the circles inscribed in the triangles $AOB$, $BOC$, $COD$, and $DOA$, respectively. Prove that the product of the perimeters of the quadrilaterals $ABCD$ and $KLMN$ is at least $4S$.

D. Yu. Burago, F. L. Nazarov

Exploration

Let the inradius of triangle $AOB$ be $r_1$, of $BOC$ be $r_2$, of $COD$ be $r_3$, and of $DOA$ be $r_4$.

The perimeter of $ABCD$ is

$$P=AB+BC+CD+DA.$$

The area decomposes as

$$S=[AOB]+[BOC]+[COD]+[DOA].$$

Since the area of a triangle equals its inradius times semiperimeter,

$$[AOB]=r_1s_1,\qquad [BOC]=r_2s_2,\qquad [COD]=r_3s_3,\qquad [DOA]=r_4s_4,$$

where $s_i$ denotes the corresponding semiperimeter.

The quadrilateral $KLMN$ is formed by the incenters of these four triangles. The first task is to understand its side lengths.

Consider two adjacent triangles, for instance $AOB$ and $BOC$. Their incenters are $K$ and $L$. Both triangles share the side $OB$. If $E$ and $F$ are the touch points of the incircles with $OB$, then the perpendicular distances of $K$ and $L$ from $OB$ are $r_1$ and $r_2$. Along $OB$ the projections of $K$ and $L$ are exactly $E$ and $F$.

A useful computation is

$$OE=s_1-AO,\qquad OF=BO-(s_2-OC).$$

Since

$$s_1=\frac{AO+BO+AB}{2},\qquad s_2=\frac{BO+OC+BC}{2},$$

one gets

$$EF=\frac{AB+BC}{2}.$$

Hence

$$KL^2=\left(\frac{AB+BC}{2}\right)^2+(r_1+r_2)^2.$$

This already yields

$$KL\ge \frac{AB+BC}{2}.$$

Analogous formulas hold for the other sides:

$$LM\ge \frac{BC+CD}{2},\quad MN\ge \frac{CD+DA}{2},\quad NK\ge \frac{DA+AB}{2}.$$

Summing,

$$p:=\operatorname{per}(KLMN)\ge P.$$

If this is true, then the desired inequality becomes a consequence of

$$P^2\ge 4S,$$

because then

$$Pp\ge P^2\ge 4S.$$

The inequality $P^2\ge 4S$ is the classical isoperimetric inequality for convex quadrilaterals, and in fact follows from the stronger fact that among quadrilaterals with fixed perimeter, the square has maximal area:

$$S\le \frac{P^2}{16}.$$

The potentially dangerous step is the computation

$$EF=\frac{AB+BC}{2}.$$

Everything rests on obtaining the side lengths of $KLMN$ correctly.

Problem Understanding

We are given a convex quadrilateral $ABCD$ of area $S$. The diagonals meet at $O$. For each of the four triangles determined by the diagonals, we take its incenter. These incenters are denoted by $K,L,M,N$.

We must prove that

$$\operatorname{per}(ABCD)\cdot \operatorname{per}(KLMN)\ge 4S.$$

This is a Type B problem. The statement is a pure inequality that must be proved.

The core difficulty is to relate the geometry of the incenters to the sides of the original quadrilateral. Once a lower bound for the perimeter of $KLMN$ in terms of the perimeter of $ABCD$ is obtained, the problem reduces to a standard area-perimeter inequality for convex quadrilaterals.

Proof Architecture

Lemma 1. If $K$ and $L$ are the incenters of triangles $AOB$ and $BOC$, then

$$KL^2=\left(\frac{AB+BC}{2}\right)^2+(r_1+r_2)^2,$$

where $r_1,r_2$ are the corresponding inradii; this follows from orthogonal projection onto the common side $OB$.

Lemma 2. Consequently,

$$KL\ge \frac{AB+BC}{2},$$

and similarly

$$LM\ge \frac{BC+CD}{2},\quad MN\ge \frac{CD+DA}{2},\quad NK\ge \frac{DA+AB}{2}.$$

Lemma 3. Summing the inequalities of Lemma 2 gives

$$\operatorname{per}(KLMN)\ge \operatorname{per}(ABCD).$$

Lemma 4. For every convex quadrilateral of perimeter $P$ and area $S$,

$$S\le \frac{P^2}{16},$$

hence

$$P^2\ge 16S\ge 4S.$$

The hardest point is Lemma 1, namely the exact computation of the distance between the projections of two neighboring incenters onto their common side.

Solution

Let

$$P=\operatorname{per}(ABCD), \qquad p=\operatorname{per}(KLMN).$$

Denote by $r_1,r_2,r_3,r_4$ the inradii of triangles $AOB$, $BOC$, $COD$, $DOA$, respectively.

Consider the neighboring triangles $AOB$ and $BOC$. Let $E$ and $F$ be the touch points of their incircles with the common side $OB$.

Since $K$ and $L$ are the incenters, the segments $KE$ and $LF$ are perpendicular to $OB$, and

$$KE=r_1,\qquad LF=r_2.$$

Because the triangles lie on opposite sides of the line $OB$, the perpendicular separation of $K$ and $L$ equals

$$r_1+r_2.$$

Let

$$s_1=\frac{AO+BO+AB}{2}, \qquad s_2=\frac{BO+OC+BC}{2}$$

be the semiperimeters of triangles $AOB$ and $BOC$.

For a triangle, the tangent lengths from the same vertex are equal. Hence on side $OB$,

$$OE=s_1-AO, \qquad BF=s_2-OC.$$

Therefore

$$\begin{aligned} EF &=OB-OE-BF\ &=OB-(s_1-AO)-(s_2-OC). \end{aligned}$$

Substituting the expressions for $s_1$ and $s_2$,

$$\begin{aligned} EF &=OB-\frac{AO+BO+AB}{2}+AO -\frac{BO+OC+BC}{2}+OC\ &=\frac{AB+BC}{2}. \end{aligned}$$

Since the projections of $K$ and $L$ onto $OB$ are $E$ and $F$, respectively,

$$KL^2=EF^2+(r_1+r_2)^2 =\left(\frac{AB+BC}{2}\right)^2+(r_1+r_2)^2.$$

Hence

$$KL\ge \frac{AB+BC}{2}.$$

Applying the same argument to the other three pairs of neighboring triangles gives

$$LM\ge \frac{BC+CD}{2},$$

$$MN\ge \frac{CD+DA}{2},$$

$$NK\ge \frac{DA+AB}{2}.$$

Adding these four inequalities,

$$\begin{aligned} p &\ge \frac{AB+BC}{2} +\frac{BC+CD}{2} +\frac{CD+DA}{2} +\frac{DA+AB}{2}\ &=AB+BC+CD+DA\ &=P. \end{aligned}$$

Thus

$$p\ge P.$$

Now use the classical isoperimetric inequality for convex quadrilaterals:

$$S\le \frac{P^2}{16}.$$

Consequently,

$$P^2\ge 16S.$$

Combining this with $p\ge P$,

$$Pp\ge P^2\ge 16S\ge 4S.$$

Therefore

$$\operatorname{per}(ABCD)\cdot \operatorname{per}(KLMN)\ge 4S.$$

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the computation of $EF$. Writing

$$OE=s_1-AO=\frac{BO+AB-AO}{2},$$

and

$$BF=s_2-OC=\frac{BO+BC-OC}{2},$$

gives

$$\begin{aligned} EF &=OB-OE-BF\ &=OB-\frac{BO+AB-AO}{2} -\frac{BO+BC-OC}{2}\ &=\frac{AO+OC-AB-BC}{2}. \end{aligned}$$

This expression is incorrect because $F$ lies between $O$ and $B$, so the quantity needed is $OF$, not $BF$. Using

$$OF=BO-BF=BO-(s_2-OC),$$

one obtains

$$EF=OF-OE=\frac{AB+BC}{2},$$

which is the correct value.

The second delicate point is the distance formula for $KL$. The projections of $K$ and $L$ onto $OB$ are $E$ and $F$. Since the two incenters lie on opposite sides of $OB$, the perpendicular separation is $r_1+r_2$, not $|r_1-r_2|$. Hence

$$KL^2=EF^2+(r_1+r_2)^2.$$

The final step uses only

$$p\ge P.$$

Even the stronger estimate

$$Pp\ge P^2\ge 16S$$

is obtained. Thus the required inequality with coefficient $4$ follows immediately.

Alternative Approaches

A different route is to compute every side of $KLMN$ exactly:

$$KL=\sqrt{\left(\frac{AB+BC}{2}\right)^2+(r_1+r_2)^2},$$

and analogous formulas for the other three sides. One may then apply Minkowski's inequality to these four vectors. This yields a quantitative lower bound for $p$ stronger than $p\ge P$.

The presented proof is preferable because it extracts only the information actually needed. After the projection computation, each side of $KLMN$ dominates a simple half-sum of two sides of $ABCD$, and the argument reduces immediately to the standard extremal area estimate for convex quadrilaterals.