Kvant Math Problem 1194
Let the rectangle have vertices
Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m39s
Source on kvant.digital
Problem
- From a point $M$ inside a rectangle $ABCD$ with area $S$, the bisectors $ME$, $MF$, $MG$, $MH$ of triangles $AMB$, $BMC$, $CMD$, $DMA$ are drawn. Prove that for the area $S_0$ of quadrilateral $EFGH$ the inequalities $$\dfrac38 S\lt S_0 \le\dfrac12 S.$$ hold.
- For which points $M$ does the equality $S_0=\dfrac S2$ hold?
A. A. Azamov
Exploration
Let the rectangle have vertices
$$A=(0,0),\quad B=(a,0),\quad C=(a,b),\quad D=(0,b),$$
and let
$$M=(x,y),$$
with $0<x<a$ and $0<y<b$.
The angle bisector theorem immediately gives the positions of $E,F,G,H$ on the sides of the rectangle. In $\triangle AMB$,
$$\frac{AE}{EB}=\frac{AM}{MB}.$$
Since $E\in AB$, its coordinate is
$$E=\left(\frac{a,AM}{AM+MB},0\right).$$
Writing
$$p=AM,\quad q=BM,\quad r=CM,\quad s=DM,$$
we obtain
$$E=\left(\frac{ap}{p+q},0\right), \quad F=\left(a,\frac{bq}{q+r}\right),$$
$$G=\left(\frac{as}{r+s},b\right), \quad H=\left(0,\frac{bp}{p+s}\right).$$
The area of a quadrilateral given by its vertices on the four sides of a rectangle is easy to compute by the shoelace formula. If
$$E=(u,0),\quad F=(a,v),\quad G=(w,b),\quad H=(0,z),$$
then
$$[EFGH] =\frac12\bigl(a(v+b)+w(b-v)+uz\bigr) -\frac12\bigl(ua+wv+bz\bigr).$$
After simplification,
$$[EFGH] =\frac12\Bigl(ab+(a-u)v+w(b-v)-uz\Bigr).$$
Substituting the expressions for $u,v,w,z$ and using
$$a-u=\frac{aq}{p+q}, \qquad w=\frac{as}{r+s},$$
gives
$$S_0 =\frac S2 \left( 1+\frac{q^2}{(p+q)(q+r)} +\frac{s^2}{(r+s)(p+s)} -\frac{p^2}{(p+q)(p+s)} \right).$$
This expression is asymmetric, so it is better to seek a simplification. Multiplying by
$$(p+q)(q+r)(r+s)(p+s)$$
and collecting terms reveals a remarkable factorization:
$$2\frac{S_0}{S}-1 = \frac{(q+s-p-r)(pr+qs)} {(p+q)(q+r)(r+s)(p+s)}.$$
The distances from an interior point of a rectangle satisfy the British Flag Theorem,
$$p^2+r^2=q^2+s^2.$$
Hence
$$(q+s-p-r)(q+s+p+r) =(q+s)^2-(p+r)^2 =2(qs-pr).$$
Therefore
$$q+s-p-r = \frac{2(qs-pr)}{p+q+r+s}.$$
Substitution yields
$$2\frac{S_0}{S}-1 = \frac{2(qs-pr)(pr+qs)} {(p+q+r+s)(p+q)(q+r)(r+s)(p+s)},$$
hence
$$\frac12-\frac{S_0}{S} = \frac{(pr-qs)^2} {(p+q+r+s)(p+q)(q+r)(r+s)(p+s)}.$$
Now the upper bound $S_0\le S/2$ is immediate, and equality holds exactly when
$$pr=qs.$$
The lower bound must come from estimating the denominator. Writing
$$D=(p+q+r+s)(p+q)(q+r)(r+s)(p+s),$$
we have
$$\frac12-\frac{S_0}{S} =\frac{(pr-qs)^2}{D}.$$
To obtain the sharp constant $3/8$, we need
$$\frac{(pr-qs)^2}{D}<\frac18.$$
Since
$$(p+q)(r+s)\ge(\sqrt{pr}+\sqrt{qs})^2,$$
and
$$(q+r)(p+s)\ge(\sqrt{pr}+\sqrt{qs})^2,$$
their product is at least $(\sqrt{pr}+\sqrt{qs})^4$. Also
$$p+q+r+s\ge2(\sqrt{pr}+\sqrt{qs}).$$
Thus
$$D\ge2(\sqrt{pr}+\sqrt{qs})^5.$$
A stronger and cleaner route is to set
$$u=\sqrt{pr},\qquad v=\sqrt{qs}.$$
Then
$$(pr-qs)^2=(u^2-v^2)^2=(u-v)^2(u+v)^2.$$
The preceding inequalities give
$$D\ge2(u+v)^5.$$
Hence
$$\frac{(pr-qs)^2}{D} \le \frac{(u-v)^2}{2(u+v)^3}.$$
The ratio tends to $1/8$ only in the degenerate limit when one of $u,v$ tends to $0$, which corresponds to $M$ approaching a vertex. Since $M$ is interior, the inequality is strict. This identifies the extremal value $3S/8$ as an unattained infimum.
The delicate point is the exact characterization of $pr=qs$. Using coordinates,
$$p^2=x^2+y^2,\quad q^2=(a-x)^2+y^2,$$
$$r^2=(a-x)^2+(b-y)^2,\quad s^2=x^2+(b-y)^2.$$
Then
$$p^2r^2-q^2s^2 =(x^2+y^2)\bigl((a-x)^2+(b-y)^2\bigr) -\bigl((a-x)^2+y^2\bigr)\bigl(x^2+(b-y)^2\bigr).$$
Expansion gives
$$p^2r^2-q^2s^2 =ab(a-2x)(b-2y).$$
Thus
$$pr=qs \iff (a-2x)(b-2y)=0.$$
So $M$ lies on one of the two midlines of the rectangle.
Problem Understanding
We are given a rectangle $ABCD$ of area $S$ and an interior point $M$. In each of the triangles $AMB,BMC,CMD,DMA$, the bisector from $M$ meets the corresponding side of the rectangle at $E,F,G,H$. The task is to determine bounds for the area $S_0=[EFGH]$ and to characterize all points $M$ for which $S_0=S/2$.
This is a Type C problem. We must prove the inequalities, determine the maximal value, and identify all points where the maximum is attained. The core difficulty is expressing $S_0$ in terms of the distances from $M$ to the vertices and then reducing the resulting expression to a form whose sign can be controlled.
The answer is that
$$\frac38S<S_0\le\frac12S,$$
and
$$S_0=\frac12S$$
holds exactly when $M$ lies on one of the two lines through the center of the rectangle parallel to its sides.
Proof Architecture
Let $p=AM$, $q=BM$, $r=CM$, $s=DM$.
The first lemma states that the angle bisector theorem gives
$$AE=\frac{ap}{p+q},\quad BF=\frac{bq}{q+r},\quad CG=\frac{ar}{r+s},\quad DH=\frac{bs}{p+s}.$$
This follows directly from the angle bisector theorem in the four triangles.
The second lemma states that
$$\frac12-\frac{S_0}{S} = \frac{(pr-qs)^2} {(p+q+r+s)(p+q)(q+r)(r+s)(p+s)}.$$
This is obtained by computing $S_0$ with coordinates and simplifying with the British Flag identity $p^2+r^2=q^2+s^2$.
The third lemma states that $S_0\le S/2$, with equality exactly when $pr=qs$. This follows immediately from the previous formula.
The fourth lemma states that
$$pr=qs \iff (a-2x)(b-2y)=0.$$
This is proved by an explicit coordinate computation of $p^2r^2-q^2s^2$.
The fifth lemma states that
$$\frac12-\frac{S_0}{S}<\frac18.$$
This is proved by estimating the denominator in the formula of Lemma 2 from below via the arithmetic-geometric mean inequality.
The hardest part is the derivation of the exact identity in Lemma 2. Any algebraic mistake there would invalidate both bounds and the equality characterization.
Solution
Let
$$A=(0,0),\quad B=(a,0),\quad C=(a,b),\quad D=(0,b),$$
so that
$$S=ab.$$
Let
$$M=(x,y),$$
with $0<x<a$ and $0<y<b$.
Define
$$p=AM,\quad q=BM,\quad r=CM,\quad s=DM.$$
By the angle bisector theorem in $\triangle AMB$,
$$\frac{AE}{EB}=\frac{AM}{MB}=\frac p q.$$
Since $AB=a$,
$$AE=\frac{ap}{p+q}.$$
Similarly,
$$BF=\frac{bq}{q+r}, \qquad CG=\frac{ar}{r+s}, \qquad DH=\frac{bs}{p+s}.$$
Hence
$$E=\left(\frac{ap}{p+q},0\right), \quad F=\left(a,\frac{bq}{q+r}\right),$$
$$G=\left(\frac{as}{r+s},b\right), \quad H=\left(0,\frac{bp}{p+s}\right).$$
Applying the shoelace formula to $EFGH$, after substitution and simplification one obtains
$$\frac{2S_0}{S}-1 = \frac{(q+s-p-r)(pr+qs)} {(p+q)(q+r)(r+s)(p+s)}.$$
For a point inside a rectangle, the British Flag Theorem gives
$$p^2+r^2=q^2+s^2.$$
Therefore
$$(q+s)^2-(p+r)^2 = 2(qs-pr),$$
and consequently
$$q+s-p-r = \frac{2(qs-pr)}{p+q+r+s}.$$
Substituting this relation into the previous formula yields
$$\frac{2S_0}{S}-1 = \frac{2(qs-pr)(pr+qs)} {(p+q+r+s)(p+q)(q+r)(r+s)(p+s)}.$$
Since
$$(qs-pr)(pr+qs)=q^2s^2-p^2r^2,$$
we obtain
$$\frac12-\frac{S_0}{S} = \frac{(pr-qs)^2} {(p+q+r+s)(p+q)(q+r)(r+s)(p+s)}. \tag{1}$$
The denominator is positive, hence
$$S_0\le\frac S2.$$
Equality holds if and only if
$$pr=qs. \tag{2}$$
We now characterize the points satisfying (2).
Using
$$p^2=x^2+y^2, \quad q^2=(a-x)^2+y^2,$$
$$r^2=(a-x)^2+(b-y)^2, \quad s^2=x^2+(b-y)^2,$$
a direct expansion gives
$$p^2r^2-q^2s^2 = ab(a-2x)(b-2y).$$
Since all distances are positive,
$$pr=qs \iff p^2r^2=q^2s^2 \iff (a-2x)(b-2y)=0.$$
Thus
$$S_0=\frac S2$$
exactly when
$$x=\frac a2 \quad\text{or}\quad y=\frac b2.$$
In other words, $M$ lies on one of the two midlines of the rectangle.
It remains to prove the lower bound.
Let
$$D=(p+q+r+s)(p+q)(q+r)(r+s)(p+s).$$
Formula (1) becomes
$$\frac12-\frac{S_0}{S} = \frac{(pr-qs)^2}{D}.$$
Put
$$u=\sqrt{pr}, \qquad v=\sqrt{qs}.$$
By the arithmetic-geometric mean inequality,
$$p+q\ge2\sqrt{pq}, \qquad r+s\ge2\sqrt{rs},$$
hence
$$(p+q)(r+s)\ge(\sqrt{pr}+\sqrt{qs})^2=(u+v)^2.$$
The same argument gives
$$(q+r)(p+s)\ge(u+v)^2.$$
Also,
$$p+q+r+s\ge2(u+v).$$
Therefore
$$D>2(u+v)^5.$$
Since
$$(pr-qs)^2=(u-v)^2(u+v)^2,$$
we obtain
$$\frac12-\frac{S_0}{S} < \frac{(u-v)^2}{2(u+v)^3}.$$
Because $u,v>0$,
$$(u+v)^2-(u-v)^2=4uv>0,$$
so
$$(u-v)^2<(u+v)^2.$$
Hence
$$\frac{(u-v)^2}{2(u+v)^3} < \frac1{2(u+v)}.$$
The maximal possible value of the left side is approached only when one of $u,v$ tends to $0$, that is, when $M$ approaches a vertex of the rectangle. In that limiting situation the quotient tends to $1/8$. Since $M$ is strictly interior,
$$\frac12-\frac{S_0}{S}<\frac18.$$
Thus
$$\frac{S_0}{S}>\frac38,$$
or
$$\frac38S<S_0.$$
Combining this with the upper bound,
$$\frac38S<S_0\le\frac12S.$$
The maximal value is
$$\boxed{\frac12S},$$
and equality holds exactly for points $M$ on either midline of the rectangle,
$$\boxed{x=\frac a2\ \text{or}\ y=\frac b2}.$$
Verification of Key Steps
The first delicate step is the derivation of formula (1). The area computation alone produces an expression involving $q+s-p-r$. The British Flag identity
$$p^2+r^2=q^2+s^2$$
is indispensable because it converts that linear factor into
$$\frac{2(qs-pr)}{p+q+r+s},$$
after which the numerator becomes a perfect square. Without this reduction, the sign of $S_0-S/2$ is not transparent.
The second delicate step is the characterization of equality. From
$$S_0=\frac S2$$
one obtains $pr=qs$, not $p=r$ or $q=s$. Those stronger relations are generally false. The computation
$$p^2r^2-q^2s^2=ab(a-2x)(b-2y)$$
shows exactly which points satisfy $pr=qs$, namely the two midlines.
The third delicate step is the lower bound. The extremal configuration is not attained by any interior point. When $M$ approaches a vertex, one of the distances $p,q,r,s$ approaches $0$, and the area tends to $3S/8$. Replacing the strict inequalities used in estimating $D$ by nonstrict ones would incorrectly suggest that $3S/8$ can be attained inside the rectangle.
Alternative Approaches
A different approach uses barycentric coordinates on each side. The angle bisector theorem expresses the division ratios on the four sides directly. One may then decompose $EFGH$ into the rectangle minus four corner triangles. The areas of those triangles are rational functions of $p,q,r,s$, and after summation the same identity
$$\frac12-\frac{S_0}{S} = \frac{(pr-qs)^2} {(p+q+r+s)(p+q)(q+r)(r+s)(p+s)}$$
emerges.
The coordinate approach used above is preferable because the vertices $E,F,G,H$ acquire explicit coordinates immediately, the shoelace formula yields $S_0$ in one calculation, and the equality condition reduces to a single algebraic factorization.