Kvant Math Problem 108
For a triangle, suppose a line intersects two sides and cuts the triangle into two parts.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m21s
Source on kvant.digital
Problem
a) Prove that the line dividing a given triangle into two polygons of equal area and equal perimeter passes through the center of the circle inscribed in the triangle.
b) Prove the analogous statement for an arbitrary polygon into which a circle can be inscribed.
Yu. I. Ionin
All-Union Mathematical Olympiad for School Students (V)
Exploration
For a triangle, suppose a line intersects two sides and cuts the triangle into two parts. Let the endpoints of the chord be $P$ and $Q$. If the two parts have equal perimeter, then the boundary of one part consists of the segment $PQ$ together with one of the two boundary arcs of the triangle joining $P$ to $Q$, while the other part consists of $PQ$ together with the complementary boundary arc.
Since the segment $PQ$ appears in both perimeters, equality of perimeters means that the two boundary arcs of the triangle joining $P$ and $Q$ have equal length. Thus the condition of equal perimeter can be translated into a statement purely about the boundary of the polygon.
For a triangle with side lengths $a,b,c$ and semiperimeter $s$, take points $P,Q$ on the boundary whose two connecting boundary arcs have equal length $s$. If the line $PQ$ also divides the area equally, then some relation between distances from the sides should emerge.
The presence of the incircle suggests expressing area through distances from the incenter. If a polygon admits an incircle of radius $r$, then every boundary segment contributes area equal to $\frac12 r$ times its length when joined to the center. Hence for any portion of the boundary, the area swept out with the center is proportional to its length. This observation immediately connects equal boundary lengths with equal areas.
The crucial point is to show that whenever a chord $PQ$ passes through the center $O$ of the inscribed circle, the two resulting boundary arcs have equal length if and only if the two resulting regions have equal area. The proportionality of area to boundary length should make this almost automatic.
For a tangential polygon, let the boundary be decomposed into two arcs $\Gamma_1,\Gamma_2$ between $P$ and $Q$. Joining $O$ to all vertices and to $P,Q$, the region bounded by $\Gamma_i$ and $OP$, $OQ$ has area $\frac r2,L_i$, where $L_i$ is the length of $\Gamma_i$. If $O$ lies on $PQ$, then the two polygons cut off by $PQ$ are exactly these two regions. Equal areas are then equivalent to $L_1=L_2$, and equal perimeters are also equivalent to $L_1=L_2$. This gives both directions at once.
The step most likely to hide an error is the identification of the two cut regions with the regions bounded by $\Gamma_i$ and the radii from $O$. One must use the fact that $O$ lies on $PQ$, so that $PQ=OP+OQ$.
Problem Understanding
We are given a line that divides a triangle into two polygons having equal area and equal perimeter. We must prove that this line passes through the incenter of the triangle.
Then we must prove the analogous statement for any polygon possessing an inscribed circle.
This is a Type B problem. The claim is already stated and must be proved.
The core difficulty is to translate the condition of equal perimeter into a statement about lengths of boundary arcs, and then to relate boundary lengths to areas by means of the inscribed circle.
Proof Architecture
Lemma 1. If a chord $PQ$ cuts a polygon into two parts and the two parts have equal perimeter, then the two boundary arcs of the polygon joining $P$ and $Q$ have equal length; this follows because the segment $PQ$ contributes equally to both perimeters.
Lemma 2. In a polygon with an inscribed circle of center $O$ and radius $r$, the area enclosed by $O$ and any boundary arc of length $L$ equals $\frac r2 L$; this follows by decomposing the region into triangles whose heights from $O$ equal $r$.
Lemma 3. If a line through $O$ meets the boundary at $P$ and $Q$, then the two regions into which the polygon is cut have equal area if and only if the corresponding boundary arcs have equal length; this follows from Lemma 2.
Lemma 4. Under the same hypothesis, the two regions have equal perimeter if and only if the corresponding boundary arcs have equal length; this follows from $PQ=OP+OQ$.
The hardest direction is the converse implication: from equal area and equal perimeter deducing that the cutting line passes through $O$.
The lemma most likely to fail under insufficient scrutiny is Lemma 3, because it requires a precise identification of the relevant regions.
Solution
Let a polygon admitting an inscribed circle be given. Denote by $O$ the center of the circle and by $r$ its radius. Part (a) will follow immediately from part (b), since every triangle possesses an incircle. We therefore prove the general statement.
Suppose a line intersects the boundary of the polygon at points $P$ and $Q$ and divides the polygon into two parts. Let $\Gamma_1$ and $\Gamma_2$ be the two boundary arcs joining $P$ and $Q$. Let their lengths be $L_1$ and $L_2$.
The perimeter of the first part is $L_1+PQ$, and the perimeter of the second part is $L_2+PQ$. Hence equality of perimeters is equivalent to
$$L_1=L_2.$$
This proves Lemma 1.
Now consider the regions bounded by $\Gamma_1$ together with the segments $OP$ and $OQ$, and by $\Gamma_2$ together with the same segments. Denote their areas by $S_1$ and $S_2$.
We claim that
$$S_i=\frac r2,L_i \qquad (i=1,2).$$
Indeed, subdivide $\Gamma_i$ at every vertex of the polygon lying on it. The resulting region is decomposed into triangles having common vertex $O$ and bases equal to the corresponding boundary segments of the polygon. Since the circle is inscribed, the distance from $O$ to every side equals $r$. Therefore each such triangle has area $\frac12 r\ell$, where $\ell$ is the length of its base. Summing over all boundary segments of $\Gamma_i$ gives
$$S_i=\frac r2,L_i.$$
This proves Lemma 2.
Assume now that the cutting line passes through $O$. Then $O$ lies on the segment $PQ$, so
$$PQ=OP+OQ.$$
The first region cut off by the line consists exactly of the region bounded by $\Gamma_1$, $OP$, and $OQ$, while the second consists exactly of the region bounded by $\Gamma_2$, $OP$, and $OQ$.
Their areas are therefore $S_1$ and $S_2$. By Lemma 2,
$$S_1=S_2 \quad\Longleftrightarrow\quad L_1=L_2.$$
By Lemma 1,
$$L_1=L_2 \quad\Longleftrightarrow\quad \text{the two parts have equal perimeter}.$$
Hence, for every line through $O$, the two conditions
$$\text{equal area} \qquad\text{and}\qquad \text{equal perimeter}$$
are equivalent.
Now suppose a line cuts the polygon into two parts of equal area and equal perimeter. By Lemma 1 we have
$$L_1=L_2.$$
From Lemma 2 it follows that the two regions bounded by $\Gamma_1$ and $\Gamma_2$ together with $OP$ and $OQ$ have equal area.
Let $T$ be the triangle $OPQ$. The area of the first cut part equals either $S_1+\operatorname{area}(T)$ or $S_1-\operatorname{area}(T)$, according to the position of $O$ relative to the line $PQ$. The area of the second cut part equals the corresponding expression with $S_2$.
Since $S_1=S_2$, the difference of the areas of the two cut parts is either
$$2,\operatorname{area}(T) \quad\text{or}\quad -2,\operatorname{area}(T).$$
The cut parts are assumed to have equal area, so this difference is zero. Hence
$$\operatorname{area}(T)=0.$$
Therefore $O$, $P$, and $Q$ are collinear. Since $P$ and $Q$ lie on the cutting line, that line passes through $O$.
Thus every line dividing a tangential polygon into two polygons of equal area and equal perimeter passes through the center of the inscribed circle.
Applying this result to a triangle yields part (a).
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the translation of equal perimeter into $L_1=L_2$. Writing the perimeters explicitly,
$$p_1=L_1+PQ,\qquad p_2=L_2+PQ.$$
Subtracting gives
$$p_1-p_2=L_1-L_2.$$
No geometric assumption is used. Equality of perimeters is exactly equivalent to equality of boundary-arc lengths.
The second delicate step is the formula
$$S_i=\frac r2 L_i.$$
A careless argument might treat $\Gamma_i$ as a smooth curve. In fact $\Gamma_i$ is a polygonal chain. Decomposing the corresponding region into triangles with vertex $O$, each triangle has height $r$ because every side of the polygon is tangent to the incircle. Summing the triangle areas gives the formula rigorously.
The third delicate step is the deduction that equal areas force $O$ to lie on $PQ$. When $O$ is not on the line, the two cut regions differ from the regions of areas $S_1$ and $S_2$ by adding the triangle $OPQ$ to one side and subtracting it from the other. Since $S_1=S_2$, the area difference between the cut regions is exactly $2\operatorname{area}(OPQ)$ in absolute value. Equality of the cut areas implies $\operatorname{area}(OPQ)=0$, hence collinearity.
Alternative Approaches
One may introduce a signed area function. Let $L_1$ and $L_2$ be the lengths of the two boundary arcs between $P$ and $Q$. Equality of perimeters gives $L_1=L_2$. For a tangential polygon, the signed area between a boundary arc and the segments joining its endpoints to the incenter equals $\frac r2$ times the arc length. Consequently the two corresponding signed areas are equal. Comparing these with the actual areas cut off by the chord $PQ$ shows that any discrepancy is exactly the area of triangle $OPQ$. Equal areas then force this triangle to degenerate.
Another approach uses the standard formula for a tangential polygon,
$$\text{Area}=rs,$$
where $s$ is the semiperimeter. After equal perimeters have been converted into equal arc lengths, one expresses each cut region as a tangential polygon together with or minus the triangle $OPQ$. The equality of areas again reduces to the vanishing of the area of $OPQ$. The method presented in the main proof avoids introducing semiperimeters and works directly with local area contributions from the incenter.