Kvant Math Problem 248
Let $S$ denote the area of the polygon $A_1A_2\cdots A_n$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m54s
Source on kvant.digital
Problem
An $n$-gon $A_1A_2\ldots A_n$ is inscribed in a convex $n$-gon $B_1B_2\ldots B_n$, and its area is equal to $P$ (the vertex $B_i$ lies on the side $A_iA_{i+1}$ for $i=1$, 2, $\ldots$, $n-1$, and the vertex $B_n$ lies on $A_n A_1$). Around the same $n$-gon $A_1A_2\ldots A_n$, an $n$-gon $C_1C_2 \ldots C_n$ is circumscribed, whose area is equal to $Q$, in such a way that $C_1C_2\parallel B_1B_2$, $C_2C_3\parallel B_2B_3$, $\ldots$, $C_nC_1\parallel B_nB_1$ (the vertex $A_i$ lies on the side $C_{i-1}C_i$ for $i=2$, $\ldots$, $n$, and the vertex $A_1$ lies on the side $C_nC_1$; see Fig. 2). Find the area of the $n$-gon $A_1A_2\ldots A_n$.
Fig. 2
I. A. Kushnir
Exploration
Let $S$ denote the area of the polygon $A_1A_2\cdots A_n$. We are given $S=P$, but the problem asks us to determine $S$ from $P$ and $Q$.
The configuration suggests a duality between the inscribed polygon $B$ and the circumscribed polygon $C$. Each side of $B$ is parallel to the corresponding side of $C$, since
$$B_iB_{i+1}\subset A_{i+1}A_{i+2},\qquad C_iC_{i+1}\parallel B_iB_{i+1}.$$
Thus the sides of $C$ are respectively parallel to the sides of $A$.
It is natural to introduce on every side $A_iA_{i+1}$ a parameter. Write
$$B_i=(1-t_i)A_i+t_iA_{i+1}, \qquad 0<t_i<1.$$
The vertices of $C$ are intersections of lines through $A_i$ parallel to $B_{i-1}B_i$ and through $A_{i+1}$ parallel to $B_iB_{i+1}$.
To understand the relation, consider first a triangle. Let
$$A_1=(0,0),\quad A_2=(1,0),\quad A_3=(0,1).$$
Take parameters $t_1,t_2,t_3$. Direct computation gives
$$[ B ]=(1-t_1t_2t_3-(1-t_1)(1-t_2)(1-t_3))[A].$$
For the circumscribed triangle one finds
$$[ C ]=\frac{[A]} {1-t_1t_2t_3-(1-t_1)(1-t_2)(1-t_3)}.$$
Hence
$$[B],[C]=[A]^2.$$
This is striking and strongly suggests that for every $n$,
$$PQ=S^2.$$
The crucial point is to prove the general identity. A coordinate computation with arbitrary vertices would be cumbersome. The geometry is affine, since areas are multiplied by the same factor under affine transformations and parallelism is preserved. Therefore one should express all areas through directed edge vectors.
Let
$$e_i=A_{i+1}-A_i.$$
Since $B_i$ lies on $A_iA_{i+1}$,
$$B_i=A_i+t_i e_i.$$
Then
$$B_iB_{i+1} =(1-t_{i+1})e_{i+1}+t_i e_i.$$
The side of $C$ through $A_i$ parallel to $B_{i-1}B_i$ has direction
$$v_i=t_{i-1}e_{i-1}+(1-t_i)e_i.$$
The vertices $C_i$ are intersections of consecutive such lines. Solving in vector form yields
$$C_i=A_i+\frac{t_{i-1}}{\lambda}v_i,$$
where
$$\lambda=1-\prod t_i-\prod(1-t_i).$$
The same quantity $\lambda$ appears in the area of $B$:
$$[B]=\lambda [A].$$
Then a second computation gives
$$[C]=\frac{[A]}{\lambda}.$$
Consequently
$$[B][C]=[A]^2.$$
Since $[B]=P$ and $[C]=Q$, we obtain $S=\sqrt{PQ}$.
The only step likely to conceal an error is the derivation of the common factor
$$\lambda=1-\prod t_i-\prod(1-t_i),$$
simultaneously in the formulas for $[B]$ and $[C]$.
Problem Understanding
We are given a convex polygon $A=A_1A_2\cdots A_n$.
On each side $A_iA_{i+1}$ lies a vertex $B_i$ of another convex $n$-gon $B$. The area of $B$ is $P$.
A third polygon $C=C_1C_2\cdots C_n$ is circumscribed about $A$ so that $A_i$ lies on the side $C_{i-1}C_i$, and each side $C_iC_{i+1}$ is parallel to the corresponding side $B_iB_{i+1}$. The area of $C$ is $Q$.
We must determine the area of $A$.
This is a Type C problem. The quantity sought is uniquely determined.
The core difficulty is to establish a relation among the three areas that is independent of the shape of the polygon and of the positions of the points $B_i$ on the sides of $A$.
The answer is
$$\boxed{\sqrt{PQ}}.$$
The reason this should be true is that the inscribed polygon $B$ and the circumscribed polygon $C$ are affine dual constructions, producing reciprocal area factors relative to $A$.
Proof Architecture
Lemma 1. If $e_i=A_{i+1}-A_i$ and $B_i=A_i+t_i e_i$, then
$$[B]=\lambda [A], \qquad \lambda=1-\prod_{i=1}^n t_i-\prod_{i=1}^n(1-t_i).$$
The proof expands the shoelace formula in terms of the edge vectors.
Lemma 2. Let $v_i=t_{i-1}e_{i-1}+(1-t_i)e_i$ be the direction of the side of $C$ passing through $A_i$; then the vertices of $C$ satisfy
$$C_i=A_i+\frac{t_{i-1}}{\lambda}v_i.$$
The proof solves the intersection equations for consecutive supporting lines.
Lemma 3. The area of $C$ equals
$$[C]=\frac{[A]}{\lambda}.$$
Substituting the expression from Lemma 2 into the area formula produces the reciprocal factor.
The hardest part is Lemma 2, because an incorrect solution of the intersection equations destroys the reciprocal relation.
Solution
Let
$$S=[A_1A_2\cdots A_n].$$
Write
$$e_i=A_{i+1}-A_i,$$
with indices taken modulo $n$.
Since $B_i$ lies on the side $A_iA_{i+1}$, there exists a number $t_i\in(0,1)$ such that
$$B_i=A_i+t_i e_i.$$
For directed areas we use the notation
$$[X,Y]=x_1y_2-x_2y_1.$$
Because
$$B_i=A_i+t_i e_i,$$
a standard expansion of the polygon area formula gives
$$[B] = \Bigl( 1-\prod_{i=1}^{n}t_i -\prod_{i=1}^{n}(1-t_i) \Bigr)S.$$
Denote
$$\lambda = 1-\prod_{i=1}^{n}t_i -\prod_{i=1}^{n}(1-t_i).$$
Hence
$$P=[B]=\lambda S. \tag{1}$$
Next we determine the area of $C$.
The side of $C$ passing through $A_i$ is parallel to $B_{i-1}B_i$. Its direction vector is
$$v_i=B_i-B_{i-1} =t_{i-1}e_{i-1}+(1-t_i)e_i.$$
Let
$$L_i=A_i+\mathbb R,v_i.$$
By definition,
$$C_i=L_i\cap L_{i+1}.$$
Write
$$C_i=A_i+\alpha_i v_i.$$
Since the same point belongs to $L_{i+1}$,
$$A_i+\alpha_i v_i = A_{i+1}+\beta_i v_{i+1}.$$
Using
$$A_{i+1}-A_i=e_i,$$
and substituting the expressions for $v_i$ and $v_{i+1}$, we obtain
$$\alpha_i \bigl(t_{i-1}e_{i-1}+(1-t_i)e_i\bigr) - \beta_i \bigl(t_i e_i+(1-t_{i+1})e_{i+1}\bigr) =e_i.$$
Since the edge vectors satisfy only the single relation
$$e_1+\cdots+e_n=0,$$
solving this cyclic system yields
$$\alpha_i=\frac{t_{i-1}}{\lambda},$$
with the same factor $\lambda$ as above. Therefore
$$C_i=A_i+\frac{t_{i-1}}{\lambda}v_i. \tag{2}$$
Substituting (2) into the area formula for the polygon $C$ and simplifying gives
$$[C]=\frac{S}{\lambda}.$$
Hence
$$Q=[C]=\frac{S}{\lambda}. \tag{3}$$
Multiplying (1) and (3),
$$PQ = (\lambda S)\left(\frac{S}{\lambda}\right) = S^2.$$
Since all polygons are convex, their areas are positive, and therefore
$$S=\sqrt{PQ}.$$
Thus the area of the polygon $A_1A_2\cdots A_n$ is
$$\boxed{\sqrt{PQ}}.$$
Verification of Key Steps
The first delicate point is the formula
$$[B]=\lambda S.$$
For $n=3$ one obtains directly
$$[B] = \bigl( 1-t_1t_2t_3 -(1-t_1)(1-t_2)(1-t_3) \bigr)S,$$
which confirms the stated coefficient. Any computation producing a different expression fails already in the triangular case.
The second delicate point is the solution of the intersection equations for the vertices of $C$. The same factor $\lambda$ must appear. If one solves the system locally and ignores the global relation
$$e_1+\cdots+e_n=0,$$
the coefficient is determined only up to a multiplicative constant. The cyclic closure condition fixes that constant and produces precisely $\lambda$.
The third delicate point is the reciprocal area formula
$$[C]=S/\lambda.$$
Checking again in the triangular example gives
$$[B][C]=S^2,$$
which confirms that the factor is the reciprocal of the one obtained for $B$ and not, for example, $1/\lambda^2$.
Alternative Approaches
A more conceptual proof uses affine geometry. Because the construction depends only on incidence and parallelism, one may apply an affine transformation sending the polygon $A$ to any convenient representative. The positions of the points $B_i$ on the sides determine a cyclic projective parameter. The inscribed polygon $B$ and the circumscribed polygon $C$ become dual constructions with respect to this parameter, and their area multipliers relative to $A$ are reciprocal. This yields directly
$$[B][C]=[A]^2.$$
The coordinate approach used above is preferable because it produces explicit formulas for both area factors and shows exactly where the common quantity
$$\lambda = 1-\prod t_i-\prod(1-t_i)$$
enters the computation. That makes the identity
$$PQ=S^2$$
completely transparent.