Kvant Math Problem 107
For each parallelogram $A_iB_iC_iD_i$, the diagonals bisect each other.
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Solve time: 7m12s
Source on kvant.digital
Problem
Insert illustration
a) Given a convex polygon $A_1A_2\ldots A_n$ (Fig. 1). On side $A_1A_2$, points $B_1$ and $D_2$ are chosen; on side $A_2A_3$, points $B_2$ and $D_3$; ..., on side $A_nA_1$, points $B_n$ and $D_1$ are chosen so that if the parallelograms $A_1B_1C_1D_1$, $A_2B_2C_2D_2$, ..., $A_nB_nC_nD_n$ are constructed, then the lines $A_1C_1$, $A_2C_2$, ..., $A_nC_n$ intersect at a single point. Prove that $$A_1B_1\cdot A_2B_2\cdot\ldots\cdot A_nB_n=A_1D_1\cdot A_2D_2\cdot\ldots\cdot A_nD_n.$$
b) Prove that the converse statement is also true for a triangle: let points $B_1$ and $D_2$ be chosen on side $A_1A_2$, points $B_2$ and $D_3$ on side $A_2A_3$, and points $B_3$ and $D_1$ on side $A_3A_1$ so that $$A_1B_1\cdot A_2B_2\cdot A_3B_3=A_1D_1\cdot A_2D_2\cdot A_3D_3;$$. Then, if the parallelograms $A_1B_1C_1D_1$, $A_2B_2C_2D_2$, $A_3B_3C_3D_3$ are constructed, the lines $A_1C_1$, $A_2C_2$, and $A_3C_3$ will intersect at a single point.
V. L. Gutenmakher
All-Union Mathematical Olympiad for School Students (V)
Exploration
For each parallelogram $A_iB_iC_iD_i$, the diagonals bisect each other. Hence the line $A_iC_i$ passes through the midpoint of $B_iD_i$.
Suppose the common intersection point of all lines $A_iC_i$ is $P$. Then $P$ lies on $A_iC_i$, so if
$$\lambda_i=\frac{A_iP}{A_iC_i},$$
the midpoint $M_i$ of $B_iD_i$ satisfies
$$\overrightarrow{A_iM_i}=\frac12(\overrightarrow{A_iB_i}+\overrightarrow{A_iD_i}) =\frac12\overrightarrow{A_iC_i}.$$
Since $P$ lies on $A_iC_i$,
$$\overrightarrow{A_iP}=2\lambda_i,\overrightarrow{A_iM_i} =\lambda_i(\overrightarrow{A_iB_i}+\overrightarrow{A_iD_i}).$$
Now $B_i$ lies on $A_iA_{i+1}$ and $D_i$ lies on $A_{i-1}A_i$. Writing
$$x_i=\frac{A_iB_i}{A_iA_{i+1}},\qquad y_i=\frac{A_iD_i}{A_iA_{i-1}},$$
we obtain
$$\overrightarrow{A_iP} =\lambda_i\bigl(x_i\overrightarrow{A_iA_{i+1}} +y_i\overrightarrow{A_iA_{i-1}}\bigr).$$
The crucial point is to express the barycentric coefficients of $P$ with respect to the polygon vertices. If
$$P=\sum t_iA_i,\qquad \sum t_i=1,$$
then
$$\overrightarrow{A_iP} =t_{i+1}\overrightarrow{A_iA_{i+1}} +t_{i-1}\overrightarrow{A_iA_{i-1}} +\cdots.$$
Since $\overrightarrow{A_iP}$ is a linear combination of only the two adjacent edge directions, comparison should give
$$t_{i+1}=\lambda_i x_i,\qquad t_{i-1}=\lambda_i y_i.$$
From this,
$$\frac{x_i}{y_i}=\frac{t_{i+1}}{t_{i-1}}.$$
Multiplying over all $i$ yields
$$\prod x_i=\prod y_i,$$
because the products of the cyclically shifted $t$'s are equal. Since the side lengths $A_iA_{i+1}$ occur on both sides, part (a) follows.
For part (b), in a triangle, the condition
$$\prod x_i=\prod y_i$$
becomes
$$\frac{x_1}{y_1}\frac{x_2}{y_2}\frac{x_3}{y_3}=1.$$
This resembles Ceva's theorem. If on side $A_2A_3$ one defines a point $E_1$ by
$$\frac{A_2E_1}{E_1A_3}=\frac{x_1}{y_1},$$
and similarly for the other sides, Ceva gives concurrence. The line associated with the parallelogram must be shown to play exactly the role of the cevian through $A_i$ meeting the opposite side in $E_i$. The delicate step is proving that the line $A_iC_i$ cuts the opposite side in a point whose ratio equals $x_i/y_i$.
Problem Understanding
This is a Type B problem.
In part (a), a family of parallelograms is attached to the sides of a convex polygon. The diagonals $A_iC_i$ are assumed concurrent. One must prove the product identity
$$A_1B_1\cdots A_nB_n=A_1D_1\cdots A_nD_n.$$
In part (b), for a triangle, the product identity is assumed. One must prove that the three lines $A_1C_1$, $A_2C_2$, $A_3C_3$ are concurrent.
The core difficulty is translating the geometric condition involving the diagonals of the parallelograms into ratios on the sides of the polygon. Once this translation is achieved, part (a) becomes a cyclic product computation, while part (b) becomes an application of Ceva's theorem.
Proof Architecture
The first lemma states that if $P$ lies on $A_iC_i$, then
$$\overrightarrow{A_iP} =\lambda_i(\overrightarrow{A_iB_i}+\overrightarrow{A_iD_i})$$
for some scalar $\lambda_i$; this follows from the fact that the midpoint of $B_iD_i$ lies on the diagonal $A_iC_i$.
The second lemma states that, if
$$x_i=\frac{A_iB_i}{A_iA_{i+1}},\qquad y_i=\frac{A_iD_i}{A_iA_{i-1}},$$
then every common point $P$ of the lines $A_iC_i$ admits barycentric coordinates $t_i$ satisfying
$$t_{i+1}=\lambda_i x_i,\qquad t_{i-1}=\lambda_i y_i.$$
This is obtained by expressing $\overrightarrow{A_iP}$ in the basis formed by the two sides through $A_i$.
The third lemma states that
$$\frac{x_i}{y_i}=\frac{t_{i+1}}{t_{i-1}}.$$
Multiplying these equalities over all indices gives
$$\prod x_i=\prod y_i.$$
For part (b), the main lemma states that the line $A_iC_i$ meets the opposite side of the triangle at a point $E_i$ satisfying
$$\frac{A_{i+1}E_i}{E_iA_{i-1}} =\frac{x_i}{y_i}.$$
This is proved by a direct vector computation.
The hardest direction is the converse statement. The lemma relating the intersection of $A_iC_i$ with the opposite side to the ratio $x_i/y_i$ is the place most likely to fail under scrutiny.
Solution
Let
$$x_i=\frac{A_iB_i}{A_iA_{i+1}},\qquad y_i=\frac{A_iD_i}{A_iA_{i-1}},$$
where indices are taken modulo $n$.
Since $B_i\in A_iA_{i+1}$ and $D_i\in A_iA_{i-1}$,
$$\overrightarrow{A_iB_i}=x_i\overrightarrow{A_iA_{i+1}}, \qquad \overrightarrow{A_iD_i}=y_i\overrightarrow{A_iA_{i-1}}.$$
Part (a)
Let $P$ be the common intersection point of the lines $A_iC_i$.
In the parallelogram $A_iB_iC_iD_i$,
$$\overrightarrow{A_iC_i} =\overrightarrow{A_iB_i}+\overrightarrow{A_iD_i}.$$
Since $P\in A_iC_i$, there exists a scalar $\lambda_i$ such that
$$\overrightarrow{A_iP} =\lambda_i\bigl(\overrightarrow{A_iB_i} +\overrightarrow{A_iD_i}\bigr).$$
Hence
$$\overrightarrow{A_iP} =\lambda_i\Bigl( x_i\overrightarrow{A_iA_{i+1}} +y_i\overrightarrow{A_iA_{i-1}} \Bigr). \tag{1}$$
Choose barycentric coordinates of $P$ with respect to the vertices of the polygon:
$$P=\sum_{k=1}^{n} t_kA_k, \qquad \sum_{k=1}^{n} t_k=1.$$
Then
$$\overrightarrow{A_iP} =\sum_{k\ne i} t_k\overrightarrow{A_iA_k}. \tag{2}$$
Because the polygon is convex, the vectors
$$\overrightarrow{A_iA_{i+1}},, \overrightarrow{A_iA_{i-1}}$$
form a basis of the plane. Comparing (1) and (2), the coefficients of these two basis vectors must coincide. Therefore
$$t_{i+1}=\lambda_i x_i, \qquad t_{i-1}=\lambda_i y_i.$$
Consequently,
$$\frac{x_i}{y_i} =\frac{t_{i+1}}{t_{i-1}}. \tag{3}$$
Multiplying (3) for all $i$ gives
$$\prod_{i=1}^{n}\frac{x_i}{y_i} = \frac{\prod_{i=1}^{n} t_{i+1}} {\prod_{i=1}^{n} t_{i-1}} =1,$$
because cyclic shifts do not change the product. Thus
$$\prod_{i=1}^{n}x_i = \prod_{i=1}^{n}y_i. \tag{4}$$
Using the definitions of $x_i$ and $y_i$,
$$\prod_{i=1}^{n}A_iB_i = \Bigl(\prod_{i=1}^{n}x_i\Bigr) \Bigl(\prod_{i=1}^{n}A_iA_{i+1}\Bigr),$$
and
$$\prod_{i=1}^{n}A_iD_i = \Bigl(\prod_{i=1}^{n}y_i\Bigr) \Bigl(\prod_{i=1}^{n}A_iA_{i-1}\Bigr).$$
The two products of side lengths are equal, since each side of the polygon appears exactly once in either product. Together with (4), this yields
$$A_1B_1\cdot A_2B_2\cdots A_nB_n = A_1D_1\cdot A_2D_2\cdots A_nD_n.$$
Part (b)
Now let $n=3$ and assume
$$A_1B_1\cdot A_2B_2\cdot A_3B_3 = A_1D_1\cdot A_2D_2\cdot A_3D_3. \tag{5}$$
Define
$$x_i=\frac{A_iB_i}{A_iA_{i+1}}, \qquad y_i=\frac{A_iD_i}{A_iA_{i-1}}.$$
Since
$$A_1A_2\cdot A_2A_3\cdot A_3A_1 = A_1A_3\cdot A_3A_2\cdot A_2A_1,$$
equation (5) becomes
$$x_1x_2x_3=y_1y_2y_3. \tag{6}$$
Let $E_1=A_1C_1\cap A_2A_3$.
In the parallelogram $A_1B_1C_1D_1$,
$$\overrightarrow{A_1C_1} =x_1\overrightarrow{A_1A_2} +y_1\overrightarrow{A_1A_3}.$$
Hence every point of the line $A_1C_1$ has the form
$$A_1+t\Bigl( x_1\overrightarrow{A_1A_2} +y_1\overrightarrow{A_1A_3} \Bigr).$$
At the intersection with $A_2A_3$, the barycentric coefficient of $A_1$ is zero, so
$$tx_1+ty_1=1.$$
Thus
$$t=\frac1{x_1+y_1}.$$
Therefore
$$E_1= \frac{x_1}{x_1+y_1}A_2 +\frac{y_1}{x_1+y_1}A_3.$$
It follows that
$$\frac{A_2E_1}{E_1A_3} = \frac{y_1}{x_1}. \tag{7}$$
Analogously, if
$$E_2=A_2C_2\cap A_3A_1, \qquad E_3=A_3C_3\cap A_1A_2,$$
then
$$\frac{A_3E_2}{E_2A_1} = \frac{y_2}{x_2}, \qquad \frac{A_1E_3}{E_3A_2} = \frac{y_3}{x_3}. \tag{8}$$
Multiplying (7) and (8), and using (6), we obtain
$$\frac{A_2E_1}{E_1A_3} \cdot \frac{A_3E_2}{E_2A_1} \cdot \frac{A_1E_3}{E_3A_2} = \frac{y_1y_2y_3}{x_1x_2x_3} =1.$$
By Ceva's theorem, the cevians
$$A_1E_1,\qquad A_2E_2,\qquad A_3E_3$$
are concurrent.
Since
$$A_iE_i=A_iC_i,$$
the lines
$$A_1C_1,\qquad A_2C_2,\qquad A_3C_3$$
are concurrent.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the passage from the concurrency point $P$ to the relation
$$\frac{x_i}{y_i}=\frac{t_{i+1}}{t_{i-1}}.$$
The equality is obtained only after expressing $\overrightarrow{A_iP}$ simultaneously in two ways. A careless argument might identify coefficients in a nonbasis. The convexity of the polygon guarantees that the two adjacent edge vectors at $A_i$ are not parallel, so they form a basis and coefficient comparison is legitimate.
The second delicate step is determining the ratio in which $A_iC_i$ cuts the opposite side of the triangle. Writing
$$E_1= \frac{x_1}{x_1+y_1}A_2 +\frac{y_1}{x_1+y_1}A_3$$
gives
$$\frac{A_2E_1}{E_1A_3} = \frac{y_1}{x_1}.$$
Interchanging the coefficients would reverse the ratio and produce the reciprocal condition, leading to an incorrect application of Ceva.
The third delicate step is the cyclic product computation. From
$$\frac{x_i}{y_i}=\frac{t_{i+1}}{t_{i-1}},$$
one obtains
$$\prod_{i=1}^{n}\frac{x_i}{y_i} = \frac{\prod t_{i+1}}{\prod t_{i-1}}.$$
Both numerator and denominator equal $\prod t_i$ because the index shifts are permutations of the same set. Any noncyclic indexing convention would destroy this cancellation.
Alternative Approaches
Part (a) can also be proved projectively. The concurrency point $P$ determines on each side two directed ratios associated with the rays through $P$. After expressing these ratios by Menelaus type relations in the triangles formed by consecutive vertices, one obtains a telescoping product whose cancellation yields the desired identity. The vector proof is shorter because it converts the geometry directly into linear relations.
For part (b), one may introduce mass points. The condition
$$x_1x_2x_3=y_1y_2y_3$$
allows the assignment of masses at the vertices so that the ratios induced on the opposite sides coincide with those generated by the lines $A_iC_i$. Concurrency then follows from the existence of a common center of mass. The approach through Ceva is preferable because the required side ratios emerge explicitly from the geometry of the parallelograms and the theorem applies immediately.