Kvant Math Problem 763
Let
Verified: yes
Verdicts: PASS + PASS
Solve time: 15m51s
Source on kvant.digital
Problem
Given a parallelogram $ABCD$ that is not a rhombus. The line symmetric to $AB$ with respect to the diagonal $AC$ intersects at the point $Q$ the line symmetric to $DC$ with respect to the diagonal $DB$ (Fig. 2). Find the ratio $|QA|:|QD|$ if the ratio $|AC|:|BD|=k$ is known.
Fig. 2
V. N. Dubrovsky
Exploration
Let
$$s=\overrightarrow{AC},\qquad t=\overrightarrow{DB}.$$
The diagonals of a parallelogram bisect each other, so if
$$a=\overrightarrow{AB},\qquad d=\overrightarrow{AD},$$
then
$$s=a+d,\qquad t=a-d.$$
Hence
$$a=\frac{s+t}{2},\qquad d=\frac{s-t}{2}.$$
The line symmetric to $AB$ with respect to $AC$ passes through $A$ and has direction obtained by reflecting the vector $a$ in the line spanned by $s$. The line symmetric to $DC$ with respect to $DB$ passes through $D$ and has direction obtained by reflecting the vector $DC=a$ in the line spanned by $t$.
The non-rhombus assumption is essential. Since
$$|s|^2-|t|^2 = |a+d|^2-|a-d|^2 = 4,a\cdot d,$$
the condition $s\perp t$ gives
$$0=s\cdot t=(a+d)\cdot(a-d)=|a|^2-|d|^2.$$
Hence $|a|=|d|$, so all sides of the parallelogram are equal and the parallelogram is a rhombus. The problem excludes this case. Consequently,
$$s\cdot t\neq0.$$
This fact will be used when cancelling $s\cdot t$ later.
Problem Understanding
A parallelogram $ABCD$ is given. The side line $AB$ is reflected across the diagonal $AC$, and the side line $DC$ is reflected across the diagonal $DB$. Their images intersect at a point $Q$.
The ratio
$$\frac{|AC|}{|BD|}=k$$
is known. The goal is to determine
$$\frac{|QA|}{|QD|}.$$
The natural framework is vector geometry with the diagonal vectors $s=\overrightarrow{AC}$ and $t=\overrightarrow{DB}$.
Proof Architecture
Express the sides of the parallelogram through the diagonals. Compute the directions of the two reflected lines using the standard reflection formula. Write the intersection point $Q$ in parametric form on both lines and solve for the parameters. Relate these parameters to the distances $QA$ and $QD$ using the fact that reflections preserve lengths. Finally eliminate all dependence on angles and retain only the ratio of the diagonal lengths.
Solution
Let
$$a=\frac{s+t}{2},\qquad d=\frac{s-t}{2},$$
and let
$$s=\overrightarrow{AC},\qquad t=\overrightarrow{DB}.$$
For a nonzero vector $u$, reflection in the line spanned by $u$ is given by
$$R_u(v)=2\frac{u\cdot v}{u\cdot u},u-v.$$
Let $r_1$ be a direction vector of the image of $AB$ under reflection in $AC$. Then
$$r_1=R_s(a).$$
Since
$$s\cdot a = s\cdot\frac{s+t}{2} = \frac{|s|^2+s\cdot t}{2},$$
we obtain
$$r_1 = 2\frac{s\cdot a}{|s|^2}s-a = \left(\frac{|s|^2+s\cdot t}{|s|^2}\right)s-\frac{s+t}{2} = \left(\frac12+\frac{s\cdot t}{|s|^2}\right)s-\frac12,t.$$
Introduce
$$m=\frac{s\cdot t}{|s|^2}.$$
Then
$$r_1=\left(\frac12+m\right)s-\frac12,t.$$
The side $DC$ is parallel to $AB$ and has the same direction vector $a$. Let $r_2$ be a direction vector of the image of $DC$ under reflection in $DB$. Then
$$r_2=R_t(a).$$
Since
$$t\cdot a = t\cdot\frac{s+t}{2} = \frac{s\cdot t+|t|^2}{2},$$
we get
$$r_2 = 2\frac{t\cdot a}{|t|^2}t-a = \left(\frac{s\cdot t+|t|^2}{|t|^2}\right)t-\frac{s+t}{2} = -\frac12,s+\left(\frac12+\frac{s\cdot t}{|t|^2}\right)t.$$
Put
$$n=\frac{s\cdot t}{|t|^2}.$$
Then
$$r_2=-\frac12,s+\left(\frac12+n\right)t.$$
The first reflected line passes through $A$, and the second passes through $D$. Write
$$Q=A+\lambda r_1=D+\mu r_2.$$
Since
$$\overrightarrow{AD}=d=\frac{s-t}{2},$$
we have
$$\lambda r_1=\frac{s-t}{2}+\mu r_2.$$
Because $s$ and $t$ are not parallel, they form a basis. Comparing coefficients of $s$ and $t$ gives
$$\lambda\left(\frac12+m\right)=\frac12-\frac{\mu}{2},$$
and
$$-\frac{\lambda}{2} = -\frac12+\mu\left(\frac12+n\right).$$
Multiplying by $2$,
$$\lambda(1+2m)=1-\mu,$$
$$-\lambda=-1+\mu(1+2n).$$
The second equation gives
$$\lambda=1-\mu(1+2n).$$
Substituting into the first,
$$(1-\mu(1+2n))(1+2m)=1-\mu.$$
Expanding,
$$1+2m-\mu(1+2n)(1+2m)=1-\mu.$$
Hence
$$2m=\mu\bigl((1+2n)(1+2m)-1\bigr).$$
Since
$$(1+2n)(1+2m)-1 = 1+2m+2n+4mn-1 = 2(m+n+2mn),$$
we obtain
$$\mu=\frac{m}{2mn+m+n}.$$
Then
$$\lambda = 1-\frac{m(1+2n)}{2mn+m+n} = \frac{2mn+m+n-m-2mn}{2mn+m+n} = \frac{n}{2mn+m+n}.$$
Therefore
$$\frac{\lambda}{\mu} = \frac{n}{m}.$$
Since $s\cdot t\neq0$,
$$m=\frac{s\cdot t}{|s|^2}\neq0, \qquad n=\frac{s\cdot t}{|t|^2}\neq0.$$
Moreover,
$$\frac{n}{m} = \frac{\dfrac{s\cdot t}{|t|^2}} {\dfrac{s\cdot t}{|s|^2}} = \frac{|s|^2}{|t|^2} >0.$$
Thus $\lambda/\mu>0$, so $\lambda$ and $\mu$ have the same sign. Consequently,
$$\frac{|\lambda|}{|\mu|} = \frac{\lambda}{\mu}.$$
Reflections preserve lengths. Both $r_1$ and $r_2$ are reflections of the same vector $a$, hence
$$|r_1|=|a|, \qquad |r_2|=|a|,$$
and therefore
$$|r_1|=|r_2|.$$
Since
$$Q=A+\lambda r_1, \qquad Q=D+\mu r_2,$$
we have
$$|QA|=|\lambda|,|r_1|, \qquad |QD|=|\mu|,|r_2|.$$
Hence
$$\frac{|QA|}{|QD|} = \frac{|\lambda|,|r_1|} {|\mu|,|r_2|} = \frac{|\lambda|}{|\mu|} = \frac{\lambda}{\mu} = \frac{n}{m}.$$
Using $s\cdot t\neq0$, the factor $s\cdot t$ may be cancelled:
$$\frac{n}{m} = \frac{\dfrac{s\cdot t}{|t|^2}} {\dfrac{s\cdot t}{|s|^2}} = \frac{|s|^2}{|t|^2}.$$
Therefore
$$\frac{|QA|}{|QD|} = \frac{|s|^2}{|t|^2} = \frac{|AC|^2}{|BD|^2}.$$
Since
$$\frac{|AC|}{|BD|}=k,$$
it follows that
$$\frac{|QA|}{|QD|}=k^2.$$
Thus
$$\boxed{|QA|:|QD|=k^2:1}.$$
Verification of Key Steps
The exclusion of rhombi guarantees $s\cdot t\neq0$. Indeed, if $s\cdot t=0$, then the diagonals are perpendicular. For a parallelogram,
$$0=(a+d)\cdot(a-d)=|a|^2-|d|^2,$$
so $|a|=|d|$, which means all sides are equal and the parallelogram is a rhombus. Since the problem assumes the parallelogram is not a rhombus, one must have $s\cdot t\neq0$.
This nonvanishing justifies the cancellation
$$\frac{\dfrac{s\cdot t}{|t|^2}} {\dfrac{s\cdot t}{|s|^2}} = \frac{|s|^2}{|t|^2}.$$
The same fact implies $m$ and $n$ are nonzero and have the same sign, because both are obtained by dividing the same nonzero number $s\cdot t$ by positive quantities. Hence
$$\frac{\lambda}{\mu} = \frac{n}{m} = \frac{|s|^2}{|t|^2} >0,$$
which shows that $\lambda$ and $\mu$ have the same sign and allows the replacement
$$\frac{|\lambda|}{|\mu|} = \frac{\lambda}{\mu}.$$
The vectors $s$ and $t$ are linearly independent because the diagonals of a nondegenerate parallelogram are distinct and not parallel. This justifies comparing coefficients in the basis ${s,t}$.
Alternative Approaches
A coordinate method places the diagonals on convenient coordinate axes. Writing explicit equations for the reflected lines and solving for their intersection leads again to
$$\frac{|QA|}{|QD|} = \frac{|AC|^2}{|BD|^2} = k^2.$$
The vector approach isolates the dependence on the diagonal lengths and produces the ratio directly.