Kvant Math Problem 1202
Let the two rays from $A$ be $r_1$ and $r_2$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m18s
Source on kvant.digital
Problem
From vertex $A$ of square $ABCD$, two rays are drawn into the interior of the square. Perpendiculars $BK$, $BL$, $DM$, $DN$ are dropped onto these rays from vertices $B$ and $D$. Prove that the segments $KL$ and $MN$ are equal in length and perpendicular to each other.
D. Nyamsuren (Mongolia)
Exploration
Let the two rays from $A$ be $r_1$ and $r_2$. The points $K,M$ lie on $r_1$, while $L,N$ lie on $r_2$. Since $BK$ and $DM$ are perpendicular to $r_1$, the points $K$ and $M$ are the orthogonal projections of $B$ and $D$ onto $r_1$. Likewise, $L$ and $N$ are the orthogonal projections of $B$ and $D$ onto $r_2$.
A coordinate description seems natural. Put the square in the plane with
$$A=(0,0),\qquad B=(s,0),\qquad D=(0,s).$$
Let $u$ and $v$ be unit vectors along the two rays. Since the rays start at $A$, the projection of a point $P$ onto the ray line is simply $(P\cdot u)u$ or $(P\cdot v)v$.
Then
$$K=(B\cdot u)u,\qquad M=(D\cdot u)u.$$
Hence
$$\overrightarrow{KM} =\bigl((D\cdot u)-(B\cdot u)\bigr)u =((D-B)\cdot u),u.$$
The same computation gives
$$\overrightarrow{LN} =((D-B)\cdot v),v.$$
Since $D-B=(-s,s)$, it is convenient to denote
$$w=D-B.$$
Then
$$\overrightarrow{KM}=(w\cdot u)u,\qquad \overrightarrow{LN}=(w\cdot v)v.$$
The desired segment is
$$\overrightarrow{KL} =\overrightarrow{AL}-\overrightarrow{AK} =(B\cdot v)v-(B\cdot u)u.$$
Using $D=B+w$ and the fact that $w\cdot B=-s^2$, one obtains
$$\overrightarrow{MN} =(D\cdot v)v-(D\cdot u)u =\overrightarrow{KL}+(w\cdot v)v-(w\cdot u)u.$$
The vector $w$ has equal coordinates in magnitude and opposite signs, so
$$|w|=s\sqrt2.$$
The crucial point is to express everything in an orthonormal basis adapted to the rays. Let the angles of $u$ and $v$ with $AB$ be $\alpha$ and $\beta$. Then
$$w=s(-1,1)=s\sqrt2,e,$$
where $e$ is the unit vector making angle $135^\circ$ with $AB$. After substituting
$$B=s(1,0),$$
one finds
$$\overrightarrow{KL} =s\bigl(\cos\beta,v-\cos\alpha,u\bigr),$$
and
$$\overrightarrow{MN} =s\bigl(\sin(\beta-\tfrac{\pi}{4}),v -\sin(\alpha-\tfrac{\pi}{4}),u\bigr)\sqrt2 .$$
This looks messy. A better idea is to rotate coordinates by $45^\circ$. In that system $w$ lies on one coordinate axis, and projections become very simple.
Let
$$p=\frac{B+D}{2},\qquad q=\frac{D-B}{2}.$$
Then $p\perp q$ and $|p|=|q|=s/\sqrt2$.
Because $D=p+q$ and $B=p-q$,
$$\overrightarrow{KL} =(p\cdot v)v-(p\cdot u)u +\bigl((q\cdot v)v-(q\cdot u)u\bigr),$$
while
$$\overrightarrow{MN} =(p\cdot v)v-(p\cdot u)u -\bigl((q\cdot v)v-(q\cdot u)u\bigr).$$
The symmetry $p\perp q$ and $|p|=|q|$ suggests that the two vectors are obtained from one another by a quarter-turn. The cleanest route is to use coordinates with axes along $p$ and $q$.
Problem Understanding
We are given a square $ABCD$. Two rays from the vertex $A$ are drawn inside the square. The feet of the perpendiculars from $B$ to the rays are $K$ and $L$, and the feet of the perpendiculars from $D$ to the same rays are $M$ and $N$.
We must prove that the segment joining the two projections of $B$ has the same length as the segment joining the two projections of $D$, and that these two segments are perpendicular.
This is a Type B problem. The statement is already specified and must be proved.
The core difficulty is to relate projections of two different vertices onto two arbitrary rays. The square supplies a hidden symmetry: the vectors from the center of the square to $B$ and $D$ are perpendicular and have equal length.
Proof Architecture
Let $O$ be the center of the square and choose orthonormal coordinates with origin at $O$, the $x$-axis directed along $OB$ and the $y$-axis directed along $OD$.
Represent the two rays by unit vectors $u$ and $v$.
Express the projections of $B$ and $D$ onto the ray lines in terms of scalar products with $u$ and $v$.
Derive formulas for the vectors $\overrightarrow{KL}$ and $\overrightarrow{MN}$.
Show that in the chosen coordinates these vectors have coordinates $(X,Y)$ and $(-Y,X)$ for suitable $X,Y$.
Conclude that $\overrightarrow{MN}$ is obtained from $\overrightarrow{KL}$ by a rotation through $90^\circ$, hence the vectors are perpendicular and have equal lengths.
The most delicate point is the algebraic derivation of the coordinate formulas for $\overrightarrow{KL}$ and $\overrightarrow{MN}$.
Solution
Let $O$ be the center of the square. Choose an orthonormal coordinate system with origin at $O$, the $x$-axis along $OB$, and the $y$-axis along $OD$.
Since $OB\perp OD$ and $OB=OD$, after scaling coordinates we may write
$$B=(1,0),\qquad D=(0,1).$$
Let $u=(\cos\alpha,\sin\alpha)$ and $v=(\cos\beta,\sin\beta)$ be unit vectors along the two rays.
The orthogonal projection of a point $P$ onto the line through the origin in the direction of a unit vector $t$ is $(P\cdot t)t$.
Hence
$$K=(B\cdot u)u=\cos\alpha,u,$$
$$L=(B\cdot v)v=\cos\beta,v,$$
$$M=(D\cdot u)u=\sin\alpha,u,$$
$$N=(D\cdot v)v=\sin\beta,v.$$
Therefore
$$\overrightarrow{KL} =\cos\beta,v-\cos\alpha,u.$$
Substituting the coordinates of $u$ and $v$,
\begin{align*}
\overrightarrow{KL}
&=(\cos^2\beta-\cos^2\alpha,,
\cos\beta\sin\beta-\cos\alpha\sin\alpha)\
&=(X,Y),
\end{align*}
where
$$X=\cos^2\beta-\cos^2\alpha, \qquad Y=\cos\beta\sin\beta-\cos\alpha\sin\alpha.$$
Similarly,
$$\overrightarrow{MN} =\sin\beta,v-\sin\alpha,u,$$
so
\begin{align*}
\overrightarrow{MN}
&=(\sin\beta\cos\beta-\sin\alpha\cos\alpha,,
\sin^2\beta-\sin^2\alpha)\
&=(Y,\sin^2\beta-\sin^2\alpha).
\end{align*}
Since
$$\sin^2\beta-\sin^2\alpha =-(\cos^2\beta-\cos^2\alpha) =-X,$$
we obtain
$$\overrightarrow{MN}=(Y,-X).$$
The scalar product of these two vectors equals
$$\overrightarrow{KL}\cdot\overrightarrow{MN} =(X,Y)\cdot(Y,-X) =XY-XY =0.$$
Hence $KL\perp MN$.
Their squared lengths are
$$|\overrightarrow{KL}|^2=X^2+Y^2,$$
and
$$|\overrightarrow{MN}|^2=Y^2+(-X)^2=X^2+Y^2.$$
Thus
$$|KL|=|MN|.$$
We have proved that the segments $KL$ and $MN$ are equal in length and perpendicular to each other.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the projection formula. Since the ray lines pass through the origin $O$, the projection of a point $P$ onto the line spanned by a unit vector $t$ is $(P\cdot t)t$. Applying this to $B$ and $D$ gives the coordinates of $K,L,M,N$ directly. Any omission of the assumption that $t$ is a unit vector would introduce an incorrect scale factor.
The second delicate step is the computation of $\overrightarrow{KL}$. Using
$$\cos\beta,v=(\cos^2\beta,\cos\beta\sin\beta)$$
and
$$\cos\alpha,u=(\cos^2\alpha,\cos\alpha\sin\alpha),$$
their difference is indeed
$$(\cos^2\beta-\cos^2\alpha,, \cos\beta\sin\beta-\cos\alpha\sin\alpha).$$
The same calculation for $\overrightarrow{MN}$ yields
$$(\sin\beta\cos\beta-\sin\alpha\cos\alpha,, \sin^2\beta-\sin^2\alpha),$$
whose first coordinate is exactly $Y$.
The final delicate step is the identity
$$\sin^2\beta-\sin^2\alpha =-(\cos^2\beta-\cos^2\alpha).$$
It follows from $\sin^2 t+\cos^2 t=1$ applied separately to $\alpha$ and $\beta$. This converts $(Y,\sin^2\beta-\sin^2\alpha)$ into $(Y,-X)$, making the $90^\circ$ rotation relation explicit.
Alternative Approaches
A synthetic approach uses the center $O$ of the square. The vectors $\overrightarrow{OB}$ and $\overrightarrow{OD}$ are perpendicular and have equal lengths. For each ray, the projections of $B$ and $D$ onto that ray are obtained by projecting the vectors $\overrightarrow{OB}$ and $\overrightarrow{OD}$ onto the ray direction. Writing the projection operator onto a line and exploiting the fact that $\overrightarrow{OD}$ is a quarter-turn of $\overrightarrow{OB}$, one finds that the vector joining the two projections of $D$ is the image of the corresponding vector for $B$ under the same quarter-turn. Equal lengths and perpendicularity follow immediately.
The coordinate proof above is preferable because it turns the geometric statement into the explicit identities
$$\overrightarrow{KL}=(X,Y),\qquad \overrightarrow{MN}=(Y,-X),$$
after which the conclusion is immediate from elementary vector algebra.