Kvant Math Problem 853

A square $ABCD$ rotates about its fixed center $O$, while a fixed line $l$ is taken as the $x$-axis.

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Verified: yes
Verdicts: PASS + PASS
Solve time: 29m33s
Source on kvant.digital

Problem

The square $ABCD$ rotates about its fixed center. Find the locus traced by the midpoint of the segment $PQ$, where $P$ is the foot of the perpendicular dropped from the point $D$ to the fixed line $l$, and $Q$ is the midpoint of the side $AB$.

Y. Tabov (Bulgaria)

Exploration

A square $ABCD$ rotates about its fixed center $O$, while a fixed line $l$ is taken as the $x$-axis. The midpoint $Q$ of $AB$ and the foot $P$ of the perpendicular from $D$ to $l$ both depend on the rotation angle $\theta$. The midpoint $M$ of $PQ$ is expressed in terms of $\theta$, and the locus is obtained by eliminating the parameter.

The configuration is handled by placing the square in a coordinate system centered at $O$, then applying a rotation matrix to all relevant points.

Problem Understanding

The square is initially centered at the origin with vertices

$A\left(-\frac a2,\frac a2\right),\quad B\left(\frac a2,\frac a2\right),\quad C\left(\frac a2,-\frac a2\right),\quad D\left(-\frac a2,-\frac a2\right).$

A rotation by angle $\theta$ about the origin is applied using

$R_\theta=\begin{pmatrix}\cos\theta&-\sin\theta\ \sin\theta&\cos\theta\end{pmatrix}.$

The midpoint $Q$ of $AB$ is rotated, the vertex $D$ is rotated, the point $P$ is obtained by orthogonal projection of $D$ onto the $x$-axis, and $M$ is the midpoint of $PQ$.

Proof Architecture

The midpoint of $AB$ before rotation is

$Q_0=\left(0,\frac a2\right).$

After rotation,

$Q=\left(-\frac a2\sin\theta,\frac a2\cos\theta\right).$

The vertex $D_0=\left(-\frac a2,-\frac a2\right)$ transforms as

$D=\left(\frac a2(\sin\theta-\cos\theta),-\frac a2(\sin\theta+\cos\theta)\right).$

The projection of $D$ onto the $x$-axis keeps the $x$-coordinate and sets the $y$-coordinate to zero, hence

$P=\left(\frac a2(\sin\theta-\cos\theta),0\right).$

The midpoint $M$ of $PQ$ is computed coordinatewise. For the $x$-coordinate,

$x_M=\frac12\left(\frac a2(\sin\theta-\cos\theta)-\frac a2\sin\theta\right)=-\frac a4\cos\theta.$

For the $y$-coordinate,

$y_M=\frac12\left(0+\frac a2\cos\theta\right)=\frac a4\cos\theta.$

Thus

$M=\left(-\frac a4\cos\theta,\frac a4\cos\theta\right).$

Solution

The parametric equations of the locus are

$x=-\frac a4\cos\theta,\qquad y=\frac a4\cos\theta.$

These relations imply the linear constraint

$x+y=0.$

The parameter $\cos\theta$ varies between $-1$ and $1$, so $y$ ranges from $-\frac a4$ to $\frac a4$. The same restriction applies to $x$ through $x=-y$.

The locus is therefore a segment of the line $y=-x$ bounded by the extreme values of $\cos\theta$.

When $\cos\theta=1$, the point is

$M\left(-\frac a4,\frac a4\right),$

and when $\cos\theta=-1$, the point is

$M\left(\frac a4,-\frac a4\right).$

Verification of Key Steps

The rotation of $Q$ follows directly from applying the rotation matrix to the midpoint of $AB$. The rotation of $D$ is obtained by applying the same matrix to $\left(-\frac a2,-\frac a2\right)$, which produces a linear combination of $\sin\theta$ and $\cos\theta$ with correct sign structure.

The projection step preserves the horizontal coordinate of $D$ because orthogonal projection onto the $x$-axis removes only the vertical component. The midpoint computation is an affine average of two correctly rotated points, which reduces both coordinates to expressions proportional to $\cos\theta$ alone after cancellation of $\sin\theta$ terms.

The resulting dependence shows that all motion collapses onto a single scalar parameter $\cos\theta$, which forces the locus to be a line segment rather than a general conic.

Alternative Approaches

The point $Q$ depends linearly on the rotated vector corresponding to the midpoint of a side of the square, while $P$ depends linearly on the rotated vertex $D$ followed by projection onto a fixed axis. Both operations preserve linear dependence on $\sin\theta$ and $\cos\theta$, but the midpoint operation eliminates the $\sin\theta$ component entirely, leaving both coordinates proportional to $\cos\theta$ with opposite signs. This reduction forces collinearity of all positions of $M$, so the image of the motion is a one-dimensional segment rather than a quadratic curve.