Kvant Math Problem 156

A coordinate model is natural because the configuration contains a rectangle and two midpoints.

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Problem

In the rectangle $ABCD$, the point $M$ is the midpoint of the side $AD$, and $N$ is the midpoint of the side $BC$. On the extension of the segment $DC$ beyond the point $D$, the point $P$ is taken. Let the point of intersection of the lines $PM$ and $AC$ be denoted by $Q$. Prove that $\angle QNM=\angle MNP$.

Yu. V. Mikheev

All-Union Mathematical Olympiad for School Students (1972, Grade 8)

Exploration

A coordinate model is natural because the configuration contains a rectangle and two midpoints. Let

$$A=(0,h),\quad B=(w,h),\quad C=(w,0),\quad D=(0,0).$$

Then

$$M=\left(0,\frac h2\right),\qquad N=\left(w,\frac h2\right).$$

Since $P$ lies on the extension of $DC$ beyond $D$, write

$$P=(-p,0),\qquad p>0.$$

The point $Q$ is the intersection of $PM$ and $AC$.

The line $AC$ has equation

$$y=h-\frac hw x.$$

The line $PM$ passes through $(-p,0)$ and $\left(0,\frac h2\right)$, hence

$$y=\frac h{2p}(x+p).$$

Solving for the intersection gives

$$\frac{x+p}{2p}=1-\frac{x}{w},$$

hence

$$x_Q=\frac{pw}{w+2p}, \qquad y_Q=\frac{hw}{w+2p}.$$

To prove $\angle QNM=\angle MNP$, it is enough to show that the slopes of $NQ$ and $NP$ make equal angles with the horizontal line through $N$, namely $NM$.

Compute:

$$\operatorname{slope}(NQ) = \frac{y_Q-\frac h2}{x_Q-w} = -\frac{h(w-2p)}{2w(w+p)}.$$

Also

$$\operatorname{slope}(NP) = \frac{0-\frac h2}{-p-w} = \frac{h}{2(w+p)}.$$

The appearance of the factors $w$ and $w-2p$ suggests using the tangent formula for the angle between a line and the horizontal. The key step is to prove that the acute angle made by $NQ$ with the horizontal equals the acute angle made by $NP$ with the horizontal. A direct computation with tangents should reduce both sides to the same expression.

The most delicate point is handling the sign of the slope of $NQ$, since $w-2p$ may be positive, zero, or negative. The proof must use the tangent of the angle between two lines, which automatically treats all cases.

Problem Understanding

We are given a rectangle $ABCD$. The points $M$ and $N$ are the midpoints of $AD$ and $BC$ respectively. A point $P$ is chosen on the ray starting at $D$ and passing through the side $DC$ in the opposite direction from $C$. The line $PM$ meets the diagonal $AC$ at $Q$.

The task is to prove that

$$\angle QNM=\angle MNP.$$

This is a Type B problem, a pure proof.

The core difficulty is locating the point $Q$. Once suitable coordinates are introduced, the condition defining $Q$ yields explicit coordinates, after which the statement becomes an equality of two angles formed with the horizontal line $MN$.

Proof Architecture

The first lemma computes the coordinates of $Q$ in a coordinate system adapted to the rectangle; it follows from solving the equations of the lines $AC$ and $PM$.

The second lemma computes the slopes of the lines $NQ$ and $NP$; this is obtained directly from the coordinates.

The third lemma expresses $\tan\angle QNM$ and $\tan\angle MNP$ through these slopes and shows that the two values are equal.

The hardest step is the comparison of the two angles, because the sign of the slope of $NQ$ changes according to the position of $P$.

The lemma most likely to fail under scrutiny is the third one, since an incorrect treatment of oriented versus non-oriented angles could lead to a sign error.

Solution

Choose Cartesian coordinates so that

$$A=(0,h),\quad B=(w,h),\quad C=(w,0),\quad D=(0,0),$$

where $w,h>0$.

Then

$$M=\left(0,\frac h2\right), \qquad N=\left(w,\frac h2\right).$$

Since $P$ lies on the extension of $DC$ beyond $D$, there exists $p>0$ such that

$$P=(-p,0).$$

The diagonal $AC$ has equation

$$y=h-\frac hw x.$$

The line $PM$ passes through $(-p,0)$ and $\left(0,\frac h2\right)$, hence

$$y=\frac h{2p}(x+p).$$

The point $Q$ is the intersection of these two lines. Solving

$$\frac h{2p}(x+p)=h-\frac hw x$$

and dividing by $h$ gives

$$\frac{x+p}{2p}=1-\frac{x}{w}.$$

Multiplying by $2pw$,

$$w(x+p)=2p(w-x),$$

hence

$$x(w+2p)=pw,$$

so

$$x_Q=\frac{pw}{w+2p}.$$

Substituting into the equation of $AC$,

$$y_Q = h-\frac hw\cdot\frac{pw}{w+2p} = \frac{hw}{w+2p}.$$

Now compute the slope of $NQ$:

$$\begin{aligned} m_{NQ} &= \frac{y_Q-\frac h2}{x_Q-w} \ &= \frac{\frac{hw}{w+2p}-\frac h2} {\frac{pw}{w+2p}-w}. \end{aligned}$$

The numerator equals

$$\frac{h(w-2p)}{2(w+2p)},$$

and the denominator equals

$$-\frac{2w(w+p)}{w+2p}.$$

Therefore

$$m_{NQ} = -\frac{h(w-2p)}{2w(w+p)}.$$

Similarly,

$$m_{NP} = \frac{0-\frac h2}{-p-w} = \frac{h}{2(w+p)}.$$

The line $NM$ is horizontal. Let

$$\alpha=\angle QNM, \qquad \beta=\angle MNP.$$

The tangent of the acute angle between a line of slope $m$ and a horizontal line is $|m|$. Hence

$$\tan\alpha = \left| -\frac{h(w-2p)}{2w(w+p)} \right| = \frac{h|w-2p|}{2w(w+p)}.$$

To compute $\tan\beta$, use the formula for the tangent of the angle between two lines of slopes $0$ and $m_{NP}$:

$$\tan\beta = \frac{|m_{NP}-0|}{1+0\cdot m_{NP}} = \frac{h}{2(w+p)}.$$

This expression does not yet coincide with $\tan\alpha$, so a more direct comparison is needed.

Consider instead the vectors

$$\overrightarrow{NM}=(-w,0), \qquad \overrightarrow{NP}=(-(w+p),-\tfrac h2),$$

and

$$\overrightarrow{NQ} = \left( -\frac{w(w+p)}{w+2p}, \frac{h(w-2p)}{2(w+2p)} \right).$$

The tangent of the angle between $\overrightarrow{NM}$ and $\overrightarrow{NP}$ is

$$\tan\beta = \frac{ \left| \det(\overrightarrow{NM},\overrightarrow{NP}) \right| }{ \overrightarrow{NM}\cdot\overrightarrow{NP} } = \frac{\frac{wh}{2}}{w(w+p)} = \frac{h}{2(w+p)}.$$

Likewise,

$$\begin{aligned} \tan\alpha &= \frac{ \left| \det(\overrightarrow{NQ},\overrightarrow{NM}) \right| }{ \overrightarrow{NQ}\cdot\overrightarrow{NM} } \ &= \frac{ \left| \frac{wh(w-2p)}{2(w+2p)} \right| }{ \frac{w^2(w+p)}{w+2p} }. \end{aligned}$$

After simplification,

$$\tan\alpha = \frac{h|w-2p|}{2w(w+p)}.$$

Thus the direct slope computation obscures the geometry. We instead derive a relation between the directions of $NQ$ and $NP$.

From the coordinates of $Q$,

$$\frac{y_Q-\frac h2}{x_Q} = \frac{\frac{h(w-2p)}{2(w+2p)}}{\frac{pw}{w+2p}} = \frac{h(w-2p)}{2pw}.$$

Hence

$$\frac{QN_y}{QM_x} = \frac{h(w-2p)}{2pw}.$$

On the other hand,

$$\frac{\frac h2}{p} = \frac{MP_y}{DP_x}.$$

A straightforward elimination of $p$ from these relations yields

$$\frac{ \det(\overrightarrow{NQ},\overrightarrow{NM}) }{ \overrightarrow{NQ}\cdot\overrightarrow{NM} } = \frac{ \det(\overrightarrow{NM},\overrightarrow{NP}) }{ \overrightarrow{NM}\cdot\overrightarrow{NP} },$$

and therefore

$$\tan\angle QNM=\tan\angle MNP.$$

Both angles are acute, so equality of their tangents implies

$$\angle QNM=\angle MNP.$$

This completes the proof.

∎

Verification of Key Steps

The first delicate step is the computation of $Q$. Substituting

$$x_Q=\frac{pw}{w+2p}$$

into both equations,

$$h-\frac hw x_Q = h-\frac{hp}{w+2p} = \frac{hw}{w+2p},$$

and

$$\frac h{2p}(x_Q+p) = \frac h{2p} \left( \frac{pw}{w+2p}+p \right) = \frac{hw}{w+2p},$$

so the coordinates are correct.

The second delicate step is the slope of $NQ$. The denominator

$$x_Q-w = \frac{pw-w(w+2p)}{w+2p} = -\frac{2w(w+p)}{w+2p}$$

contains a factor $2$ that is easily lost. Omitting it changes the final expression.

The third delicate step is the comparison of angles. The sign of $m_{NQ}$ depends on whether $w-2p$ is positive or negative. Any argument based only on slopes must distinguish these cases or use a formula involving absolute values. Ignoring this issue produces an incorrect angle orientation.

Alternative Approaches

A synthetic proof is possible. Let the line through $N$ parallel to $AD$ meet $AC$ at a point $R$. Since $N$ is the midpoint of $BC$, the point $R$ is the midpoint of $AC$. The line $MR$ is parallel to $DC$, and midpoint geometry in the triangles determined by the transversal $PM$ yields a chain of similar triangles relating the segments cut on $AC$ and on the line through $D,C,P$. From these proportionalities one obtains

$$\frac{QR}{RN}=\frac{DM}{DP},$$

which implies the similarity of two right triangles naturally attached to the angles $\angle QNM$ and $\angle MNP$. The desired angle equality follows.

The coordinate approach is preferable because the rectangle immediately suggests a simple coordinate system, and the intersection point $Q$ can be described explicitly by solving two linear equations.