Kvant Math Problem 1269

Let $a=AB$ and $c=AC$.

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Problem

A triangle $ABC$ is given in the plane. The line $p$ is parallel to the line $AB$ and is at a distance $AC$ from it, in such a way that the strip bounded by these two lines ($p$ and $AB$) contains no interior points of the triangle $ABC$. The line $q$ is parallel to the line $AC$ and is at a distance $AB$ from it, in such a way that the strip bounded by these two lines ($q$ and $AC$) contains no interior points of the triangle $ABC$. The lines $p$ and $q$ intersect at the point $L$. Prove that the line $AL$ passes through the midpoint of $BC$.

Ya. Koval'

Exploration

Let $a=AB$ and $c=AC$. The line $p$ is parallel to $AB$ and at distance $c$ from it. Since the strip between $AB$ and $p$ contains no interior points of the triangle, $p$ lies on the side of $AB$ opposite to $C$.

Similarly, $q$ is parallel to $AC$ and at distance $a$ from it, lying on the side of $AC$ opposite to $B$.

The condition involves distances equal to side lengths. This suggests introducing coordinates adapted to the sides through $A$.

Take $A=(0,0)$. Let the $x$-axis lie along $AB$, and let $AC$ make an angle $\gamma=\angle BAC$ with the $x$-axis. Then

$$B=(a,0), \qquad C=(c\cos\gamma,c\sin\gamma).$$

Since $AB$ is the line $y=0$, the line parallel to it at distance $c$ on the side opposite to $C$ is

$$p:\ y=-c.$$

The line $AC$ has equation

$$-\sin\gamma,x+\cos\gamma,y=0.$$

A parallel line at distance $a$ has equation

$$-\sin\gamma,x+\cos\gamma,y=\pm a.$$

To determine the sign, substitute $B=(a,0)$; the value is $-a\sin\gamma<0$. Hence the side opposite to $B$ corresponds to the positive sign:

$$q:\ -\sin\gamma,x+\cos\gamma,y=a.$$

Intersecting $q$ with $p$ gives

$$-\sin\gamma,x-c\cos\gamma=a,$$

hence

$$x=-\frac{a+c\cos\gamma}{\sin\gamma}, \qquad y=-c.$$

The midpoint of $BC$ is

$$M=\left(\frac{a+c\cos\gamma}{2},\frac{c\sin\gamma}{2}\right).$$

The coordinates of $L$ are proportional to those of $M$:

$$L=-\frac{2}{\sin\gamma},M.$$

Thus $A,M,L$ are collinear.

The only delicate point is choosing the correct side for the parallel line $q$ and checking that the sign in its equation is positive.

Problem Understanding

A triangle $ABC$ is given. Through a construction depending on the lengths $AB$ and $AC$, two lines $p$ and $q$ are drawn parallel to $AB$ and $AC$ respectively, each lying outside the triangle. Their intersection is $L$. The task is to prove that the line $AL$ passes through the midpoint of $BC$.

This is a Type B problem, a pure proof.

The core difficulty is translating the geometric condition describing on which side of the original lines the parallels lie. Once that is done correctly, the statement becomes a straightforward coordinate computation.

Proof Architecture

Introduce coordinates with $A=(0,0)$, $B=(a,0)$, and $C=(c\cos\gamma,c\sin\gamma)$.

Show that the line $p$ has equation $y=-c$, because it is parallel to $AB$ and lies at distance $c$ on the side opposite to $C$.

Show that the line $q$ has equation $-\sin\gamma,x+\cos\gamma,y=a$, because it is parallel to $AC$, at distance $a$ from it, and lies on the side opposite to $B$.

Compute the coordinates of $L=p\cap q$.

Compute the midpoint $M$ of $BC$.

Prove that the position vectors of $L$ and $M$ are proportional, which implies that $A$, $M$, and $L$ are collinear.

The most delicate lemma is the determination of the correct equation of $q$, since choosing the wrong sign places the line on the wrong side of $AC$.

Solution

Let

$$a=AB,\qquad c=AC,\qquad \gamma=\angle BAC.$$

Choose Cartesian coordinates so that

$$A=(0,0),\qquad B=(a,0),$$

and the positive $x$-axis is directed from $A$ to $B$. Then

$$C=(c\cos\gamma,c\sin\gamma).$$

Since $AB$ is the line $y=0$, and $C$ lies above it because $c\sin\gamma>0$, the line parallel to $AB$ at distance $c$ on the opposite side is

$$p:\ y=-c.$$

The line $AC$ passes through the origin and has equation

$$-\sin\gamma,x+\cos\gamma,y=0.$$

For a line

$$-\sin\gamma,x+\cos\gamma,y=t,$$

parallel to $AC$, the distance from $AC$ equals $|t|$, because

$$\sqrt{(-\sin\gamma)^2+(\cos\gamma)^2}=1.$$

Hence every line parallel to $AC$ and at distance $a$ from it has equation

$$-\sin\gamma,x+\cos\gamma,y=\pm a.$$

At the point $B=(a,0)$ the left-hand side equals

$$-a\sin\gamma<0.$$

Therefore $B$ lies in the half-plane

$$-\sin\gamma,x+\cos\gamma,y<0.$$

Since $q$ must lie on the side of $AC$ opposite to $B$, its equation is

$$q:\ -\sin\gamma,x+\cos\gamma,y=a.$$

Let $L=p\cap q$. Substituting $y=-c$ into the equation of $q$ gives

$$-\sin\gamma,x-c\cos\gamma=a,$$

so

$$x=-\frac{a+c\cos\gamma}{\sin\gamma}.$$

Thus

$$L= \left( -\frac{a+c\cos\gamma}{\sin\gamma}, -c \right).$$

Let $M$ be the midpoint of $BC$. Then

$$M= \left( \frac{a+c\cos\gamma}{2}, \frac{c\sin\gamma}{2} \right).$$

Multiplying the coordinates of $M$ by the scalar

$$-\frac{2}{\sin\gamma}$$

yields

$$-\frac{2}{\sin\gamma}M = \left( -\frac{a+c\cos\gamma}{\sin\gamma}, -c \right) = L.$$

Hence the position vector of $L$ is a scalar multiple of the position vector of $M$. Consequently the points $A=(0,0)$, $M$, and $L$ are collinear.

Since $M$ is the midpoint of $BC$, the line $AL$ passes through the midpoint of $BC$.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is locating $p$. The strip bounded by $AB$ and $p$ contains no interior point of the triangle. Since the interior of the triangle near vertex $C$ lies on the same side of $AB$ as $C$, the line $p$ must be drawn on the opposite side. In the chosen coordinates, that side is $y<0$, giving $p:y=-c$ rather than $y=c$.

The second delicate step is choosing the sign in the equation of $q$. The family of parallels at distance $a$ from $AC$ is

$$-\sin\gamma,x+\cos\gamma,y=\pm a.$$

Evaluating at $B$ gives a negative value. Thus $B$ lies in the negative half-plane determined by $AC$. The required line must be in the opposite half-plane, which corresponds to the equation with right-hand side $+a$. Choosing $-a$ would place the line on the same side as $B$ and violate the hypothesis.

The final step is the collinearity test. The midpoint is

$$M= \left( \frac{a+c\cos\gamma}{2}, \frac{c\sin\gamma}{2} \right),$$

while

$$L= \left( -\frac{a+c\cos\gamma}{\sin\gamma}, -c \right).$$

Both coordinates are multiplied by the same factor $-2/\sin\gamma$, so $L$ lies on the line through $A$ and $M$. No additional geometric argument is required.

Alternative Approaches

A synthetic proof can be obtained by introducing vectors based at $A$. Let $\mathbf{u}$ and $\mathbf{v}$ be unit vectors along $AB$ and $AC$. Since $p$ is obtained from $AB$ by translating it through the vector $-c\mathbf{n}_1$, where $\mathbf{n}_1$ is a unit normal to $AB$, and $q$ is obtained from $AC$ by translating it through the vector $a\mathbf{n}_2$, where $\mathbf{n}_2$ is a unit normal to $AC$, one can solve for the intersection point $L$ in the basis ${\mathbf{u},\mathbf{v}}$. The resulting expression shows that

$$\overrightarrow{AL} = -\frac{1}{\sin\gamma} \left( \overrightarrow{AB} + \overrightarrow{AC} \right).$$

Since the midpoint $M$ of $BC$ satisfies

$$\overrightarrow{AM} = \frac12 \left( \overrightarrow{AB} + \overrightarrow{AC} \right),$$

the vectors $\overrightarrow{AL}$ and $\overrightarrow{AM}$ are proportional.

The coordinate method is preferable because the distance conditions immediately translate into simple linear equations, and the midpoint computation becomes completely transparent.