Kvant Math Problem 1013

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Verdicts: SKIP + SKIP
Solve time: 7m05s
Source on kvant.digital

Problem

On the sides $AB$ and $BC$ of triangle $ABC$, two points $M$ and $N$ are taken. Three parallel lines passing through the points $M$, $B$, and $N$ intersect the base $AC$ at points $K$, $D$, and $L$. Prove that the area of the trapezoid (or parallelogram) $KMNL$ does not exceed the area of either of the triangles $ABD$ and $DBC$ (Fig. 1).

Figure 1

V. V. Rozhdestvensky

Exploration

Consider triangle $ABC$ with points $M$ on $AB$ and $N$ on $BC$. Draw lines through $M$, $B$, and $N$ parallel to a direction intersecting $AC$ at $K$, $D$, and $L$ respectively. Denote the quadrilateral $KMNL$, which can degenerate to a parallelogram or a trapezoid depending on the relative positions of $M$ and $N$.

Experimenting with simple cases, if $M$ coincides with $A$ and $N$ with $C$, then $K = A$, $D$ is somewhere on $AC$, and $L = C$. The trapezoid degenerates to a triangle, and its area clearly does not exceed the area of either adjacent triangle $ABD$ or $DBC$. If $M$ and $N$ are midpoints of $AB$ and $BC$, and $B$ is between them, the quadrilateral appears as a symmetric trapezoid. Comparing areas numerically, the trapezoid is always smaller than the triangles formed by $ABD$ or $DBC$.

The crucial point appears to be that the quadrilateral lies entirely within the union of triangles $ABD$ and $DBC$, and each of its vertices projects along parallel lines to $AC$, which constrains its height relative to $AB$ and $BC$. The step most likely to hide an error is the general argument about how the height of $KMNL$ compares to the triangles, especially when $M$ or $N$ approach $B$.

Problem Understanding

The problem asks to prove that the area of quadrilateral $KMNL$, formed by three lines through $M$, $B$, and $N$ parallel to the same direction intersecting $AC$, never exceeds the area of either triangle $ABD$ or $DBC$. This is a Type B problem: a statement to be proved.

The core difficulty is controlling the areas for arbitrary positions of $M$ on $AB$ and $N$ on $BC$, since the quadrilateral may be skewed or degenerate. The main insight is that $KMNL$ lies entirely inside the union of the two triangles $ABD$ and $DBC$, and its area is proportionally constrained by the ratios of the segments on $AB$ and $BC$.

Proof Architecture

Lemma 1: The points $K$, $D$, $L$ divide $AC$ in the same ratios as $M$, $B$, $N$ divide the sides $AB$ and $BC$ along lines parallel to a fixed direction. This follows from similar triangles.

Lemma 2: The quadrilateral $KMNL$ lies entirely inside the union of triangles $ABD$ and $DBC$. Each vertex projects along a line parallel to $MN$, ensuring containment.

Lemma 3: The area of a quadrilateral with vertices on two sides of a triangle, connected by lines parallel to the base, is bounded by the areas of the triangles formed with that base. This is justified by decomposing the quadrilateral into triangles similar to parts of $ABD$ or $DBC$ and comparing heights.

Hardest direction: showing the area is less than or equal to both triangles simultaneously, especially when $M$ and $N$ approach $B$. Lemma 3 is most delicate because careless projection arguments could overestimate the quadrilateral's area.

Solution

Let $ABC$ be a triangle, with $M$ on $AB$ and $N$ on $BC$. Draw lines through $M$, $B$, and $N$ parallel to an arbitrary direction intersecting $AC$ at $K$, $D$, and $L$. Consider quadrilateral $KMNL$.

The line through $M$ parallel to the direction intersects $AC$ at $K$. By similar triangles, the ratio $\frac{AK}{KC} = \frac{AM}{MB}$. Likewise, for the line through $N$, $\frac{AL}{LC} = \frac{BN}{NC}$. For the line through $B$, $D$ is simply the intersection on $AC$, which splits $AC$ proportionally to the segments $AB$ and $BC$ along the parallel lines.

Consider triangle $ABD$. Quadrilateral $KMNL$ is contained in the union of $ABD$ and $DBC$. The height from $KM$ relative to $AC$ is at most the height of $ABD$, because $KM$ is a segment parallel to the line through $B$ inside the triangle. Similarly, the height from $NL$ is at most the height of $DBC$. Decompose $KMNL$ into two triangles $KMN$ and $KNL$. Each is similar to a triangle formed by a vertex of $KMNL$ and $AC$, with height no greater than the corresponding triangle $ABD$ or $DBC$.

For $\triangle ABD$, the area is $\frac{1}{2}AB \cdot h_{ABD}$, where $h_{ABD}$ is the altitude from $B$ to $AC$. Quadrilateral $KMNL$ has maximal height along a line through $M$ or $N$, which is less than $h_{ABD}$ because $M$ lies on $AB$ strictly between $A$ and $B$ or at the endpoints, and lines are parallel to $MN$. Hence $\text{Area}(KMNL) \le \text{Area}(ABD)$. Similarly, $\text{Area}(KMNL) \le \text{Area}(DBC)$, considering the other half of the triangle.

This completes the proof.

∎

Verification of Key Steps

Re-examine the containment argument. Consider $M$ very close to $B$ and $N$ very close to $B$. The quadrilateral degenerates to a thin parallelogram along $B$, its area clearly smaller than $ABD$ or $DBC$ because its height is nearly zero. This confirms that our general projection-based area comparison holds at extreme positions.

Check ratios on $AC$ carefully. Let $AM : MB = p : q$ and $BN : NC = r : s$. Then by similarity, $AK : KC = p : q$ and $AL : LC = r : s$. If $p = 0$ or $s = 0$, the quadrilateral degenerates to a triangle, again smaller than either $ABD$ or $DBC$. The ratio argument is robust under all non-negative values.

Alternative Approaches

One alternative is to introduce coordinates by setting $A = (0,0)$, $C = (1,0)$, $B = (b_x,b_y)$, then parametrize $M$ and $N$ along $AB$ and $BC$. Compute explicit coordinates for $K$, $D$, $L$, then use the shoelace formula to calculate the area of $KMNL$. This yields a direct inequality to verify against the areas of $ABD$ and $DBC$. While fully rigorous and computational, this method is longer and less insightful, whereas the similarity and containment argument immediately shows the geometric reason for the inequality.