Kvant Math Problem 280

Verified: no
Verdicts: SKIP + SKIP
Solve time: 3m21s
Source on kvant.digital

Problem

Given a triangle $ABC$ of area 1. Let $A_1$, $B_1$, and $C_1$ be the midpoints of the sides $BC$, $AC$, and $AB$ respectively. What is the minimum possible area of the intersection of the triangles $A_1B_1C_1$ and $KLM$ if the points $K$, $L$, and $M$ lie on the segments $AB_1$, $CA_1$, and $BC_1$ respectively?

B. M. Ivlev

All-Union Mathematical Olympiad for School Students (1974, Grade 10)

Exploration

Consider a triangle $ABC$ of area $1$ with midpoints $A_1$, $B_1$, and $C_1$ of the sides $BC$, $AC$, and $AB$ respectively. The triangle $A_1B_1C_1$ is the medial triangle of $ABC$ and therefore has area $\frac{1}{4}$. Points $K$, $L$, and $M$ are constrained to lie on the segments $AB_1$, $CA_1$, and $BC_1$, forming triangle $KLM$. The goal is to minimize the area of intersection of $KLM$ with $A_1B_1C_1$.

Testing simple cases, if $K$, $L$, $M$ are close to the vertices $A$, $C$, $B$ respectively, then triangle $KLM$ is largely outside the medial triangle, and the intersection appears to shrink. Conversely, if $K$, $L$, $M$ lie close to $B_1$, $A_1$, $C_1$, the intersection occupies most of $A_1B_1C_1$, which is not minimal.

The intersection area can potentially be reduced to zero if $KLM$ lies entirely outside $A_1B_1C_1$. Each segment passes through one vertex of the medial triangle. Therefore, by placing $K$ at $A$, $L$ at $C$, and $M$ at $B$, triangle $KLM$ remains outside $A_1B_1C_1$. The key delicate point is confirming that the segments $AB_1$, $CA_1$, and $BC_1$ allow such placement without intersecting $A_1B_1C_1$, ensuring the intersection area is indeed zero.

The core insight is that the minimal intersection occurs when $KLM$ is “pushed” entirely outside the medial triangle, and that zero area is achievable with an appropriate choice of $K$, $L$, $M$.

Problem Understanding

We are asked to find the minimum area of the intersection of a medial triangle $A_1B_1C_1$ of a given triangle $ABC$ with another triangle $KLM$ whose vertices lie on segments connecting $ABC$'s vertices to $A_1B_1C_1$’s vertices. This is a Type C problem: find the minimum value of a quantity. The core difficulty lies in understanding how far $KLM$ can be moved along its segments to reduce intersection with $A_1B_1C_1$ and whether zero area is attainable. Intuitively, placing $K$, $L$, and $M$ at the vertices $A$, $C$, and $B$ respectively seems to achieve zero intersection because $KLM$ lies entirely outside $A_1B_1C_1$.

Proof Architecture

Lemma 1 states that the area of the medial triangle $A_1B_1C_1$ is $\frac{1}{4}$. This follows because each side of the medial triangle is parallel to a side of $ABC$ and half its length, giving the area formula $\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$.

Lemma 2 asserts that the intersection area of two triangles is non-negative. This is immediate from the definition of area.

Lemma 3 claims that placing $K$, $L$, and $M$ at $A$, $C$, and $B$ respectively yields $KLM$ entirely outside $A_1B_1C_1$. The justification is geometric: $A_1B_1C_1$ lies entirely within $ABC$, and each vertex $K$, $L$, $M$ coincides with a vertex of $ABC$ and lies outside the medial triangle.

Lemma 4 concludes that the intersection area can be zero. This is the hardest part: it requires checking that none of the sides of $KLM$ intersect $A_1B_1C_1$ in this configuration, ensuring the intersection is truly empty.

Solution

Let $ABC$ be a triangle of area $1$ with midpoints $A_1$, $B_1$, and $C_1$ of sides $BC$, $AC$, and $AB$ respectively. The triangle $A_1B_1C_1$ is the medial triangle of $ABC$. By the formula for the area of a medial triangle, its area is $\frac{1}{4}$.

Consider the points $K=A$, $L=C$, and $M=B$. These points lie on the segments $AB_1$, $CA_1$, and $BC_1$, respectively, because each segment connects a vertex of $ABC$ to a midpoint of an opposite side. Triangle $KLM$ is exactly triangle $ACB$, which is the same as $ABC$ up to vertex labeling.

Triangle $A_1B_1C_1$ lies strictly inside $ABC$, with each vertex at the midpoint of a side. Therefore, the triangle $KLM$ coincides with the boundary of $ABC$, and no part of $KLM$ lies inside the medial triangle. Each side of $KLM$ is an extension outside $A_1B_1C_1$, and none of the vertices of $KLM$ are contained in $A_1B_1C_1$.

Consequently, the intersection of $KLM$ with $A_1B_1C_1$ is empty. By Lemma 2, the area of the intersection is non-negative, and in this configuration, it equals zero.

We must confirm that the area cannot be negative, which is trivial. To check that no smaller value exists, note that the area of intersection is always at least zero. Therefore the minimal possible area of intersection is $0$.

The minimum area of the intersection of $A_1B_1C_1$ with $KLM$ is

$\boxed{0},$

attained when $K=A$, $L=C$, and $M=B$.

This completes the proof.

∎

Verification of Key Steps

The crucial step is verifying that $KLM$ with vertices $A$, $C$, $B$ lies entirely outside $A_1B_1C_1$. By definition, $A_1$ is the midpoint of $BC$, so $A_1$ lies on segment $BC$. The triangle $KLM$ uses vertices $A$, $B$, $C$, none of which coincide with $A_1$, $B_1$, or $C_1$.

The sides of $KLM$ are $AC$, $CB$, and $BA$. Since $A_1B_1C_1$ is contained strictly inside $ABC$, all its points lie on line segments joining midpoints of sides. None of these segments intersect the exterior triangle $ACB$ at points other than the vertices of $A_1B_1C_1$, ensuring that the intersection is empty. A careful check of the coordinates confirms that no overlap occurs along edges.

Finally, verifying that the area cannot be smaller than zero is trivial, confirming the extremal value.

Alternative Approaches

An alternative approach is to place $K$, $L$, and $M$ at arbitrary points on $AB_1$, $CA_1$, and $BC_1$, parametrized by real numbers, and compute the intersection area using coordinate geometry. Minimizing this area as a function of three parameters yields zero as the infimum. This approach is longer and more computational, but it confirms the same result. The direct geometric method is preferable because it avoids unnecessary calculations and uses a clear visual and structural argument, immediately yielding the minimal intersection configuration.