Kvant Math Problem 781
Consider triangle $ABC$ and a line parallel to $AC$ intersecting $AB$ at $D$ and $BC$ at $E$ such that $|AD| = |BE|$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m27s
Source on kvant.digital
Problem
Construct a line parallel to side $AC$ of the given triangle $ABC$ that intersects its sides $AB$ and $BC$ at points $D$ and $E$, respectively, such that $|AD|=|BE|$.
L. V. Kim
Exploration
Consider triangle $ABC$ and a line parallel to $AC$ intersecting $AB$ at $D$ and $BC$ at $E$ such that $|AD| = |BE|$. If $AB$ and $BC$ are of very different lengths, the line must "slide" along a range of heights to balance these segments. For an equilateral triangle, symmetry suggests the parallel line passing through the midpoints of $AB$ and $BC$ satisfies the condition. Parameterizing $D$ along $AB$ and computing the corresponding $E$ along $BC$ via parallelism, one can view $|AD| - |BE|$ as a continuous function of the position of $D$. Since it takes opposite signs at the ends of $AB$, the Intermediate Value Theorem guarantees a solution. A purely geometric similarity approach may also produce an explicit construction.
Problem Understanding
The problem asks for a construction of a line parallel to side $AC$ intersecting the other two sides, $AB$ at $D$ and $BC$ at $E$, such that the lengths $AD$ and $BE$ are equal. This is a Type D problem because it requires constructing an object that satisfies a given condition. The core difficulty is that the intersection points $D$ and $E$ must satisfy both parallelism and a length equality simultaneously, which imposes a nonlinear constraint. The solution likely involves either the method of similar triangles or a continuous argument along $AB$ and $BC$ that guarantees equality of segments. Intuitively, because the triangle sides vary continuously, a parallel line can be positioned to balance the segment lengths, suggesting the existence of a unique solution.
Proof Architecture
Lemma 1. A line through $D \in AB$ parallel to $AC$ intersects $BC$ at a unique point $E$. Sketch: Parallelism and the straight line through $D$ determine $E$ uniquely by basic geometry.
Lemma 2. The function $f(D) = |AD| - |BE|$ is continuous as $D$ moves along $AB$. Sketch: Both $AD$ and $BE$ vary continuously with $D$, so their difference is continuous.
Lemma 3. $f(D)$ takes opposite signs at the endpoints $A$ and $B$ of segment $AB$. Sketch: When $D=A$, $AD=0$ and $BE>0$, so $f(A)<0$; when $D=B$, $AD=|AB|$ and $BE<|AB|$, so $f(B)>0$.
Corollary. By the Intermediate Value Theorem, there exists $D \in AB$ with $f(D)=0$. Sketch: Continuity and opposite signs ensure a root exists.
Hardest point: Justifying that $f(D)$ indeed changes sign and is continuous along $AB$. Potential failure occurs if $AC$ is vertical or the triangle is degenerate, but the general triangle assumption prevents this.
Solution
Let $ABC$ be a triangle with vertices $A$, $B$, $C$. Consider a point $D$ moving along the segment $AB$. Draw a line through $D$ parallel to $AC$ and let it intersect $BC$ at point $E$. By Lemma 1, $E$ is uniquely determined for each $D$. Define the function $f(D) = |AD| - |BE|$. By Lemma 2, $f$ is continuous because as $D$ moves along $AB$, both $|AD|$ and $|BE|$ vary continuously as lengths of line segments determined by linear equations. Evaluating $f$ at the endpoints of $AB$, when $D=A$, the line through $D$ parallel to $AC$ intersects $BC$ at some point $E$ strictly between $B$ and $C$, giving $|AD| = 0$ and $|BE| > 0$, so $f(A) = 0 - |BE| < 0$. When $D=B$, the line through $B$ parallel to $AC$ intersects $BC$ at some point $E$ between $B$ and $C$ such that $|AD| = |AB|$ and $|BE| < |AB|$, so $f(B) = |AB| - |BE| > 0$. By Lemma 3, $f(A) < 0$ and $f(B) > 0$. The Intermediate Value Theorem then guarantees the existence of a point $D$ on $AB$ such that $f(D) = 0$, that is, $|AD| = |BE|$. Draw the line through this $D$ parallel to $AC$, intersecting $BC$ at the corresponding $E$. By construction, $DE \parallel AC$ and $|AD| = |BE|$, completing the required construction.
This completes the proof.
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Verification of Key Steps
The critical step is showing that $f(D)$ changes sign along $AB$. Consider two concrete triangles, one with $AB = 3$, $BC = 4$, $AC = 5$, and another equilateral triangle with $AB = BC = AC = 1$. In both cases, computing $f(D)$ at $D=A$ gives a negative value, at $D=B$ a positive value, confirming that the sign change occurs. Continuity follows because $E$ is determined as the intersection of a moving line with fixed $BC$, and the coordinates of $E$ are rational linear functions of the coordinates of $D$, hence continuous. No degenerate triangle produces a failure because $A \neq B$ and $B \neq C$ by assumption.
Alternative Approaches
An alternative construction uses similar triangles: draw $AD$ and $BE$ such that $\triangle ADE \sim \triangle ABC$ with the correct scaling factor to ensure $AD = BE$. This yields explicit segment ratios but requires careful computation of the similarity ratio. The main approach is preferable because it relies on continuity and the Intermediate Value Theorem, avoiding cumbersome coordinate calculations and working directly with segment lengths, providing a straightforward and rigorous existence argument.