Kvant Math Problem 346

Place the square in a coordinate system with algebraic convenience so that perpendicularity can be tested by a dot product condition.

kvantmathematicsolympiad

Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m29s
Source on kvant.digital

Problem

Point $K$ is the midpoint of side $AB$ of square $ABCD$, and point $L$ divides diagonal $AC$ in the ratio $3:1$ (Fig. 1). Prove that angle $KLD$ is a right angle.

Figure 1

Yu. G. Bogaturov, 10th grade student (Kutaisi)

Exploration

Place the square in a coordinate system with algebraic convenience so that perpendicularity can be tested by a dot product condition. Let $A$ be the origin, $AB$ lie on the $x$-axis, and $AD$ on the $y$-axis. Then $A(0,0)$, $B(1,0)$, $C(1,1)$, $D(0,1)$ describes a unit square without loss of generality.

The midpoint $K$ of $AB$ becomes $(\tfrac12,0)$. The point $L$ lies on the diagonal $AC$, hence has the form $(t,t)$. The ratio $AL:LC=3:1$ determines $t=\tfrac34$.

The claim that $\angle KLD$ is right translates to $(\overrightarrow{LK})\cdot(\overrightarrow{LD})=0$. Computing these vectors gives a direct algebraic condition that can be verified exactly.

The central step is confirming that the two resulting direction vectors are orthogonal after substitution of $t=\tfrac34$.

Problem Understanding

This is a Type B problem, requiring a proof that a specific angle is a right angle in a geometric configuration involving a square, a midpoint, and a division point on a diagonal.

The core difficulty lies in translating the geometric configuration into a form where perpendicularity can be verified rigorously without relying on diagrammatic intuition. The expected outcome is that $\angle KLD=90^\circ$.

Proof Architecture

The first lemma establishes coordinates for a square and expresses $K$ and $L$ in these coordinates. This is justified by rigid motions preserving angles and ratios.

The second lemma computes the condition $AL:LC=3:1$ in terms of the parameter $t$ on the diagonal $AC$, yielding $t=\tfrac34$.

The third lemma translates the right angle condition into orthogonality of vectors $\overrightarrow{LK}$ and $\overrightarrow{LD}$ via the dot product criterion.

The final step evaluates this dot product explicitly and verifies it equals zero. The most delicate point is the correct interpretation of the ratio on the diagonal and the correct computation of vector components.

Solution

Let the square $ABCD$ be placed in the coordinate plane so that $A(0,0)$, $B(1,0)$, $C(1,1)$, and $D(0,1)$. This choice preserves all incidence relations and angle measures.

The midpoint $K$ of $AB$ has coordinates $K\left(\tfrac12,0\right)$.

Point $L$ lies on segment $AC$, hence $L$ has coordinates $(t,t)$ for some $t\in(0,1)$.

The condition $AL:LC=3:1$ implies

$$\frac{AL}{AC}=\frac{3}{4}.$$

Since $AC$ is parameterized by $(t,t)$ with $t$ measuring proportional distance from $A(0,0)$ to $C(1,1)$, it follows that $t=\tfrac34$. Hence $L\left(\tfrac34,\tfrac34\right)$.

To prove $\angle KLD$ is right, it suffices to show that $\overrightarrow{LK}\perp \overrightarrow{LD}$, that is,

$$\overrightarrow{LK}\cdot \overrightarrow{LD}=0.$$

We compute

$$\overrightarrow{LK}=K-L=\left(\tfrac12-\tfrac34,,0-\tfrac34\right)=\left(-\tfrac14,,-\tfrac34\right),$$

and

$$\overrightarrow{LD}=D-L=\left(0-\tfrac34,,1-\tfrac34\right)=\left(-\tfrac34,,\tfrac14\right).$$

Their dot product equals

$$\overrightarrow{LK}\cdot \overrightarrow{LD} =\left(-\tfrac14\right)\left(-\tfrac34\right)+\left(-\tfrac34\right)\left(\tfrac14\right) =\tfrac{3}{16}-\tfrac{3}{16}=0.$$

Therefore $\overrightarrow{LK}\perp \overrightarrow{LD}$, which implies $\angle KLD=90^\circ$.

This completes the proof. ∎

Verification of Key Steps

The identification of the parameter $t=\tfrac34$ follows directly from interpreting the ratio on a segment as a proportional division of the parameter along a straight line. Since $AC$ is represented by $(t,t)$, distances along it scale linearly with $t$, so the fraction of the whole segment from $A$ to $L$ equals $t$.

The vector computations are verified independently by recomputing each coordinate difference: subtraction in both components yields consistent expressions for $\overrightarrow{LK}$ and $\overrightarrow{LD}$.

The orthogonality check reduces entirely to arithmetic in rational numbers, and cancellation occurs between equal magnitude opposite-sign terms, ensuring no hidden geometric assumption is used.

Alternative Approaches

A synthetic approach avoids coordinates by using properties of midpoints and angle chasing in similar triangles formed by the diagonal of the square. One can express $K$ as the midpoint of $AB$ and compare triangles $ALK$ and $ADL$ using similarity induced by the square’s symmetry. This route is more sensitive to auxiliary constructions and angle tracking, while the coordinate method reduces the problem to a single explicit orthogonality computation with no auxiliary geometric choices.