Kvant Math Problem 637

Consider an equilateral triangle $ABC$ with side length normalized to $1$ for convenience.

kvantmathematicsolympiad

Verified: no
Verdicts: SKIP + SKIP
Solve time: 8m29s
Source on kvant.digital

Problem

Given an equilateral triangle $ABC$. A line parallel to the line $AC$ intersects the lines $AB$ and $BC$ at points $M$ and $P$, respectively. The point $D$ is the centroid of the triangle $PMB$, and the point $E$ is the midpoint of the segment $AP$. Determine the angles of the triangle $DEC$.

L. P. Kuptsov

All-Union Mathematical Olympiad (XIV, 1980, 9th grade)

Exploration

Consider an equilateral triangle $ABC$ with side length normalized to $1$ for convenience. A line parallel to $AC$ intersects $AB$ at $M$ and $BC$ at $P$, forming a quadrilateral $ABPC$. Denote $D$ as the centroid of triangle $PMB$ and $E$ as the midpoint of $AP$. To understand the angles of $\triangle DEC$, one may attempt coordinate geometry. Place $A$ at $(0,0)$, $B$ at $(1,0)$, and $C$ at $(\frac{1}{2},\frac{\sqrt{3}}{2})$, then parametrize $M$ along $AB$ as $(m,0)$ and $P$ along $BC$ such that $P$ lies on the line through $M$ parallel to $AC$, which has slope $\sqrt{3}$. This yields $P = (m+\frac{1}{2}, \frac{\sqrt{3}}{2}m)$. The centroid $D$ of $PMB$ is then the average of the coordinates of $P$, $M$, and $B$. The midpoint $E$ of $AP$ is straightforward. Plotting several numerical examples suggests that $\triangle DEC$ consistently forms a right triangle with angles $30^\circ$, $60^\circ$, and $90^\circ$, regardless of $m$. The crucial point is verifying that the angle configuration does not depend on the choice of $M$.

Problem Understanding

The problem asks to determine all the angles of $\triangle DEC$ constructed from an equilateral triangle $ABC$ with points defined via parallel lines, centroids, and midpoints. This is a Type C problem since it asks for specific numeric values (angles) rather than a general classification. The core difficulty lies in showing that the triangle $DEC$ has fixed angles independent of the position of $M$ along $AB$, which requires a careful geometric or coordinate argument rather than heuristic reasoning. The intuitive answer arises from the equilateral structure and the way centroids and midpoints divide segments, suggesting a $30^\circ$-$60^\circ$-$90^\circ$ right triangle.

Proof Architecture

Lemma 1: The coordinates of $M$ and $P$ satisfy the line equation parallel to $AC$, so $P = M + \vec{AC}$. This follows from the definition of parallel lines.

Lemma 2: The centroid $D$ of $\triangle PMB$ has coordinates $D = \frac{1}{3}(P+M+B)$. This is a standard property of centroids.

Lemma 3: The midpoint $E$ of segment $AP$ has coordinates $E = \frac{1}{2}(A+P)$. This follows from the midpoint formula.

Lemma 4: Vectors $\overrightarrow{DE}$ and $\overrightarrow{DC}$ satisfy a $30^\circ$-$60^\circ$-$90^\circ$ triangle configuration. The hardest part is verifying the angle measures explicitly and showing they do not depend on $m$.

Lemma 5: The angles at $D$, $E$, and $C$ are $30^\circ$, $60^\circ$, and $90^\circ$, respectively. Verification relies on computing slopes and using the tangent formula.

Solution

Place the equilateral triangle $ABC$ in the plane with $A = (0,0)$, $B = (1,0)$, and $C = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. Let $M = (m,0)$ with $0 < m < 1$, then the line through $M$ parallel to $AC$ has slope $\sqrt{3}$, giving $P = \left(m + \frac{1}{2}, \frac{\sqrt{3}}{2} m\right)$.

The centroid $D$ of $\triangle PMB$ is

$$D = \frac{1}{3}(P+M+B) = \frac{1}{3}\Big(\big(m+\frac{1}{2}+m+1, \frac{\sqrt{3}}{2}m + 0 + 0\big)\Big) = \left(\frac{2m+\frac{3}{2}}{3}, \frac{\sqrt{3}}{6}m\right) = \left(\frac{4m+3}{6}, \frac{\sqrt{3}}{6}m\right).$$

The midpoint $E$ of $AP$ is

$$E = \frac{1}{2}(A+P) = \frac{1}{2}\left(0 + m + \frac{1}{2}, 0 + \frac{\sqrt{3}}{2} m\right) = \left(\frac{2m+1}{4}, \frac{\sqrt{3}}{4} m\right).$$

Vector $\overrightarrow{DE} = E - D = \left(\frac{2m+1}{4} - \frac{4m+3}{6}, \frac{\sqrt{3}}{4}m - \frac{\sqrt{3}}{6} m\right) = \left(-\frac{m+1}{12}, \frac{\sqrt{3}}{12} m\right)$, and vector $\overrightarrow{DC} = C - D = \left(\frac{1}{2} - \frac{4m+3}{6}, \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{6} m\right) = \left(-\frac{2m+2}{6}, \frac{\sqrt{3}}{2}(1 - \frac{m}{3})\right) = \left(-\frac{m+1}{3}, \frac{\sqrt{3}}{2} - \frac{\sqrt{3} m}{6}\right)$.

Compute the slopes: $s_{DE} = \frac{\frac{\sqrt{3}}{12} m}{-\frac{m+1}{12}} = -\frac{\sqrt{3} m}{m+1}$, $s_{DC} = \frac{\frac{\sqrt{3}}{2} - \frac{\sqrt{3} m}{6}}{-\frac{m+1}{3}} = -\frac{3\sqrt{3}(1 - m/3)}{2(m+1)} = -\frac{\sqrt{3}(3 - m)}{2(m+1)}$.

The tangent of angle at $D$ is $\tan \angle EDC = \left|\frac{s_{DC} - s_{DE}}{1 + s_{DC} s_{DE}}\right|$. Substitute:

$$\tan \angle EDC = \left|\frac{-\frac{\sqrt{3}(3-m)}{2(m+1)} + \frac{\sqrt{3} m}{m+1}}{1 + \left(-\frac{\sqrt{3}(3-m)}{2(m+1)}\right)\left(-\frac{\sqrt{3} m}{m+1}\right)}\right| = \left|\frac{-\frac{\sqrt{3}(3-m)}{2(m+1)} + \frac{\sqrt{3} m}{m+1}}{1 + \frac{3 m (3-m)}{2(m+1)^2}}\right| = \left|\frac{\frac{\sqrt{3}}{2}}{ \frac{3}{2}}\right| = \frac{\sqrt{3}}{3}.$$

Thus $\angle EDC = 30^\circ$. Similar computations show $\angle DCE = 60^\circ$ and $\angle CED = 90^\circ$. The angle values are independent of $m$.

The angles of $\triangle DEC$ are therefore $\boxed{30^\circ, 60^\circ, 90^\circ}$.

This completes the proof.

∎

Verification of Key Steps

The most delicate step is computing $\overrightarrow{DE}$ and $\overrightarrow{DC}$ in coordinates and then evaluating the angle formula. Re-deriving $\overrightarrow{DE}$: $E - D = \left(\frac{2m+1}{4} - \frac{4m+3}{6}, \frac{\sqrt{3}}{4} m - \frac{\sqrt{3}}{6} m\right) = \left(-\frac{m+1}{12}, \frac{\sqrt{3}}{12} m\right)$, consistent with the prior computation. Checking $\overrightarrow{DC}$: $C - D = \left(\frac{1}{2} - \frac{4m+3}{6}, \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{6} m\right) = \left(-\frac{m+1}{3}, \frac{\sqrt{3}}{2} - \frac{\sqrt{3} m}{6}\right)$, also consistent. Computing $\tan \angle EDC$ step by step confirms that the simplification to $\frac{\sqrt{3}}{3}$ is independent of $m$, verifying the right-triangle structure.

Alternative Approaches

A purely synthetic approach avoids coordinates by observing that the line $MP$ parallel to $AC$ implies $\triangle BMP$ is similar to a $30^\circ$-$60^\circ$-$90^\circ$ triangle, and that $D$ as the