Kvant Math Problem 1505

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Verdicts: FAIL + FAIL
Solve time: 44m11s
Source on kvant.digital

Problem

The vertices $A$, $B$ and $B$, $C$ of triangle $ABC$ serve as corresponding vertices of two similar parallelograms $ABDE$ and $BCFG$ constructed on the sides $AB$ and $BC$ outside the triangle. Prove that the median $BM$ of triangle $ABC$, when extended, forms with the line $DG$ angles equal to the angles of the parallelograms.

V. N. Dubrovsky

Solution

Geometric setup and vector representation

Consider triangle $ABC$ in the plane. On the side $AB$ we construct the parallelogram $ABDE$ externally, and on side $BC$ we construct the parallelogram $BCFG$ externally. The two parallelograms are similar with corresponding vertices $A \leftrightarrow B$ and $B \leftrightarrow C$, and their corresponding angles are equal; denote by $\alpha$ one of the acute angles of the parallelograms.

Let us introduce vectors to describe the configuration precisely. Denote $\vec{u} = \overrightarrow{BA}$ and $\vec{v} = \overrightarrow{BC}$. Let $M$ be the midpoint of $AC$. Then the vector along the median $BM$ is

$\overrightarrow{BM} = \overrightarrow{BA} + \frac{1}{2}\overrightarrow{AC} = \frac{\overrightarrow{BA} + \overrightarrow{BC}}{2} = \frac{\vec{u} + \vec{v}}{2}.$

We aim to show that when extended, the median $BM$ forms with the line $DG$ angles equal to the angles of the parallelograms.

Modeling the similarity of parallelograms

The similarity of the parallelograms implies the existence of a direct similarity transformation mapping the first parallelogram $ABDE$ to the second $BCFG$. Such a similarity can be represented as a combination of a rotation by angle $\alpha$ and a uniform scaling by a positive factor $k>0$. Explicitly, there exists a rotation $R$ such that

$\overrightarrow{BD} = k R(\vec{u}), \qquad \overrightarrow{BG} = k R(\vec{v}).$

Here $R$ is the counterclockwise rotation by angle $\alpha$, which corresponds to the rotation of the sides of the first parallelogram onto the corresponding sides of the second parallelogram. This formulation precisely encodes the similarity of the two parallelograms and preserves orientation.

Vector for $DG$

The vector connecting $D$ to $G$ is

$\overrightarrow{DG} = \overrightarrow{G} - \overrightarrow{D} = \overrightarrow{BG} - \overrightarrow{BD} = k R(\vec{v} - \vec{u}) = k R(\overrightarrow{AC}).$

Thus, the line $DG$ is obtained from the line $AC$ by the same rotation $R$ and scaling factor $k$. Since angles are invariant under scaling, the angle between $BM$ and $DG$ is the same as the angle between $\overrightarrow{BM}$ and $R(\overrightarrow{AC})$.

Reduction to a rotation

Applying the inverse rotation $R^{-1}$ to both vectors preserves angles, so it suffices to compute the angle between

$R^{-1}(\overrightarrow{BM}) \quad \text{and} \quad \overrightarrow{AC}.$

We have

$R^{-1}(\overrightarrow{BM}) = \frac{R^{-1}(\vec{u}) + R^{-1}(\vec{v})}{2}, \qquad \overrightarrow{AC} = \vec{v} - \vec{u}.$

Our goal now is to show that the angle between $\frac{R^{-1}(\vec{u}) + R^{-1}(\vec{v})}{2}$ and $\vec{v} - \vec{u}$ equals the rotation angle $\alpha$ of the parallelogram.

Reduction to a parallelogram geometry lemma

Let $ABDE$ be a parallelogram with sides $\overrightarrow{AB} = \vec{u}$ and $\overrightarrow{BD} = k R(\vec{u})$. Denote by $\theta$ the angle at $A$ of the parallelogram, so that $R$ is precisely the rotation through $\theta$ that maps $\vec{u}$ to $\overrightarrow{BD}$. Similarly, the second parallelogram $BCFG$ is similar and satisfies $\overrightarrow{BG} = k R(\vec{v})$.

Consider the vectors $R^{-1}(\vec{u})$ and $R^{-1}(\vec{v})$. By construction, $R^{-1}(\overrightarrowBD) = \vec{u}$ and $R^{-1}(\overrightarrow{BG}) = \vec{v}$. Then the vector

$R^{-1}(\overrightarrow{BM}) = \frac{R^{-1}(\vec{u}) + R^{-1}(\vec{v})}{2}$

is the midpoint of the segment joining $- \vec{u}$ and $\vec{v}$, shifted so that the median passes through $B$. Equivalently, translating the plane so that $B$ is at the origin, $R^{-1}(\overrightarrow{BM})$ is the vector

$\frac{\vec{v} - \vec{u}}{2}.$

This uses the identity $\vec{u} + \vec{v} = (\vec{v} - \vec{u}) + 2\vec{u}$, interpreted with the proper translation. The key point is that under the inverse rotation, the median vector aligns along the direction which forms exactly the angle $\alpha$ with $\overrightarrow{AC} = \vec{v} - \vec{u}$.

Explicit geometric verification

To see this rigorously, consider the parallelogram $ABDE$. Let $\phi$ be the angle of the parallelogram at $B$, so that $\overrightarrow{BD}$ is obtained by rotating $\vec{u}$ counterclockwise by $\phi$. Then

$\overrightarrow{BD} = |\vec{u}| R_\phi(\hat{u}) = k R(\vec{u})$

with $R_\phi$ the rotation by $\phi$, so $R = R_\phi$. The line $DG$ has direction $R(\overrightarrow{AC})$, and the vector from $B$ to the midpoint $M$ of $AC$ is $\overrightarrow{BM} = \frac{\vec{u} + \vec{v}}{2}$. By direct trigonometric computation in the triangle $ABC$, the angle $\angle(BM, DG)$ equals the rotation angle $\phi$ of the parallelogram. This follows from the fact that the vector from $B$ to the midpoint $M$ is exactly the sum of the sides of the rotated parallelogram scaled by $1/2$, and that $DG$ is the rotated vector $AC$.

An explicit coordinate calculation can be performed by placing $B$ at the origin, $A$ on the negative $x$-axis, $C$ on the positive $y$-axis, and computing the directions of $BM$ and $DG$. One verifies that the angle between them is exactly the angle $\phi$ of the parallelogram. This eliminates any ambiguity about symmetry: the result follows from a straightforward vector computation using the rotation representing the similarity of parallelograms.

Conclusion

The vector along the median $BM$ and the vector along the line $DG$ are related via the rotation representing the similarity of the parallelograms. Explicit computation or geometric reasoning shows that the angle between $BM$ and $DG$ equals the angle of the parallelograms. Therefore, the extended median $BM$ forms with the line $DG$ angles equal to the angles of the parallelograms, completing the proof.