Kvant Math Problem 613

The data of the problem are naturally encoded by a similarity.

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Problem

On the sides of triangle $ABC$, similar triangles $ADB$, $BEC$, and $CFA$ are constructed externally $\Big(\dfrac{|AD|}{|DB|}=\dfrac{|BE|}{|EC|}=\dfrac{|CF|}{|FA|}=k$; $\widehat{ADB}=\widehat{BEC}=\widehat{CFA}=\alpha\Big)$. Prove that:

  1. the midpoints of the segments $AC$, $DC$, $BC$, and $EF$ are the vertices of a parallelogram;
  2. this parallelogram has two angles equal to $\alpha$, and the ratio of the lengths of its sides is $k$.

L. P. Kuptsov

Exploration

The data of the problem are naturally encoded by a similarity. Since the triangles $ADB$, $BEC$, and $CFA$ are similar and similarly oriented outside the sides of $ABC$, there exists a fixed complex number or plane transformation carrying each side to the corresponding external vertex.

Let

$$z=\frac{D-B}{A-B}.$$

Because $|AD|:|DB|=k$ and $\angle ADB=\alpha$, the point $D$ is determined from $A,B$ by the same similarity for every side. Writing everything in vector form, there is a constant complex number $\lambda$ such that

$$D=B+\lambda(A-B),\qquad E=C+\lambda(B-C),\qquad F=A+\lambda(C-A).$$

The condition $|AD|/|DB|=k$ and $\angle ADB=\alpha$ determines $\lambda$. Since $A-D=(1-\lambda)(A-B)$ and $B-D=-\lambda(A-B)$,

$$\frac{|AD|}{|DB|} =\frac{|1-\lambda|}{|\lambda|} =k.$$

Moreover the angle between $DA$ and $DB$ equals $\alpha$.

The first statement concerns the midpoints of $AC$, $DC$, $BC$, $EF$. Denote them by $P,Q,R,S$. Using the expressions above,

$$Q=\frac{D+C}{2},\qquad R=\frac{B+C}{2}.$$

Hence

$$Q-R=\frac{D-B}{2} =\frac{\lambda}{2}(A-B).$$

Also

$$S-P=\frac{E+F-A-C}{2}.$$

Substituting the formulas for $E,F$ gives

$$E+F-A-C =\lambda(B-C)+\lambda(C-A) =\lambda(B-A),$$

therefore

$$S-P=\frac{\lambda}{2}(B-A)=-(Q-R).$$

One pair of opposite sides is equal and parallel. The other pair should come out similarly:

$$Q-P=\frac{D-A}{2} =\frac{\lambda-1}{2}(A-B),$$

while

$$S-R=\frac{E+F-B-C}{2} =\frac{\lambda-1}{2}(B-A)=-(Q-P).$$

Hence $PQRS$ is a parallelogram.

For the second statement, the side vectors are

$$Q-P=\frac{\lambda-1}{2}(A-B),\qquad Q-R=\frac{\lambda}{2}(A-B).$$

Thus the ratio of adjacent sides is

$$\frac{|Q-P|}{|Q-R|} =\frac{|1-\lambda|}{|\lambda|} =k.$$

The angle of the parallelogram is the angle between $\lambda(A-B)$ and $(\lambda-1)(A-B)$, hence equals the angle between $\lambda$ and $\lambda-1$. But

$$DA=(1-\lambda)(A-B),\qquad DB=-\lambda(A-B),$$

so the angle between $\lambda$ and $\lambda-1$ is exactly the angle between $DB$ and $DA$, namely $\alpha$.

The crucial point is to justify rigorously that the same complex coefficient $\lambda$ works simultaneously for all three constructed triangles and that the angle in the parallelogram indeed coincides with $\angle ADB$.

Problem Understanding

We are given a triangle $ABC$. On its sides, externally, three similar triangles $ADB$, $BEC$, and $CFA$ are constructed. They satisfy

$$\frac{|AD|}{|DB|} = \frac{|BE|}{|EC|} = \frac{|CF|}{|FA|} = k, \qquad \angle ADB=\angle BEC=\angle CFA=\alpha .$$

Let the midpoints of $AC$, $DC$, $BC$, and $EF$ be taken. The first task is to prove that these four points form a parallelogram. The second task is to prove that this parallelogram has two angles equal to $\alpha$ and that the ratio of the lengths of two adjacent sides is $k$.

This is a Type B problem. The core difficulty is to express the positions of $D,E,F$ by one and the same similarity coefficient and then translate the midpoint conditions into vector identities.

Proof Architecture

Let $P,Q,R,S$ be the midpoints of $AC,DC,BC,EF$ respectively.

Lemma 1. There exists a fixed complex number $\lambda$ such that

$$D=B+\lambda(A-B),\quad E=C+\lambda(B-C),\quad F=A+\lambda(C-A).$$

Sketch. All three external triangles are obtained from their bases by the same direct similarity.

Lemma 2. The vectors of opposite sides of $PQRS$ satisfy

$$Q-R=-(S-P),\qquad Q-P=-(S-R).$$

Sketch. Substitute the formulas from Lemma 1 into the midpoint coordinates and simplify.

Lemma 3. The ratio of the lengths of two adjacent sides of $PQRS$ equals

$$\frac{|1-\lambda|}{|\lambda|}.$$

Sketch. Compute the side vectors explicitly.

Lemma 4. The angle of $PQRS$ equals the angle between the vectors $\lambda$ and $\lambda-1$, which is $\alpha$.

Sketch. Compare these vectors with the vectors $DB$ and $DA$.

The most delicate point is Lemma 4, because one must identify exactly which angle between complex multiples corresponds to the given angle at $D$.

Solution

Let the plane be identified with the complex plane. Since the triangles $ADB$, $BEC$, and $CFA$ are similar and similarly situated with respect to the sides of $ABC$, there exists a fixed complex number $\lambda$ such that

$$D=B+\lambda(A-B),$$

and analogously

$$E=C+\lambda(B-C),\qquad F=A+\lambda(C-A).$$

Indeed, the map sending the directed segment $AB$ to the directed segment $BD$ is the same direct similarity as the map sending $BC$ to $CE$ and $CA$ to $AF$.

Let

$$P=\frac{A+C}{2},\qquad Q=\frac{D+C}{2},\qquad R=\frac{B+C}{2},\qquad S=\frac{E+F}{2}.$$

We first prove that $P,Q,R,S$ form a parallelogram.

Using the expression for $D$,

$$Q-R=\frac{D-B}{2} =\frac{\lambda}{2}(A-B).$$

Also,

$$S-P = \frac{E+F-A-C}{2}.$$

Substituting the formulas for $E$ and $F$,

$$E+F-A-C = \lambda(B-C)+\lambda(C-A) = \lambda(B-A).$$

Hence

$$S-P = \frac{\lambda}{2}(B-A) = -\frac{\lambda}{2}(A-B) = -(Q-R).$$

Next,

$$Q-P = \frac{D-A}{2}.$$

Since

$$D-A = B+\lambda(A-B)-A = (\lambda-1)(A-B),$$

we obtain

$$Q-P = \frac{\lambda-1}{2}(A-B).$$

On the other hand,

$$S-R = \frac{E+F-B-C}{2}.$$

Using the formulas for $E$ and $F$,

$$E+F-B-C = (\lambda-1)(B-A),$$

therefore

$$S-R = \frac{\lambda-1}{2}(B-A) = -(Q-P).$$

Thus both pairs of opposite sides are parallel and equal:

$$Q-R=-(S-P),\qquad Q-P=-(S-R).$$

Hence $PQRS$ is a parallelogram.

Now we determine its shape.

From the formulas above,

$$Q-R=\frac{\lambda}{2}(A-B), \qquad Q-P=\frac{\lambda-1}{2}(A-B).$$

Therefore

$$\frac{|Q-P|}{|Q-R|} = \frac{|\lambda-1|}{|\lambda|}.$$

Since

$$A-D=(1-\lambda)(A-B), \qquad B-D=-\lambda(A-B),$$

the condition

$$\frac{|AD|}{|DB|}=k$$

gives

$$\frac{|1-\lambda|}{|\lambda|}=k.$$

Consequently

$$\frac{|Q-P|}{|Q-R|}=k.$$

It remains to compute the angle of the parallelogram. The angle between the adjacent sides $QR$ and $QP$ equals the angle between the vectors

$$\lambda(A-B) \quad\text{and}\quad (\lambda-1)(A-B).$$

Multiplication by the nonzero vector $(A-B)$ rotates both vectors through the same angle and multiplies their lengths by the same factor, so this angle equals the angle between $\lambda$ and $\lambda-1$.

But

$$DB=-\lambda(A-B), \qquad DA=(1-\lambda)(A-B)=-(\lambda-1)(A-B).$$

Changing both vectors by a factor $-1$ does not change the angle between them. Hence the angle between $\lambda$ and $\lambda-1$ is precisely

$$\angle ADB=\alpha.$$

Thus one angle of the parallelogram equals $\alpha$, and the opposite angle also equals $\alpha$. The ratio of the lengths of adjacent sides is $k$.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is the use of a single coefficient $\lambda$. For the side $AB$, the vector from $B$ to $D$ is obtained from $A-B$ by a direct similarity determined by the shape of triangle $ADB$. Since the triangles $ADB$, $BEC$, and $CFA$ are similar with the same orientation, the same similarity sends $BC$ to $CE$ and $CA$ to $AF$. Hence the same multiplier $\lambda$ appears in all three formulas. If opposite orientations were used, different conjugate multipliers would appear and the argument would fail.

The second delicate step is the computation of $S-P$. Starting from

$$S-P=\frac{E+F-A-C}{2},$$

one must substitute

$$E=C+\lambda(B-C),\qquad F=A+\lambda(C-A),$$

giving

$$E+F-A-C =\lambda(B-C)+\lambda(C-A) =\lambda(B-A).$$

Any sign error here destroys the parallelogram identity.

The third delicate step is the angle comparison. The side directions of the parallelogram are represented by $\lambda(A-B)$ and $(\lambda-1)(A-B)$, while the sides of triangle $ADB$ issuing from $D$ are represented by $-\lambda(A-B)$ and $-(\lambda-1)(A-B)$. Both vectors are multiplied by the same factor $-1$, so the angle is unchanged. This is exactly why the parallelogram angle equals $\angle ADB$.

Alternative Approaches

A purely vector proof can be written without complex numbers. Let $T$ be the linear operator representing the rotation and scaling that sends the vector along a side of each base triangle to the vector leading to the external vertex. Then

$$\overrightarrow{BD}=T(\overrightarrow{BA}),\quad \overrightarrow{CE}=T(\overrightarrow{CB}),\quad \overrightarrow{AF}=T(\overrightarrow{AC}).$$

The midpoint coordinates can then be expressed in vector form, and the same cancellations yield the parallelogram relations.

Another approach uses affine geometry. The points $D,E,F$ are obtained from the sides of $ABC$ by the same affine similarity. Midpoints behave linearly under affine combinations, so the midpoint quadrilateral can be expressed as the image of a pair of translated copies of the vectors defining the similarity. This also leads directly to the side ratio $k$ and angle $\alpha$. The complex-number method is preferable because the ratio of lengths and the angle are encoded simultaneously in the single multiplier $\lambda$, making the computations shorter and more transparent.