Kvant Math Problem 561

Verified: yes
Verdicts: PASS + PASS
Solve time: 11m32s
Source on kvant.digital

Problem

Two triangles $A_1B_1C_1$ and $A_2B_2C_2$, whose areas are $S_1$ and $S_2$, are positioned so that the rays $A_1B_1$ and $A_2B_2$, $B_1C_1$ and $B_2C_2$, $C_1A_1$ and $C_2A_2$ are parallel but directed oppositely. Find the area of the triangle whose vertices are the midpoints of the segments $A_1A_2$, $B_1B_2$, $C_1C_2$.

L. P. Kuptsov

Exploration

The condition says that corresponding sides are parallel, but the directions are reversed. In vector form this means

$$\overrightarrow{A_1B_1}=-\lambda,\overrightarrow{A_2B_2},\qquad \overrightarrow{B_1C_1}=-\lambda,\overrightarrow{B_2C_2},\qquad \overrightarrow{C_1A_1}=-\lambda,\overrightarrow{C_2A_2}$$

for some positive number $\lambda$. Since the three side vectors of a triangle satisfy

$$\overrightarrow{A_1B_1}+\overrightarrow{B_1C_1}+\overrightarrow{C_1A_1}=0,$$

the same factor $\lambda$ must occur for all three pairs of corresponding sides. Thus the triangles are homothetic with opposite orientation.

Let

$$M_A=\frac{A_1+A_2}{2},\qquad M_B=\frac{B_1+B_2}{2},\qquad M_C=\frac{C_1+C_2}{2}$$

be the midpoints. The desired triangle is $M_AM_BM_C$.

A natural computation is

$$\overrightarrow{M_AM_B} =\frac{\overrightarrow{A_1B_1}+\overrightarrow{A_2B_2}}2.$$

Since $\overrightarrow{A_2B_2}=-(1/\lambda)\overrightarrow{A_1B_1}$,

$$\overrightarrow{M_AM_B} =\frac{\lambda-1}{2\lambda},\overrightarrow{A_1B_1}.$$

The same factor multiplies every side. Hence $M_AM_BM_C$ is similar to $A_1B_1C_1$, with similarity coefficient

$$\left|\frac{\lambda-1}{2\lambda}\right|.$$

Its area should therefore be

$$S=\frac{(\lambda-1)^2}{4\lambda^2}S_1.$$

Now express $\lambda$ through $S_1,S_2$. Since the triangles are similar,

$$S_1=\lambda^2S_2.$$

Therefore

$$S=\frac{(\sqrt{S_1}-\sqrt{S_2})^2}{4}.$$

A quick check. If $S_1=S_2$, then $\lambda=1$, the triangles are translates with reversed orientation, and the three midpoint segments collapse to one point; the formula gives $0$. If $S_2=0$ formally, the formula gives $S_1/4$, matching the midpoint triangle of a single triangle. The expression is plausible.

The step most likely to hide an error is proving that the same proportionality factor $\lambda$ applies to all three pairs of corresponding sides.

Problem Understanding

We are given two triangles $A_1B_1C_1$ and $A_2B_2C_2$ with areas $S_1$ and $S_2$. Each side of the first triangle is parallel to the corresponding side of the second triangle, but the directions of the corresponding side vectors are opposite.

Let $M_A,M_B,M_C$ be the midpoints of $A_1A_2,B_1B_2,C_1C_2$. We must find the area of triangle $M_AM_BM_C$ in terms of $S_1$ and $S_2$.

This is a Type C problem, since a numerical quantity must be determined.

The core difficulty is extracting from the parallelism condition that the two triangles are similar with opposite orientation, and then computing the scale factor of the midpoint triangle.

The answer should be

$$\boxed{\frac{(\sqrt{S_1}-\sqrt{S_2})^2}{4}}.$$

Proof Architecture

First lemma: there exists a positive number $\lambda$ such that

$$\overrightarrow{A_1B_1}=-\lambda\overrightarrow{A_2B_2},\quad \overrightarrow{B_1C_1}=-\lambda\overrightarrow{B_2C_2},\quad \overrightarrow{C_1A_1}=-\lambda\overrightarrow{C_2A_2}.$$

Sketch: write separate positive proportionality constants and use the relation that the sum of the side vectors of a triangle is zero.

Second lemma: the triangles are similar and

$$S_1=\lambda^2S_2.$$

Sketch: corresponding sides have common ratio $\lambda$.

Third lemma: for the midpoint triangle,

$$\overrightarrow{M_AM_B} =\frac{\lambda-1}{2\lambda},\overrightarrow{A_1B_1},$$

and analogous formulas hold for the other sides.

Sketch: express midpoint vectors and substitute the relation from the first lemma.

Fourth lemma: $M_AM_BM_C$ is similar to $A_1B_1C_1$ with coefficient

$$\left|\frac{\lambda-1}{2\lambda}\right|.$$

Sketch: all three side vectors are multiplied by the same factor.

The most delicate lemma is the first one, because different proportionality factors must be shown impossible.

Solution

Let

$$M_A,\quad M_B,\quad M_C$$

be the midpoints of $A_1A_2$, $B_1B_2$, $C_1C_2$, respectively.

Since the rays $A_1B_1$ and $A_2B_2$ are parallel and oppositely directed, there exists a positive number $p$ such that

$$\overrightarrow{A_1B_1}=-p,\overrightarrow{A_2B_2}.$$

Similarly,

$$\overrightarrow{B_1C_1}=-q,\overrightarrow{B_2C_2}, \qquad \overrightarrow{C_1A_1}=-r,\overrightarrow{C_2A_2},$$

for some positive numbers $q,r$.

The side vectors of each triangle satisfy

$$\overrightarrow{A_1B_1}+\overrightarrow{B_1C_1} +\overrightarrow{C_1A_1}=0,$$

and

$$\overrightarrow{A_2B_2}+\overrightarrow{B_2C_2} +\overrightarrow{C_2A_2}=0.$$

Substituting the first three relations into the first identity gives

$$p,\overrightarrow{A_2B_2} +q,\overrightarrow{B_2C_2} +r,\overrightarrow{C_2A_2}=0.$$

Since

$$\overrightarrow{C_2A_2} =-\overrightarrow{A_2B_2}-\overrightarrow{B_2C_2},$$

we obtain

$$(p-r)\overrightarrow{A_2B_2} +(q-r)\overrightarrow{B_2C_2}=0.$$

The vectors $\overrightarrow{A_2B_2}$ and $\overrightarrow{B_2C_2}$ are not parallel because $A_2B_2C_2$ is a triangle. Hence

$$p-r=0,\qquad q-r=0.$$

Therefore

$$p=q=r=\lambda$$

for some $\lambda>0$, and

$$\overrightarrow{A_1B_1}=-\lambda\overrightarrow{A_2B_2},$$

$$\overrightarrow{B_1C_1}=-\lambda\overrightarrow{B_2C_2},$$

$$\overrightarrow{C_1A_1}=-\lambda\overrightarrow{C_2A_2}.$$

Thus the triangles are similar with similarity ratio $\lambda$, so

$$S_1=\lambda^2S_2.$$

Now compute the sides of the midpoint triangle. Using position vectors,

$$\overrightarrow{M_AM_B} =\frac{(B_1+B_2)-(A_1+A_2)}2 =\frac{\overrightarrow{A_1B_1}+\overrightarrow{A_2B_2}}2.$$

Since

$$\overrightarrow{A_2B_2} =-\frac1\lambda,\overrightarrow{A_1B_1},$$

we get

$$\overrightarrow{M_AM_B} =\frac{\lambda-1}{2\lambda},\overrightarrow{A_1B_1}.$$

The same calculation yields

$$\overrightarrow{M_BM_C} =\frac{\lambda-1}{2\lambda},\overrightarrow{B_1C_1},$$

$$\overrightarrow{M_CM_A} =\frac{\lambda-1}{2\lambda},\overrightarrow{C_1A_1}.$$

Hence $M_AM_BM_C$ is similar to $A_1B_1C_1$ with similarity coefficient

$$\left|\frac{\lambda-1}{2\lambda}\right|.$$

If $S$ denotes the area of $M_AM_BM_C$, then

$$S=\frac{(\lambda-1)^2}{4\lambda^2}S_1.$$

Using $S_1=\lambda^2S_2$,

$$S =\frac{(\lambda\sqrt{S_2}-\sqrt{S_2})^2}{4} =\frac{(\sqrt{S_1}-\sqrt{S_2})^2}{4}.$$

Therefore the required area is

$$\boxed{\frac{(\sqrt{S_1}-\sqrt{S_2})^2}{4}}.$$

Verification of Key Steps

The first delicate step is proving that the three proportionality factors coincide. Writing

$$\overrightarrow{A_1B_1}=-p\overrightarrow{A_2B_2},\quad \overrightarrow{B_1C_1}=-q\overrightarrow{B_2C_2},\quad \overrightarrow{C_1A_1}=-r\overrightarrow{C_2A_2},$$

and using the triangle identities produces

$$(p-r)\overrightarrow{A_2B_2} +(q-r)\overrightarrow{B_2C_2}=0.$$

Because the two displayed vectors are linearly independent, the coefficients must vanish. Any argument that merely states that the side ratios are equal without using linear independence leaves a gap.

The second delicate step is the computation of $\overrightarrow{M_AM_B}$. The midpoint formula gives

$$\overrightarrow{M_AM_B} =\frac{\overrightarrow{A_1B_1}+\overrightarrow{A_2B_2}}2.$$

Substituting

$$\overrightarrow{A_2B_2} =-\frac1\lambda\overrightarrow{A_1B_1}$$

produces

$$\frac{\lambda-1}{2\lambda}\overrightarrow{A_1B_1}.$$

A sign error here would change the final factor from $(\lambda-1)^2$ to $(\lambda+1)^2$.

The third delicate step is replacing $\lambda$ by the areas. Since similar triangles have area ratio equal to the square of the similarity ratio,

$$\lambda=\sqrt{\frac{S_1}{S_2}},$$

with $\lambda>0$. Substituting this positive square root yields

$$\frac{(\lambda-1)^2}{4\lambda^2}S_1 =\frac{(\sqrt{S_1}-\sqrt{S_2})^2}{4}.$$

Using the negative square root would be inconsistent with $\lambda>0$.

Alternative Approaches

An affine-coordinate approach is possible. Choose coordinates so that $A_2B_2C_2$ is represented by vectors $u,v$. The parallelism condition implies that $A_1B_1C_1$ is represented by $-\lambda u,-\lambda v$. Writing coordinates of the six vertices and then the three midpoints gives side vectors of the midpoint triangle directly as $\frac{\lambda-1}{2}u$ and $\frac{\lambda-1}{2}v$. The area follows from the determinant formula.

The vector approach used above is preferable because it extracts the common factor $\lambda$ from the geometric condition in a coordinate-free way and converts the entire problem into a short similarity computation.