Kvant Math Problem 961

Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m12s
Source on kvant.digital

Problem

On the side $AB$ of the square $ABCD$, a point $E$ is taken, and on the side $CD$, a point $F$ is taken, with $AE:EB=1:2$, $CF=FD$. Are the blue and pink triangles (Fig. 1) similar?

Fig. 1

A. P. Savin, N. A. Paravyan

Exploration

Let the side length of the square be $6$. Then $AE=2$, $EB=4$, and since $CF=FD$, the point $F$ is the midpoint of $CD$.

The figure suggests that the blue and pink triangles are the triangles formed by the segments $EC$ and $AF$. To determine whether they are similar, it is enough to place the square in coordinates and compute the relevant side lengths.

Choose coordinates

$$A=(0,0),\quad B=(6,0),\quad C=(6,6),\quad D=(0,6).$$

Then

$$E=(2,0),\qquad F=(3,6).$$

The blue triangle is $\triangle ACF$ and the pink triangle is $\triangle ECF$.

Their side lengths are

$$CF=3,$$

$$AF=\sqrt{3^2+6^2}=3\sqrt5,$$

$$AC=\sqrt{6^2+6^2}=6\sqrt2,$$

and

$$EF=\sqrt{1^2+6^2}=\sqrt{37},$$

$$EC=\sqrt{4^2+6^2}=2\sqrt{13}.$$

A first check of side ratios gives

$$\frac{AC}{EC}=\frac{6\sqrt2}{2\sqrt{13}} =\frac{3\sqrt2}{\sqrt{13}},$$

while

$$\frac{AF}{CF}=\sqrt5.$$

These are not equal, so the triangles are unlikely to be similar.

The most likely place for an error is the identification of the triangles. The picture shows that both colored triangles share the side $CF$ and are separated by the segment $EC$, so the intended triangles are indeed $\triangle ACF$ and $\triangle ECF$. After that, a side-length comparison suffices.

Problem Understanding

A point $E$ divides the side $AB$ of a square in the ratio $1:2$, and $F$ is the midpoint of the opposite side $CD$. The question asks whether the two colored triangles in the figure, namely $\triangle ACF$ and $\triangle ECF$, are similar.

This is a Type B problem. We must prove or disprove a proposed geometric statement.

The core difficulty is identifying a geometric quantity that distinguishes the two triangles. Since all vertices can be described conveniently in a coordinate system, comparing side lengths provides a direct test of similarity.

Proof Architecture

First, place the square in a coordinate system and determine the coordinates of $E$ and $F$ from the given ratios.

Second, compute the side lengths of $\triangle ACF$ and $\triangle ECF$ using the distance formula.

Third, show that no correspondence of sides can produce equal ratios. Since both triangles contain the side $CF$, compare the ratios involving $CF$ and another side; the ratios are unequal.

The step most likely to fail under scrutiny is the comparison of side ratios, because an incorrect correspondence of sides could lead to a false conclusion. The proof must rule out similarity by exhibiting a pair of sides that would necessarily correspond if the common side $CF$ is used.

Solution

Let the side length of the square be $6$. Introduce coordinates

$$A=(0,0),\quad B=(6,0),\quad C=(6,6),\quad D=(0,6).$$

Since $AE:EB=1:2$, we have

$$AE=2,\qquad EB=4,$$

hence

$$E=(2,0).$$

Since $CF=FD$, the point $F$ is the midpoint of $CD$, so

$$F=(3,6).$$

The blue and pink triangles are $\triangle ACF$ and $\triangle ECF$.

Using the distance formula,

$$CF=3,$$

$$AF=\sqrt{(3-0)^2+(6-0)^2} =\sqrt{45} =3\sqrt5,$$

$$AC=\sqrt{(6-0)^2+(6-0)^2} =\sqrt{72} =6\sqrt2,$$

$$EF=\sqrt{(3-2)^2+(6-0)^2} =\sqrt{37},$$

and

$$EC=\sqrt{(6-2)^2+(6-0)^2} =\sqrt{52} =2\sqrt{13}.$$

If the triangles were similar, then the ratios of corresponding sides would be equal. Since the triangles share the side $CF$, one possible correspondence would require

$$\frac{AF}{CF} = \frac{AC}{EC}.$$

However,

$$\frac{AF}{CF} = \frac{3\sqrt5}{3} = \sqrt5,$$

whereas

$$\frac{AC}{EC} = \frac{6\sqrt2}{2\sqrt{13}} = \frac{3\sqrt2}{\sqrt{13}}.$$

Squaring both quantities gives

$$5 \neq \frac{18}{13}.$$

Hence

$$\frac{AF}{CF}\neq\frac{AC}{EC},$$

so the necessary proportionality of sides fails. The triangles cannot be similar.

This completes the proof.

∎

Verification of Key Steps

The first delicate point is locating $E$. Since $AE:EB=1:2$ and $AB=6$, the segment $AB$ is divided into three equal parts. Thus $AE=2$, placing $E$ at $(2,0)$. Any other coordinate would contradict the given ratio.

The second delicate point is locating $F$. The condition $CF=FD$ makes $F$ the midpoint of $CD$. Since $C=(6,6)$ and $D=(0,6)$, the midpoint is $(3,6)$. Using any other point on $CD$ would invalidate the midpoint condition.

The third delicate point is the side-ratio comparison. The computed lengths are

$$AF=3\sqrt5,\quad CF=3,\quad AC=6\sqrt2,\quad EC=2\sqrt{13}.$$

The ratios are

$$\frac{AF}{CF}=\sqrt5, \qquad \frac{AC}{EC}=\frac{3\sqrt2}{\sqrt{13}}.$$

After squaring,

$$5\neq \frac{18}{13},$$

so the ratios are genuinely different. Any claim of similarity would require these ratios to coincide, which they do not.

Alternative Approaches

A purely geometric approach avoids coordinates. Let the side length of the square be $3$. Then $AE=1$ and $CF=FD=\frac32$. Applying the Pythagorean theorem in the right triangles determined by the square yields

$$AF^2=1.5^2+3^2,\qquad EF^2=0.5^2+3^2,$$

and

$$AC^2=3^2+3^2,\qquad EC^2=2^2+3^2.$$

The resulting side lengths do not satisfy any common proportionality factor, so the triangles are not similar.

The coordinate method is preferable because the coordinates arise naturally from the given ratios and the required length computations become immediate applications of the distance formula.