Problem 1353

I. solution. The two triangles are congruent because the lengths of their sides are equal in pairs: $A B=E F=15, B C=D E=$ $\sqrt{250}, C A=F D=5$ units; in the first triangle, the vertices $A, B, C$ correspond to $F, E, D$ in the second.

The traversal of the vertices of the triangles in the order $A, B, C$ and $F, E, D$ is in opposite directions, i.e., the former is counterclockwise and the latter is clockwise. - Therefore, the two triangles can be transformed into each other by congruence transformations, and among these, there must be an axial reflection, because a rotation and translation in the plane of the drawing do not change the direction of traversal.

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Figure 1

If there is a line $t$ that meets the conditions of the problem, let the reflection of the $F E D$ triangle on this line be $F^{\prime} E^{\prime} D^{\prime}$. Thus, the midpoint $F_{0}$ of the segment $F F^{\prime}$ is on $t$. The same is true for the midpoint $F_{1}$ of the segment $F A$, because $F_{0} F_{1}$ is the midline of the triangle $F F^{\prime} A$ parallel to $F^{\prime} A$, and $F^{\prime} A$, after the reflection and translation, is parallel to $t$; thus, we indeed reach $F_{1}$ from $F_{0}$ along $t$. Therefore, the line of the sought axis can be obtained by connecting $F_{1}$ with the midpoint of another pair of vertices, for example, the midpoint $E_{1}$ of the segment $E B$.

The coordinates of $F_{1}$ are $(11 ; 10.5)$, and those of $E_{1}$ are $(20.6 ; 3.3)$, from which the equation of $t$ is:

$$y-10.5=\frac{-7.2}{9.6}(x-11)=-\frac{3}{4}(x-11) ; \quad 3 x+4 y-75=0$$

Thus, $F_{0}$, as the foot of the perpendicular from $F$ to $t$: $F_{0}(13 ; 9)$, and the coordinates of $F^{\prime}$, denoted by $u$ and $v$,

$$\frac{u+22}{2}=13, \quad u=4, \quad \frac{v+21}{2}=9, \quad v=-3 ; \quad F^{\prime}(4 ;-3)$$

The magnitude of the translation after the reflection is $F^{\prime} A=5$ units, the direction is parallel to $t$, and during the translation, the abscissas of the points decrease. We can also say: the $X$-component of the translation is -4 units, the $Y$-component is +3 units. Indeed, the reflections of $E$ and $D$ on $t$ are $E^{\prime}(19 ;-3)$ and $D^{\prime}(4 ; 2)$, and from these, the above translation takes us to $B$ and $C$.

If we reflect $A B C$ and translate the reflection to $F E D$, then the axis and the magnitude of the translation are the same, but the direction is opposite to the previous one, and during the translation, the abscissas of the points increase.

Kajcsos Zsolt (Szombathely, Nagy Lajos g. IV. o. t.)

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II. solution. Translate one of the two congruent triangles, for example, $F E D-\mathrm{t}$, so that one of its vertices, for example, $F$, coincides with the corresponding vertex of the $A B C$ triangle, $A$. Denote the vertices of the translated triangle by $F^{*}=A, E^{*}, D^{*}$ (Figure 2). Reflect the $A B C$ triangle on the bisector of the angle between the corresponding sides, for example, the bisector $f$ of $\angle B A E^{*}$. The triangle $A B C$ transforms into $F^{*} E^{*} D^{*}$ (and only with this reflection). Now translate $F^{*} E^{*} D^{*}$ in a direction perpendicular to $f$ so that $E^{*}$ lands on the perpendicular projection $E_{1}$ of $E$ on the line $B E^{*}$, and the new positions of $D^{*}$ and $F^{*}$ are $D_{1}$ and $F_{1}$. Then, on the one hand, the triangle $F_{1} E_{1} D_{1}$ is the reflection of the triangle $A B C$ on the perpendicular bisector $t$ of the segment $B E_{1}$, which is parallel to $f$, and on the other hand, translating $F_{1} E_{1} D_{1}$ so that $E_{1}$ lands on $E$, the triangle transforms into $F E D$, since the corresponding sides of the two triangles are parallel and equal. This translation is perpendicular to $E_{1} B$, hence parallel to $t$. Thus, we have found a reflection and a translation that meet the requirements of the problem.

Given the data of the problem, $E^{*}$ and $D^{*}$ can be obtained by reducing the abscissas by 22 and the ordinates by 21: $E^{*}(4.2 ;-14.4), D^{*}(-4.8 ;-1.4)$. The bisector of $\angle B A E^{*}$ is the line connecting $A$ with the midpoint $A_{1}(9.6 ;-7.2)$ of the segment $B E^{*}$, because the triangle $B A E^{*}$ is isosceles; its slope is $-7.2 / 9.6=-3 / 4$. Now, the point $E_{1}$ is the intersection of the line through $B$ with slope $-1 /(-3 / 4)=4 / 3$ and the line through $E$ with slope $-3 / 4$, the lines $4 x-3 y=60$ and $3 x+4 y=105$: $E_{1}(-22.2 ; 9.6)$, so the $X$ and $Y$ components of the translation are -4 and 3. The axis of reflection is the line through the midpoint $(18.6 ; 4.8)$ of the segment $B E_{1}$ with slope $-3 / 4$, its equation is: $3 x+4 y-75=0$.

Balla Katalin (Budapest, Radnóti M. gyak. g. IV. o. t.)

Note. The first part of the solution only used the fact that the two triangles are congruent and have opposite traversal directions. Thus, we have generally shown that one of two such triangles can always be transformed into the other by a reflection and a translation parallel to the reflection axis.