Translation, Rotation and Reflection

In one sentence

Translation, rotation and reflection have one thing in common: the figure moves house but does not change shape. Every segment stays the same length, and every angle stays the same size.

So when the conditions of a problem are scattered all over the figure (two segments you need to add are far apart, three segments you need to compare do not make a triangle), move one piece across unchanged so that they meet.

Reflection: carry points from one side of a line to the other. A broken path can then be straightened.
Rotation: turn about the shared vertex of “two equal segments sharing a vertex”. One side lands on the other.
Translation: slide along a set of parallel lines. Two segments that were not joined now meet end to end.

Predict first

In the figure above, $A$ is $2$ from line $l$ and $B$ is $3$ from $l$ . Where is the point $P$ that makes $A P + P B$ shortest?

Directly below

A

Exactly halfway between the two feet of the perpendicularsBetween the two feet, closer to the foot below

A

Choose an answer first; then you can see the result.

Stop 1 · Reflection: straighten the broken path (Vol. 4 · inscribed triangles)

$A (- 3, 2)$ and $B (3, 3)$ are on the same side of line $l$ , and $P$ is on $l$ . What is the smallest possible value of $A P + P B$ ?

Example: two points on the same side of a line, shortest broken path

Step 1 / 4

A broken path is hard to compare
$A P$ and $P B$ are the two pieces of a broken path. As $P$ moves, one gets longer and the other shorter, and it is hard to see when the sum is smallest.

$A P + P B = ?$
Reflect A across
Swap in a new broken path
Straighten it

Light paths in triangles, bouncing billiard balls, the inscribed triangle with the smallest perimeter: they all use this one move. Each reflection removes one bend from the path.

Stop 2 · Rotate by 60°: equilateral triangles (Vol. 4 · equilateral triangles)

$P$ is a point inside equilateral $△ A B C$ with $P A = 5$ , $P B = 12$ , $P C = 13$ . Find $∠ A P B$ .

$5^{2} + 1 2^{2} = 1 3^{2}$ , so there must be a right angle hiding somewhere. But $P A$ , $P B$ , $P C$ are like three scattered chopsticks: they do not form a triangle.

Example: PA = 5, PB = 12, PC = 13

Step 1 / 5

Find equal sides
In an equilateral triangle $B A = B C$ , with a $6 0^{\circ}$ angle between them. So rotating $6 0^{\circ}$ about $B$ carries $B C$ exactly onto $B A$ .

$P A = 5, P B = 12, P C = 13$
Rotate △BPC across
PB has been moved out
PC has moved across too
Add the two angles

Stop 3 · Rotate by 90°: the half angle in a square (Vol. 4 · squares)

Square $A B C D$ has side $6$ . $E$ is on $B C$ , $F$ is on $C D$ , $∠ E A F = 4 5^{\circ}$ , and $B E = 2$ . Find $E F$ .

Predict first

First drag $E$ in the figure below and watch the readings. How is $E F$ related to $B E + D F$ ?

E F

is always longer

E F

is always shorterThey are always equal

Choose an answer first; then you can see the result.

Example: square of side 6, BE = 2, ∠EAF = 45°

Step 1 / 5

Three segments in three places
$B E$ is on the bottom side, $D F$ on the right side, and $E F$ runs across the middle. To prove $E F = B E + D F$ , first join $B E$ and $D F$ into one segment.

$B E + D F = ? E F$
Rotate 90° about A
Check collinearity first
A pair of congruent triangles
Work it out

Stop 4 · Translation: put the diagonals end to end (Vol. 4 · congruent triangles)

In trapezoid $A B C D$ , $A D ∥ B C$ , the diagonals satisfy $A C ⊥ B D$ , $A C = 6$ and $B D = 8$ . Find the sum of the two bases, and the area of the trapezoid.

Example: a trapezoid with perpendicular diagonals

Step 1 / 4

The conditions do not reach
The area formula needs “top base + bottom base” and the height, and neither is given. What is given is two crossing diagonals.

$A C = 6, B D = 8, A C ⊥ B D$
Translate one diagonal
Top base plus bottom base, ready-made
The area moves across too

Drag $D$ in the figure: the diagonals are no longer perpendicular, but “trapezoid area $=$ area of $△ B D C^{'}$ ” still holds. The translation carries areas and lengths across; it does not depend on that right angle.

Practice

Exercise 1Basic

Points $A$ and $B$ are on the same side of line $l$ , at distances $2$ and $4$ from $l$ , and the feet of their perpendiculars are $8$ apart. $P$ is on $l$ . What is the smallest possible value of $A P + P B$ ?

Exercise 2Advanced

Square $A B C D$ has side $15$ . $E$ is on $B C$ , $F$ is on $C D$ , $∠ E A F = 4 5^{\circ}$ , and $B E = 3$ . Find $E F$ .

Exercise 3Challenge

$P$ is a point inside equilateral $△ A B C$ with $P A = 6$ , $P B = 8$ , $P C = 10$ . How many degrees is $∠ A P B$ ? (Enter just the number.)

In one sentence

Stop 1 · Reflection: straighten the broken path (Vol. 4 · inscribed triangles)

Example: two points on the same side of a line, shortest broken path

Stop 2 · Rotate by 60°: equilateral triangles (Vol. 4 · equilateral triangles)

Example: PA = 5, PB = 12, PC = 13

Stop 3 · Rotate by 90°: the half angle in a square (Vol. 4 · squares)

Example: square of side 6, BE = 2, ∠EAF = 45°

Stop 4 · Translation: put the diagonals end to end (Vol. 4 · congruent triangles)

Example: a trapezoid with perpendicular diagonals

Practice

Mistake Museum · Exhibits from this stop

Rotating and then adding straight away

Rotating even though the two sides are not equal

Stops on this line