Floor‑function bounding with $\lfloor\sqrt{n}\rfloor$
Solution.
We need to count the number of sequences of three transformations that return triangle $T$ to its original position.
Let’s analyze each transformation:
- Rotation by $90°$ counterclockwise: $(x,y) \to (-y,x)$
- Rotation by $180°$ counterclockwise: $(x,y) \to (-x,-y)$
- Rotation by $270°$ counterclockwise: $(x,y) \to (y,-x)$
- Reflection across $x$-axis: $(x,y) \to (x,-y)$
- Reflection across $y$-axis: $(x,y) \to (-x,y)$
The original triangle $T$ has vertices $(0,0)$, $(4,0)$, and $(4,3)$.
Step 1: Find transformations that return $T$ to itself
Let’s check which single transformations return $T$ to itself:
Rotation by $180°$: $(0,0) \to (0,0)$, $(4,0) \to (-4,0)$, $(4,3) \to (-4,-3)$
- This does NOT return $T$ to itself.
Reflection across $x$-axis: $(0,0) \to (0,0)$, $(4,0) \to (4,0)$, $(4,3) \to (4,-3)$
- This does NOT return $T$ to itself.
Reflection across $y$-axis: $(0,0) \to (0,0)$, $(4,0) \to (-4,0)$, $(4,3) \to (-4,3)$
- This does NOT return $T$ to itself.
Rotation by $90°$: $(0,0) \to (0,0)$, $(4,0) \to (0,4)$, $(4,3) \to (-3,4)$
- This does NOT return $T$ to itself.
Rotation by $270°$: $(0,0) \to (0,0)$, $(4,0) \to (0,-4)$, $(4,3) \to (3,-4)$
- This does NOT return $T$ to itself.
Step 2: Find sequences of two transformations
Let’s check sequences of two transformations:
Identity transformation: The composition of any transformation with its inverse gives the identity.
- Rotation by $90°$ followed by rotation by $270°$ = Identity
- Rotation by $180°$ followed by rotation by $180°$ = Identity
- Reflection across $x$-axis followed by reflection across $x$-axis = Identity
- Reflection across $y$-axis followed by reflection across $y$-axis = Identity
Step 3: Find sequences of three transformations
For a sequence of three transformations to return $T$ to itself, the composition of all three must be the identity transformation.
Let’s denote the transformations as:
- $R_{90}$: Rotation by $90°$
- $R_{180}$: Rotation by $180°$
- $R_{270}$: Rotation by $270°$
- $R_x$: Reflection across $x$-axis
- $R_y$: Reflection across $y$-axis
We need to find all ordered triples $(A, B, C)$ such that $A \circ B \circ C = \text{Identity}$.
This means $A \circ B = C^{-1}$.
Step 4: Count the valid sequences
Since we have 5 transformations and each has an inverse, we need to count the number of ways to choose $A$ and $B$ such that $A \circ B$ is one of the 5 transformations.
For each choice of $A$ (5 choices), there is exactly one choice of $B$ such that $A \circ B$ equals any given transformation.
However, we need to be more systematic. Let’s use the fact that:
- Rotations: $R_{90} \circ R_{270} = R_{180} \circ R_{180} = \text{Identity}$
- Reflections: $R_x \circ R_x = R_y \circ R_y = \text{Identity}$
For three transformations to compose to identity, we need:
- Either all three are the same transformation (and it’s an involution)
- Or the first two compose to the inverse of the third
Let’s count systematically:
Case 1: All three transformations are the same
- $R_{180} \circ R_{180} \circ R_{180} = R_{180}$ (not identity)
- $R_x \circ R_x \circ R_x = R_x$ (not identity)
- $R_y \circ R_y \circ R_y = R_y$ (not identity)
Case 2: First two compose to the inverse of the third
For $R_{90}$:
- $R_{90} \circ R_{270} \circ R_{180} = \text{Identity} \circ R_{180} = R_{180}$ (not identity)
- $R_{90} \circ R_{180} \circ R_{270} = R_{270} \circ R_{270} = R_{180}$ (not identity)
Wait, let me recalculate this more carefully.
Actually, let’s use a different approach. We know that:
- $R_{90} \circ R_{270} = \text{Identity}$
- $R_{180} \circ R_{180} = \text{Identity}$
- $R_x \circ R_x = \text{Identity}$
- $R_y \circ R_y = \text{Identity}$
For three transformations to compose to identity, we need the first two to compose to the inverse of the third.
The inverses are:
- $R_{90}^{-1} = R_{270}$
- $R_{180}^{-1} = R_{180}$
- $R_{270}^{-1} = R_{90}$
- $R_x^{-1} = R_x$
- $R_y^{-1} = R_y$
So we need:
- $A \circ B = R_{270}$ and $C = R_{90}$
- $A \circ B = R_{180}$ and $C = R_{180}$
- $A \circ B = R_{90}$ and $C = R_{270}$
- $A \circ B = R_x$ and $C = R_x$
- $A \circ B = R_y$ and $C = R_y$
For each case, we need to count how many pairs $(A,B)$ satisfy $A \circ B = \text{target}$.
Since we have 5 transformations, for each target transformation, there are exactly 5 pairs $(A,B)$ such that $A \circ B = \text{target}$ (one for each choice of $A$).
Therefore, the total number of valid sequences is $5 \times 5 = 25$.
Answer: $\boxed{25}$