Question

a 144° rotation maps a polygon p onto itself. which of the following could be polygon p? select all that apply. an equilateral triangle a square a rectangle that is not a square a regular pentagon a regular hexagon a regular octagon

Explanation:

To determine which polygon is mapped onto itself by a \(144^\circ\) rotation, we use the property of regular polygons: the minimum angle of rotational symmetry of a regular \(n\)-sided polygon is \(\frac{360^\circ}{n}\). A rotation of \(144^\circ\) must be a multiple of this minimum angle for the polygon to map onto itself.

Step 1: Recall the formula for rotational symmetry

For a regular \(n\)-sided polygon, the angle of rotational symmetry is \(\frac{360^\circ}{n}\) (and its multiples). We need to check if \(144^\circ\) is a multiple of \(\frac{360^\circ}{n}\), or equivalently, if \(n\) divides \(\frac{360^\circ}{144^\circ}=\frac{5}{2}\) (wait, actually, we should check if \(144^\circ\) is a multiple of \(\frac{360^\circ}{n}\), so \(\frac{144^\circ}{\frac{360^\circ}{n}}=\frac{144n}{360}=\frac{2n}{5}\) must be an integer. So \(2n\) must be divisible by \(5\), or \(n\) must be a multiple of \(\frac{5}{2}\), but since \(n\) is an integer, \(n\) must be a multiple of \(5\) (because \(2\) and \(5\) are coprime). Wait, let's re-express:

We need \(\frac{360^\circ}{n} \times k = 144^\circ\) for some integer \(k\), so \(k=\frac{144n}{360}=\frac{2n}{5}\). So \(k\) must be an integer, so \(2n\) must be divisible by \(5\), which implies \(n\) must be a multiple of \(5\) (since \(2\) and \(5\) are coprime). Wait, no: \(2n\) divisible by \(5\) means \(n\) divisible by \(5/ gcd(2,5)=5\). So \(n\) must be a multiple of \(5\)? Wait, let's test with each polygon:

- **Equilateral triangle (\(n=3\))**: Rotational symmetry angle is \(120^\circ\). \(144^\circ\) is not a multiple of \(120^\circ\) (since \(144/120=1.2\), not integer). So no.
- **Square (\(n=4\))**: Rotational symmetry angle is \(90^\circ\). \(144/90=1.6\), not integer. No.
- **Rectangle (not square, \(n=4\))**: Rotational symmetry angle is \(180^\circ\). \(144\) is not a multiple of \(180\) (144/180=0.8, not integer). No.
- **Regular pentagon (\(n=5\))**: Rotational symmetry angle is \(72^\circ\) (since \(360/5=72\)). Now, \(144/72=2\), which is an integer. So a rotation of \(144^\circ\) (which is \(2 \times 72^\circ\)) maps the regular pentagon onto itself. Yes.
- **Regular hexagon (\(n=6\))**: Rotational symmetry angle is \(60^\circ\). \(144/60=2.4\), not integer. No.
- **Regular octagon (\(n=8\))**: Rotational symmetry angle is \(45^\circ\). \(144/45=3.2\), not integer. No.

Wait, but wait, maybe my initial approach is wrong. Let's think again: a regular polygon has rotational symmetry for any rotation that is a multiple of \(\frac{360^\circ}{n}\). So we need \(144^\circ\) to be a multiple of \(\frac{360^\circ}{n}\), i.e., \(\frac{144^\circ}{\frac{360^\circ}{n}} = \frac{144n}{360} = \frac{2n}{5}\) must be an integer. So \(\frac{2n}{5}\) must be integer, so \(n\) must be a multiple of \(5/ gcd(2,5)=5\). So \(n\) must be a multiple of \(5\). So only the regular pentagon (\(n=5\)) satisfies this, because \(n=5\): \(\frac{2*5}{5}=2\), which is integer. Let's check:

- Regular pentagon: rotational symmetry angle is \(72^\circ\) (360/5=72). \(72^\circ \times 2 = 144^\circ\), so a \(144^\circ\) rotation is two times the basic rotational symmetry angle, so it maps the pentagon onto itself.

Now let's check the other options:

- Equilateral triangle: rotational symmetry angles are \(120^\circ, 240^\circ\). \(144^\circ\) is not a multiple of \(120^\circ\), so no.
- Square: rotational symmetry angles are \(90^\circ, 180^\circ, 270^\circ\). \(144^\circ\) is not a multiple of \(90^\circ\), so no.
- Rectangle (not square): rotational symmetry angle is \(180^\circ\). \(144^\circ\) is not a multiple of \(180^\circ\), so no.
- Regular hexagon: rotational symmetry angles are \(60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ\). \(144^\circ\) is not a multiple of \(60^\circ\), so no.
- Regular octagon: rotational symmetry angles are \(45^\circ, 90^\circ, 135^\circ, 180^\circ, 225^\circ, 270^\circ, 315^\circ\). \(144^\circ\) is not a multiple of \(45^\circ\) (144/45=3.2), so no.

So only the regular pentagon works.

Answer:

D. a regular pentagon