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How to construct for rotating a point around a given point with an angle $\alpha$

Writer Sophia Terry
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Explain the construction steps of rotating a point around a given point for a given angle $\alpha$.

I know that we have to translate the point of rotation to origin and then

$$x' = x\cos{\alpha} - y \sin{\alpha}$$$$y' = y\cos{\alpha} + x \sin{\alpha}$$Then we have to add the given point of rotation to $x'$ and $y'$. But I have a confusion that it may not be the construction step of rotating. Can you please tell me that if construction is based on drawing figures using rulers and protractors? And if that is true, what will be the construction steps? Thanks in advance.

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2 Answers

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You cannot do it for all angles with only an unmarked ruler and a compass.

Lemma: An angle of $10^\circ$ is not constructible

Proof: Suppose otherwise. Let $\angle XYZ=10^{\circ}$, where $Y$ is the origin and the distance from $X$ to the origin is $1$. Additionally, suppose $Z$ is on the x-axis. Then the distance from $X$ to the x-axis must be constructible, but this distance is precisely $\sin(10^\circ)$. We know that $\sin(10^\circ)$ isn't constructible, since it has minimal polynomial $8x^3-6x+1$, contradiction.

Now, suppose that we can rotate a point $A$ with angle $10^{\circ}$ about $P$, mapping $A$ to $B$. Then $\angle APB=10^{\circ}$ was constructed, impossible by our lemma.


Now, to do it for some whole number of degrees $\alpha$ given that it's possible to do it for that $\alpha$.

  • First, draw the circle centered at $P$ passing through $A$ with the compass
  • Construct the angle $\alpha$.
  • Copy the angle $\alpha$ over such that $\angle APC=\alpha$
  • Let $B$ be the intersection of the circle and $PC$, then $B$ is a degree $\alpha$ rotation of $A$ about $P$.

The main idea here is that this problem is essentially the problem of constructing an angle. However, not all angles are constructible (as I showed for $10^\circ$). Using the same argument for $10^\circ$, we get the following:

An angle $\alpha$ is constructible if and only if $\sin\alpha$ is a constructible number

Now, whether a number is constructible or not is a highly researched problem (and probably out of the scope of what you are asking). You can do some googling, it boils down to the degree of the minimal polynomial of that number. However, see Chebyshev Polynomials! This helps us find a polynomial that $\sin\alpha$ is a root of.

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You mentioned protractor. If a protractor is allowed, life is a lot easier because we can easily construct $\alpha$. Hence using the method in the second section of this answer, we can solve the problem. However, the problem of "constructing" something generally refers to only using a compass and an unmarked ruler, since these were the only 100% accurate tools in the times of the Greek.

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Assume that you need to move a point (7,4) from position B to another point F. There is $ \alpha = 90^{\circ}$ anti-clockwise rotation on a circle 5 units radius centered at A(2,3) and next 9 units displacement to the right.

Pierce the compass needle at point A (2,3). Set needle to pencil radial distance 5 units when using the compass. Draw quarter circle using a protractor. You got to C. Now displace from C to F towards right direction along the arrow as shown.

All the above is a practical way to achieve point F (10,8) on graph paper.

All these steps are can be included in a math notation using trigonometry and vectors.. symbolically.

$$x' = x\cos{\alpha} - y \sin{\alpha}+9$$$$y' = y\cos{\alpha} + x \sin{\alpha}+0$$

If you choose symbolic option your answer will be obtained symbolically or analytically and fast.. by using a computer. It represents what you did on the graph paper without using a school compass box used only at beginning stage.

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