Drawing a tangent to a circle at a given point in just 3 ruler-and-compass constructions

euclidean-geometrygeometric-construction

A friend of mine recently introduced me to an interesting app called Euclidea that challenges you to complete geometric constructions like the ones we learnt in high school. The app has successfully made me feel quite silly about myself, for I am unable to solve this elementary problem:

Given a circle and a point on its circumference, construct the tangent to the circle at that point using ruler-and-compass constructions.

The difficulty I am facing is in finding an optimal solution: I want to accomplish this using as few constructions as possible. So, drawing a line with the ruler would count as one construction, and drawing a circle with the pair of compasses would count as one construction.

I am able to do it in four constructions like this:

But, apparently there is a solution using just three constructions. And, after several weeks of trying (and failing), I have decided to ask for help. Can anyone help me see the light?

Best Answer

Pick an arbitrary point $A$ on the circle, close to $P$, the point of tangency. (0 moves)
Draw a circle centred at $A$ passing through $P$ and intersecting the circle again at $B$. (1 move)
Draw a circle centred at $P$ passing through $B$. Let it intersect the circle centred at $A$ at $C$. (2 moves)
Draw $PC$. (3 moves)

Variant 1

@Blue has provided a solution, excavated here from the depths of the discussions above:

Variant 1

Construct a point $Q \in$ the given external line $\ell$ [0L, 0E].
Construct $\bigcirc P$ with radius $PQ$ [1L, 1E].
- Let point $Q' := \bigcirc P \cap \ell \neq Q$.
Construct $\bigcirc Q$ with radius $PQ$ [2L, 2E].
- Let points $S, S' := \bigcirc P \cap \bigcirc Q$; S is the 'upper' intersection WLOG.
- $m\angle PQS = 60° \implies m\overset{\mmlToken{mo}{⏜}}{PS\,} = 60°$.
Construct $\overline{Q'S'}$ [3L, 3E].
- Resulting $\angle QQ'S'$ inscribes $\overset{\mmlToken{mo}{⏜}}{PS\,} \implies m\angle QQ'S' = 30°$
Construct $\ell'$, the perpendicular bisector of $\overline{Q'S'}$ [4L, 6E].
- Let $U := \ell' \cap \ell$.
- Let $V := \ell' \cap \overline{Q'S'}$.
- By construction, $\angle UVQ'$ is a right angle.
- Therefore, $m\angle VUQ' = 60°$.
- Moreover, $\ell'$ perpendicularly bisects a chord $\overline{Q'S'}$ of a circle $\bigcirc P \implies \ell'$ contains a diameter of $\bigcirc P \implies \ell'$ passes through $P$.
- By construction, $\angle VUQ' = \angle PUQ = 60°$, QEF.

Also, Euclidea gives additional points for constructing all possible versions of each geometric object in question. In this case, two possible angles can be produced. Below for completeness, therefore, is the Variant 1-type construction that produces both angles.

Complete Variant 1

Variant 2

Variant 2 has been solved by @Blue above. A simple angle-chasing proof of that construction is also above. For what it's worth, though, here is an analytical proof:

4L4E Figure

Set $PQ = radius(\bigcirc P) = radius(\bigcirc Q) := 1$.
- $m\angle PRS = 60°$
Define a right-handed coordinate system.
- $Q := (0, 0)$.
- $R := (-1, 0)$.
Let $\theta := 2 m\angle PQU$.
- $P = (cos(2\theta), sin(2\theta))$.
- $S = (cos(2\theta + 60°), sin(2\theta + 60°))$.
Calculate $RS^2 = radius^2(\bigcirc R)$. $$ \begin{align} RS^2 & = (x_S - x_R)^2 + (y_S - y_R)^2 \\ & = (cos(2\theta + 60°) - (-1))^2 + (sin(2\theta + 60°) - (0))^2 \\ & = (cos^2(2\theta + 60°) + 2cos(2\theta + 60°) + 1) + sin^2(2\theta + 60°) \\ & = 2 + 2cos(2\theta + 60°) ) \\ RS^2 & = 4cos^2(\theta + 30°) \\ \end{align} $$
Determine the equation of $\bigcirc R$. $$ \begin{align} RS^2 & = (x - x_R)^2 + (y - y_R)^2 \\ 4cos^2(\theta + 30°) & = (x + 1)^2 + y^2 \\ \end{align} $$
Determine the equation of $\bigcirc P$. $$ \begin{align} PQ^2 & = (x - x_P)^2 + (y - y_P)^2 \\ 1 & = (x - cos(2\theta))^2 + (y - sin(2\theta))^2 \\ \end{align} $$
Find the coordinates of $T$, the other intersection of $\bigcirc P$ and $\bigcirc R$.
$$ \left\{ \begin{aligned} 4cos^2(\theta + 30°) & = (x_T + 1)^2 + y_T^2 \\ 1 & = (x_T - cos(2\theta))^2 + (y_T - sin(2\theta))^2 \\ \end{aligned} \right. $$ WolframAlpha solves this system as $T := \bigcirc P \cap \bigcirc R = \left(cos(2\theta) - \frac{1}{2}, sin(2\theta) - \frac{\sqrt{3}}{2} \right)$.
Find the slope $s$ of $\overline{PT}$. $$ \begin{align} s & = \frac{y_T - y_P}{x_T - x_P} = \frac{\left(sin(2\theta) - \frac{\sqrt{3}}{2}\right) - sin(2\theta)}{\left(cos(2\theta) - \frac{1}{2}\right) - cos(2\theta)} \\ s & = \sqrt{3} \end{align} $$
The angle of $\overline{PT}$ with respect to the positive x-axis is $atan(s) = 60°$. QED. $\square$

Best Answer

Related Solutions

Euclidean Geometry – Inscribing a Square in a Circle in Seven Compass-and-Straightedge Steps

Euclidean Geometry – Construct 60° Angle Through Point in Four Compass-and-Straightedge Steps

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