a vertex and two parallel lines are given. construct an equilateral triangle whose other vertices lie on two parallel lines this is a question from the app euclidia. i tried to make equilateral triangle but one vertex is not touching the parallel line. please provide a solutionand how the construction works
How to construct an equilateral triangle whose one vertex is given and other two vertices lie on two parallel lines
euclidean-geometrygeometric-constructiongeometryplane-geometrytriangles
Related Solutions
Suppose we have centroid $M = (x_0,\ y_0)$ and vertex $A=(x_1,\ y_1)$.
First let us center the triangle at the origin with shifted vertex $$A' = (x_1',\ y_1') = (x_1 - x_0,\ y_1 - y_0)$$ The other vertices will be reached from this one by a rotation about the origin $120^\circ$ clockwise and counter-clockwise. The counter-clockwise rotation matrix is $$R_{120^\circ} = \begin{pmatrix}\cos120^\circ & -\sin120^\circ \\ \sin120^\circ & \cos120^\circ\end{pmatrix} = \begin{pmatrix}-\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}$$ with the clockwise rotation matrix as $$R_{-120^\circ} = \begin{pmatrix}\cos120^\circ & \sin120^\circ \\ -\sin120^\circ & \cos120^\circ\end{pmatrix} = \begin{pmatrix}-\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}$$ Your vertices are then $$B' = \begin{pmatrix}-\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}\begin{pmatrix}x_1' \\ y_1'\end{pmatrix}=\begin{pmatrix}-\frac{1}{2}x_1' - \frac{\sqrt{3}}{2}y_1' \\ \frac{\sqrt{3}}{2}x_1' - \frac{1}{2}y_1'\end{pmatrix}$$
$$C' = \begin{pmatrix}-\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}\begin{pmatrix}x_1' \\ y_1'\end{pmatrix}=\begin{pmatrix}-\frac{1}{2}x_1' + \frac{\sqrt{3}}{2}y_1' \\ -\frac{\sqrt{3}}{2}x_1' - \frac{1}{2}y_1'\end{pmatrix}$$ Adding $M$ to each coordinate shifts back the triangle to the original spot.
EUCLIDEA IOS PROBLEM 2.8 / ANDROID 5.7 | 5E SOLUTION REQUIRED
Given Two parallel lines ${a}$ (for 'above') and ${b}$ ('below'), separated by a distance $2D$.
Goal Construct line ${m}$ (for 'middle') parallel to and equidistant from both given lines, i.e., separated by a distance $D$ from both $a$ and $b$.
Constraints Only an unmarked straightedge and a non-rusty collapsible compass (i.e., a compass that, while on paper, can have any radius but which cannot maintain said radius when not on paper) can be used. Use only from the construction steps below a number of steps that together require just five elementary steps; the fifth step must construct the goal line itself.
- Construct a point: 0 elementary steps (E).
- Mark the intersection of two curves with a point: 0E.
- Construct a new line (*or line segment or ray): 1E.
- Extend a given line segment (*or ray): 1E.
- Construct a circle (non-rusty collapsible compass): 1E.
- Construct the perpendicular bisector of a line segment: 3E.
- Construct a new line perpendicular to an old line: 3E.
Hints from Euclidea (including Twitter support) A known 5E solution employs five elementary steps in the order: circle, circle, line, line, line. The first circle is twice the radius of the second one. Homothety is used in the known solution.
SOLUTION Please refer to the figure below. Apologies for not having labels on this figure.
Construct two arbitrary points $O$ and $P$ on $b$ separated by a distance $OP > D$ (just eyeball it) [OE running total].
Construct circle $P(O)$ centered on point $P$ and with radius $OP$ [1E running total].
- $P(O)$ intersects $b$ at a new point $Q$.
Construct $Q(O)$ [2E running total].
- $Q(O)$ also intersects $a$ at two points $L$ (for 'left') and $R$ ('right').
Construct $\overleftrightarrow{OL}$ [3E running total].
- $P(O)$ intersects $\overleftrightarrow{OL}$ at $O$ and another point $M_L$.
Construct $\overleftrightarrow{OR}$ [4E running total].
- $P(O)$ intersects $\overleftrightarrow{OR}$ at $O$ and another point $M_R$.
Construct the desired line $m = \overleftrightarrow{M_{L}M_{R}}$, shown in yellow in the figure below [5E final total].
PROOF OF THE SOLUTION
Given that $OP$ is chosen such that $OQ$ is larger than the gap $2D$ between the two give lines, there exist exactly two points $L$ and $R$ of intersection between $Q(O)$ and $a$.
Homothety is then used in this construction. First, we define a homothetic point ${O}$ on one of the lines, with $b$ used here without loss of generality. $Q(O)$ is constructed to have twice the radius of $P(O)$. Hence, taking $O$ as the homothetic point, also known as the center of dilation, we can view $P(O)$ as the half-fold 'dilation' of $Q(O)$. This means that every line through $O$ intersects $P(O)$ and $Q(O)$ at corresponding points, with the point on $P(O)$ half as far from $O$ as its corresponding point on $Q(O)$ is.
More specifically, the point $M_L$ on $P(O)$ that corresponds to the point $L$ on $P(O)$ can be found by intersecting $\overleftrightarrow{OL}$ and $P(O)$. Point $M_R$ can also be found analogously using $R$.
Homothety preserves parallelism. Therefore, line $m := \overleftrightarrow{M_LM_R}$ is parallel to both $a$ and $b$, as desired.
Given that the homothety involved here involves a half-fold dilation of $Q(O)$ to $P(O)$, we know that the gap between $a$ and $m$ is also half of the gap between $a$ and $b$. This is equivalent to saying that $m$ is equidistant from both $a$ and $b$, as desired. $\blacksquare$
Best Answer
Let $A$ be a given point and $a$ and $b$ be given lines.
Also, let $R_{A}^{\alpha}$ be a rotation around $A$ by $\alpha$.
Now, let $$R_{A}^{60^{\circ}}(a)\cap b=\{C\}$$ and let $$R_{A}^{-60^{\circ}}(\{C\})=\{B\}.$$ Thus, $\Delta ABC$ is an equilateral triangle.
Can you end it now?
There are number of cases.