Draw an Equilateral Triangle Inscribed in a Circle

Inscribing an equilateral triangle in a circumvolve

Alignments to Content Standards: Yard-CO.D.thirteen

Job

Suppose we are given a circle of radius $r$. The goal of this task is to construct an equilateral triangle whose three vertices prevarication on the circumvolve. Suppose $\overline{AB}$ is a diameter of the circle. Draw a circle with center $A$ and radius $r$ and label the 2 points of intersection of the circles $P$ and $Q$ equally pictured below:

1. Testify that $thou(\angle ABP) = 30$ and $m(\angle ABQ) = 30$.
2. Show that $\triangle PBQ$ is an equilateral triangle inscribed in the given circle.

IM Commentary

Amalgam an equilateral triangle inscribed in a circle requires creativity and a adept knowledge of geometry and/or trigonometry. The arroyo taken in the solution hither uses the fact that $\sin{xxx} = \frac{i}{ii}$. Hence in order to construct a 30 caste bending it is sufficient to exist able to construct a right triangle whose hypotenuse has twice the length of a leg. Students will need to know that $\bending APB$ is correct since it is inscribed in a circle with i ray along a bore of the circle. The ratio of $(2:1)$ for the hypotenuse to the leg is arranged by using the diameter of a circle for the hypotenuse and the radius of a congruent circle for the leg.

This job implements many of import ideas from geometry including trigonometric ratios, important facts nearly triangles, and reflections if students choose to view $\triangle ABQ$ as the reflection of $\triangle ABP$ almost $\overleftrightarrow{AB}$. As a result, it is recommended that this job exist undertaken relatively late in the geometry curriculum. Information technology could too exist given in open ended course , that is requesting to produce an inscribed equilateral triangle within a given circle without suggesting an approach or providing a moving-picture show. An alternative method begins by noting that if $O$ is the eye of the original circumvolve, then $\triangle AOP$ is equilateral (all three sides are radii of congruent circles). We tin can continue drawing circles, with centers $P$ and $Q$ and each time we go another equilateral triangle. The end event volition be a regular hexagon inscribed inside the circle: joining every other vertex of this hexagon gives an equilateral triangle inscribed in the circumvolve. A variant of this argument, focused only on the hexagon, is presented in http://www.illustrativemathematics.org/illustrations/786.

Work on this task can develop skills related to MP1, Make Sense of Problems and Persevere in Solving Them, and as well MP3, Construct Viable Arguments and Critique the Reasoning of Others. In the first instance, the argument is complex and requires putting together several abstract arguments which need to be related to the construction of the desired equilateral triangle. Considering of this complexity, it is ideally suited for grouping piece of work so that students tin combine their strengths and skills.

Solution

1. We know that $|AP| = r$ because it is a radius of our auxiliary circle. We also know that $|AB| = 2r$ since it a diamteter of our given circle of radius $r$:

   

   Thus $$ \frac{|AP|}{|AB|} = \frac{1}{two}. $$ Since $\triangle ABP$ is inscribed in a circumvolve and side $\overline{AB}$ is a bore of this circle, this ways that $\angle APB$ is a correct angle. Side $\overline{AP}$ is the side opposite angle $ABP$. Therefore $$ \sin{\angle ABP} = \frac{|AP|}{|AB|} = \frac{1}{2}. $$ This means that $\bending ABP$ is a xxx degree bending.

   Nosotros tin can apply the same argument to testify that $\angle ABQ$ is a thirty caste angle. Alternatively, since $\overleftrightarrow{AB}$ contains the centers of both circles, reflection about $\overleftrightarrow{AB}$ maps the two circles to themselves and interchanges $P$ and $Q$. This means that reflection nigh $\overleftrightarrow{AB}$ maps $\angle ABP$ to $\bending ABQ$. Since reflections preserve bending measurements, this means that $m(\bending ABQ) = m(\angle ABP) = 30$.

2. We know that $m(\bending PBQ) = k(\angle ABQ) + g(\angle ABP)$. We have found that the angles $ABQ$ and $ABP$ each measure thirty degrees and so $m(\bending PBQ) = 60$. We also know that $\triangle PBQ$ is isosceles with $\overline{Atomic number 82} = \overline{QB}$. This is truthful because reflection near $\overleftrightarrow{AB}$ maps $\overline{PB}$ to $\overline{QB}$ and vice versa. Base angles in an isosceles triangle are congruent so $$ thou(\bending BPQ) = g(\angle BQP). $$ We also know that $$ chiliad(\angle PBQ) + thou(\angle BPQ) + m(\angle BQP) = 180. $$ Using the fact that $grand(\angle PBQ) = threescore$ we find that $$ m(\bending BPQ) + chiliad(\bending BQP) = 120. $$ Since these two angles are congruent nosotros find that $thou(\angle BPQ) = m(\bending BQP) = 60$. Since all three angles in $\triangle PBQ$ are lx degree angles it is an equilateral triangle. The vertices $P, B, Q$ all lie on the circumvolve so $\triangle PBQ$ is an equilateral triangle inscribed in the given circle. It is pictured beneath with all of the auxiliary constructions removed:

   

Inscribing an equilateral triangle in a circle

Suppose we are given a circumvolve of radius $r$. The goal of this task is to construct an equilateral triangle whose three vertices lie on the circumvolve. Suppose $\overline{AB}$ is a diameter of the circumvolve. Draw a circle with middle $A$ and radius $r$ and label the 2 points of intersection of the circles $P$ and $Q$ as pictured below:

1. Show that $m(\angle ABP) = xxx$ and $m(\bending ABQ) = thirty$.
2. Show that $\triangle PBQ$ is an equilateral triangle inscribed in the given circle.

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Source: https://tasks.illustrativemathematics.org/content-standards/tasks/1557

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