Euclidea 4.4

Euclidea 4.4 Equilateral Triangle in Circle


Inscribe an equilateral triangle in the circle using the given point A as a vertex. The center of the circle is not given

Goal: 5L, 6E 

Solution 5L:



 * 1) Center $$ {\rm A} $$, arbitrary radius, draw circle to intersect $$ \odot {\rm O} $$  at points $$ {\rm B} $$ and $$ {\rm C} $$;
 * 2) Center $$ {\rm B} $$, radius $$\overline{\rm BA}$$, draw circle to intersect $$\odot A$$  at points $$ {\rm D} $$ and $$ {\rm E} $$;
 * 3) Perpendicular bisector of $$ \overline{\rm CD}$$ passes through $$ {\rm A} $$
 * 4) Perpendicular bisector of $$ \overline{\rm CE}$$ passes through $$ {\rm A} $$
 * 5) Connect the interestions of the two bisectors with $$ \odot {\rm O} $$

'''Why it works? '''

We need to make the following auxilary lines as shown, where $$ \overline{\rm FG}$$ is the tangent line of $$ \odot {\rm O} $$ at point $$ {\rm A} $$, and we have used the center $$ {\rm O} $$ even though it is not used in the solution. The key step is to show that $$ \angle HAO = 30^{\circ}$$.



To see that, we compute by brute force

$$ \begin{align} \angle CAH &= \frac{1}{2} ( 180^{\circ} - \angle DAB - \angle BAG - \angle CAF) \\ &= \frac{1}{2}( 180^{\circ} - 60^{\circ} - 2 \angle CAF) \end{align} $$

$$ \angle HAO = 90^{\circ}- \angle CAF - \angle CAH = 30^{\circ}$$

Similarly, the other bisector is at the desired angle too.

Solution 6E:
Each of the following steps costs 1E:


 * 1) Center on an arbitrary point $$ {\rm B} $$ on $$ \odot {\rm O} $$, draw a circle to intersect  $$ \odot {\rm O} $$ at $$ {\rm A} $$, and another point $$ {\rm C} $$;
 * 2) Center $$ {\rm A} $$, radius $$ \overline{\rm AC} $$, intersect $$ \odot {\rm B} $$ at $$ {\rm D} $$;
 * 3) Center $$ {\rm D} $$, radius $$ \overline{\rm DA} $$, intersect $$ \odot {\rm A} $$ at $$ {\rm E} $$ and at $$ {\rm F} $$;
 * 4) Connect $$ \overline{\rm AF} $$;
 * 5) Connect $$ \overline{\rm AE} $$;
 * 6) Connect the intersections of the above two lines with  $$ \odot {\rm O} $$ to complete the triangle.



Why it works?

Like always, the key step is at the beginning of this procedure. After step 2, we need to show that point $$ {\rm D} $$ is at the right place. For that, we need auxilary lines as follows. Again, we used the center $$ {\rm O} $$ in the proof, but not in the solutions. We will show below that $$ \angle OAD = 90^{\circ}$$:



This is true because

$$ \angle AOB = 2 \angle ACB = 2 \angle CAB $$

and since $$ \triangle ACD $$ and $$ \triangle OAB $$ are both isosceles triangle, we have

$$ \angle CAB = \angle DAB $$

and

$$ 90^{\circ} = \frac{1}{2} \angle AOB + \angle OAB = \angle CAB+\angle OAB = \angle DAB + \angle OAB = \angle OAD $$

After this everything is easy. We have, in the solution figure, two equal sized circles $$ \odot {\rm A}$$ and $$ \odot {\rm D}$$ intersecting, which makes triangles $$ \triangle AED $$ and $$ \triangle AFD $$ both equilateral. This puts $$ \overline{\rm AE} $$ and $$ \overline{\rm AF}$$ both in the desired directions.