I chose an arbitrary length r and constructed two circles with centers in M and N of radius r and marked their intersections with the given circle. Then I looked at the angle formed by the center of the given circle and those two intersections. I found the bisector of that angle and marked the intersection of the bisector and the circle with S. Finally, I found points A and B on the given circle such that triangles ASO and BS0 are equilateral, where O is the center of the given circle. However, I get that at such construction, |AM| is only approximately equal to |BN|. How can I solve this problem?
The points M and N and the circle k are given. Construct an equilateral triangle ABC inscribed in circle k such that |AM|=|BN|.
constructive-mathematicseuclidean-geometrygeometry
Related Solutions
A geometric proof for part a.
From the OP:
We need to prove that the new equilateral triangle is essentially a rotation of the original one, about an axis passing through center of its circumcircle perpendicular to its plane.
I think it's easier to think of it as a reflection. In the diagram below
- $GH$ is the diameter of the circumcircle that is perpendicular to $AM$.
- It's given that $BN$ and $CP$ are parallel to $AM$.
- Hence, points $M,N,P$ are reflections of points $A,B,C$ respectively in the line $GH$.
- Hence, $\triangle MNP$ is a reflection of $\triangle ABC$ through the line $GH$.
Edit, to add part b.
By way of the reflective symmetry about $GH$ and the rotational symmetry of the equilateral triangles we can note that $AP=BN=CM$, and that $AN=BM=CP$. From this we can draw a system of parallel lines to show that $AM=BN+CP$.
From @Blue's answer to my Mathematics StackExchange question:
Construct $\bigcirc A$ through $O$.
- Let $P_1$ and $P_2$ be the points where $\bigcirc A$ meets $\bigcirc O$.
Construct $\bigcirc P_1$ through $P_2$.
- Let $C$ be the (other) point where $\bigcirc P_1$ meets $\bigcirc O$.
Construct $\overleftrightarrow{OP_1}$.
- Let $Q_1$ and $Q_2$ be the points where $\overleftrightarrow{OP_1}$ meets $\bigcirc P_1$.
Construct $\overleftrightarrow{CQ_1}$.
- Let $B$ be the point where $\overleftrightarrow{CQ_1}$ meets $\bigcirc O$.
- Note that we have constructed $\overleftrightarrow{BC}$.
Construct $\overleftrightarrow{CQ_2}$.
- Let $D$ be the point where $\overleftrightarrow{CQ_2}$ meets $\bigcirc O$.
- Note that we have constructed $\overleftrightarrow{CD}$.
Construct $\overleftrightarrow{AB}$.
- Construct $\overleftrightarrow{AD}$.
Square $\square ABCD$, with constructed edge-lines, is inscribed in $\bigcirc O$. (Proof that the quadrilateral is, in fact, a square, is left as an exercise to the reader.)
Edit. Having been asked to elaborate on the square ...
As of Step 2, we know $\triangle P_1 P_2 C$ is equilateral and that $\overline{OA}$ is on the perpendicualr bisector of side $\overline{P_1 P_2}$. Therefore, $\overline{AC}$ is a diameter of $\bigcirc O$, and we have that $\angle ABC$ and $\angle ADC$ (for point $D$ constructed later) are right angles by Thales' Theorem.
As of Step 4, as observed by Jan and Tristan in the comments, $\overline{Q_1 Q_2}$ is a diameter of $\bigcirc{P_1}$, so $\angle O_1 C Q_2$ is a right angle. Therefore, $\square ABCD$ is at least a rectangle.
Now, define $a := |\overline{OA}|$, so that $|\overline{P_1P_2}| = a\sqrt{3}$ and $|\overline{OQ_1}| = a( 1 + \sqrt{3})$. Since $\angle AOQ_1 = 60^\circ$, if we let $R$ be the foot of the perpendicular from $Q_1$ to $\overleftrightarrow{OA}$, then $|\overline{OR}| = \frac{a}{2}( 1 + \sqrt{3})$ and $$|\overline{Q_1R}| = \frac{a \sqrt{3}}{2}(1+\sqrt{3}) = \frac{a}{2}(3 + \sqrt{3}) = a + |\overline{OR}| = |\overline{CR}|$$ Thus, $\angle Q_1 C R = 45^\circ$ and we may conclude that $\square ABCD$ is a square. $\square$
Best Answer
Rotate point $N$ by $120°$ counterclockwise about the center $O$ of the circle, to $N'$ (this can be done by constructing three circles, as in figure below). If the perpendicular bisector of $MN'$ intersects the given circle at $A$, then $AM=AN'$. Finally, rotate $A$ by $120°$ clockwise about $O$, to $B$: we have then $AM=AN'=BN$ (by the congruence of triangles $OAN'$ and $OBN$), as requested.
Both intersections of the perpendicular bisector give a solution, and the initial rotation can also be clockwise: hence we can have up to four distinct solutions.