Minimum operations to find tangent to circle

Question

I've been playing the game Euclidea 3, and I can't really wrap my mind around one of the minimal solutions:

https://www.youtube.com/watch?v=zublg6ZevKo&feature=youtu.be&t=9

The object is to get a tangent line on right of the blue circle with only 3 operations. The minimum solution involves only 2 compass operations (shown above as the gray circles) and the final operation is a straight edge through the point on the blue circle and the intersection of the two compass operations.

My question is: how does this work? How do these two compass operations always result in a point tangential to the blue circle's point?

Answer 1

In the construction above, we are given the circle centred on $O$ (black) and the point $P$. We pick an arbitrary point $A$ on the circle and construct two more circles (in light grey) which intersect in $B$ and $C$; a segment of the tangent is then $PC$. $Z$ is a point on the black circle, on the other side of the chord $PB$ from $A$, and is solely a component of our proof: that $\angle OPC=90^\circ$. Line segments of the same length have the same colour (except the ones incident to $Z$).

Say $\angle BPA=x$. Since $\triangle PAB$ and $\triangle CAP$ are congruent and isosceles, $\angle APC=x$.

$\angle PAB$ is $180^\circ-2x$, and since quadrilateral $PABZ$ is cyclic, $\angle PAB+\angle BZP=180^\circ$, or $\angle BZP=2x$. By the inscribed angle theorem, $\angle BOP=4x$.

Since $\triangle BOP$ is isosceles, $\angle OPB=\frac12(180^\circ-4x)=90^\circ-2x$. Then $$\angle OPC=\angle OPB+\angle BPA+\angle APC=90^\circ-2x+x+x=90^\circ$$ Hence $PC$ is tangent to the black circle at $P$.

Answer 2

(1) $\angle MXN = 90^0$. (Angles in semi-circle)

(2) $\angle RTN = 90^0$. (Line of centers is perpendicular to common chord)

(1) and (2) imply XM // TR.

From (2), we have $\triangle RTN \cong \triangle RTX$. Then, $\angle 1 = \angle 2$.

Also, $\angle 2 = \angle 3$ , $\angle 2 = \angle 4$ and $\angle 4 = \angle 5$.

Therefore, $\angle 5 = \angle 1$. This means MRN is tangent to the blue circle by “converse of angles in alternate segment”.

Answer 3

The original circle had center $A$ and $B$ on the boundary, and all of the red line segments are radii. Point $C$ on that circle was arbitrarily chosen, and the circle with center $C$ and $B$ on its boundary (with green radii) meets the original circle at $D$. Finally, the circle whose center is $B$ and has $D$ on its boundary is drawn (with blue radii), and $E$ is where that circle meets the second circle.

Claim: $\angle ABE$ is a right angle.

Proof:

• $C$ is equidistant from $B$ and $D$, and so is $A$. Therefore, the perpendicular bisector of $\overline{BD}$ (i.e. the locus of all points equidistant from $B$ and $D$) is $\overleftrightarrow{AC}$ and thus $\triangle BCF$ is a right triangle.
• $\angle DBC\cong\angle CBE$ as they are corresponding angles in the two congruent blue-green-green triangles.
• $\angle ABC\cong \angle BCA$, as they are the base angles in an isosceles triangle.
• Therefore, $m\angle ABE=m\angle ABC+\angle CBE = m\angle BCF + m\angle FBC=90^\circ$