Euclidea 7.8

Euclidea 7.8 Circle Tangent to Three Lines
Construct a circle that is tangent to the three given lines. Two of the lines, $$\overline{\rm AC}, \overline{\rm BD}$$, are parallel.



The 4L solution to this problem is trivial. However, the 6E solution is perhaps the hardest in the game to this point. So let's talk about that.

The first steps are easy,

1. Draw circle $$ \odot{\rm A} $$ with radius $$\overline{\rm AB}$$ to intersect at point $$ {\rm C} $$, and connect line $$ \overline{\rm BC} $$. Here we are starting to make a rhombus, with vertices $$ {\rm A, B, C} $$. With $$ \overline{\rm BC} $$ as the diagonal of the rhombus, all we need to do is to find the middle point of $$ \overline{\rm BC} $$, which is the center of the desired circle. And then we can drop a perpendicular line, say, to $$ \overline{\rm AC} $$, and use that as the radius, then we are done. The problem is, like always, that we don't have the budget to do that.



The next step is the amazing part.

2. Draw circle $$ \odot{\rm C} $$ with radius $$\overline{\rm CB}$$;

3. Find the intersecting point $$ {\rm D} $$ between circle $$ \odot{\rm A} $$ and line $$\overline{\rm AC} $$, draw circle $$ \odot{\rm D} $$ with radius $$\overline{\rm DC} $$ to intersect $$ \odot{\rm C}  $$ at $$ {\rm E, F} $$.

The claim now is that line $$ \overline{\rm EF} $$ is the desired line: it is perpendicular to $$ \overline{\rm AC} $$, which is trivial by the symmetry, but much more importantly, it passes through the middle point of $$ \overline{\rm BC} $$. Thus, we not only have found the middle point, which is the center of the desired circle but also can use the distance between that to line $$ \overline{\rm AC} $$ as the radius to finish the problem.

So why would $$ \overline{\rm EF} $$ go through the middle point of $$ \overline{\rm BC} $$? This is quite amazing, since the radius of $$ \odot{\rm A} $$ and the radius of $$ \odot{\rm C} $$ do not have any relation. Let's name the intersecting point of $$ \overline{\rm AC} $$ and $$ \overline{\rm EF} $$ as $$ {\rm G} $$. An equivalent statement is that if we connect $${\rm B} $$ and $$ {\rm H} $$, the two intersecting points between $$ \odot{\rm A} $$ and $$ \odot{\rm C} $$, and call the intersecting point between $$ \overline{\rm AC} $$ and $$ \overline{\rm BH} $$ point $$ {\rm J} $$, then $$ \overline{\rm CG} = \overline {\rm GJ} $$. This is a very non-trivial statement. To see that more clearly, I will start from scratch and state this as a Lemma. I do not have a really clearn proof of this, but I have one that works.



In this new picture, consider a line $$ \overline{\rm AB} $$. We draw a circle $$ \odot{\rm A} $$ with radius $$ \overline{\rm AB} $$. Let's call this radius $$ {\rm r} $$. Now we draw a circle $$ \odot{\rm B} $$ with a different radius $$ {\rm R} $$. Let the intersections between the two circles be $$ {\rm F, G}$$ and let $$ \overline{\rm FG}$$ intersect $$ \overline{\rm AB} $$ at $$ {\rm H} $$. Now we also draw circle $$ \odot{\rm E} $$ with radius $$ \overline{\rm EB} = 2r $$. This intersects $$ \odot{\rm B} $$ at $$ {\rm C, D} $$, and line $$ \overline{\rm CD} $$ intersects $$ \overline{\rm AB} $$ at $$ {\rm J} $$.

To make our life easier, let's name the lengths of line segments as

$$ \overline{\rm AH} = a $$

$$ \overline{\rm HJ} = b $$

$$ \overline{\rm JB} = c $$

$$ \overline{\rm HF} = d $$

$$ \overline{\rm JD} = e $$

and we would like to show that $$ b=c $$

To see that, we have a bunch of Pythagorean relations:

$$(1). \hspace{1cm}    a^2+ d^2 = r^2 $$

$$(2). \hspace{1cm}    (b+c)^2+ d^2 = R^2 $$

$$(3). \hspace{1cm}    c^2+ e^2 = R^2 $$

$$(4). \hspace{1cm}    (r+a+b)^2+ e^2 = (2r)^2 $$

and

$$(5). \hspace{1cm} a+b+c = r $$

We have here 5 equations and 7 unknowns. But fortunately, we can get something out of it. Use (5) to replace $$ (b+c) $$ in (2) by $$ (r-a) $$, and then subtract (1), we have $$ (r-a)^2 -a^2 = R^2 - r^2$$, which simplifies to

$$ a = \frac{2r^2 - R^2}{2r} $$.

Similarly replace $$ (r+a+b) $$ in (4) by $$ (2r - c) $$ and subtract (3) to get $$ (2r - c)^2 - c^2 = (2r)^2 - R^2 $$, which simplifies to

$$ c = \frac[R^2}{4r} $$

Now we plug these two in (5) to solve that $$ d = \frac{R^2}{4r} = c $$, which is the desired result.