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I saw a question that was nearly the same as this, but I couldn't understand the answers.

enter image description here

Assume that everything that seems to be tangent should be tangent, and that everything that appears to be a radius is a radius.

This question is based on a proof for circles with radii a, b, and c: (1/sqrt(a))=(1/sqrt(b))+(1/sqrt(c))

I already got the two tangent circles to a line, and I'm trying to construct the small circle in between.

I was given the following hint: Find the square root of the radii of all circles.

I found it by doing the following. This example is for length a:

  • Drew any length x
  • Drew a+x and constructed a line perpendicular to the point where a and x meet ("line j")
  • Found the midpoint of a+x, set the compass length to 0.5(a+x), and drew a semicircle
  • Found the point at which the semicircle and line j. Extended that point to each of the ends of segment a+x to make a right triangle.
  • Used the geometric mean to find the square root of each radius.

How can I use the square root of the radius? How do I construct the rest of it?

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Here is a construction of mine that takes a square root but uses some other tricks to reduce the number of steps in the overall construction. There probably is a better construction but this is my best so far. This version is closer to the graphic you have in your question and combines everything into one graphic.

I will take it as already shown that

$$\frac 1{\sqrt{a}}+\frac 1{\sqrt{b}}=\frac 1{\sqrt{c}}$$

In the diagram, your two given circles are drawn in red ($a'$ with center $A$ and radius $AP=a$) and blue ($b'$ with center $B$ and radius $BP=b$). Point $P$ is the intersection of segment $\overline{AB}$ with the circles.

enter image description here

Construct semicircle $d$ between points $A$ and $B$, and point $D$ on semicircle $d$ so that line $\overleftrightarrow{DP}$ is perpendicular to $\overline{AB}$. We then have

$$DP=\sqrt{ab}$$

as can be shown by considering the tree right triangles formed by points $A,B,P,$ and $D$.

In the next stage, construct ray $h$ to bisect angle $BPD$ and point $F$ at its intersection with segment $\overline{BD}$, and drop a perpendicular from point $F$ to segment $\overline{BP}$ intersecting at point $G$. Length $FG=PG$ is the reciprocal of the sum of the reciprocals of $BP$ and $DP$. By construction and the formula at the start of this answer, we have

$$FG=PG=\sqrt{bc}$$

In the next stage, construct quarter-circle $f$ with center $P$ and radius $PG$, point $H$ at the intersection of quarter-circle $f$ with line $\overleftrightarrow{DP}$, and point $J$ on segment $\overline{PH}$ such that segment $\overline{GJ}$ is parallel to segment $\overline{BH}$. By similar triangles we then have $\frac{GJ}{BH}=\frac{PJ}{PH}$, which means that

$$PJ=c$$

which is the desired radius of the third circle tangent to the line.

In the last stage, construct a semicircle with center $P$ and radius $PJ$, point $N$ the intersection of that semicircle with segment $AP$, point $O$ the intersection of that semicircle with segment $BP$, the circular arc $o$ with center $A$ and radius $AO$, the circular arc $n$ with center $B$ and radius $BN$, and point $C$ the intersection of arcs $n$ and $o$. Draw the circle $c'$ with center $C$ and radius length $PJ$, and you now have that third circle!

Draw the line tangent to any two of those circles, which you say you can do, and you are all done. I hope you can see in my diagram that the three circles do indeed have the same tangent line $t$.

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The distance between the base points is (by Pythagoras) $$\sqrt{(a+b)^2-(a-b)^2}=2\sqrt{ab} $$ hence we obtain the condition $$ 2\sqrt {ac}+2\sqrt{bc}=2\sqrt{ab}$$ and hence dividing by $2\sqrt{abc}$ $$ \frac1{\sqrt b}+\frac1{\sqrt a}=\frac1{\sqrt c}$$ Thus if you can use this equation determine $c$ from $a,b$, you can construct the figure by constructing the triangle of the three centers, which is determined by its sides $b+c,a+c,a+b$.

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  • $\begingroup$ How can I construct 1/sqrt(a)? $\endgroup$ – Rex May 10 '15 at 13:13

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