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These are distinct (their radii might differ or not) semicircles on the $x-axis$ as depicted below. I want to show that there exist no such four Euclidean semicircles such that the intersections occurs at a right angle as in the figure. How can show that algebraically?

Fig 1

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    $\begingroup$ Maybe start by fixing one semicircle and parameterizing all the semicircles that make a right angle intersection with it? Do you require the centers of the semicircles to fall on the $x$-axis as the diagram suggests? $\endgroup$
    – hardmath
    Sep 13, 2020 at 14:31
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    $\begingroup$ Fix the leftmost and the rightmost semicircles. Try to determine how to draw another semicircle that makes 90 degree angles with the fixed 2. You will find that there only exists 1 possibility for said semicircle. $\endgroup$ Sep 13, 2020 at 15:25

2 Answers 2

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Here are three proofs, from the simplest to the more elaborate: 1) using hyperbolic geometry, 2) using analytical geometry, 3) using the Apollonius configuration.

  1. With hyperbolic geometry:

Using the [Poincaré half-plane model] (http://pi.math.cornell.edu/~mec/Winter2009/Mihai/section6.html), the upper part of the figure displays an "ordinary" quadrilateral whose sum of angles is $2 \pi$ which is impossible (this sum of angles being twice the sum of angles of a triangle, itself less than $\pi$ as is well known for hyperbolic geometry).

  1. Algebraic proof:

Let us give indices $k=1...4$ to the circles from left to right in the order of their leftmost point, with resp. equations:

$$x^2+y^2-2a_kx-2b_ky+c_k=0$$

Let us recall that a necessary (and in fact sufficient) condition for circles with equations:

$$\begin{cases}x^2+y^2-2ax-2by+c&=&0\\x^2+y^2-2a'x-2b'y+c'&=&0\end{cases}\tag{0}$$ to be orthogonal is

$$2(aa'+bb')=c+c'$$

(Reference: https://mathworld.wolfram.com/OrthogonalCircles.html).

Please note that $(a,b)$ and $(a',b')$ are the coordinates of the centers of these circles. Here, the $b,b'$ components are zero because the centers of all these circles belong to the $x$ axis.

The four orthogonality conditions give rise to the following system:

$$\begin{cases}2a_1a_2&=&c_1+c_2&(A)\\2a_1a_3&=&c_1+c_3&(B)\\2a_3a_4&=&c_3+c_4&(C)\\2a_2a_4&=&c_2+c_4&(D)\end{cases}$$

Combining these equations under the form (A)-(B)+(C)-(D), one gets:

$$(a_1-a_4)(a_3-a_2)=0$$

As $a_1=a_4$ is an excluded case (according to the figure), we must have $a_2=a_3$.

Replacing now $a_3$ by $a_2$ into relationship (B), and comparing with (A), we get $c_2=c_3$ which would mean that the 2nd and 3rd circle, having $a_2=a_3$ and $c_2=c_3$, would have the same equation, therefore would be identical, contradicting the fact that all the circles are assumed distinct.

  1. Using the Apollonius configuration:

[Following the suggestion of Alan Abraham] Let us consider the pencil of circles $\lambda C_1+(1-\lambda) C_4$ of circles generated by the two leftmost and rightmost circles ; the generic circle of this pencil has equation:

$$x^2+y^2-2(\lambda a_1+(1-\lambda)a_4)x+(\lambda c_1+(1-\lambda)c_4)=0$$

(with all their centers on the $x$-axis). In the case of the given figure, $C_1 \cap C_4 = \{A,A'\}$ ; therefore the pencil above is the set of red circles in the figure below:

enter image description here

Fig. 1: the two families of orthogonal circles.

The set of circles orthogonal to $C_1$ and $C_4$ is a certain set of circles (in blue on figure above) with their centers on line $AA'$ and the property that any point on $AA'$ is the center of a well determined blue circle ; as we are looking for a center on the $x$ axis, there is only one such circle. Let it be circle $C_2$. In this case, there is no place for a circle $C_3$ different from $C_2$ with its center on the $x$ axis...

Remark: if $C_1 \cap C_4=\emptyset$, just exchange the roles of blue and red circles.

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Let's simplify the problem by an inversion. Specifically, let's invert about a right-angle intersection between one of the two semicircles.

Those two semicircles (or rather the circles they represent) becomes a pair of perpendicular lines; without loss of generality, they're the $x$-axis and the $y$-axis.

The line in your diagram becomes a circle perpendicular to both axes, which means its center must be at the origin; without loss of generality, it's the unit circle.

For the remaining two semicircles (or rather the circles they represent):

  • One must be perpendicular to the $x$-axis, which means its center has coordinate $(a,0)$ for some $a$. Without loss of generality $a>0$.
  • The other must be perpendicular to the $y$-axis, which means its center has coordinate $(0,b)$ for some $b$. Without loss of generality $b>0$.
  • Both must be perpendicular to the unit circle. If the circle centered at $(a,0)$ intersects the unit circle at $(x,y)$, then the three points $(0,0)$, $(a,0)$, and $(x,y)$ form a right triangle with hypotenuse $a$ whose other two sides are the radii. So the radius of this circle is $\sqrt{a^2-1}$. Similarly, the circle centered at $(0,b)$ has radius $\sqrt{b^2-1}$.

For a circle with radius $\sqrt{a^2-1}$ to be orthogonal to a circle with radius $\sqrt{b^2-1}$, the distance between their centers must be $$ d = \sqrt{(\sqrt{a^2-1})^2 + (\sqrt{b^2-1})^2} = \sqrt{a^2 + b^2 - 2}. $$ But we already know that the distance between their centers $(a,0)$ and $(0,b)$ is $\sqrt{a^2+b^2} \ne \sqrt{a^2+b^2-2}$, contradiction.

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  • $\begingroup$ [+1] Very interesting. I would simplify the final development by saying that the unit circle $U$ and the circle $C$ with center $(a,0)$ define a pencil $P$ of circles (the set of circles passing through points of intersection $(d,e)$ and $(d,-e)$ with $d \ne 0$ of circles $U$ and $C$). Let $P^{\perp}$ (to which the circle with center $(0,b)$ should belong) be the orthogonal pencil of $P$; the centers of the circles belonging to pencil $P^{\perp}$ have the same abscissa $d \ne 0$ defined above): contradiction with the fact that the abscissa is $0$... $\endgroup$
    – Jean Marie
    Sep 14, 2020 at 12:23

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