The solution will find point P on the line connecting points A and B.

In the original problem there is a line that has points A and B on the same side of it. The point P has to be on that line, not on AB

Thanks.

Golland

Eta zadacha iz 1-go tura olimpiadi 1937 goda. Pomechena zvezdochkoi, i teper ponyatno pochemu. Ne dumau chto bi shkolniki v 1937 godu znali analiticheskuu geometriu. Vidimo est kakoi-to truk dlya reshenia etoi zadachi, no v Kiseleve ego ne naidesh.

Ya pomestil echo odnu zadachu na postroenie, no ona zateryalas
in the High school bin. Dan treugolnik i tochka vne ego.
Cherez tochku provesti pryamuu kotoraya razobyet treugolnik na 2 ravnie ploschadi.

Eto ne iz olimpiadi. Eto obobchenie zadachi o pryamougolnom torte
iz kotorogo virezan malenkii pryamougolnik i etot tort nado odnim razrezom podelit popolam. (nado porezat cherez tsentri)

esli tochka P dana na storone treugolnika to soedinit ee s protivopolozhnoi vershinoi, opustit medianu iz etoi vershini, cherez osnovanie mediani provesti pryamuu palallelnuu pervoi linii.ona peresechet druguu storonu v tochke S. PS iskomaya linia.

No esli P vne treugolnika to kak togda postroit? Chto to ne poluchaetsya.

Zaranee spasibo.

Golland

Given a line L and 2 points A, B on the same side of the line, construct a point P on the line L that has a given sum d of distances from the points A and B.

Even students not familiar with analytical geometry (especially when they expect to do well in a math competition) should know that the locus of points with a constant sum d of distances from 2 given points A, B is an ellipse with focal points A and B. Generally, the ellipse would intersect the given line L in 2 different points P and Q (if the line is close enough), which would solve the problem. However, an ellipse cannot be constructed with a ruler and a compass.

From my classes in descriptive geometry, I remembered that although we cannot construct an ellipse with a ruler and a compass, we certainly can construct as many points of the ellipse as we wish. Moreover, descriptive geometry deals with projections of objects, including cylinders, cones, etc. Even students not familiar with descriptive geometry should realize that the projection of a circle is generally an ellipse. There must be a plane (tilted with respect to the plane in which the line L and the points A, B are given), from which a circle projects to our ellipse. If we can figure out the circle center and radius, we can construct it with a compass. Since any line projects as a line, there is a line L' in such plane that projects to the given line L. Moreover, this line intersects the circle and the intersections P' and Q' project to the points P and Q, the solutions of our problem.

In fact, there is an infinite number of such planes and we should choose one in a way to make the solution as simple as possible. What immediately comes to mind is to tilt our plane by some angle around the axis of rotation AB. In this case, we do not have to project the points A, B (not that we need their projections). We know the circle center (midpoint O of the line segment AB) and if we can figure out the length of the ellipse primary axis, we also know the circle radius. For the line L' that projects into the given line L, we need 2 points (unless the line L is parallel to the line segment AB, in which case we need only 1 point). If the line L intersects the line AB, the intersection S is one of the points (because it is on the axis of rotation AB). The only remaining task is to choose another arbitrary point R on the line L and find the corresponding point R' of the line L'.

We have the ellipse focal points A, B and we have to find the lengths of the ellipse primary and secondary axes a and b. Moving the point P to the line AB, say to the point D right of the focal point B, we have

d = AD + BD = (AO + OD) + (OD - BO) = (AO - BO) + 2a = 2a
a = d/2

The primary axis CD of the ellipse can be easily constructed. Moving the point P to the normal to the line AB erected at the midpoint O, say to the point E below the line segment AB, we have an isosceless DABE or 2 right angle triangles DAOE and DBOE. From the DAOE:

AO² + OE² = (AB/2)² + b² = (d/2)²
b = 1/2·Ö(d² - AB²)

The isosceles DABE and consequently, the secondary axis EF of the ellipse can be also easily constructed. Then we draw a circle centered at the midpoint O of the line segment AB and with radius a = d/2 = CD/2. We have 1 point of the line L' - the intersection S of the line L and the axis of rotation AB. To find the second point R', we project the intersection R of the line L with the ellipse secondary axis EF (extended if necessary). The ratio of the ellipse secondary and primary axes b/a gives us the ratio OR/OR':

b/a = OR/OR'

Transfer the ellipse secondary axis b = OE to the axis of rotation AB - the line segment OG. Connect points G and R. Draw a parallel to the line GR through the point C. The parallel intersects the ellipse secondary axis EF (extended) at the point R'. Draw the line L' by connecting the points S and R'. The line L' intersects the circle at the points P' and Q'. Drop normals from the points P' and Q' to the axis of rotation AB. The normals intersect the line L at the points P and Q, the solutions of our problem. Apart from the ellipse main axes CD and EF and the desired points P, Q, we did not have to construct any other points of the ellipse.

Since I am tilting the plane around the secondary axis of the ellipse, the circle center O' º O and the circle radius equals to the secondary ellipse axis b = OE = OF = EF/2. All I have to do is to project one point S of the line L to find one point S' of the line L', both lines being parallel to the axis of rotation. The ratio of the ellipse primary and secondary axes a/b gives me the ratio OS/OS':

a/b = OS/OS'

Transfer the primary ellipse axis on the axis of rotation EF: OG = OD = a. Draw a parallel to the line DG through the point F. The parallel intersects the line AB at the point S'. Erect normal to the line AB at the point S' - this is the line L'. The line L' intersects the circle at the points P' and Q'. Drop normals from these points to the axis of rotation and extend them in the opposite direction. The normals intersect the line L at the points P and Q, the solutions of our problem.

Recently I came across a different solution to this problem.

If we draw a circle C with radius of the given length and the center in point A and reflect the other point B in the given line to B', then we can construct a circle X passing through the B, B' and touching the circle C. The center of circle X is on the given line
and is the solution to the problem.

This solution does not require the special knowledge of ellipse
characteristics.

Now my stumbling block is the inscribed circle. How is it possible
to find the 3rd point where the inscribed circle touches the
circle C ? I hope it is rather simple, but for me it turned out to be a problem.

G

You did not offer any proof that the 2 circles solve the given problem. But since it is rather trivial to see, I will skip it. The original problem is reduced to the following problem:

Given a circle a with the center A and radius d (the sum of distances from the line l to the points A and B) and 2 points B and C inside the circle, construct a circle p through the 2 points B and C and touching the circle a. This is a degenerate Apollonius problem (which is why you are having difficulties). Normal Apollonius problem is to construct a circle touching 3 given circles. Any of the 3 given circles can be degenerated to a point or to a line. For example, if all 3 circles are degenerated to points, the single solution is the triangle circumcircle. If all 3 circles are degenerated to lines, the 4 solutions are the triangle incircle and 3 excircles. As a rule, degeneracy simplifies the solution. See Apollonius Problem at https://www.cut-the-knot.org/Curriculum/Geometry/Apollonius.shtml.

In the spirit of solving Apollonius problems without generating a nasty quadratic equation, the idea is to invert the given circle a in some other circle w such that the inverted circle a' degenerates to a line. The points B and C are also inverted into points B' and C' and the problem is reduced to constructing a circle p' through the inverted points B' and C' touching the line a'. The circle a will degenerate into a line a' when inverted in a circle w with the center O on the circle a. You can easily prove it and you can also play with the inversion tool at https://www.cut-the-knot.org/Curriculum/Geometry/InversionDemo.shtml.

Given a line l and 2 points A and B on the same side of the line, construct a point P such that the sum of distances AP and BP equals to a given length D.

1. Draw a circle a with the center A and radius d.

2. Reflect the point B in the line l. Point C is the reflected point.

3. Select a point O on the circle a. To make things as easy as possible, extend the line BC until it intersects circle a at the point O. Then both the inverted points B' and C' will be on the line BC. Draw a circle w with the center O and an arbitrary radius. Again, to make things as easy as possible, choose radius OB. Then the reflection B' of the point B in the circle w is identical with the point B and we only have to invert the point C.

4. Circle w intersects the circle a at points G and H. Line a' º GH is the inversion of circle a in the circle w.

5. Invert point C in the circle w. The inverted point C' is on the ray OC and

OC·OC' = OB²

Erect a normal to the line OC º BC at the point O. The normal intersects the circle w at the point E. Transfer the line segment OC to the normal OE - to the line segment OD. Connect points D and B. Draw a line through the point E parallel to the line BD. The parallel intersects the ray OC at the point C', the reflection of the point C in the circle w.

6. Construct the blue circles p' and q' through the points B' º B and C' and touching the line a' (see below). Label the circle centers are P' and Q', respectively.

7. Circles p' and q' intersect the circle w at some points I did not label - these are the 3rd points of the red circles p and q. However, you do not have to construct the red circles, just connect the point O with the points P' and Q' and extend until the rays OP' and OQ' intersect the given line l at points P and Q, respectively. These 2 points are solutions of the original problem and also centers of the red circles p and q, respectively, solving the degenerate Apollonius problem.

The remaining problem is the construction of a circle (say p) through 2 given points (say B and C) and touching a given line (say a).

1. Construct axis d of the line segment BC. The axis intersects the line a at the point O.

2. Construct line b by doubling the angle between the axis d and the line a.

3. Draw an arbitrary circle p' with the center S' on the axis d and touching both the lines a and b. I chose to erect a normal at the point H', the intersection of the line BC and line a. The normal intersects the axis d at the point S', center of the circle p'. Radius if this arbitrary circle is S' H'.

4. Draw a ray OC and extend until it intersects the circle p' at the points C' and C". Connect points S' and C'. Draw a parallel to the line S'C' through the point C. The parallel intersects the axis d at the point S, center of the desired circle p. Drop a normal to the line a from the point S and label the heel of this normal H. SH is radius of the desired circle p. The 2nd solution (circle q - not drawn) is constructed by using the point C" instead of C'.

In conclusion, I would say that the solution is much more complex than what I did before with the ellipse. I am even tempted to say that the ellipse construction is a preferred solution of this degenerate Apollonius problem. And one more thing. Please, Golland, do not give up so easily. The true joy of geometry is in solving the problem, not just in seeing the solution (like in everything else).

Regards, Vladimir

Sorry, this step is totally unnecessary and can be skipped. It is just how I visualized the solution.

Thanks for the detailed explanation of the Appolonius approach to
the solution of the restated original problem. It is rather involved,
I must agree with you. But his solution was for 4 circles (three
given and one touching solution). Here we have only one given.

As a "true" mathematician you brought the problem to an existing
solution. And though it is a working solution,
I am convinced that there should be a simpler one, the one that
also does not rely on the ellipse.

For the following reason: if the original problem is restated for
difference of distances (not the sum) given, the solution is still a
center of a circle on the given line (only this time the touching is
external).

How would you construct it using your method when instead of an ellipse we have hyperbola?

There could be a link in construction between this problem (when the
difference of distances is given - hyperbola ) and the problem of cutting the triangle into 2 equal areas by a line through a given point.

I looked up "hyperbola" on mathworld.com and found that our friend Appolonious proved a theorem that area of the triangle made up by an
angle and the tangent to a hyperbola that has sides of angle as its asymptotes is invariant. So, by selecting proper hyperbola (the side
of the deltoid ) whose focal point is a function of the length of 2
triangle sides , we can make that invariant area equal to 1/2 of the
given triangle. Another interesting point : the point where tangent line touches the hyperbola is the middle point of the tangent line between the triangle sides it crosses. That should help in construction.

G.

The ellipse solution and the 2 different solutions of a degenerate Apollonius problem are not working solutions, they are simple and beautiful solutions. And I did not just reduced the problem to an existing solution - I recognized the type of the problem and used a method yielding solution for this type of problems (BTW, I am a physicist, not a mathematician). I will show you what a working solution is: using analytical geometry. To make problem as simple as possible, put the point A at the coordinate origin and point B on the x-axis. Equation of the circle centered at A and with radius d is

x² + y² = d²

Denote C the reflection of the point B reflected in the line l. Coordinates of the point B are (x_B, 0). Coordinates of the point C are (x_C, y_C). They can be either given or calculated from the coordinates of the given point B and the equation of the given line l. Equation of the circle through the points B and C and touching the other circle is

(x - x_P)² + (y - y_P)² = r²

where (x_P, y_P) are unknown coordinates of the point P that solves the problem and r is an unknown radius of this circle. Since points B and C are on this circle, we have

(1) (x_B - x_P)² + y_P² = r²

(2) (x_C - x_P)² + (y_C - y_P)² = r²

Denote T the point of contact of the 2 circles with unknown coordinates (x_T, y_T). Since the point T is on both circles, we have

(3) x_T² + y_T² = d²

(4) (x_T - x_P)² + (y_T - y_P)² = r²

These are 4 equations for 5 unknowns r, x_T, y_T, x_P, and y_P. The 5th equation is

(5) y_T/x_T = y_P/x_P

because the point T is on the ray AP. Eliminating 4 unknowns (say x_T, y_T, y_P, and r), you get a nasty quadratic equation for the remaining unknown x_P, which you can solve and, in principle, construct a line seqment of the length x_P using a ruler and compass. Yuck.

>For the following reason: if the original problem is restated
>for difference of distances (not the sum) given, the solution
>is still a center of a circle on the given line (only this time the
>touching is external).
>
> How would you construct it using your method when instead
>of an ellipse we have hyperbola?

I do not know the solution off hand. But I made the following statement at the beginning of the ellipse solution: Although we cannot construct an ellipse with a ruler and compass, we certainly can construct as many points of the ellipse as we wish. It is true for a hyperbola as well. The difference would be that I cannot not visualize the problem using the parallel projection of a circle. I may even try the hyperbola solution.

>There could be a link in construction between this problem
>(when the difference of distances is given - hyperbola) and the
>problem of cutting the triangle into 2 equal areas by a
>line through a given point.

Probably, certainly worth trying.

>I looked up "hyperbola" on mathworld.com and found that our friend
>Appolonious proved a theorem that area of the triangle made up by
>an angle and the tangent to a hyperbola that has sides of angle
>as its asymptotes is invariant. So, by selecting proper hyperbola
>(the side of the deltoid) whose focal point is a function of the
>length of 2 triangle sides, we can make that invariant area equal to
>1/2 of the given triangle. Another interesting point: the point
>where tangent line touches the hyperbola is the middle point of
>the tangent line between the triangle sides it crosses. That should
>help in construction.

Well, Euclidean geometry is an ancient enterprise, so it is not so easy to come up with something new. But the hyperbola theorem poped up in my analysis of the "cut the triangle area in half" problem following the construction.

Regards, Vladimir

This may be finished off more quickly with a similar inversive idea.

(adopting your notation...)

If we invert the plane w.r.t. the circle S, centre B passing through B', p must invert to a straight line, tangent to the inverse of a. Such a line is easy to construct. More precisely:

(...and style, which I like)

1. Construct a', the inverse of a w.r.t. S.

2. Construct lines l_1 and l_2 through B', tangent to a'.

3. Let p_1 and p_2 be the inverses of l_1 and l_2 with respect to S. They are the required circles.

The above constructions are all standard, if a little arduous with ruler and compasses.

Thankyou

sfwc
<><

(BTW, what does sfwc stand for?). This is a beautiful twist and worth of presenting a construction. 2 cautionary comments though:

1. While inversion of a circle in a circle that intersects it is rather trivial, inversion of a circle in a circle that does not intersect it is a little harder to grasp. In fact, I did not do it right in my 1st construction attempt and then wondered why the resulting inscribed circles did not touch the big one.

2. The construction tends to get crowded inside the inversion circle which can be rather small, so it is a good idea to start with a large exterior circle.

1. Draw a circle a with the center A and radius d. The circle intersects the line AB at points E and F.

2. Reflect the point B in the line l. Point B' is the reflected point.

3. Draw a circle w iwith the center B and radius r = BB'. The circle intersects the line AB at points C and D. This is the inversion circle.

4. Invert the circle a in the circle w. Center of the inverted circle a' will be on the line AB. We will invert the points E and F into points E' and F', respectively.

r = BB' = BC = BD
BE·BE' = r² = BC²
BF·BF' = r² = BD²

Erect a normal to the line AB at the point B. The normal intersects the inversion circle w at the point K. Transfer the line segments BE and BF on the normal BK - to the line segments BG and BH, respectively. Connect points G and C and draw a parallel to the line GC through the point K. The parallel intersects the line AB at the point E', a reflection of the point E in the circle w. Similarly, connect points H and D and draw a parallel to the line HD through the point K. The parallel intersects the line AB at the point F', a reflection of the point F in the circle w. Point A' is the midpoint of the line segment E'F'. Draw the circle a' centered at A' and with the radius A'E' = A'F' = E'F'/2.

5. Find 2 tangents p' and q' from the point B' to the inverted circle a'. A tangent is normal to the circle radius at the point of contact. Construct the points of contact S' and T': Draw a circle t centered at the midpoint O of the line segment A'B'. Circle t intersects the circle a' at the points S' and T'. Tangents p' and q' are the desired circles p and q inverted in the circle w. The circles p and q degenerate to lines p' and q' because they both pass through the point B, the center of the inversion circle w.

6. Tangents p' and q' intersect the inversion circle w at the points M and N, respectively. Since the points M and N are on the inversion circle w, they are identical with their own reflections. Therefore, they are on the circles p and q, respectively, the solutions of the problem.

7. Circles p and q are circumcircles of isosceles triangles DMBB' and DNBB', respectively. The 2 triangles are isosceles because each has 2 sides equal to the radius of the inversion circle w. Centers P and Q of the circumcircles p and q are both on the line l, the axis of the line segment BB'. To find the centers P and Q, construct axes of the line segments MB' and NB' by dropping normals from the point B to these line segments (remember, DMBB' and DNBB' are isosceles). The normals intersect the line l at the points P and Q, respectively.

8. This last step is a bonus, it is not necessary to complete the construction. Draw the rays AP and AQ and extend them until they intersect the circle a at points S and T. These are the points of contact of the circles p and q with the circle a.

Regards, Vladimir

Well, my original nickname was shrew (the origins of this are lost in the mists of time), this was 'shortened' to smallfurrywoodlandcreature and then re-shortened to sfwc.

Sorry about the nasty inversion construction. I tend to 'chunk' my constructions, and inversion is a standard 'chunk'. However, I am convinced that sketchpad would benefit from a standard, inbuilt inversion function.

Thankyou

sfwc
<><

PS That was a little too chatty for a maths forum. Sorry Alex. If you don't want to post this, that's fine (but please let me know so I can try to contact Vladimir by some other means). For more chatty stuff, I may be found at (strangely enough) sfwc@hotmail.com.

While I have nothing against Apolonious solution ( it is a piece
of beauty ), the use of it in given problem is not warranted.
Like using cannon to shoot down birds.

the problem: a circle C and two points A and B inside. Find a
point on the common tangent to circle C and circle through A and
B and touching C.

We will find a point X where this tangential crosses the AB line.

Additional construction is any circle Y passing through A and B.

It's center is on symmetry line for A and B.

Circle Y will cross circle C in points M and N.

Point X where AB and MN cross is the solution. From this point we

can draw 2 tangentials to circle C. Hence the original problem has

2 solutions.

Another solution (this one will result in the actual point of touch):

Call the solution circle S.

We know that any 2 circles have center of symmetry. Construct

an additional circle (like in the first solution, but start with

construction of chord A'B' of any length having the same line of

sym. as AB and draw a circle Y through A' and B'). Find center of

sym. C and Y. Find the center of sym. Y and X (use chord AB and

and A'B' ). A line through those centers of sym. will cross the

given circle C at the point where it touches S.

There are 2 solutions for there are 2 centers of sym. for 2 circles:

always on the line through the centers, but one between the centers

and the other on the side of the smaller circle.

I have to thank you guys for forcing me to look it all up in the

geometry book. When I placed the problem on the site I hoped that

someone with knowhow and passion for both geometry and teaching

will explain it all to me without my "wasting" all that time.

It turned out not to be wasted, for I learned about this pretty

powerful technique of additional circle !

G.

> We will find a point X where this tangential crosses the AB line.
> Additional construction is any circle Y passing through A and B.
> It's center is on symmetry line for A and B. Circle Y will cross
> circle C in points M and N. Point X where AB and MN cross is the
> solution. From this point we can draw 2 tangentials to circle C.
> Hence the original problem has 2 solutions.

This is not correct. First, you did not offer any proof (or at least an explanation) why the above construction should solve the problem. Second, when I tried to perform the construction, it immediately became clear that the 2 blue tangents e and h and their points of contact U, V with the circle a (centered at A and with radius d) are symmetrical with respect to the line AB, while the 2 red circles p and q solving the problem, their centers P, Q, and their points of contact S, T with the circle a are not.

I have not checked the 2nd solution yet. When you are trying to learn a theory and you do not completely solve some example problems (i.e., when you stop before the solution is totally unquestionable), it may produce a false sense of understanding (this is a personal experience). Don't worry, I made my share of blunders, the most recent and exquisite one has been kindly deleted by Alex. Please, Alex, do not restore it!!!.

Regards, Vladimir

At the beginning of my message I stated the problem and I repeat:

*********************************************************
the problem: a circle C and two points A and B inside. Find a
point on the common tangent to circle C and circle through A and
B and touching C.
************************************************************

the problem stated this way does not mention any lines. We get the line as the symmetry axis for A and B.

The points A and B here are the points B and B' (the reflection of
B in the given line) from the original problem. The circle here has its center in C (compared to original problem center A ).

You are right. I should have named points the same. It is confusing.
But now I hope there is no confusion any more.

G.

Suppose a line l and a point B not on the line are given. Reflect the point B in the line l, label it C and forget about the line l (i.e., erase the line).

Suppose 2 different points B and C are given. Construct the axis l of the line segment BC and forget about the point C (i.e., erase this point).

You get the same result when the other point is everywhere labeled B'.

Please, with this in mind, go through my replies to your 2 solutions. You should find your 1st solution incorrect, no question about that. As for the 2nd solution, you can either clear up the ambiguities I pointed out or you can find it also incorrect. But until you do one of these, I cannot offer you any more help (not a personal decision, but an impossibility).

Regards, Vladimir

Please, open up!

In your analysis you place the center of the given circle in one of the two points in question (A). That is wrong. The center of the
given circle is somewhere else, and the two points A and B are inside that circle (we create a symmetry line for A and B while circle center is in O).

Thanks for your helping me in solving this difficult construction
problem that occupies my mind for almost a year now.

I liked your projective solution using ellipse property.

I'm still interested in projective (or any other) construction in
case of hyperbola (not ellipse).

Lots and lots of thanks.

G.

P.S. The second "solution" is for the case when the resulting
circle touches two crossing lines instead of passing through 2
points. I made WRONG conclusion that it was possible to create
the center of symmetry for the arbitrary circle and the
solution circle using the two points of the solution circle and
the centers line. If it were possible, then the 3 central symmetry
centers theorem could have been used.

If there are 3 circles then the 3 centers of symmetry of each pair
are collinear. Unfortunately this neat theorem can not be utilized
in this case.

Now it works and you did it yourself. Not only should you feel like a million bucks, hopefully, you won't get stuck the next time around. Still, you did not offer any proof or explanation of why the construction should work, which is a serious deficiency. Do not skip this step.

So, why does the construction work? It uses the concept of a power p of a point B to a circle w centered at a point Y and with a radius r

p = YB² - r²

and of a radical line, the locus of points with the same power to 2 non-concentric circles. If the 2 circles intersect, their radical line is particulary simple to draw - it is the line connecting their 2 intersection points. Now, the radical line of the circles p and q, the solutions of our problem, is the line r º BC. When we draw another arbitrary circle w through the points B and C (its center Y being on the line l, the axis of symmetry of the line segment BC), it'shares the same radical line r with the desired circles p and q. We should draw this arbitrary circle in such a way that it intersects the circle a centered at the point A and with the radius d at some points M and N. This makes the radical line s º MN of these 2 circles (a and w) easy to construct. The 2 radical lines s and r intersect at the point X. Consequently, point X has the same power to all 4 circles. Now we found the point with the same power to the 3 circles of interest a, p, and q, and we can forget about the circle w. Construct the 2 tagents e and h from the point X to the circle a by finding midpoint O of the line segment AX and drawing a circle centered at the midpoint O and with radius OA = OX = AX/2. This circle intersects the circle a at the points U and V, the 2 points of contact with the tangents. Power of the point X to the 3 circles a, p, and q is the same

p = AX² - d² = PX² - r_P² = QX² - r_Q²

where r_P and r_Q are radii of the desired circles p and q, respectively. For the point of contact U we get

p = AX² - AU² = PX² - PU² = UX²

and similarly for the point of contact V we get

p = AX² - AV² = QX² - QV² = VX²

Recalling Pythagoras' theorem, this implies an existence of a pair of right-angle triangles DAXU and DPXU with a common side UX for the point of contact U and a pair of and right-angle triangles DAXV and DQXV with a common side VX for the point of contact V. Consequently, the tangent q is a tangent to both the circles a and p with a common point of contact U. Similarly, the tangent h is a tangent to both the circles a and q with a common point of contact V. Now the centers P and of the desired circles p and q, respectively, can be found simply by drawing rays AU and AV, which intersect the line l at the desired points P and Q, respectively.

I would arrange the 4 solutions to the original problem (a) by their simplicity and (b) by their universality:

Simplicity:
1. solution with an ellipse
2. solution with radical lines
3. 2 solutions of degenerate Apollonius problem

Universality:
1. 2 solutions of degenerate Apollonius problem
2. solution with radical lines
3. solution with an ellipse

As for the problem similar to the original one, namely to find a point P on a given line l with a constant difference d of distances from 2 given points A and B, it can be solved using a hyperbola. The idea is similar to solving the original problem using an ellipse, namely, to tilt the plane around the axis AB by an appropriate angle, transforming the general hyperbola to a normal hyperbola (a hyperbola with its primary and secondary axes a and b equal to each other and consequently, with perpendicular asymptotes). While a circle (i.e., a normal ellipse) can be constructed with a compass, a normal hyperbola cannot. Of course, I can proceed with some other construction steps once I have the normal hyperbola, but I am still searching for steps simpler than those I found so far. It may take some time.

Regards, Vladimir

Let M' and N' be the intersections of any line l through X with a. Then:

XM'*XN' = XM*XN (Intersecting Chords Thm.)
= XB*XC (Intersecting Chords Thm.)
=> M', N', B and C are cyclic (Converse to the Intersecting Chords Thm.)

In particular, each tangent from X to a meets a at a double point PP' say. Then the concyclicity condition derived means that P, P' B and C are cyclic. To put it another way, there is a circle w through B and C and tangent to a at PP' as required.

Thankyou

sfwc
<><

PS The idea of double points may be new to you, but basically you consider a point as a pair of points with infinit'simal distance between them but direction of a line through them known. (so if they lie on a circle, that line must be a tangent line to the circle). most Theorems still apply, as in the above proof. My most wierd application involved a double point at infinity.

It'seems to me that there is no real need to perform inversion
of the Apollonius type to solve construction problems of the following kind:

construct a circle touching
a) 3 given circles,
b) 2 circles and a line
c) 2 circles and a point
........................
all the way to 3 points.

In some of these cases (when one of the given forms is a circle)
we should be able to construct it using only the edge.

Then in what kind of geometry construction problems do we need to actually perform the inversion?

G.

Many works by Apollonius have been lost - see Apollonius of Perga at
https://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Apollonius.html)
- but it does not appear that he new about the inversion.
He certainly knew about radical lines - see Apollonius' Theorem at
https://www.maths.gla.ac.uk/~wws/cabripages/inversive/newapollonius.html
Since he solved his own problem, inversion is not necessary to solve the Apollonius problem, it is just a handy tool.

Alex, please, we need you. Do you kow what kind of geometry problems reqiure the use of inversion in their solutions?

Regards, Vladimir

I do not know if you are aware of the, I believe, Carleman's solution:

Form three right cones with the three circles as bases. The the three intersect at a point. With this point as the apex form another right cone looking in the opposite direction. Its intersection with the plane of the circles solves the problem.

>
>Alex, please, we need you. Do you kow what kind of geometry
>problems reqiure the use of inversion in their
>solutions?
>
I am tremendously busy right now. Let postpone this till the weekend.

Just as an afterthought. I did not have time to follow thoroughly the thread, but what does the question mean? Inversion can be performed with compass and straightedge. Thus, inversion is a convenience, not a necessity.

Now, on an administrative note: I am closing this thread which gew too long for the forum software to handle comfortably. Please start a new one.

Comment:
The condition "on the same side of the line" is an unnecessary limitation both for the ellipse and for the hyperbola problems. However, this condition guarantees the existence of 2 different solutions for the hyperbola problem. On the other hand, the opposite condition would have guaranteed the existence of 2 different solutions for the ellipse problem.

When solving the ellipse problem, we constructed a circle and a line l' in a plane tilted with respect to the given plane around the axis AB, which projected to the given line l and an ellipse intersecting the line at the points P and Q, the solutions of our problem. A circle is an ellipse with the primary and secondary axes equal to each other and it can be constructed with a compass. The initial idea for the solution of the hyperbola problem is similar to the ellipse problem solution - to perform the construction in a plane tilted with respect to the given plane around the axis AB, from which a normal hyperbola (i.e., a hyperbola with the primary and secondary axes equal to each other) and a line l' project to the given line l and a hyperbola intersecting the line at the points P and Q, the solutions of our problem. Although we cannot construct the normal hyperbola with a compass, we get another benefit: the normal hyperbola has asymptotes perpendicular to each other. We will use these asymptotes as an x-y coordinate system, express the equation of the line l' and the normal hyperbola in these coordinates, find a quadratic equation for the 2 intersections of this line with the normal hyperbola, and solve this equation with a ruler and compass. The point is that the normal hyperbola in such coordinate system has a very simple equation and, consequently, our quadratic equation will also be simple.

First, we have to find the primary and secondary axes a and b of the hyperbola. Moving the points P or Q to the line AB (i.e., to the points C and D, respectively), we get

d = BC - AC = (BO + OC) - (AO - OC) = (BO - AO) + 2a = 2a
a = d/2

d = AD - AD = (AO + OD) - (BO - OD) = (AO - BO) + 2a = 2a
a = d/2

The primary axis CD of the hyperbola can be easily constructed by drawing a circle a centered at the point A and with the radius d, which intersects the line AB at the point D between the points A and B. Similarly, we could construct a circle b centered at the point B and with the radius d (not drawn), which would intersect the line AB at the point C between the points A and B.

We cannot move the points P or Q to the secondary axis of the hyperbola, like we did with the ellipse. For the ellipse, the secondary axis b was given by the equation

a² = c² + b² (ellipse)

while for the hyperbola the secondary axis b is given by a similar equation

c² = a² + b² (hyperbola)

In both cases, c = AB/2 is half of the distance separating the focal points or the distance of each focal point from the midpoint of the line segment AB. The secondary axis of the hyperbola is therefore constructed using the Pythagoras' theorem: We draw a circle d centered at the point D and with the radius c = AB/2 = BO = AO. The circle intersects the normal to the line AB erected at the midpoint O at the points E and F and the secondary axis of the hyperbola is b = EO = FO.

Now we tilt the plane by an appropriate angle to transform the hyperbola into a normal hyperbola with the same primary axis and with perpendicular asymptotes. So we just draw the 2 asymptotes x and y, both making the angle 45º with the line AB - the primary axis of both hyperbolas. To project the given line l into the line l' in the new plane, we need 2 points of this line. One is the point S, the intersection of the given line l with the axis of rotation AB. To get the 2nd point, we project the intersection R of the hyperbola secondary axis with the given line l to the point R'. The ratio of the hyperbola secondary and primary axes b/a gives us the ratio OR/OR':

b/a = OR/OR'

Transfer the hyperbola secondary axis b = OE to the axis of rotation AB - the line segment OG. Connect points G and R. Draw a parallel to the line GR through the point C. The parallel intersects the hyperbola secondary axis EF (extended) at the point R'. Draw the line l' by connecting the points S and R'. The line l' intersects the normal hyperbola asymptotes x and y at the points U and V, respectively.

Now we have to resort to analytical geometry. Since the above construction is already crowded, we will use another sketch:

The equation of the line l' in the x-y coordinate system is

(1) y = v/u·x + v

where u = OU and v = OV have already been constructed. The equation of the normal hyperbola is

(2) xy = a²/2

This is easily seen by considering equation of the normal hyperbola in the original X-Y coordinate system (i.e., the 2 hyperbola branches left and right of the origin O), where the axes X º CD and and Y º EF coincide with its primary and secondary axes:

X² - Y² = a²

Rotating the coordinate axes X and Y by the angle f = -45º (to coincide with axes x and y) and taking into account sin(-45º) = -Ö2/2, cos(-45º) = Ö2/2:

X = x·cosf - y·sinf
Y = x·sinf + y·cosf

X = x·Ö2/2 + y·Ö2/2
Y = -x·Ö2/2 + y·Ö2/2

X² - Y² = 2xy = a²

Eliminating y from the 2 equations (1) and (2), we get a quadratic equation for the x-coordinate of the points P' and Q':

v/u·x + v = a²/(2x)
x² + u·x - a²·u/(2v) = 0

To simplify things, we select the primary axis of the hyperbola as the unit length (a = 1). The circle w centered at the origin O and with the radius a becomes the unit circle intersecting the x and y axes at the points C', D' and E', F', respectively. The above quardatic equation becomes

x² + u·x - u/(2v) = 0

Solving for the x-coordinate we get

x = -u/2 ± Ö{u²/4 + u/(2v)}

Denote w² = u/(2v) > 0. We have to construct line segments u/2 and w = Ö{u/(2v)}. The square root will be a hypotenuse of a right angle triangle with sides u/2 and w. Adding this hypotenuse to or subtracting it from the line segment u/2, we will get the 2 solutions for the x-coordinates of the points P and Q. Proceed as follows:

1. Find the midpoint W of the line segment u = OU. This gives us the length u/2.

2. Connect the points W and V and draw a parallel to the line WV through the point F'. The parallel intersects the x-axis at the point X. This gives us the length u/(2v).

3. The line segment XD' equals to u/(2v) + 1. Find the midpoint H of the line segment XD' and construct the circle x centered at the midpoint H and with the radius HX = HD' = XD'/2. The circle x intersects the y-axis at the point Y. This gives us the length w = OY = Ö{u/(2v)}.

4. The hypothenuse YW of the right angle triangle DOYW with the sides u/2 = OW and w = OY = Ö{u/(2v)} gives us the length Ö(u²/4 + w²) = Ö{u²/4 + u/(2v)}.

5. Add this hypothenuse to the length u/2 = OW or substract it from the length u/2 = OW by constructing the circle z centered at the point W and with the radius OY equal to this hypothenuse. The circle z intersects the x-axis at the points M and N. By a pure coincidence, the point N is very close to the point H - disregard this coincidence.

6. Erect normals m and n to the x-axis at the points M and N, respectively. The 2 normals intersect the line l' at the points P' and Q', the projections of the points P and Q that solve the problem.

7. Return to the first sketch (the one with the hyperbolas). Drop normals from the points P' and Q' to the axis of rotation AB. The 2 normals intersect the given line l at the points P and Q, the solutions of our problem.

Huh. Done. It is easy to make promises...

In fact your construction proves the method. If you insist, it is OK
to use point A as center. Then the solution cicle passes through A, B
and touches circle alpha in symmetrical points U and V.

The red circles are solutions to a slightly different problem.
But if you want a solution for points B and C on your chart then
the arbitrary circle has to go through B and C. It will cross
circle alpha in appropriate points and line through these two points and BC will cross in a new point Z'.

G.

What chord has the same line of symmetry as the line segment AB? I say none. Or is AB a chord through the points A and B, point A being the center of circle C ?

> Find center of sym. C and Y. Find the center of sym. Y and X
> (use chord AB and A'B').

What is X and how do I use chords AB and A'B' ?

> A line through those centers of sym. will cross the given
> circle C at the point where it touches S. There are 2 solutions
> for there are 2 centers of sym. for 2 circles: always on the line
> through the centers, but one between the centers and the other
> on the side of the smaller circle.

Please could you clear up these ambiguities and/or give more details ?

Regards, Vladimir

The second "solution" is off the table now. It flew away, for
it was full of hot air and wishful thinking. Sorry.

G

"Given a line l and 2 points A and B such that the line segment AB does not intersect the line l. With a ruler and compass, find a point P on the line l such that the sum of distances PA and PB to the 2 given points A and B equals to a given length d."

It is true that the condition "such that the line segment AB does not intersect the line l" is an unnecessary limitation of the problem.