Correctly constructed sketches are available by clicking on the links in quotes or the diagrams shown. If you have trouble linking to Sketchpad, then go to [GSP Help].
Part 1: Sketch tracing P as A moves (see end of Part 1): "Monge1a.gsp"
and a sketch of a tangent to to a circle through an external point: "Monge1b.gsp"
Part 2: Sketch tracing m as O1 and O2 move (see end of Part II): "Monge2a.gsp"
and a sketch of an external tangent to two circles: "Monge2b.gsp"
Part 3: Sketch of the three pairs of external tangents to three circles. This should illustrate Monge's Theorem, as in 3.5.: "Monge3.gsp"
Part 4: Sketch of dilating a polygon by marked ratio: "Monge4a.gsp"
Part 4: Sketch of the composition of two dilations: "Monge4b.gsp"
Mathematical Fact: Proof: Construction:
Mathematical Fact: Construction:
Monge's Theorem:
The dilation that maps A onto A' exists but is not unique.
The center of dilation can be any point C on the line AA'. If C is not between A and A', the scaling factor of the dilation would be CA' / CA. If C is between A and A', the scaling factor of the dilation would be - CA' / CA. The scaling factor could be any real number depending on the location of C. (The scaling factor is only equal to 1 if A' = A and only equal to 0 if A' = A = C.)
Given four points A, A', B, and B', does there exist a dilation
such that A' is the image of A and B' is the image of B? If not, what
are conditions for such a dilation to exist? Is such a dilation unique?
Explain.
In general, such a dilation does not exist.
As was just stated above, the center of dilation that maps A onto A' is on the line AA'. Likewise, the center of dilation that maps B onto B' is on the line BB'. So for one dilation to do both, the center of dilation can only be the intersection point of line AA' and line BB'. Call this point of intersection C. For this to occur, line AA' and line BB' cannot be parallel; in addition the scaling factor to map A onto A' is determined to be CA' / CA (or - CA' / CA). Likewise, the scaling factor to map B onto B' is determined to be CB' / CB (or - CB' / CB). These two conditions can simultaneous occur only when CA' / CA = CB' / CB, which is quite a stringent condition.
When are two circles related by a dilation? Is such a dilation
unique? Relate this to Part 2.
Two circles with different radii are always related by a unique dilation. The center of dilation is the center of symmetry discovered in Problem 3 (Part 2).
Consider two circles centered at A and A' of different radii with an external tangent constructed. Let T and T' be the points of tangency of the circles centered at A and A' respectively. Let C be the point of intersection of line AA' and line TT'. Radii AT and A'T' are parallel so angle(CAT) = angle(CA'T'). Obviously angle(ACT) = angle(A'CT'). Thus triangles ACT and A'CT' are similar by a scaling factor equal to CA' / CA = CT' / CT. Note that this meets the stringent condition in the previous section. Thus there is a unique dilation centered at C with a scaling factor CA' / CA such that A' is the image of A and T' is the image of T.
In fact if we go back to Problem 3 (Part 2), we could take any points P and P' on the respective circles such that AP and A'P' are parallel. With the same arguement just given for points T and T', the same unique dilation centered at C with a scaling factor CA' / CA would make A' the image of A and P' the image of P. Since all points of one circle are the dilation of all points of the other circle, the two circles are related by a unique dilation. (This treats two inverse dilations is a single unique dilation, though they could be considered two dilations with reciprocal scaling factors.) The center of the dilation is referred to as the center of symmetry.
We discovered from observations that the composition of two dilations (centered at C1 and C2) is a third dilation, and this resulting figure is also a dilation of the first figure.
It's center of dilation, call is C12, is collinear with C1 and C2. In order to
understand why the new center is collinear with the others, construct points C1 and C2
and the line containing them. Then construct a point P on this line. After
two successive dilations of P, the first image, P', and the final image, P'', were
both points on the line, so therefore C12 (the center of the
composition) must also be on this line.
Note: The composition of two dilations is usually another dilation. However, if the scaling factors of the two dilations are reciprocals of each other, then the composition of the dilations is a translation, not a dilation.
Explanation: (more formal than above) Let C12 be the center of d12. Assume that C12 is not on line C1C2. Since P'' is the image of P under dilation d12, we know that P, P'', and C12 are collinear from the first part of Problem 5 (Part 4 of Monge's Theorem Investigation). But since P and P'' are already on line C1C2, then C12 is also on line C1C2. This is a contradiction to the assumption. Thus all three centers of dilation must be collinear.
From the last part of Problem 5, we can view any pair of circles (with different radii) as related by a dilation. The center of dilation is the point of intersection of the two external tangents. Consider three circles, c1, c2, and c3 with non-equal radii. The composition of dilations taking c1 to c2 and taking c2 to c3 gives the dilation from c1 to c3. Thus the centers of dilation (where the external tangents intersect) are collinear from Problem 6.
Given any hexgaon inscribed in a circle such that opposite sides aren't parallel, extend the sides. Construct the three points where the opposite extended sides intersect. These three points of intersection are collinear. The sketch of this is available"here. [GSP Help]
The point of tangency lies on the circle with diameter OQ.
Let O be the center of the circle and Q be the point the tangent line should go through. Since the tangent line and a radius are perpendicular first consider two more general perpendicular lines: one passing through O and the other passing through Q. Let P be the point of intersection of these lines. Consider OQ to be the diameter of a circle. All triangles with vertices O, Q, and any other point on this circle is necessarily a right triangle. Triangle OPQ is such a right triangle so P must lie on the circle with diameter PQ.
Construct the midpoint M of segment OQ. Then construct a circle centered at M with radius MO. Mark a point P (there are two) where this new circle intersects the circle centered at O. P is the point of tangency. Finally, construct the line containing points P and Q.
Given two circles, there is a common intersection point for the set of lines containing the points on the circles which are on parallel radii. (See figure below.)
Let O1 and O2 be the center of the circles to construct an external tangent line to. An external tangent to both circles must be perpendicular to radii of both circles. Thus these two radii must be parallel to each other. Using this idea, consider (and construct) two parallel lines, each containing a radius from one of the circles. Mark the point of intersection of each line and their respective circles. Note that there are four points of intersection - two for each line. Select only one point of intersection for each so that the two points are on the "same side" of their respective circles. (This is so that we may study external tangents rather than internal tangents.) Construct the line m containing these two points of intersection. Remember that the parallel lines constructed could contain any pair of radii. Observe that no matter which parallel lines were chosen, all possible lines m always intersects the line containing O1 and O2 at the same point.
Construct the line containing O1 and O2. Also construct the point Q where line containing O1 and O2 intersects with a line m. Since all lines m pass through Q, the line tangent to the circles must also pass through Q. Thus we can construct the desired tangent line by constructing a tangent to the circle centered at O1 (or O2) through point Q by the method described in Problem 2 (Part 1 of Monge's Theorem Investigation). This line will be an external tangent to the circles.
Consider any three circles in the plane with different radii. For each pair of circles the external tangents intersect at a point. The three points of intersection are collinear.
Let d1 and d2 denote dilations centered at C1 and C2 respectively such that the scaling factors of the dilations are not reciprocals of each other. Then the composition of d1 followed by d2 is a dilation, call it d12. Take any point P on the line C1C2. Dilating P by d1 will give an image P' which is collinear with P and C1. This was shown in the first part of Problem 5 (Part 4 of Monge's Theorem Investigation). In other words P' is on line C1C2. Likewise, dilating P' by d2 will give an image P'' which is is on line C1C2. So dilating P by d12 gives the image P'' which is is on line C1C2.
General Sketchpad questions related to Monge's Theorem.
![[HOME]](/pix/home.gif)