Download 9 Harmonic Points

Figure 9.1: Menelaus’ Theorem
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9.1
Harmonic Points
The Theorems and Menelaus and Cevá
Theorem 9.1 (Theorem of Menelaus). Let a transversal cut the three sides of
ABC, extended if necessary, in the three points X, Y, Z. One orders the six segments
as they occur on a closed curve circumventing the triangle: AX, XB, BY, Y C, CZ and
finally ZA. Note this curve can be drawn without lifting the pencil. That ordering
understood, one gets
AX BY CZ
·
·
= −1
XB Y C ZA
Proof of Menelaus’ Theorem. Draw the parallel to triangle side AB through vertex C.
Let G be its intersection point with the transversal. Similar triangles with common
vertex Y and opposite parallel side GC and XB yield the proportion
(9.1)
|Y C|
|GC|
=
|XB|
|Y B|
Similar triangles with common vertex Z and opposite parallel side GC and XA yield
the proportion
(9.2)
|GC|
|ZC|
=
|XA|
|ZB|
Dividing (9.1) over (9.2) yields
|XA|
|Y C| |ZB|
=
·
|XB|
|Y B| |ZC|
Now rearranging yields the assertion.
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Figure 9.2: The area of the middle triangle is one seventh of the total area.
Problem 9.1. All three sides of an equilateral triangle are trisected. Three transversals
are drawn from the vertices to one of the partition points of the opposite side, in such a
way to obtain a smaller equilateral triangle in the middle. Assume that the area of the
original triangle is 21. Find the area of the triangle in the middle. Which fraction of
the total area is covered by the middle triangle?
Answer. Since the total area is given to be 21, the areas of the three triangles adjacent to
the left side add up to 7. Hence the areas x of the middle triangle, y of the quadrilaterals,
and z of the small triangles in the corners satisfy the two equations
x + 3y + 3z = 21
y + 2z = 7
A third equation among the three unknowns x, y, z can be obtained from Menelaus’
theorem. We use the left triangle BCD and the transversal AU . Starting at vertex
B, we enumerate the six segments along the sides of the triangle cut by the transversal
and obtain
BU CV DA
·
·
= −1
U C V D AB
because of the areas of the
The first and third ratio are given. The second ratio is y+z
z
triangles left to the side CD. Hence we obtain
1 y + z −1
·
·
= −1
2
z
3
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Figure 9.3: What is the area of the triangle in the middle?
and hence y+z = 6z. Plugging y = 5z into the first two equations, we obtain z = 1, y = 5
and finally x = 3. Hence the area of the middle triangle is one seventh of the total area.
Definition 9.1. Four points A, B, X, Y on a line are called harmonic, if the two points
X and Y divide segment AB from inside and outside in the same ratio. In other words,
the cross ratio
AX · BY
(AB, XY ) =
= −1
BX · AY
for the directed segments.
Remark. To remember the correct appearance of the points on top and bottom of the
fraction, you can use the diagram
A B X→Y
↓ ↑
B A X→Y
Theorem 9.2 (Theorem of the harmonic quadrilateral). Take any quadrilateral
ABCD. Let a pair of opposite sides AB and CD intersect in point X, and the other
pair of opposite sides in point Y . Let the diagonals intersect in point M . Let the line
Y M intersect side AB in point X . Then the four points A, B, X, X are harmonic
points.
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Figure 9.4: The Theorem of Cevá
Theorem 9.3 (Theorem of Cevá). Let M be a point inside (or outside) of triangle
ABC. Let the three lines through M and the vertices of the triangle cut the opposite
sides in the three points X, Y, Z. One orders the six segments XA, XB, Y B, Y C, ZC
and ZA such that their midpoints occur in that order on a 360◦ around the triangle.
That ordering understood, one gets
|XA| |Y B| |ZC|
·
·
=1
|XB| |Y C| |ZA|
Proof. take triangle ABC from the figure of the theorem of Cevá, and let the two line
AY and BZ intersect in point M .
Now draw line ZY and use it as a transversal for the theorem of Menelaus. Let
X be the intersection of the transversal with the extended side AB. The theorem of
Menelaus yields
|X A| |Y B| |ZC|
·
·
= −1
|X B| |Y C| |ZA|
One needs still to draw line CM . Let X be its intersection point with line AB. One
gets harmonic quadrilateral based on ABY Z. Hence A, B, X, X are four harmonic
points.
|AX| · |BX |
(AB, XX ) =
= −1
|BX| · |AX |
The two equations together imply
|XA| |Y B| |ZC|
·
·
=1
|XB| |Y C| |ZA|
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Figure 9.5: The harmonic quadrilateral
as to be shown.
Problem 9.2. Use the Theorem of Cévà to prove: The three segments from the vertices
of a triangle to the touching points of its in-circle intersect at one point.
Answer.
9.2
The circle of Apollonius
Proposition 9.1. Given are two circles which have the diameters AB and XY lying
on one line, say in the order A ∗ X ∗ B ∗ Y . Let point P be an intersection point of the
two circles.
The three angles ∠AP X, ∠XP B and ∠BP Y are 45◦ if and only if the two circles
intersect orthogonally.
Proof. In the figure on page 519, the angles at vertices A and O are α and 2α, the
latter being an exterior angle at the top of an isosceles triangle. Similarly, the angles
at vertices Y and Q are y and 2y, respectively. The angle sum in the triangles OP Q
and AP Y yield
∠OP Q = 180◦ − 2α − 2y
∠AP Y = 180◦ − α − y
∠OP Q − 90◦ = 2(∠AP Y − 135◦ )
Hence ∠OP Q is a right angle if and only if α + y = 45◦ if and only if ∠AP Y = 135◦ . In
this case, the two circles intersect perpendicularly. Too, all three angles ∠AP X, ∠XP B
and ∠BP Y are 45◦ since by Thales’ theorem ∠XP Y = ∠AP B = 90◦ .
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Figure 9.6: The Gergonne point.
Figure 9.7: Three angles of 45◦ .
−→
Problem 9.3. Explain why the ray T Y bisects the angle ∠U T V . Find and mark in the
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Figure 9.8: How three angles of 45◦ are obtained.
Figure 9.9: Find angles α, 90◦ − α and 90◦ − 2α.
figure above: angle α in red, angles 90◦ − α in blue, angles 90◦ − 2α in green.
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Figure 9.10: The points on Apollonius circle have all same ratio of distances from points C
and G.
Figure 9.11: The points on Apollonius circle have all same ratio of distances from A and
B, and the segments AX and BX appear under equal angles.
For any four harmonic points A ∗ X ∗ B ∗ Y , a circle with the harmonic division
points X and Y as diameter XY , is called an Apollonius circle.
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Theorem 9.4 (Apollonius’ Theorem). All points on an Apollonius circle over the
division point X, Y of segment AB have the same ratio of distance from A and B. From
any point on the Apollonius circle, the segments AX and BX appear under congruent
angles.
Let now C be any point not on line AB. We mark the three adjacent angles u, v, v with
vertex C and points A, X, B, Y on their sides, and finally the angle u , such that the
four angles add up to two right angles. The following statements are equivalent:
(1) u ∼
= v: the two segments AX and XB are seen from C appearing under congruent
angles.
(1’) u ∼
= v : the two segments AY and BY are seen from C appearing under supplementary angles. 43
(2)
|AC|
|AX|
=
|BC|
|BX|
(2’)
|AY |
|AC|
=
|BC|
|BY |
(3) Point C lies on the Thales’ circle over the harmonic division points X and Y .
Lemma 9.1.
sin v |AX|
|AC|
=
|BC|
sin u |BX|
|AC|
sin v |AY |
(9.4)
=
|BC|
sin u |BY |
AX · BY
sin u · sin v (9.5)
=−
(AB, XY ) =
sin v · sin u
BX · AY
Proof. Let x = ∠AXC and y = ∠AY C. We use the sin theorem several times and get
(9.3)
(9.6)
(9.7)
(9.8)
(9.9)
|AC|
sin x
|BC|
sin x
|AC|
sin y
|BC|
sin y
|AX|
sin u
|BX|
=
sin v
|AY |
=
sin u
|BY |
=
sin v =
Note we have used that the sin of supplementary angles are equal. By division we get
formulas as claimed.
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more exactly, one has to say ”angles adding up to two right”
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”(1) → (2)”: Under the assumption u = v, formula (9.3) yields, as to be shown
|AX|
|AC|
=
|BC|
|BX|
”(2) → (1)”: Under the latter assumption, formula (9.3) yields, sin u = sin v. Since
u + v < 2R, the two angles u and v cannot be supplementary. Hence we get u = v, as
to be shown.
”(2) ↔ (2 )”: Follows at once from the assumption that A, B, X, Y are harmonic points.
We see with a similar argument that (1) ↔ (2 ). Hence all four statements (1),(1’),(2)
and (2’) are equivalent.
”(1) → (3)”: Assume u = v. By the reasoning above, we conclude that u = v . Hence
−−→
−−→
the rays CX and CY are interior and exterior angular bisectors of the angle ∠ACB.
As seen earlier (see section 9.6 about the in-circle and ex-circles), these bisectors are
perpendicular to each other. By the converse Thales’ theorem 32, the vertex C lies on
the circle with diameter XY , which is an Apollonius circle.
”(3) → (1)”: Assume that point C lies on an Apollonius circle with diameter XY . By
Thales’ theorem, angle ∠XCY is a right angle, and hence u + u and v + v are right
angles. From formulas 9.3 and 9.4 we conclude
(9.10)
(9.11)
sin v
cos v
sin v =
=
sin u
sin u
cos u
tan u = tan v
Both u and v are acute angles. Hence we conclude u = v, as to be shown.
Problem 9.4. Explain an easy construction of the forth harmonic point with three points
A ∗ X ∗ B given. Use a Thales circle over diameter AB. Confirm that the construction
is valid.
Construction. Let the given segment XY be a diameter of circle δ. At the given point U
inside the segment XY , we erect the perpendicular, and let S and T be the intersection
points of the perpendicular with this circle. The tangent at T intersect the line XY at
the forth harmonic point V .
Reason of validity. To check that X, Y, U, V are harmonic points we prove that the ray
−→
T Y bisects the angle ∠U T V . Indeed, by Euclid III.32, the angle between a chord and
the tangent at its endpoint is congruent to the circumference angle of this chord. Hence
∠ST Y ∼
= ∠Y T V . The converse argument is easily checked, too.
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Figure 9.12: Harmonic points XY, U V from Apollonius circle.
Remark. As follows easily from the leg theorem, U and V are inverse images of each
other. The circle with diameter U V contains a pair of inverse points. As well known
(from the basics for Poincaré’s disk model) it is thus orthogonal to the circle δ.
Furthermore, the V T S has Y as center of its in-circle, and X as center of its
ex-circle. Once more, this implies that (XY, U V ) = −1.
Proposition 9.2. Given are two circles which have the diameters AB and XY lying
on one line, say in the order A ∗ X ∗ B ∗ Y . Let point P be an intersection point of the
two circles.
(9.12)
(AB, XY ) = − tan2 ∠AP X
Corollary 49 (Thales’ second donkey). Equivalent are:
(a) Two circles intersect perpendicularly.
−→ −→ −−→ −−→
(b) the four rays P Y , P A, P X, P B from an intersection point P of the two circle make
three angles of 45◦ .
(c) Their two diameters on the line through their center are four harmonic points.
Problem 9.5. Prove this corollary, ready to be engraved on the tomb of my former high
school math teacher.
9.3
An application to electrostatics
Problem 9.6. Near a conducting sphere of diameter AB = 3 m, a point charge of
Q Coulomb is placed, say at point Y with distance BY = 1 m from the conducting
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Figure 9.13:
Figure 9.14: The attraction by a conducting sphere is equivalent to the attraction by a
mirror charge.
sphere. We want to calculate the attraction force between the point charge and the
sphere.
On any conducting boundary, the total potential has to be constant. To assure the
potential to vanish on the surface of the conducting sphere, we determine an appropriate
mirror charge. The electric field is equal to the field produced by the given point charge
together with the field of the mirror charge Q . Thus outside of the conducting sphere,
its effect is equivalent to the effect of the mirror charge.
The location and strength of the mirror charge is determined by means of the boundary condition
Q
Q
+
=0
|CY | |CX|
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which needs to hold for any point C on the surface of the sphere. Convince yourself that,
because of Apollonius theorem, the mirror charge Q is located at the forth harmonic point
X.
(a) Calculate the distances |AX| and |BX| for the forth harmonic point X.
(b) Use the boundary condition to determine the mirror charge.
(c) Calculate the attraction force F from Coulomb’s law
F =
Q Q
|XY |2
Answer. (a) The forth harmonic point satisfies
|AY |
|AX|
=
=4
|BX|
|BY |
since |AY | = |AB| + |BY | = 4 m and |BY | = 1 m are given. With |AB| = 3 m,
we get
4
1
|AX| = 3 m and |BX| = 3 m
5
5
(b) The total potential from the two charges needs to satisfy the boundary condition
for any point C on the conducting surface. This turns out to be achievable because
of Apollonius theorem. Indeed, with any point C on the conducting sphere, we
get
Q
|CX|
|BX|
3
− =
=
=
Q
|CY |
|BY |
5
(c) The distance of the charge to the mirror charge is
3
8
|XY | =
+1 m= m
5
5
We get the attraction force F from Coulomb’s law
F =
Q Q
3 Q2
15 Q2
Coulomb m−2
=
−
=
−
|XY |2
5 (8/5)2
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9.4
The perspective view
Figure 9.15: Albrecht Dürer 1530
We want to construct the perspective view of objects, lying mainly in a horizontal
xy-plane as they appear in a vertical xz-plane. I call the horizontal xy-plane the object
plane and the vertical xz-plan the picture plane.
Problem 9.7. Suppose you know that the given quadrilateral is a perspective image of
a square in a horizontal plane. We have to
• reconstruct the horizon
• reconstruct the position of the viewer’s eye.
To this end, we use the half-plane above the horizon and assume it to be rotated by a
right angle around the horizon towards the viewer, out of the vertical picture plane.
Answer.
• The extensions of the opposite sides AB and CD meet at point G, the
opposite sides BC and DA meet at point H. The line GH is the horizon.
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Figure 9.16: From were does a square look like this?
• The position of the viewer’s eye is reconstructed in the half-plane above the horizon. The diagonals BD and AC intersect the horizon in the points I and J,
respectively. The viewer’s eye lies at the intersection of the Thales’ circles above
the segments GH and IJ.
The light rays from the points G and H on the horizon (or points nearby, in
reality) enter the viewer’s eye at right angles. Hence the viewer’s eye lies on a
Thales circle with diameter GH.
Similarly, light rays from the points I and J on the horizon enter the viewer’s eye
at right angles. Hence the viewer’s eye lies on a Thales circle with diameter IJ,
too.
Problem 9.8. As in problem 9.7, we reconstruct a square in a horizontal object plane
from its perspective image in the picture plane. Reconstruct the horizon, the position of
the viewer’s eye, and finally the object square in the object plane.
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Figure 9.17: Reconstruction of the position of the eye.
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Figure 9.18: Reconstruction of the eye position and the object square from its perspective
image.
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Figure 9.19: Tiling of a plane.
Problem 9.9. Explain how the tiling of the plane with translations of the original square
is constructed in the perspective view.
Why can there no contradictions arise? Give an example of the little Theorem of
Desargues applied to two triangles, which shows that an incidence holds as expected.
Figure 9.20: The mapping by perspective in coordinates.
The following problem gives the mapping of perspective in three-dimensional Cartesian coordinates. The eye of the viewer is positioned behind the picture plane at point
E, say with coordinates (0, −f, h). The length f is the distance of the eye from the
picture plane. In the set up of photography it is the focus length. The coordinate h is
determined by the height of the eye above the object plane.
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For any given point P = (x, y, z), we want to determine its image Q = (x , 0, z ) on
the picture plane. The three points E, P and Q lie on the line given by the light ray
from the object to the eye of the viewer. From the parametric representation of a line
in three dimensional space, we get
Q = tE + (1 − t)P
⎡ ⎤
⎡ ⎤
⎡ ⎤
x
x
0
⎣ 0 ⎦ = t ⎣−f ⎦ + (1 − t) ⎣y ⎦
z
h
z
with a real parameter t varying along the line.
Problem 9.10. Determine parameter t as a function of y. Get the coordinates x and
z as a function of x, y, z.
y
Answer. The second component of the system gives 0 = −tf + (1 − t)y. Hence t = f +y
f
and 1 − t = f +y
. We plug this solution into the equations for the first and third
component to get
fx
f +y
hy + f z
z = th + (1 − t)z =
f +y
x = (1 − t)x =
Problem 9.11. Use the result
fx
f +y
hy + f z
z =
f +y
x =
to determine the image of a ray y = mx, z = 0 with x ≥ 0. Which point Q∞ on the
horizon do you get in the limit x → ∞.
Answer. The image of the given ray, and the point on the horizon are
fx
f + mx
hmx
z =
f + mx
x =
lim x =
x→∞
f
m
lim z = h
x→∞
Hence the line z = h is the horizon.
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