Figure 9.1: Menelaus’ Theorem

9 Harmonic Points

9.1 The Theorems and Menelaus and Ceva

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 segmentsas they occur on a closed curve circumventing the triangle: AX,XB,BY, Y C,CZ andfinally ZA. Note this curve can be drawn without lifting the pencil. That orderingunderstood, one gets

AX

XB· BYY C

· CZZA

= −1

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 commonvertex Y and opposite parallel side GC and XB yield the proportion

(9.1)|GC||XB| =

|Y C||Y B|

Similar triangles with common vertex Z and opposite parallel side GC and XA yieldthe proportion

(9.2)|GC||XA| =

|ZC||ZB|

Dividing (9.1) over (9.2) yields

|XA||XB| =

|Y C||Y B| · |ZB|

|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 transversalsare drawn from the vertices to one of the partition points of the opposite side, in such away to obtain a smaller equilateral triangle in the middle. Assume that the area of theoriginal triangle is 21. Find the area of the triangle in the middle. Which fraction ofthe 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 tothe 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 vertexB, we enumerate the six segments along the sides of the triangle cut by the transversaland obtain

BU

UC· CVV D

· DAAB

= −1

The first and third ratio are given. The second ratio is y+zz

because of the areas of thetriangles left to the side CD. Hence we obtain

1

2· y + zz

· −1

3= −1

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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 = 5and 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 pointsX and Y divide segment AB from inside and outside in the same ratio. In other words,the cross ratio

(AB,XY ) =AX · BYBX · AY = −1

for the directed segments.

Remark. To remember the correct appearance of the points on top and bottom of thefraction, 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 otherpair of opposite sides in point Y . Let the diagonals intersect in point M . Let the lineYM intersect side AB in point X ′. Then the four points A,B,X,X ′ are harmonicpoints.

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Figure 9.4: The Theorem of Ceva

Theorem 9.3 (Theorem of Ceva). 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 oppositesides in the three points X, Y, Z. One orders the six segments XA,XB, Y B, Y C, ZCand ZA such that their midpoints occur in that order on a 360◦ around the triangle.That ordering understood, one gets

|XA||XB| · |Y B|

|Y C| · |ZC||ZA| = 1

Proof. take triangle �ABC from the figure of the theorem of Ceva, and let the two lineAY and BZ intersect in point M .

Now draw line ZY and use it as a transversal for the theorem of Menelaus. LetX ′ be the intersection of the transversal with the extended side AB. The theorem ofMenelaus yields

|X ′A||X ′B| · |Y B|

|Y C| · |ZC||ZA| = −1

One needs still to draw line CM . Let X ′ be its intersection point with line AB. Onegets harmonic quadrilateral based on �ABY Z. Hence A,B,X,X ′ are four harmonicpoints.

(AB,XX ′) =|AX| · |BX ′||BX| · |AX ′| = −1

The two equations together imply

|XA||XB| · |Y B|

|Y C| · |ZC||ZA| = 1

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Figure 9.5: The harmonic quadrilateral

as to be shown.

Problem 9.2. Use the Theorem of Ceva to prove: The three segments from the verticesof 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 lyingon one line, say in the order A ∗X ∗B ∗ Y . Let point P be an intersection point of thetwo circles.

The three angles ∠APX,∠XPB and ∠BPY are 45◦ if and only if the two circlesintersect orthogonally.

Proof. In the figure on page 519, the angles at vertices A and O are α and 2α, thelatter being an exterior angle at the top of an isosceles triangle. Similarly, the anglesat vertices Y and Q are y and 2y, respectively. The angle sum in the triangles �OPQand �APY yield

∠OPQ = 180◦ − 2α− 2y

∠APY = 180◦ − α− y∠OPQ− 90◦ = 2(∠APY − 135◦)

Hence ∠OPQ is a right angle if and only if α+ y = 45◦ if and only if ∠APY = 135◦. Inthis case, the two circles intersect perpendicularly. Too, all three angles ∠APX,∠XPBand ∠BPY are 45◦ since by Thales’ theorem ∠XPY = ∠APB = 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−→TY bisects the angle ∠UTV . 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 Cand G.

Figure 9.11: The points on Apollonius circle have all same ratio of distances from A andB, and the segments AX and BX appear under equal angles.

For any four harmonic points A ∗ X ∗ B ∗ Y , a circle with the harmonic divisionpoints 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 thedivision point X, Y of segment AB have the same ratio of distance from A and B. Fromany point on the Apollonius circle, the segments AX and BX appear under congruentangles.Let now C be any point not on line AB. We mark the three adjacent angles u, v, v′ withvertex C and points A,X,B, Y on their sides, and finally the angle u′, such that thefour 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 congruentangles.

(1’) u′ ∼= v′: the two segments AY and BY are seen from C appearing under supple-mentary angles. 43

(2)|AC||BC| =

|AX||BX|

(2’)|AC||BC| =

|AY ||BY |

(3) Point C lies on the Thales’ circle over the harmonic division points X and Y .

Lemma 9.1.

|AC||BC| =

sin v

sin u

|AX||BX|(9.3)

|AC||BC| =

sin v′

sin u′|AY ||BY |(9.4)

(AB,XY ) =AX · BYBX · AY = −sin u · sin v′

sin v · sin u′(9.5)

Proof. Let x = ∠AXC and y = ∠AY C. We use the sin theorem several times and get

|AC|sin x

=|AX|sin u

(9.6)

|BC|sin x

=|BX|sin v

(9.7)

|AC|sin y

=|AY |sin u′

(9.8)

|BC|sin y

=|BY |sin v′

(9.9)

Note we have used that the sin of supplementary angles are equal. By division we getformulas as claimed.

43more 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

|AC||BC| =

|AX||BX|

”(2) → (1)”: Under the latter assumption, formula (9.3) yields, sinu = sin v. Sinceu + v < 2R, the two angles u and v cannot be supplementary. Hence we get u = v, asto 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 areperpendicular to each other. By the converse Thales’ theorem 32, the vertex C lies onthe circle with diameter XY , which is an Apollonius circle.

”(3) → (1)”: Assume that point C lies on an Apollonius circle with diameter XY . ByThales’ theorem, angle ∠XCY is a right angle, and hence u + u′ and v + v′ are rightangles. From formulas 9.3 and 9.4 we conclude

sin v

sin u=

sin v′

sin u′=

cos v

cosu(9.10)

tan u = tan v(9.11)

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 pointsA ∗X ∗B given. Use a Thales circle over diameter AB. Confirm that the constructionis valid.

Construction. Let the given segment XY be a diameter of circle δ. At the given point Uinside the segment XY , we erect the perpendicular, and let S and T be the intersectionpoints of the perpendicular with this circle. The tangent at T intersect the line XY atthe forth harmonic point V .

Reason of validity. To check that X, Y, U, V are harmonic points we prove that the ray−→TY bisects the angle ∠UTV . Indeed, by Euclid III.32, the angle between a chord andthe tangent at its endpoint is congruent to the circumference angle of this chord. Hence∠STY ∼= ∠Y TV . The converse argument is easily checked, too.

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Figure 9.12: Harmonic points XY,UV from Apollonius circle.

Remark. As follows easily from the leg theorem, U and V are inverse images of eachother. The circle with diameter UV contains a pair of inverse points. As well known(from the basics for Poincare’s disk model) it is thus orthogonal to the circle δ.

Furthermore, the �V TS has Y as center of its in-circle, and X as center of itsex-circle. Once more, this implies that (XY,UV ) = −1.

Proposition 9.2. Given are two circles which have the diameters AB and XY lyingon one line, say in the order A ∗X ∗B ∗ Y . Let point P be an intersection point of thetwo circles.

(AB,XY ) = − tan2∠APX(9.12)

Corollary 49 (Thales’ second donkey). Equivalent are:

(a) Two circles intersect perpendicularly.

(b) the four rays−→PY ,

−→PA,

−−→PX,

−−→PB 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 highschool math teacher.

9.3 An application to electrostatics

Problem 9.6. Near a conducting sphere of diameter AB = 3m, a point charge ofQ Coulomb is placed, say at point Y with distance BY = 1m 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 amirror charge.

sphere. We want to calculate the attraction force between the point charge and thesphere.

On any conducting boundary, the total potential has to be constant. To assure thepotential to vanish on the surface of the conducting sphere, we determine an appropriatemirror charge. The electric field is equal to the field produced by the given point chargetogether 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 bound-ary condition

Q

|CY | +Q′

|CX| = 0

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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 pointX.

(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

|AX||BX| =

|AY ||BY | = 4

since |AY | = |AB| + |BY | = 4m and |BY | = 1m are given. With |AB| = 3m,we get

|AX| = 4

53m and |BX| = 1

53m

(b) The total potential from the two charges needs to satisfy the boundary conditionfor any point C on the conducting surface. This turns out to be achievable becauseof Apollonius theorem. Indeed, with any point C on the conducting sphere, weget

−Q′

Q=

|CX||CY | =

|BX||BY | =

3

5

(c) The distance of the charge to the mirror charge is

|XY | =(3

5+ 1

)m =

8

5m

We get the attraction force F from Coulomb’s law

F =Q′Q

|XY |2 = −3

5

Q2

(8/5)2= −15Q2

64Coulombm−2

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9.4 The perspective view

Figure 9.15: Albrecht Durer 1530

We want to construct the perspective view of objects, lying mainly in a horizontalxy-plane as they appear in a vertical xz-plane. I call the horizontal xy-plane the objectplane and the vertical xz-plan the picture plane.

Problem 9.7. Suppose you know that the given quadrilateral is a perspective image ofa 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 aright 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, theopposite 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 hori-zon. 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 abovethe segments GH and IJ .

The light rays from the points G and H on the horizon (or points nearby, inreality) enter the viewer’s eye at right angles. Hence the viewer’s eye lies on aThales circle with diameter GH.

Similarly, light rays from the points I and J on the horizon enter the viewer’s eyeat 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 planefrom its perspective image in the picture plane. Reconstruct the horizon, the position ofthe 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 perspectiveimage.

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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 squareis constructed in the perspective view.

Why can there no contradictions arise? Give an example of the little Theorem ofDesargues 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 Carte-sian coordinates. The eye of the viewer is positioned behind the picture plane at pointE, say with coordinates (0,−f, h). The length f is the distance of the eye from thepicture plane. In the set up of photography it is the focus length. The coordinate h isdetermined 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′) onthe picture plane. The three points E,P and Q lie on the line given by the light rayfrom the object to the eye of the viewer. From the parametric representation of a linein three dimensional space, we get

Q = tE + (1 − t)P⎡⎣x

′

0z′

⎤⎦ = t

⎡⎣ 0

−fh

⎤⎦+ (1 − t)

⎡⎣xyz

⎤⎦

with a real parameter t varying along the line.

Problem 9.10. Determine parameter t as a function of y. Get the coordinates x′ andz′ as a function of x, y, z.

Answer. The second component of the system gives 0 = −tf + (1 − t)y. Hence t = yf+y

and 1 − t = ff+y

. We plug this solution into the equations for the first and thirdcomponent to get

x′ = (1 − t)x =fx

f + y

z′ = th+ (1 − t)z = hy + fz

f + y

Problem 9.11. Use the result

x′ =fx

f + y

z′ =hy + fz

f + y

to determine the image of a ray y = mx, z = 0 with x ≥ 0. Which point Q∞ on thehorizon do you get in the limit x → ∞.

Answer. The image of the given ray, and the point on the horizon are

x′ =fx

f +mxlimx→∞

x′ =f

m

z′ =hmx

f +mxlimx→∞

z′ = h

Hence the line z′ = h is the horizon.

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