MST Topics in History of Mathematics Euclid’s Elements and the Works of Archimedes

Paul Yiu

Department of Mathematics Florida Atlantic University

Summer 2014

July 9 7

Miquel’s theorem

A

Z

Y

P

B X C

A

Z

Y

P

B X C 8

Fermat point

(1) For an arbitrary point P on the minor arc BC of the circumcircle of an equilat- eral triangle ABC, AP = BP + CP. A

Q

B C

P

Proof. If Q is the point on AP such that PQ = PC, then triangle CPQ is equilat- eral since . Note that ∠ACQ = ∠BCP. Thus, ACQ ≡BCP by the test. From this, AQ = BP, and AP = AQ + QP = BP + CP. (2) Consider a triangle ABC whose angles are all less than 120◦. Construct equilateral triangles BCX, CAY , and ABZ outside the triangle. 9

Theorem (Fermat point). If equilateral triangles BCX, CAY , and ABZ are erected on the sides of triangle ABC, then the lines AX, BY , CZ concur at a point F called the Fermat point of triangle ABC. If the angles of triangle ABC are all less than 120◦ (so that F is an interior point), then the segments AX, BY , CZ have the same length AF + BF + CF.

Y

A

Z

F

B C

X Let F be the intersection of the circumcircles of the equilateral triangles. Note that ∠CFA = ∠AF B = 120◦. It follows that ∠AF B = 120◦, and F lies on the circumcircle of the equilateral triangle BCX as well. Therefore, ∠XFC = ∠XBC =60◦, and ∠XFC + ∠CFA = 180◦. The three points A, F , X are collinear. Thus, AX = AF + FX = AF + BF + CF by (1) above. Similarly, B, F , Y are collinear, so do C, F , Z, and BY , CZ are each equal to AF + BF + CF. 10

Napoleon’s Theorem

Theorem (Napoleon). If equilateral triangles BCX, CAY , and ABZ are erected on the sides of triangle ABC, the centers of these equilateral triangles form another equilateral triangle.

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A

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B C

X Chapter 6

Euclid’s Elements, Book IV

A 2 Euclid’s Elements, Book IV

6.1 Definitions

Definitions. (IV.1). A rectilineal figure is said to be inscribed in a rectilineal figure when the respective angles of the inscribed figure lie on the respective sides of that in which it is inscribed. (IV.2). Similarly a figure is said to be circumscribed about a figure when the respective sides of the circumscribed figure pass through the respective angles of that about which it is circumscribed. 6.1 Definitions 3

Definitions. (IV.3). A rectilineal figure is said to be inscribed in a circle when each angle of the inscribed figure lies on the circumference of the circle. (IV.4). A rectilineal figure is said to be circumscribed about a circle when each side of the circumscribed figure touches the circumference of the circle.

Definitions. (IV.5). Similarly a circle is said to be inscribed in a figure when the circumference of the circle touches each side of the figure in which it is inscribed. (IV.6). A circle is said to be circumscribed about a figure when the circumference of the circle passes through each angle of the figure about which it is circumscribed.

Definition (IV.7). A straight line is said to be fitted into a circle when its extremi- ties are on the circumference of the circle. 4 Euclid’s Elements, Book IV

6.2 Some basic constructions

Euclid (IV.1). Into a given circle to fit a straight line equal to a given straight line which is not greater than the diameter of the circle. 6.2 Some basic constructions 5

Euclid (IV.2). In a given circle to inscribe a triangle equiangular with a given triangle.

B E

F

C

D G

A H 6 Euclid’s Elements, Book IV

Euclid (IV.3). About a given circle to circumscribe a triangle equiangular with a given triangle. (IV.4). In a given triangle to inscribe a circle.

A

G

E I

B F C 6.2 Some basic constructions 7

Euclid (IV.5). About a given triangle to circumscribe a circle.

A A A D E

D E B C

D E

B C I I I

B C

Corollary. When the center of the circle falls within the triangle, the triangle is acute-angled; when the center falls on a side, the triangle is right-angled; and when the center of the circle falls outside the triangle, the triangle is obtuse-angled. 6

Simson line theorem

Z

A P

Y

O

B X C

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A P

Y

O

B X C 5

The orthocenter

Theorem. The three altitudes of a triangle are concurrent.

C A B

Y

Z H

B X C

A

Proof. Let ABC be a given triangle. Through each vertex construct a line parallel to its opposite side. These three parallel lines bound a larger triangle ABC. Note that ABCB and ACBC are both parallelograms since each has two pairs of parallel sides. It follows that BA = BC = AC and A is the midpoint of BC.

Consider the altitude AX of triangle ABC. Seen in triangle ABC, this line is the perpendicular bisector of BC since it is perpendicular to BC through its midpoint A. Similarly, the altitudes BY and CZ of triangle ABC are perpendicular bisectors of CA and AB. As such, the three lines AX, BY , CZ concur at a point H. This is called the orthocenter of triangle ABC. 8 Euclid’s Elements, Book IV

6.3 Construction of square and regular hexagon

Euclid (IV.6). In a given circle to inscribe a square. (IV.7). About a given circle to circumscribe a square.

A G A F

E I B D B D

C K C H Construct two perpendicular diameters. The endpoints of the diameters are the vertices of a square inscribed in the circle. (IV.6). The perpendiculars at these endpoints to the respective diameters bound a square which circumscribes the circle (IV.7). 6.3 Construction of square and regular hexagon 9

Euclid (IV.8). In a given square to inscribe a circle. (IV.9). In a given square to circumscribe a circle. 10 Euclid’s Elements, Book IV

Euclid (IV.15). In a given circle to inscribe an equilateral and equiangular hexagon.

D

C E

G

B F

A 6.4 Construction of regular pentagon 11

6.4 Construction of regular pentagon

Euclid (IV.10). To construct an isosceles triangle having each of the angles at the base double of the remaining one.

B

C

D A

E

Given: A segment AB. To construct: Triangle ABD such that ∠B = ∠D =2∠A. Construction: Construct (by II.11) a point C on AB such that the square on AC is equal to the rectangle contained by AB and CB, a point D such that AD = AB and BD = CA. To prove: In triangle ABD, ∠B = ∠D =2∠A. Proof : Since AB = AD, ∠B = ∠D. The rectangle contained by BA and BC is equal to the square on BD (by construc- tion). Therefore, BD is tangent to the circle ACD at D. (III.37) From this, ∠DAC = ∠CDB, (III.32) ∠BCD = ∠CAD + ∠ADC = ∠CDB + ∠ADC = ∠ADB = ∠ABD. Therefore, CD = BD = CA, and ∠ADC = ∠DAC. Therefore, ∠D = ∠ADC + ∠CDB =2∠A. 12 Euclid’s Elements, Book IV

Euclid (IV.11). In a given circle to inscribe an equilateral and equiangular pen- tagon. (IV.12). About a given circle to circumscribe an equilateral and equiangular pen- tagon. (IV.13). In a given pentagon, which is equilateral and equiangular, to inscribe a circle. (IV.14). About a given pentagon, which is equilateral and equiangular, to circum- scribe a circle.

A

F

B E

G H

C D 6.4 Construction of regular pentagon 13

Euclid (IV.16). In a given circle to inscribe a fifteen-angled figure which shall be both equilateral and equiangular.

A

B

E

C D

Corollary to IV.16. And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be cir- cumscribed about the circle a fifteen-angled figure which is equilateral and equian- gular.