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March 12th, 2017, 07:12 PM   #1
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Examine the special relationship between the single and triple angles of an arbitrary

Examine the special relationship between the single and triple angles of an arbitrary angle to infer the feasibility of using classical geometric construction to trisect an angle

De-Shi Chiu

Chu Dong Junior High School, Hsinchu County 306, Taiwan (R.O.C)

Corresponding author: De-Shi Chiu, Chu Dong Junior High School, No. 8, Ln. 308, Shigangzi, Guanxi Township, Hsinchu County 306, Taiwan (R.O.C). Tel.: +886-937-517-503
(作者:邱得時 中華民國 台灣 30645新竹縣關西鎮石光里308巷8號)
E-mail: cds12345678@yahoo.com.tw

Abstract. Examine the special relationship between the single and triple angles of an arbitrary angle to infer the feasibility of using classical geometric construction to trisect an angle.
Trisecting a line segment by using classical geometric construction was previously believed to be impossible because 1 cannot be perfectly divided by 3. However, using classical geometric construction to trisect a line segment is feasible. Hence, conclusions vary depending on the logic used to infer the proposition.
Although French mathematician Pierre Wantzel proved that trisecting an arbitrary angle by using only a compass and straightedge was impossible, a different logic can be applied to infer the feasibility of the classical method. Specifically, the special relationship between the single and triple angles of an arbitrary angle can be used to infer that classical geometric construction can be used to trisect any angle. The inference method is as follows. First, find (prove) the properties (special relationship) between the single and triple angles of the arbitrary angle. Then draw an auxiliary graph for the classical geometric construction to find (prove) its properties (special relationship). Finally, use the properties of the single and triple angles of the arbitrary angle and the properties of the auxiliary graph to prove the feasibility of the classical geometric construction, namely, that an arbitrary angle can be trisected using only a compass and straightedge.

Mathematics Subject Classification.
51 GEOMETRY
Keywords.
Classical; geometry; construction trisect; angle.

Content to be proved:

Derive the particular features (particular relationships) of an angle and its triple angle.
Construct an auxiliary graph (following a classical geometric construction) and demonstrate its particular features.
Integrate the particular features of an angle and its triple angle as well as those of the auxiliary graph, prove the existence of common points between the two types of graph, and apply the feature of common points in the classical geometric construction for trisecting a given angle of 0°−180°.

Construction

Current studies refute that an angle can be trisected using geometric construction because no studies have been found to be capable of doing so. However, an arbitrary angle ∠An has three equal angles ∠Bn (=3∠An). Hence, it cannot be refuted that three equal angles∠An must exist for∠Bn. This fact cannot be overlooked and is also the key basis of my paper.
The innovativeness of this paper can be ascertained from geometrical perspectives as follows:

Explains the peculiar relationship between an arbitrary angle and its three equal angles.
Construction 1 is consistent with the auxiliary figures of classical geometric constructions and illustrates the characteristics of auxiliary figures.
The characteristics of auxiliary figures corresponding to the peculiar relationship between an arbitrary angle and its three equal angles are determined, and a geometric construction of an angle trisection that adheres with classical geometric construction is constructed.

The geometric construction for trisecting a given angle AOB (0° ≤ ∠AOB ≤ 180°) is as follows (Figure 1):

Given an angle AOB (0° ≤ ∠AOB ≤ 180°).
Construct a straight line L that passes through point O and bisects ∠AOB.
Construct a straight line M that passes through point O and is perpendicular to line L.
Let point 1 on M ⃡ make ¯O1 = ¯OA = ¯OB.
Let point 2 on L ⃡ make ¯21 = 2¯O1.
Construct ¯2B (connecting point 2 and point B) that intersects with M ⃡ at point 6.
Let point 7 on ¯2B make ¯27 = ¯21.
Let point 8 on L ⃡ make ¯67 = ¯28.
Draw ¯8A and ¯8B to make ∠A8B = 1/3 ∠AOB.
With the center at point O and radius as ¯OA (= ¯O1), draw an arc AB that intersects with ¯OB at point B. Construct a (OC) ⃗ parallel to ¯8A and intersecting with arc AB at point C, and construct a (OD) ⃗ parallel to ¯8B and intersecting with arc AB at point D. Subsequently, ∠AOC = ∠COD = ∠DOB = 1/3 ∠AOB.

Particular features (particular relationships) between an angle and its triple angle

When ∠BoOBn = 3∠BoCnBn, the specific features (i.e., specific relationship) between the two angles are as follows (Figure 2):

Construct a straight line L perpendicular to and intersecting with straight line M at point O. Given a point Bo on L ⃡, with point O as center and (〖OB〗_o ) ̅ as radius, construct an arc that intersects with L ⃡ at point Bo and with M ⃡ at point B90. Given a point Bn on arc B0B90 and given a point Cn on L ⃡, draw (C_n B_n ) ̅ that intersects with M ⃡ at point En and construct (〖OB〗_n ) ̅ (Figure 2).

If ∠BoOBn = 3∠BoCnBn, then (C_n E_n ) ̅ = 2(OB_o ) ̅; if (C_n E_n ) ̅ = 2(OB_o ) ̅, then ∠BoOBn = 3∠BoCnBn, where (OB_o ) ̅ = (OB_n ) ̅.

If ∠BoOBn = 3∠BoCnBn, then (C_n E_n ) ̅ = 2(OB_o ) ̅
Proof. Locate the circumcenter point Fn of right △CnOEn (on the hypotenuse (C_n E_n ) ̅). Because point Fn is the circumcenter of right △CnOEn, draw (OF_n ) ̅ and (OB_n ) ̅.

Given (OF_n ) ̅ = (F_n C_n ) ̅ = (F_n E_n ) ̅ = 1/2 (C_n E_n ) ̅, then (C_n E_n ) ̅ = 2(OF_n ) ̅ = 2(F_n C_n ) ̅.
According to a., (OF_n ) ̅ = (F_n C_n ) ̅ in △CnFnO, and therefore, ∠OCnFn = ∠FnOCn.
According to b., △CnOEn, △CnFnO, and Figure 2, ∠OFnEn = ∠OCnFn + ∠FnOCn = 2∠OCnFn, and therefore, ∠OFnEn = 2∠OCnFn = 2∠BoCnBn.
According to Figure 2 and the known equation ∠BoOBn = 3∠BoCnBn, ∠BoOBn = 3∠BoCnBn = ∠BoCnBn + ∠OBnCn. Subsequently, ∠OBnCn = 2∠BoCnBn.
Based on c. and d., in △FnOBn, ∠OFnEn = 2∠OCnFn = 2∠BoCnBn, and ∠OBnCn = 2∠BoCnBn. Therefore, 2∠BoCnBn = ∠OFnEn = ∠OBnCn, namely 2∠BoCnBn = ∠OFnBn = ∠OBnFn. Subsequently, (OF_n ) ̅ = (OB_n ) ̅ = (OB_o ) ̅.
According to a. and e., (C_n E_n ) ̅ = 2(OF_n ) ̅= 2(F_n C_n ) ̅, and (OF_n ) ̅ = (OB_n ) ̅ = (OB_o ) ̅. Subsequently, (C_n E_n ) ̅ = 2(OF_n ) ̅ = 2(OB_n ) ̅ = 2(OB_o ) ̅, and therefore, (C_n E_n ) ̅ = 2(OB_o ) ̅. (proven)

Consequently, if ∠BoOBn = 3∠BoCnBn, then the equation of (C_n E_n ) ̅ = 2(OB_o ) ̅ is true.

If (C_n E_n ) ̅ = 2(OB_o ) ̅, then ∠BoOBn = 3∠BoCnBn
Proof. Given that point Fn is the circumcenter of right △CnOEn (on the hypotenuse (C_n E_n ) ̅):

(OF_n ) ̅ = (F_n C_n ) ̅ = (F_n E_n ) ̅ = 1/2 (C_n E_n ) ̅ = (OB_o ) ̅ = (OB_n ) ̅, and thus, (C_n E_n ) ̅ = 2(OF_n ) ̅ = 2(F_n C_n ) ̅.
In △CnFnO, according to a. and Figure 2, (OF_n ) ̅ = (F_n C_n ) ̅. Therefore, ∠OCnFn = ∠FnOCn = 1/2∠OFnEn = 1/2∠OFnBn, and ∠BoCnBn = ∠OCnFn = 1/2∠OFnEn = 1/2∠OFnBn. Thus, 2∠BoCnBn = ∠OFnBn.
According to △FnOBn , a., and b., (OF_n ) ̅ = (F_n C_n ) ̅ = (F_n E_n ) ̅ = 1/2 (C_n E_n ) ̅ = (OB_o ) ̅ = (OB_n ) ̅, and thus, (OF_n ) ̅ = (OB_n ) ̅. Therefore, ∠OBnFn = ∠OFnBn = 2∠BoCnBn.
According to c., and Figure 2, ∠BoOBn = ∠BoCnBn + ∠OBnCn = 2∠BoCnBn + ∠BoCnBn. That is, ∠BoOBn = 3∠BoCnBn. (proven)

Therefore, if (C_n E_n ) ̅ = 2(OB_o ) ̅, then ∠BoOBn = 3∠BoCnBn is true.

Continuing the argument in 1., if ∠BoOBn = 3∠BoCnBn, then (C_n E_n ) ̅ = 2(OBo) ̅; if (C_n E_n ) ̅ = 2(OBo) ̅, then ∠BoOBn = 3∠BoCnBn, where (OBo) ̅ = (OBn) ̅. The equation ∠BoOBn = 3∠BoCnBn can be expressed as follows (0° = ∠BoOBn ≤ ∠BoOBn = 3∠BoCnBn ≤ ∠BoOB90 = 90°; Figure 3):

Draw a straight line L perpendicular to and intersecting with a straight line M at point O. Given a point Bo on L ⃡, with the center at point O and radius as (OB_o ) ̅, construct an arc intersecting with L ⃡ at points Bo and b0 and with M ⃡ at point B90. Mark a point En on M ⃡ and a point Cn on L ⃡ to make (C_n E_n ) ̅ = 2(OB_0 ) ̅. Maintain the length of (C_n E_n ) ̅ = 2(OB_o ) ̅ and adjust point En on (OB_90 ) ̅ and point Cn on L ⃡. Thereby, (C_n E_n ) ̅ intersects with arc b0B90 at point bn; when point En on (C_n E_n ) ̅ arrives at point B90 (i.e., point E90), making (C_90 B_90 ) ̅ = (C_90 E_90 ) ̅ = 2(OB_o ) ̅ and intersect with arc b0B90 at point b90. In this case, point Cn arrives at point C90. When point En arrives at point O (i.e., E0), (C_0 O) ̅ = (C_0 E_0 ) ̅ = 2(OB_o ) ̅ and intersects with arc b0B90 at point b0, and point Cn is equivalent to point C0. In addition, the angle turns from ∠OCnE0 (i.e., ∠OCnO) = 0° into ∠OCnE90 (i.e., ∠OCnB90) = 30°. That is, with a fixed segment (C_n E_n ) ̅ = 2(OB_0 ) ̅, point En moves on (OB_90 ) ̅ (i.e., (OE_90 ) ̅), and point Cn moves on the (C_0 C_90 ) ̅ on L ⃡. Let (C_n E_n ) ̅ be a ray(C_n E_n ) ⃗, where (C_n E_n ) ̅ = 2(OB_90 ) ̅. With point Cn moving on the (C_0 C_90 ) ̅ on L ⃡ and point En on (OB_90 ) ̅ (i.e., (OE_90 ) ̅), (C_n E_n ) ⃗ intersects with arc B0B90 at point Bn. If point En on (C_n E_n ) ⃗ moves from point O (i.e., point E0) to point B90 (i.e., point E90) on (OB_90 ) ̅ (i.e., (OE_90 ) ̅), then point Cn moves from point C0 to point C90 on the (C_0 C_90 ) ̅ on L ⃡, and (C_n E_n ) ⃗ intersects with arc B0B90 at all the points on the arc (namely, (C_n E_n ) ⃗ passes through every point linking B0 to point B90 on arc B0B90).

Interconnected (Ob_o ) ̅ and (OB_n ) ̅ (Figure 3)

a. When (C_n E_n ) ⃗ is regarded as (C_n b_n ) ⃗ (point bn on (C_n E_n ) ⃗; point Cn on (C_0 C_90 ) ̅), then (C_n E_n ) ⃗ intersects with arc b0B90 at point bn, with (OB_90 ) ̅ (i.e., (OE_90 ) ̅) at point En, and with arc B0B90 at point Bn. If (C_n b_n ) ̅ = (OB_n ) ̅ = (Ob_0 ) ̅ = (OB_0 ) ̅, then ∠BoOBn = 3∠BoCnBn.
Proof.
Because (C_n b_n ) ̅ = (OB_n ) ̅ = (Ob_0 ) ̅ = (OB_0 ) ̅, ∠OBnbn = ∠ObnEn = 2∠OCnEn = 2∠BOCnBn. That is, ∠OBnCn = ∠OBnbn = ∠ObnEn = 2∠OCnEn = 2∠BoCnBn. Thus, ∠OBnCn = 2∠BOCnBn.
According to I and Figure 3, ∠BoOBn = ∠OBnCn + ∠BoCnBn = 2∠BoCnBn + ∠BoCnBn = 3∠BoCnBn. (proven)

Therefore, (C_n b_n ) ⃗ (point bn is on (C_n E_n ) ⃗) intersects with arc b0B90 at point bn, with (OB_90 ) ̅ (i.e., (OE_90 ) ̅) at point En, and with arc B0B90 at point Bn. When (C_n b_n ) ̅ = (OB_n ) ̅ = (Ob_0 ) ̅ = (OB_0 ) ̅, then ∠BoOBn = 3∠BoCnBn.

When (C_n E_n ) ⃗ is equivalent to (C_90 E_90 ) ⃗, (C_n b_n ) ⃗ becomes the (C_90 b_90 ) ⃗ that intersects with arc b0B90 at point b90. That is, (C_n b_n ) ⃗ intersects with arc b0B90 at the points ranging from b0 to b90, and the range of ∠b0Cnbn is 0° = ∠b0C0b0 ≤ ∠b0Cnbn ≤ ∠b0C90b90 = 30°.
Because any two points can make a line, any given point bn on arc b0b90 can be connected with the point on L ⃡ to make a segment. Maintain the length of (C_n b_n ) ̅ = (OB_0 ) ̅, and point bn on arc bnb90 and point Cn on L ⃡ can make (C_n b_n ) ̅. Subsequently, point bn departs from point b0 on arc b0b90, which connects with point C0 on L ⃡ to make (C_0 b_0 ) ̅ = (OB_0 ) ̅, to point bn that connects with Cn on L ⃡ to make (C_n b_n ) ̅ = (OB_0 ) ̅, and finally, to point b90 that connects with point C90 on L ⃡ to make (C_90 b_90 ) ̅ = (OB_0 ) ̅. When ((C_n b_n ) ̅ = (OB_0 ) ̅)point Cn on (C_n b_n ) ⃗ moves on (C_0 C_90 ) ̅ and point bn on arc b0b90, (C_n b_n ) ⃗ intersects with arc b0b90 at any given point on the arc (i.e., (C_n b_n ) ⃗ passes through every point on arc b0b90). Thus, when (C_n b_n ) ⃗ intersects with arc b0b90 at point bn, the range of the included angle b0Cnbn formed by (C_n b_n ) ⃗ and L ⃡ can be expressed as 0° = ∠b0C0b0 ≤ ∠b0C1b1 ≤… ≤ ∠b0Cn-1bn-1 ≤ ∠b0Cnbn ≤… ≤ ∠b0C 90-1b90-1 ≤ ∠b0C90b90 = 30°.
According to b. and c., (C_n b_n ) ⃗ intersects with arc b0b90 at any given point on the arc (i.e., (C_n b_n ) ⃗ passes through every point on arc b0b90). When (C_n b_n ) ⃗ intersects with arc b0b90 at point bn, the range of the included ∠b0Cnbn formed by (Cnbn) ⃗ and L ⃡ can be expressed as 0° = ∠b0C0b0 ≤ ∠b0C1b1 ≤… ≤ ∠b0Cn-1bn-1 ≤ ∠b0Cnbn ≤… ≤ ∠b0C90-1b90-1 ≤ ∠b0C90b90 = 30°. Because (C_n b_n ) ⃗ is equivalent to (C_n E_n ) ⃗ (Figure 3), (C_n E_n ) ⃗ intersects with arc b0b90 at any given point on the arc (i.e., (C_n E_n ) ⃗ passes through every point on arc b0b90). When (C_n E_n ) ⃗ intersects with arc b0b90 at point bn, the range of the included ∠b0CnEn formed by (C_n E_n ) ⃗ and L ⃡ can be expressed as 0° = ∠b0C0b0 = ∠b0C0E0 ≤ ∠b0C1E1 ≤… ≤ ∠b0Cn-1En-1 ≤ ∠b0CnEn ≤… ≤ ∠b0C90-1E90-1 ≤ ∠b0C90E90 = ∠b0C90B90 = 30°.
According to c. and d., (C_n E_n ) ̅ on (C_n E_n ) ⃗ is equal to 2(OB_0 ) ̅. Point Cn moves on (C_0 C_90 ) ̅ on L ⃡, point En moves on (OB_90 ) ̅ (i.e., (OE_90 ) ̅), and (C_n E_n ) ⃗ intersects with arc b0b90 at any given point on the arc (i.e., (C_n E_n ) ⃗ passes through every point on arc b0b90). When (C_n E_n ) ⃗ intersects with arc b0b90 at point bn, the range of the included ∠b0CnEn formed by (C_n E_n ) ⃗ and L ⃡ can be expressed as 0° = ∠b0C0b0 = ∠b0C0E0 ≤ ∠b0C1E1 ≤… ≤ ∠b0Cn-1En-1 ≤ ∠b0CnEn ≤… ≤ ∠b0C90-1E90-1 ≤ ∠b0C90E90 = ∠b0C90B90 = 30°. Furthermore, (C_n E_n ) ⃗ intersects with arc B0B90 at point Bn. If point En on (C_n E_n ) ⃗ departs from point O (i.e., point E0) to point B90 (i.e., point E90) on (OB_90 ) ̅ (i.e., (OE_90 ) ̅), and point Cn from point C0 to point C90 on the (C_0 C_90 ) ̅ on L ⃡, then (C_n E_n ) ⃗ intersects with arc b0b90 at any given point bn. The range of the included ∠b0CnEn formed by (C_n E_n ) ⃗ and L ⃡ can be expressed as 0° = ∠b0C0b0 = ∠b0C0E0 = ∠B0C0E0 ≤ ∠b0Cnbn = ∠B0CnEn ≤ ∠b0C90E90 = ∠b0C90B90 = ∠B0C90B90 = 30°. According to Section A.1. and Figure 3, if (C_n E_n ) ̅ = 2(OB_o ) ̅, then ∠BoOBn = 3∠BOCnBn, where ∠BOCnBn = ∠b0Cnbn. Thus, (C_n E_n ) ⃗ intersects with arc B0B90 at point Bn (i.e., (C_n E_n ) ⃗ passes through every point on arc b0b90). When point Bn connects with point O to form (OB_n ) ̅, ∠BoOBn = 3∠BOCnBn, and 0° = ∠b0C0b0 = ∠b0C0E0 = ∠B0C0E0 ≤ ∠b0Cnbn = ∠B0CnEn ≤ ∠b0C90E90 = ∠b0C90B90 = ∠B0C90B90 = 30°. Therefore, 0° = ∠BOOB0 ≤ ∠BOOBn = 3∠BOCnBn ≤ ∠BOOB90 = 90° (i.e., one ∠BOCnBn must have a corresponding ∠BOOBn = 3∠BOCnBn). Thus, (C_n E_n ) ⃗ intersects with arc B0B90 at every point on the arc (i.e., (C_n E_n ) ⃗ passes through all points ranging from B0 to B90 on the arc B0B90), and the included angle formed by every (C_n E_n ) ⃗ and L ⃡ is different.

2.2. Continuing the argument in 2.1, if the center O is connected with a given point Bn on arc B0B90, then a specific feature (relationship) emerges as follows: ∠BOOBn = 3∠BOCnBn and (C_n E_n ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r = (2radius)(Figure 3).

Let (center; radius) = (Cn; (C_n E_n ) ̅) refer to using point Cn as the center and (C_n E_n ) ̅ as the radius to construct an arc that intersects with L ⃡ at point Kn.

Use (Cn; (C_n E_n ) ̅) to draw an arc, where (C_n E_n ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r. As point En moves on (OB_90 ) ̅ (i.e., (OE_90 ) ̅) and point Cn on (C_0 C_90 ) ̅, (C_n E_n ) ̅ intersects with L ⃡ at point Kn, and (C_n E_n ) ⃗ intersects with arc B0B90 at point Bn. Thus, (C_n E_n ) ̅ = (C_n K_n ) ̅ = (C_n K_0 ) ̅ + (K_0 K_n ) ̅ = (C_n O) ̅ + (OK_n ) ̅ = 2(OB_n ) ̅ = 2r (thus, point O = point K0).
Use (C0; (C_0 O) ̅) to construct an arc, where (C_0 E_0 ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, and point E0 (i.e., point O) and point C0 are located on (C_0 C_90 ) ̅. The constructed arc intersects with L ⃡ at point K0 (i.e., point O). Thus, (C_0 E_0 ) ̅ = (C_0 K_0 ) ̅ = (C_n O) ̅ (point O = point K0 or point E0).
Use (C90;(C_90 E_90 ) ̅) to construct an arc, where (C_90 E_90 ) ̅ = 2 (OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, and point E90 (i.e., point B90) and point C90 are located on (C_0 C_90 ) ̅. The constructed arc intersects with L ⃡ at point K90. Thus, (C_90 E_90 ) ̅ = (C_90 B_90 ) ̅ = (C_90 K_90 ) ̅ = (C_90 K_0 ) ̅ + (K_0 K_90 ) ̅ = (C_90 O) ̅ + (OK_90 ) ̅.
According to a., b., c., and Figure 3, when (Cn; (C_n E_n ) ̅) is used to draw an arc, (C_n E_n ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, point En moves on (OB_90 ) ̅ (i.e., (OE_90 ) ̅), point Cn moves on (C_0 C_90 ) ̅, and the constructed arc intersects with L ⃡ at point Kn. Thus, 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r = (C_n E_n ) ̅ = (C_n K_n ) ̅ = (C_n K_0 ) ̅ + (K_0 K_n ) ̅ = (C_n O) ̅ + (OK_n ) ̅ (point O = point K0). Namely,
2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r = (C_n E_n ) ̅ = (C_n K_n ) ̅ = (C_0 K_0 ) ̅ = (C_1 K_0 ) ̅ + (K_0 K_1 ) ̅ = … = (C_n K_0 ) ̅ + (K_0 K_n ) ̅ = (C_(n-1) K_0 ) ̅ + (K_0 K_(n-1) ) ̅ = … = (C_(90-1) K_0 ) ̅ + (K_0 K_(90-1) ) ̅ = (C_90 K_0 ) ̅ + (K_0 K_90 ) ̅ (i.e., 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r = (C_n K_n ) ̅ = (C_0 O) ̅ = (C_1 O) ̅ + (OK) ̅ = … = (C_n O) ̅ + (OK_n ) ̅ = (C_(n-1) O) ̅ + (OK_(n-1) ) ̅ = … = (C_(90-1) 0) ̅ + (OK_(90-1) ) ̅ = (C_90 O) ̅ + (OK_90 ) ̅; point O = point K0 =point E0).
The hypotenuse (C_n E_n ) ̅ of right △CnOEn is fixed, and therefore, the length of (C_n O) ̅ decreases as that of (OE_n ) ̅ increases (Figure 4).
When (Cn; (C_n E_n ) ̅) is used to draw an arc, (C_n E_n ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, point En moves on (OB_90 ) ̅ (i.e., (OE_90 ) ̅), point Cn moves on (C_0 C_90 ) ̅, and the constructed arc intersects with L ⃡ at point Kn. Thus, 2(OB_0 ) ̅ = (C_n E_n ) ̅ = (C_n K_n ) ̅ = (C_(n-1) K_(n-1) ) ̅ = (C_(n-1) E_(n-1) ) ̅ (Figure 4). The right △CnOEn and △Cn-1OEn-1 show that, when (OC_n ) ̅ ≤ (OC_(n-1) ) ̅, (OE_(n-1) ) ̅ must be shorter than or equal to (OEn) ̅ for both hypotenuses (C_n E_n ) ̅ and (C_(n-1) K_(n-1) ) ̅ to be equal in length (both points En and En-1 are located on M ⃡).
(C_90 K_0 ) ̅ ≤ (C_(90-1) K_0 ) ̅ ≤ … ≤ (C_n K_0 ) ̅ ≤ (C_(n-1) K_0 ) ̅ ≤ … ≤ (C_1 K_0 ) ̅ ≤ (C_0 K_0 ) ̅ (i.e., (C_90 O) ̅ ≤ (C_(90-1) O) ̅ ≤ … ≤ (C_n O) ̅ ≤ (C_(n-1) O) ̅ ≤ … ≤ (C_1 O) ̅ ≤ (C_0 O) ̅; point O = point K0 = point E0).
(OO) ̅ = (K_0 K_0 ) ̅ ≤ (K_0 K_1 ) ̅ ≤ … ≤ (K_0 K_(n-1) ) ̅ ≤ (K_0 K_n ) ̅ ≤ … ≤ (K_0 K_(90-1) ) ̅ ≤ (K_0 K_90 ) ̅ (or (OO) ̅ = (OK_0 ) ̅ ≤ (OK_1 ) ̅ ≤ … ≤ (OK_(n-1) ) ̅ ≤ (OK_n ) ̅ ≤ … ≤ OK_(90-1) ≤ (OK_90 ) ̅; point O = point K0 = point E0).
According to II, when (Cn; (C_n E_n ) ̅) is used to draw an arc, (C_n E_n ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, point En moves on (OB_90 ) ̅ (i.e., (OE_90 ) ̅), point Cn moves on (C_0 C_90 ) ̅, and point Kn at which the constructed arc intersects with L ⃡ is the only point of intersection and presents a certain order.
According to a., b., c., and d., with the center at a given point En on (OE_90 ) ̅ (i.e., (OB_90 ) ̅ or (E_0 E_90 ) ̅), use (Cn; (C_n E_n ) ̅) to construct an arc, where (C_n E_n ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, point En moves on (OB_90 ) ̅ (i.e., (OE_90 ) ̅), point Cn moves on (C_0 C_90 ) ̅, and point Kn at which the constructed arc intersects with L ⃡ is the only point of intersection and presents a certain order. That is, (OO) ̅ = (K_0 K_0 ) ̅ ≤ (K_0 K_1 ) ̅ ≤ … ≤ (K_0 K_(n-1) ) ̅ ≤ (K_0 K_n ) ̅ ≤ ... ≤ (K_0 K_(90-1) ) ̅ ≤ (K_0 K_90 ) ̅ (or (OO) ̅ = (OK_0 ) ̅ ≤ (OK_1 ) ̅ ≤ … ≤ (OK_(n-1) ) ̅ ≤ (OK_n ) ̅ ≤ … ≤ (OK_(90-1) ) ̅ ≤ (OK_90 ) ̅; point O = point K0 = point E0). In addition, (C_n E_n ) ⃗ intersects with arc B0B90 at point Bn, and every (C_n E_n ) ⃗ intersects with arc B0B90 at a different point Bn.

2.3. Let (center; radius) = (Cn; (C_n K_n ) ̅) refer to using Cn as the center and (C_n K_n ) ̅ as the radius to construct an arc that intersects with M ⃡ at point En. Use (Cn; (C_n K_n ) ̅) to construct an arc, where (C_n K_n ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, point Cn moves on (C_90 C_0 ) ̅, point Kn moves on (K_0 K_90 ) ̅ (i.e., (OK_90 ) ̅; point O = point K0 = point E0), and the constructed arc intersects with M ⃡ at point En. Thus,

Use (C90; (C_90 K_90 ) ̅) to draw an arc, where (C_90 K_90 ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, and therefore, (C_90 O) ̅ + (OK_90 ) ̅ = (C_90 K_90 ) ̅ = (C_90 E_90 ) ̅ = (C_90 B_90 ) ̅.
Use (C0; (C_0 K_0 ) ̅) to construct an arc, where (C_0 K_0 ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, and therefore, (C_0 O) ̅ = (C_0 K_0 ) ̅ = (C_0 E_0 ) ̅.
Use (Cn; (C_n K_n ) ̅) to construct an arc, where (C_n K_n ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r.
The right △CnOEn and right △Cn-1OEn-1 show that, when (OC_n ) ̅ ≤ (OC_(n-1) ) ̅, (OE_(n-1) ) ̅ must be shorter than or equal to (OE_n ) ̅ for both hypotenuses (C_n E_n ) ̅ and (C_(n-1) K_(n-1) ) ̅ to be equal in length (both points En and En-1 are located on M ⃡; Figure 4). Thus, point En at which the arc intersects with M ⃡ demonstrates the following relationships: (OO) ̅ ≤ (OE_0 ) ̅ ≤ (OE_1 ) ̅ ≤ … ≤ (OE_(n-1) ) ̅ ≤ (OE_n ) ̅ ≤ … ≤ (OE_(90-1) ) ̅ ≤ (OE_90 ) ̅ = (OB_90 ) ̅ = (OB_0 ) ̅.
According to a., b., and c., when (Cn; (C_n K_n ) ̅) is used to draw an arc, (C_n K_n ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, point Cn moves on (C_90 C_0 ) ̅, point Kn moves on (K_0 K_90 ) ̅ (i.e., (OK_90 ) ̅; point O = point K0 = point E0), and point En at which the constructed arc intersects with M ⃡ is the only point of intersection and presents a certain order. Furthermore, (C_n E_n ) ⃗ intersects with arc B0B90 at point Bn, and every (C_n E_n ) ⃗ intersects with arc B0B90 at a different point Bn.
According to a., b., c., and d., with a given point Kn on (OK_90 ) ̅ (i.e., (〖K_0 K〗_90 ) ̅) as the center, when (Cn; (C_n K_n ) ̅) is used to construct an arc, (C_n K_n ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, point Kn moves on (OK_90 ) ̅ (i.e., (K_0 K_90 ) ̅), point Cn moves on (C_90 C_0 ) ̅, and point En at which the constructed arc intersects with M ⃡ is the only point of intersection and presents a certain order.
(OO) ̅ = (E_0 E_0 ) ̅ ≤ (E_0 E_1 ) ̅ ≤ … ≤ (E_0 E_(n-1) ) ̅ ≤ (E_0 E_n ) ̅ ≤ … ≤ (E_0 E_(90-1) ) ̅ ≤ (E_0 E_90 ) ̅ (or (OO) ̅ = (OE_0 ) ̅ ≤ (OE_1 ) ̅ ≤ … ≤ (OE_(n-1) ) ̅ ≤ (OE_n ) ̅ ≤ … ≤ (OE_(90-1) ) ̅ ≤ (OE_90 ) ̅ = (OB_90 ) ̅ = (OB_0 ) ̅; point O = point E0 =point K0).
According to I, (C_n E_n ) ⃗ intersects with arc B0B90 at point Bn, every (C_n E_n ) ⃗ intersects with arc B0B90 at a different point Bn, and the points of intersection exhibit a certain order. According to I, the following relational expression can be derived: ∠OC90O = ∠OC90E0 ≤ ∠OC90E1 ≤ … ≤ ∠OC90En ≤ ∠OC90En-1 ≤ … ≤ ∠OC90E90-1 ≤ ∠OC90E90. Therefore, every (C_n E_n ) ⃗ intersects with arc B0B90 at a different point Bn, which is the only point of intersection and exhibits a certain order.

According to 2.2 e. and 3.3 e. in 2., the following statements can be derived:

3.1. (C_n E_n ) ⃗ intersects with arc B0B90 at point Bn, and every (C_n E_n ) ⃗ intersects with arc B0B90 at a different point Bn.

3.2. Given a point En on (OE_90 ) ̅ (i.e., (OB_90 ) ̅ or (〖E_0 E〗_90 ) ̅), use (Cn; (C_n E_n ) ̅) to draw an arc, where (C_n E_n ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, point En moves on (OB_90 ) ̅(i.e.,(OE_90 ) ̅), point Cn moves on (〖C_0 C〗_90 ) ̅, and the constructed arc intersects with L ⃡ at point Kn, which is the only point of intersection and exhibits a specific order.

3.3. Given a point Kn on (OK_90 ) ̅ (i.e., (K_0 K_90 ) ̅), use (Cn; (C_n K_n ) ̅) to construct an arc, where (C_n K_n ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, point Kn moves on (OK_90 ) ̅ (i.e., (K_0 K_90 ) ̅), point Cn moves on (〖C_0 C〗_90 ) ̅, and the arc intersects with M ⃡ at point En, which is the only point of intersection and exhibits a specific order.

a. Given a point En on (OE_90 ) ̅ (i.e., (OB_90 ) ̅ or (〖E_0 E〗_90 ) ̅), use (Cn; (C_n E_n ) ̅) to construct an arc, where (C_n E_n ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, point En moves on (OB_90 ) ̅ (i.e., (OE_90 ) ̅), point Cn moves on (〖C_0 C〗_90 ) ̅, and the constructed arc intersects with L ⃡ at point Kn. Given a point Kn on (OK_90 ) ̅ (i.e., (K_0 K_90 ) ̅), use (Cn; (C_n K_n ) ̅) to construct an arc, where (C_n K_n ) ̅ = 2(OB_0 ) ̅ = 2(OB_n ) ̅ = 2r, point Kn moves on (OK_90 ) ̅ (i.e., (K_0 K_90 ) ̅), point Cn moves on (〖C_0 C〗_90 ) ̅, and the constructed arc intersects with M ⃡ at point En. The intersection points Kn and En were the same in number and present specific orders (because both constructions form single points of intersection that exhibit specific orders).
b. The two aforementioned constructions produce the same number of intersection points as that of point Bn, at which (C_n E_n ) ⃗ intersects with (B_0 B_90 ) ̅ because the constructed arc has the same number of intersection points with L ⃡ and M ⃡. Furthermore, because (C_n E_n ) ⃗ intersects with M ⃡ at point En, the number of intersection points between (C_n E_n ) ⃗ and M ⃡ is equal to that of the intersection point Bn between (C_n E_n ) ⃗ and arc B0B90, and the intersection point Bn between every (C_n E_n ) ⃗ and arc B0B90 exhibits a specific order (because the included angle formed by every (C_n E_n ) ⃗ and L ⃡ is different).
c. The numbers of intersection points Kn (between the arc and L ⃡) and En (between the arc and M ⃡) produced by the two constructions in a. are the same as that of the intersection point Bn between (C_n E_n ) ⃗ and arc B0B90 described in b., and all of these points exhibit a specific order. (These relationships can be derived from the concepts of surjective and injective functions as well.) This states the specific features (relationships) between an angle and its triple angle.

Auxiliary graph and its specific features (Figure 5)

Construct a straight line L perpendicular to and intersecting with a straight line M at point O. Given a point Bo on L ⃡, with the center at point O and the radius as (OB_o ) ̅, construct an arc intersecting L ⃡ at point B0 and with M ⃡ at point B90. Given a point C90 on L ⃡ to make (C_90 B_90 ) ̅ = 2(OB_0 ) ̅, point C90 can be connected with a given point Bn on arc B0B90 to produce (C_90 B_n ) ̅ that intersects with M ⃡ at point en.

1.1. Because any two points can make a line, when point C90 is connected with a given point Bn on arc B0B90 to make (C_90 B_n ) ̅ that interests with M ⃡ at point en, the numbers of intersection points Bn (between (C_90 B_n ) ̅ and arc B0B90) and en (between (C_90 B_n ) ̅ and M ⃡) are the same because all points Bn and en are on (C_90 B_n ) ̅.

1.2. Thus, the intersecting order of (C_90 B_n ) ̅ with M ⃡ at point en, with (C_90 B_n ) ̅ as the line segment intersecting with arc B0B90 at points ranging from B0, Bn, to B90 (point O = point e0; point B90 = point e90). That is,

(OO) ̅ = (e_0 e_0 ) ̅ ≤ (e_0 e_1 ) ̅ ≤ … ≤ (e_0 e_(n-1) ) ̅ ≤ (e_0 e_n ) ̅ ≤ … ≤ (e_0 e_(90-1) ) ̅ ≤ (e_0 e_90 ) ̅ = (e_0 B_90 ) ̅ = (OB_0 ) ̅ (or (OO) ̅ = (Oe_0 ) ̅ ≤ (Oe_1 ) ̅ ≤ … ≤ (Oe_(n-1) ) ̅ ≤ (Oe_n ) ̅ ≤ … ≤ (Oe_(90-1) ) ̅ ≤ (Oe_90 ) ̅ = (OB_90 ) ̅ = (OB_0 ) ̅).
According to a, the included angles formed by (C_90 B_n ) ̅ and L ⃡ are different. That is, ∠OC90O = ∠OC90e0 ≤ ∠OC90e1 ≤ … ≤ ∠OC90en-1 ≤ ∠OC90en ≤ … ≤ ∠OC90e90-1 ≤ ∠OC90e90.

1.3. Use (C90; (C_90 e_n ) ̅) to construct an arc, where point en moves on (e_0 e_90 ) ̅ (or (OB_90 ) ̅), and the constructed arc intersects with L ⃡ at point hn (in this case, point O = point e0 = point h0; point B90 = point e90). Thus, (〖C_90 e〗_n ) ̅ = (〖C_90 h〗_n ) ̅ = (〖C_90 h〗_0 ) ̅ + (〖h_0 h〗_n ) ̅ = (C_90 O) ̅ + (Oh_n ) ̅.

According to 1.2 and the right △C90e0en (or right △C90Oen) in Figure 6, where (〖C_90 e〗_0 ) ̅ (or (C_90 O) ̅) is fixed, the following relational expression can be derived: (C_90 O) ̅ = (〖C_90 e〗_0 ) ̅ ≤ (C_90 e_1 ) ̅ ≤ … ≤ (C_90 e_(n-1) ) ̅ ≤ (C_90 e_n ) ̅ ≤ … ≤ (C_90 e_(90-1) ) ̅ ≤ (C_90 e_90 ) ̅ = (C_90 B_90 ) ̅ = 2(OB_0 ) ̅.
According to a. and the equation of (C_90 e_n ) ̅ = (C_90 h_n ) ̅ = (〖C_90〗_ h_0 ) ̅ + (h_0 h_n ) ̅ = (C_90 O) ̅ + (Oh_n ) ̅, the following relational expression can be derived: (OO) ̅ = (e_0 h_0 ) ̅ ≤ (e_0 h_1 ) ̅ ≤ … ≤ (e_0 h_(n-1) ) ̅ ≤ (e_0 h_n ) ̅ ≤ … ≤ (e_0 h_(90-1) ) ̅ ≤ (e_0 h_90 ) ̅ (or (OO) ̅ = (Oh_0 ) ̅ ≤ (Oh_1 ) ̅ ≤ … ≤ (Oh_(n-1) ) ̅ ≤ (Oh_n ) ̅ ≤ … ≤ (Oh_(90-1) ) ̅ ≤ (Oh_90 ) ̅).
According to a. and b., let (center; radius) = (C90; (C_90 e_n ) ̅) refer to using point C90 as the center and (C_90 e_n ) ̅ as the radius to construct an arc that intersects with L ⃡ at point hn. When (C90; (C_90 e_n ) ̅) is used to construct an arc, point en moves on (e_0 e_90 ) ̅ (or (OB_90 ) ̅), and the arc intersects with L ⃡ at a single point hn in a certain order.
According to a., b., and c., given a point en on (e_0 e_90 ) ̅ (or (OB_90 ) ̅), use (C90; (C_90 e_n ) ̅) to construct an arc that intersects with L ⃡ at a single point hn with a specific order. In addition, (C_90 e_n ) ̅ intersects with arc B0B90 at a single point with a specific order.

2. Let (center; radius) = (C90; (C_90 h_n ) ̅) refer to using point C90 as the center and (C_90 h_n ) ̅ as the radius to construct an arc that intersects with M ⃡ at point en. Subsequently, use (C90; (C_90 h_n ) ̅) to construct an arc, where point hn moves on (h_0 h_90 ) ̅ (or (Oh_90 ) ̅), the constructed arc intersects with M ⃡ at point en (in this case, point O = point e0 and h0; point B90 = point e90), and (C_90 e_n ) ⃗ intersects with arc B0B90 at point Bn. Thus, (C_90 h_n ) ̅ = (C_90 e_n ) ̅ = (C_90 h_0 ) ̅ + (h_0 h_n ) ̅ = (C_90 O) ̅ + (Oh_n ) ̅ (Figure 6).

a. According to the equation of (C_90 h_n ) ̅ = (C_90 e_n ) ̅ = (C_90 h_0 ) ̅ + (h_0 h_n ) ̅ = (C_90 O) ̅ + (Oh_n ) ̅, 1.b ((OO) ̅ = (e_0 h_0 ) ̅ ≤ (e_0 h_1 ) ̅ ≤ … ≤ (e_0 h_(n-1) ) ̅ ≤ (e_0 h_n ) ̅ ≤ … ≤ (e_0 h_(90-1) ) ̅ ≤ (e_0 h_90 ) ̅, or (OO) ̅ = (Oh_0 ) ̅ ≤ (Oh_1 ) ̅ ≤ … ≤ (Oh_(n-1) ) ̅ ≤ (Oh_n ) ̅ ≤ … ≤ (Oh_(90-1) ) ̅ ≤ (Oh_90 ) ̅), and because (〖C_90 e〗_0 ) ̅ in the right △C90e0en (or right △C90Oen) is fixed (in this case, point O = point e0; point B90 = point e90), the following results can be derived:
(OO) ̅ = (h_0 h_0 ) ̅ = (e_0 e_0 ) ̅ ≤ (e_0 e_1 ) ̅ ≤ … ≤ (e_0 e_(n-1) ) ̅ ≤ (e_0 e_n ) ̅ ≤ … ≤ (e_0 e_(90-1) ) ̅ ≤ (e_0 e_90 ) ̅ = (OB_90 ) ̅ = (OB_0 ) ̅ (or (OO) ̅ = (h_0 h_0 ) ̅ = (e_0 e_0 ) ̅ ≤ (Oe_1 ) ̅ ≤ … ≤ (Oe_(n-1) ) ̅ ≤ (Oe_n ) ̅ ≤ … ≤ (Oe_(90-1) ) ̅ ≤ (Oe_90 ) ̅ = (OB_90 ) ̅ = (OB_0 ) ̅).
(C_90 O) ̅ = (C_90 h_0 ) ̅ = (〖C_90 e〗_0 ) ̅ ≤ (C_90 e_1 ) ̅ ≤ … ≤(C_90 e_(n-1) ) ̅ ≤ (C_90 e_n ) ̅ ≤ … ≤ (C_90 e_(90-1) ) ̅ ≤ (C_90 e_90 ) ̅ =(C_90 B_90 ) ̅ = 2(OB_0 ) ̅.
According to a., when (C90; (C_90 h_n ) ̅) is used to construct an arc, point hn moves on (h_0 h_90 ) ̅ (or (Oh_90 ) ̅) and the arc intersects with M ⃡ at point en (in this case, point O = point e0 = point h0; point B90 = point e90), and (C_90 e_n ) ⃗ intersects with arc B0B90 at point Bn.
Use (C90; (C_90 h_n ) ̅) to construct an arc, where point hn moves on (h_0 h_90 ) ̅ (or (Oh_90 ) ̅) and the arc intersects with M ⃡ at a single point en with a specific order because (OO) ̅ = (h_0 h_0 ) ̅ = (e_0 e_0 ) ̅ ≤ (e_0 e_1 ) ̅ ≤ … ≤ (e_0 e_(n-1) ) ̅ ≤ (e_0 e_n ) ̅ ≤ … ≤ (e_0 e_(90-1) ) ̅ ≤ (e_0 e_90 ) ̅ = (OB_90 ) ̅ = (OB_0 ) ̅ (or (OO) ̅ = (h_0 h_0 ) ̅ = (e_0 e_0 ) ̅ ≤ (Oe_1 ) ̅ ≤ … ≤ (Oe_(n≤1) ) ̅ ≤ (Oe_n ) ̅ ≤ … ≤ (Oe_(90-1) ) ̅ ≤ (Oe_90 ) ̅ = (OB_90 ) ̅ = (OB_0 ) ̅).
(C_90 e_n ) ⃗ intersects with arc B0B90 at a single point Bn with a specific order (because (OO) ̅ = (h_0 h_0 ) ̅ = (e_0 e_0 ) ̅ ≤ (e_0 e_1 ) ̅ ≤ … ≤ (e_0 e_(n-1) ) ̅ ≤ (e_0 e_n ) ̅ ≤ … ≤ (e_0 e_(90-1) ) ̅ ≤ (e_0 e_90 ) ̅ = (OB_90 ) ̅ = (OB_0 ) ̅ or (OO) ̅ = (h_0 h_0 ) ̅ = (e_0 e_0 ) ̅ ≤ (Oe_1 ) ̅ ≤ … ≤ (Oe_(n≤1) ) ̅ ≤ (Oe_n ) ̅ ≤ … ≤ (Oe_(90-1) ) ̅ ≤ (Oe_90 ) ̅ = (OB_90 ) ̅ = (OB_0 ) ̅. Thus, all the included angles formed by every (C_90 e_n ) ⃗ and L ⃡ are different, and therefore, (C_90 e_n ) ⃗ intersects with M ⃡ at a single point en and with arc B0B90 at a single point Bn with a specific order.)
c. Thus, according to b., given a point hn on (h_0 h_90 ) ̅ (or (Oh_90 ) ̅), use (C90; (C_90 h_n ) ̅) to construct an arc that intersects with M ⃡ at a single point en with a specific order (in this case, point O = point e0 = point h0; point B90 = point e90) and (C_90 e_n ) ⃗ intersects with arc B0B90 at a single point Bn with a specific order.

3. Based on 1.d. and 2.c.

3.1. Given a point en on (Oe_90 ) ̅ (i.e., (OB_90 ) ̅ or (e_0 e_90 ) ̅), use (C90; (C_90 e_n ) ̅) to construct an arc, point en moves on (OB_90 ) ̅ (i.e., (Oe_90 ) ̅), and point C90 moves on (C_90 B_90 ) ̅ ((C_90 B_90 ) ̅ = 2(OB_0 ) ̅). The number of point hn at which the constructed arc intersects with L ⃡, the number of the given point hn on (Oh_90 ) ̅ (i.e., (h_0 h_90 ) ̅), and the number of point en at which the arc constructed using (C90; (C_90 e_n ) ̅) when point hn falls on (Oh_90 ) ̅ (i.e., (h_0 h_90 ) ̅) and point C90 falls on (C_90 B_90 ) ̅ ((C_90 B_90 ) ̅ = 2(OB_0 ) ̅) are the same with a specific order (because the two constructions produce single intersection points with a specific order).

3.2. The number of intersection point Bn at which (C_90 e_n ) ⃗ intersects with arc B0B90 is the same as that for the two types of construction in 3.1 because the constructed arc intersects with L ⃡ and M ⃡ at the same number of points hn and en, respectively. The number of intersection point en is the same as that of the intersection point Bn between (C_90 e_n ) ⃗ and arc B0B90 (or between (C_90 B_n ) ̅ and arc B0B90). (C_90 e_n ) ⃗ intersects with arc B0B90 at a single point Bn with a specific order as well (because every (C_90 e_n ) ⃗ forms a different included angle with L ⃡).

3.3. According to 3.1 and 3.2, the intersection points hn (between the arc and L ⃡) and en (between the arc and M ⃡) in the two constructions in (1) and the intersection point Bn (between (C_90 e_n ) ⃗ and arc B0B90) in 2.b. are the same in number and present a specific order (these relationships can be derived from the concepts of surjective and injective functions). This is the specific feature of the auxiliary graph.

C. Integrate the specific features of an angle and its triple angle with those of the auxiliary graph, prove that the points in the two graphs are common, and apply these features in the classical geometric construction for trisecting a given angle of 0°–180° (Figure 7).

1. Construct a straight line L perpendicular and intersecting with M ⃡ at point O. Given a point Bo on L ⃡, with the center at point O and the radius as (OB_o ) ̅, draw an arc that intersects with L ⃡ at point Bo and with M ⃡ at point B90. Locate a point C90 and point C0 to satisfy (C_0 O) ̅ = (C_90 B_90 ) ̅ = 2(OB_0 ) ̅. Subsequently locate a point En on M ⃡ (on (OB_90 ) ̅) and point Cn on L ⃡ (on (〖C_0 C〗_90 ) ̅) to satisfy (C_n E_n ) ̅ = 2(OB_0 ) ̅ and the intersection of (C_n E_n ) ⃗ and arc B0B90 at point Bn. Construct (C_90 B_n ) ̅ that intersects with M ⃡ at point en and construct (OB_n ) ̅.

1.1. Please refer to Figure 7.

According to A.2.1.e., (C_n E_n ) ⃗ can intersect with arc B0B90 at every point on the arc (i.e., (C_n E_n ) ⃗ passes through all the points ranging from B0 to B90 on arc B0B90) with a specific order. Certainly, (C_n E_n ) ⃗ intersects with arc B0B90 and M ⃡ at the same number of points Bn and En, respectively, with a specific order.
Point C90 can be connected with any given point Bn on arc B0B90 (i.e., (C_90 B_n ) ̅ intersects with every point on arc B0B90), and (C_90 B_n ) ̅ intersects with M ⃡ at point en with a specific order. Furthermore, (C_90 B_n ) ̅ intersects with arc B0B90 and with M ⃡ at the same number of points Bn and en, respectively, with a specific order.
According to a. and b., given a point Bn on arc B0B90, construct (C_90 B_n ) ̅ that intersects with M ⃡ at point en. That is, (C_n E_n ) ⃗ intersects with arc B0B90 at any given point Bn (i.e., passes through point Bn) and intersects with M ⃡ at point En. Connect center O with point Bn to draw (OB_n ) ̅. Thus, ∠BoOBn = 3∠BOCnBn (Figure 7).

1.2. According to A.3.3.c., the numbers of intersection points Kn (between the arc and L ⃡) and En (between the arc and M ⃡) produced by the two constructions in a. are the same as that of the intersection point Bn between (C_n E_n ) ⃗ and arc B0B90 described in b., and all of these points exhibit a specific order. According to B.3.3, the intersection points hn (between the arc and L ⃡) and en (between the arc and M ⃡) in the two constructions in (1) and the intersection point Bn (between (C_90 e_n ) ⃗ and arc B0B90) are the same in number and present a specific order. All of these intersection points are the same in number and present a specific order because the numbers of the intersection points formed by the two constructions are the same as that of intersection point Bn on arc B0B90.

2. According to 1.1 and 1.2 in 1., the following results can be derived (Figure 7):

2.1. [Given a point En on M ⃡ (on (OB_90 ) ̅) and a point Cn on L ⃡ (on (〖C_0 C〗_90 ) ̅), the number of intersection point Kn at which the arc constructed using (Cn; (C_n E_n ) ̅) intersects with L ⃡] = [Given a point Kn on L ⃡ (on (OK_90 ) ̅) and a point Cn on L ⃡ (on (〖C_0 C〗_90 ) ̅), the number of intersection point En at which the arc constructed using (Cn; (C_n K_n ) ̅) intersects with M ⃡] = [Given a point en on M ⃡ (on (OB_90 ) ̅) and given a point C90 on L ⃡, that satisfies (C_90 B_90 ) ̅ = 2(OB_0 ) ̅, the number of point hn at which the arc constructed using (C90; (C_90 e_n ) ̅) intersects with L ⃡] = [Given a point hn on L ⃡ (on (Oh_90 ) ̅) and a point C90 on L ⃡ that satisfies (C_90 B_90 ) ̅ = 2(OB_0 ) ̅, the number of point en at which the arc constructed using (C90; (C_90 h_n ) ̅) intersects with M ⃡] = [the number of point Bn at which (C_90 e_n ) ⃗ (point C90 on L ⃡ and point en on (OB_90 ) ̅) intersects with arc B0B90] = [ the number of points at which (C_n E_n ) ⃗ (point Cn on (〖C_0 C〗_90 ) ̅ and point En on (OB_90 ) ̅) intersects with arc B0B90] = [the number of points at which (C_n E_n ) ⃗ (point Cn on (〖C_0 C〗_90 ) ̅ and point En on (OB_90 ) ̅) intersects with arc B0B90] = [the number of points at which (C_90 B_n ) ̅ intersects with arc B0B90, with point C90 on L ⃡ ((C_90 B_90 ) ̅ = 2(OB_0 ) ̅) and point Bn as any given point on arc B0B90], and all of these intersection points present a specific order (points E0, K0, h0, e, and O are the same point).

Subsequently, as (C_90 B_90 ) ̅ = (C_90 E_90 ) ̅ = (C_90 K_90 ) ̅ = (C_90 e_90 ) ̅ = (C_90 h_90 ) ̅ = 2(OB_0 ) ̅ = (C_90 O) ̅ + (OK_90 ) ̅ = (C_90 O) ̅ + (Oh_90 ) ̅. Thus, (OK_90 ) ̅ = (Oh_90 ) ̅.

2.2. According to 2.1, the number of intersection point Kn between the arc constructed using (Cn; (C_n E_n ) ̅) and L ⃡ is the same as that of intersection point hn between the arc constructed using (C90; (C_90 e_n ) ̅) and L ⃡ ((OK_90 ) ̅ = (Oh_90 ) ̅). All of the intersection points start from point O on L ⃡ and end at point K90 (or point h90 because (OK_90 ) ̅ = (Oh_90 ) ̅) on L ⃡, and the numbers of intersection points are the same and with a specific order (Figure 7).

3. According to 2.1 and 2.2 in 2, the following results can be derived (Figure 7):

3.1. Given a point Kn on L ⃡ (on (OK_90 ) ̅), use (Cn; (C_n K_n ) ̅) to construct an arc that intersects with M ⃡ at a single point En with a specific order. Given a point hn on L ⃡ (on (Oh_90 ) ̅), use (C90; (C_90 h_n ) ̅) to construct an arc that intersects with M ⃡ at a single point en with a specific order. All of the intersection points start from the point O on L ⃡ (i.e., point K0 or h0) and end at point K90 on L ⃡ (i.e., point h90 because (OK_90 ) ̅ = (Oh_90 ) ̅), with the same numbers of intersection points and a specific order. If (OK_90 ) ̅ = (Oh_90 ) ̅, point Kn and point hn must share the common point (overlap). That is, (OK_n ) ̅ = (Oh_n ) ̅.

3.2. According to 3.1, the following results can be derived (Figure 7): Every point Bn on arc B0B90 (all points ranging from B0 to B90 on the arc) involves two geometric constructions. First, with point en being where (C_90 B_n ) ̅ intersects with M ⃡), use (C90; (C_90 e_n ) ̅) to construct an arc intersecting with L ⃡ at point hn (i.e., point Kn). Second, with (C_n E_n ) ⃗ intersecting with arc B0B90 at point Bn, use (Cn; (C_n E_n ) ̅) to construct an arc intersecting with L ⃡ at point hn (i.e., point Kn). That is, using the same point Bn on arc B0B90 for the aforementioned constructions results in shared points hn and Kn on L ⃡.

4. According to 2. C. 3., and (Figure 7), points Kn and hn are common. The proof of the classical geometric construction for trisecting a given angle is as follows:

Given a point Bn on arc B0B90, connect point Bn with the point C90 on L ⃡ to construct (C_90 B_n ) ̅ that intersects with M ⃡ at point en. With the center at point C90 and the radius as (C_90 e_n ) ̅, construct an arc intersecting with L ⃡ at point hn.
With the center at point C90 and the radius as (C_90 B_90 ) ̅, construct an arc intersecting with L ⃡ at point K90 and with (C_90 B_n ) ̅ at point W.
According to Figure 7, 2(OB_0 ) ̅ = (C_90 W) ̅ = (C_90 e_n ) ̅ + (e_n W) ̅ = (C_90 h_90 ) ̅ = (C_90 h_n ) ̅ + (h_n h_90 ) ̅. Because (C_90 e_n ) ̅ = (C_90 h_n ) ̅, (e_n W) ̅ = (h_n h_90 ) ̅.
According to Figure 7, 2(OB_0 ) ̅ = (C_n E_n ) ̅ = (C_n K_n ) ̅ = (C_n C_90 ) ̅ + (C_90 K_n ) ̅.
According to 3.4. and because points Kn and hn are common, 2(OB_0 ) ̅ = (C_90 W) ̅ = (C_90 e_n ) ̅ + (e_n W) ̅ = (C_90 h_90 ) ̅ = (C_90 h_n ) ̅ + (h_n h_90 ) ̅ = (C_n E_n ) ̅ = (C_n K_n ) ̅ = (C_n C_90 ) ̅ + (C_90 K_n ) ̅ = (C_n C_90 ) ̅ + (C_90 h_n ) ̅.
According to 3. and 5., (e_n W) ̅ = (h_n h_90 ) ̅, and 2(OB_0 ) ̅ = (C_n C_90 ) ̅ + (C_90 K_n ) ̅ = (C_n C_90 ) ̅ + (C_90 h_n ) ̅ = (C_90 h_n ) ̅ + (h_n h_90 ) ̅ = (C_90 e_n ) ̅ + (e_n W) ̅ = (C_90 h_n ) ̅ + (e_n W) ̅.
According to 6, (C_n C_90 ) ̅ + (C_90 h_n ) ̅ = (C_90 h_n ) ̅ + (e_n W) ̅. Thus, (C_n C_90 ) ̅ = (e_n W) ̅. (proven)


Figure Captions

FIGURE 1 Purpose of the diagram: To illustrate an angle being divided into three equivalent parts.

FIGURE 2 Purpose of the diagram: To show the relationship between an angle and a triple angle. (a) If ∠BoOBn = 3∠BoCnBn, then (C_n E_n ) ̅ = 2(OB_o ) ̅. and (b) If (C_n E_n ) ̅ = 2(OB_o ) ̅, then ∠BoOBn = 3∠BoCnBn, where (OB_o ) ̅ = (OB_n ) ̅

FIGURE 3 Purpose of the diagram: To demonstrate the special relationship between an angle and a triple angle. If ∠BoOBn = 3∠BoCnBn, then (C_n E_n ) ̅ = 2(OBo) ̅; if (C_n E_n ) ̅ = 2(OB_o ) ̅, then ∠BoOBn = 3∠BoCnBn, where (OB_o ) ̅ = (OB_n ) ̅. The equation ∠BoOBn = 3∠BoCnBn can be expressed as follows (0° = ∠BoOBn ≤ ∠BoOBn = 3∠BoCnBn ≤ ∠BoOB90 = 90°): (a) (C_n E_n ) ̅ = 2(OB_o ) ̅, where point C_n moves along L ⃡ and point E_n moves along M ⃡; (C_n E_n ) ⃗, which intersects M ⃡ at point En, intersects arc B0B90 at Bn and (b) Point C_n ((C_n E_n ) ̅ = 2(OB_o ) ̅, where 〖point C〗_n moves along L ⃡) is the center of the circle and (C_n E_n ) ̅ is the radius, forming an arc intersecting L ⃡ at point Kn

FIGURE 4 Purpose of the diagram: To demonstrate the special relationship between an angle and a triple angle. (a) Point C_n ((C_n E_n ) ̅ = 2(OB_o ) ̅, where point C_n moves along L ⃡) is the center of the circle and (C_n E_n ) ̅ is the radius, forming an arc intersecting M ⃡ at 〖point E〗_n (where (C_n E_n ) ̅ = 2(OB_o ) ̅). and (b) Line (C_n E_n ) ⃗ intersects M ⃡ at 〖point E〗_n and arc B0B90 at point Bn

FIGURE 5 Purpose of the diagram: To show the special relationship in the auxiliary graph. (a) Point C90 is the center of the circle and (C_90 e_n ) ̅ is the radius, forming an arc intersecting M ⃡ at 〖point e〗_n and L ⃡ at 〖point h〗_n (where (C_90 e_n ) ̅= 2(OB_o ) ̅) and (b) Line (C_90 e_n ) ⃗ intersects M ⃡ at point e_n and arc B0B90 at point Bn

FIGURE 6 Purpose of the diagram: To illustrate the special relationship in the auxiliary graph. (a) Point C90 is the center of the circle and (C_90 h_n ) ̅ is the radius, forming an arc intersecting M ⃡ at 〖point e〗_n and L ⃡ at 〖point h〗_n (where (C_90 h_n ) ̅ = 2(OB_o ) ̅). (b) Line (C_90 e_n ) ⃗ intersects M ⃡ at point e_n and arc B0B90 at point Bn

FIGURE 7 Purpose of the diagram: To show the special relationship in the combined auxiliary graph and the special relationship between an angle and a triple angle. (a) Point C_n (where (C_n E_n ) ̅ = 2(OBo) ̅ and point C_n moves along L ⃡) is the center of the circle and (C_n E_n ) ̅ is the radius, forming an arc intersecting L ⃡ at point Kn; 〖point C〗_n (where (C_n E_n ) ̅ = 2(OBo) ̅, and 〖point C〗_n moves along L ⃡) is the center of the circle and (C_n K_n ) ̅ is the radius, forming an arc intersecting M ⃡ at 〖point E〗_n (where (C_n K_n ) ̅ = 2(OB_o ) ̅); and (C_n E_n ) ⃗ intersects arc B0B90 at point Bn, (b) Point C90 is the center of the circle and (C_90 e_n ) ̅ is the radius, forming an arc intersecting M ⃡ at 〖point e〗_n and L ⃡ at point h_n (where (C_90 e_n ) ̅ = 2(OB_o ) ̅); (C_90 e_n ) ⃗ intersects M ⃡ at point e_n and arc B0B90 at point Bn; point C90 is the center of the circle and (C_90 h_n ) ̅ is the radius, forming an arc intersecting M ⃡ at point e_n and L ⃡ at 〖point h〗_n (where (C_90 h_n ) ̅ = 2(OB_o ) ̅); and (C_90 e_n ) ⃗ intersects M ⃡ at 〖point e〗_n and arc B0B90 at point Bn, (c) The result generated by combining Figures a and b; because Figures a and b were in the same area ((Oh_90 ) ̅ = (OK_90 ) ̅), the number of corresponding intersections is identical (i.e., arc B0B90). Similarly, the number of corresponding intersections within arc B0B90 is identical ((Oh_90 ) ̅ = (OK_90 ) ̅) and (d) Combining Figures a and b produces the following result: (C_n C_90 ) ̅ + (C_90 h_n ) ̅ = (C_90 h_n ) ̅ + (e_n W) ̅. Thus, (C_n C_90 ) ̅ = (e_n W) ̅






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(Figure 7)
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Deshichiu is offline  
 
March 12th, 2017, 08:18 PM   #2
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Do you understand the term "neusis construction"?
skipjack is offline  
March 13th, 2017, 02:56 AM   #3
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2 comments which may help.

1. Nobody will even look at your work unless you typeset it using Latex. This is a critical tool for writing modern mathematics for better or worse.

2. Unless your work can find a flaw in the existing theorem which guarantees that this construction is impossible, you will not find any traction. Without refuting this theorem, it is simply assumed that your construction is flawed and nobody will waste the time finding your mistake and trying to convince you of it.

It might help to know that the proof of this theorem is not geometric. It more or less goes like this:

It can be proved that every constructable value using only compass and straightedge must lie in a sequence of quadratic extension over the field of rational numbers. At the same time, basic trigonometric formulas tell us that trisecting an arbitrary angle is equivalent to computing roots of a polynomial whose splitting field is degree 3 over the rationals. Group theory tells us that these are not compatible so that some angles can't possibly lie in extensions of both degree 2 and 3 so there must be some angles that aren't constructable.

If you don't know this material, I suggest you spend more time reading than trying to do constructions.
Thanks from Joppy
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