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多邊形的尋短

歐翰青

Abstract

In the field of geometry, the shortest path problem has always been an old, popular and rich applications issue. From the earliest to the river then go back and put out the fire to modern telecommu- nications network systems, whether it is set up rail lines, facilities, relief route planning, it all has applications. This work is focusing on how to find the shortest circumference of the inscribed polygon within a polygon. The subjects we’re discussing are whether the solu- tion exists or not, solvability conditions, mapping methods and what graphics it will approach if no solution. Research method is to use the Mirror method or as light reflection principle, which commonly used in geometry, and classified into odd and even side polygon various situations to discuss respectively.

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2.1 有 有 有解 解 解情 情 情況 況 況

2.1.1 銳銳銳角角角三三三角角角形形形:

1. 設 D, E, F分別為 BC, AC, AB 上的任意點, 將原三角形對其三邊依序 鏡射五次, 並延長 BC, B00C00, B00000C0000可得圖 1 (0的數目意義請見附錄 .B):

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=2(DE+FE+ED) = DF+FE0+E0D00+D00F000+F000E0000+E0000D0000

≥ DD00+D00D0000 ≥DD0000. 四邊形 AC0PB中,

∠P0PC =360− (180− ∠AC0B00) − (180− ∠ABC) − ∠C0AB

= ∠B+ ∠C− ∠A = ∠PP0B00000

⇒ BC//B00000C0000 (內錯角相等), 又 DC =D0000C0000

⇒ DCD0000C0000 為平行四邊形, DD0000 =CC0000(定值)

⇒ DD0000 為定值不受 D 點移動之影響, 且為 D, D0000 兩點間可能的最 短距離.

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⇒D, F, E0, D00, F000, E0000, D0000 會共線⇒ 4FED周長之兩倍

=DF+FE0+E0D00+D00F000+F000E0000+E0000D0000 =DD0000

∠FDB = ∠EDC,∠DEC = ∠FEA,∠AFE = ∠DFB 時, 4FED 為此三角形內接周長最短三角形. 同理, 此證明法在凸 n 邊形(n = 2k+1, k∈ N)時也是可行的, 設 n 個頂點依序為 a1, a2,· · ·, an, 只要對

ana1鏡射 2n−1 次,並用稍後證明的引理 2.2 來證明 anan1//an0(2n1)an10(2n2)

(即三角形的 BC//B00000C0000), 其他的步驟一概相同, 同時可知奇數邊形 時若解存在則必為唯一.

2. 求求求證證證. 作三角形三邊之垂足並將之連接, 所得的三角形會符合定理 2.1 的條件, 也就是 ∠FDB = ∠EDC,∠DEC = ∠FEA,∠AFE = ∠DFB.

∵ HD ⊥BC, HF ⊥BA

∴ HDBF 共圓, 令

∠FDB= ∠1,∠FHB = ∠2,∠EDC = ∠3,∠EHC = ∠4,

1

1

_

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∠FDB = ∠EDC,∠DEC= ∠FEA,∠AFE= ∠DFB, 證畢.

2.1.2 圓圓圓內內內接接接四四四邊邊邊形形形－－－圓圓圓心心心在在在四四四邊邊邊形形形內內內的的的情情情況況況:

ABCD 為圓內接四邊形, 且圓心在四邊形中, 連接兩對角線, 令其交點為 E, 過 E 分別對四邊垂線, 與 AB, BC, CD, DA 的交點分別為 FGHI, 連接 FGH I. 我們宣稱此四邊形為其最小內接四邊形之一, 而其內最小四邊形有 無限多個, 只要在 AB 上特定的範圍內任取一點 F0, 過其作 FG, FI 之平行線 分別交於 G0, I0,再過 G0, I0作 GH, IH 之平行線必與 CD 交於一點 H0, 四邊 形 F0G0H0I0即為另一解, 如圖 4 :

∠I HD 時, 四邊形 FGH I 會最小.

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CD //←−−→

C00D000(這點稍 後證明), 四邊形 FGHI 的周長≥ HH000 =CC00 (定值), 易知當 HIF0G00H000 共線時, 四邊形 FGHI 為過 H 的內接周長最小四邊形, 而事實上 H 只要在 邊上的一定範圍內內接周長最小四邊形的周長都是一樣的, 即是這種情況有 無限多個解, 這點稍後證明.

∠AIF= ∠DI H,∠AFI = ∠BFG,∠FGB= ∠HGC

· · ·, an 只要對 ana1鏡射 n−1 次再重複上述步驟即可.

∠AIF = ∠DI H,∠AFI = ∠BFG,

∠FGB= ∠HGC,∠GHC = ∠I HD, 如圖 6:

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∵ ∠AFE= ∠EI A =90

⇒ AIEF共圓

∴ ∠1 = 1 2

IE

= ∠3 ,同理∠2= 12

GE

= ∠4, 又∠3= 12

_

CD

= ∠4

⇒ ∠1= ∠3= ∠4= ∠2

⇒ ∠1= ∠2

⇒90− ∠1=90− ∠2

∴ ∠AIF = ∠DI H,

⇒以此作法所做出之內接四邊形周長最小.

CD與←−−→

C00D000 會形成一夾角, 令其 為∠X, 如圖 7:

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CD與←−−→

C00D000 會之夾角: 由於 XC00B0AD 會形成一凹五邊形, 由 n 邊形內角和= (n−2)180可得

∠X+ ∠C+ (360− ∠B) + ∠A+ (180− ∠D) = 5400

⇒ ∠X = (∠B+ ∠D) − (∠A+ ∠C), 若←→

CD與←−−→

C00D000 之交點在另一邊, 則

∠X = (∠A+ ∠C) − (∠B+ ∠D), 故我們可將∠X 寫為

∠X =|∠A− ∠B+ ∠C− ∠D|.

∠X = |∠k1− ∠k2+ ∠k3− ∠k4+ · · · + (−1)n1∠kn1+ (−1)nkn|.

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∠X+ (180− ∠kn) + ∠k1+ (360− ∠k2) + ∠k3+ (360− ∠k4) + · · ·

= (n+1−2)180

⇒ ∠X = −∠k1+ ∠k2− ∠k3+ ∠k4− · · · − (−1)n1∠kn1− (−1)n∠kn. 若兩線之交點在另一邊, 則

∠X = ∠k1− ∠k2+ ∠k3− ∠k4+ · · · + (−1)n1∠kn1+ (−1)n∠kn, 故可將∠X寫為

∠X = ∠k1− ∠k2+ ∠k3− ∠k4+ · · · + (−1)n1∠kn1+ (−1)n∠kn

. 如果我們將一個圓內接四邊形鏡射, 如圖 9:

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CD與 ←−−→

C00D000 之夾角為|∠A− ∠B+ ∠C− ∠D| 但由 於四邊形 ABCD 為一圓內接四邊形

⇒ ∠A+ ∠C= ∠B+ ∠D, 由引理 2.2 可知←→

CD與←−−→

C00D000 的夾角為|∠A− ∠B+ ∠C− ∠D| =0

⇒←→

CD//←−−→

C00D000,

⇒H000H JO為平行四邊形

⇒ JO= HH000.

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2.1.3 凸凸凸 n 邊邊邊形形形 (n=2k, kN)－－兩兩組間角角和和相相等等且且卡卡點點未未卡卡到到的的情情況況:

(a6, a60(5)

),再過卡點

a1, a20, a300· · ·an0(n1)

a1, a20, a300· · ·a60(5) , 作 anan0(n1)a6a60(5)之平行線, 可得圖 10 (圖以六邊形為例):

ana1, a1a20,· · · , an10(n2)an0(n1)

a6a1, a1a20,· · · , a50(4)a60(5) 上的交點鏡射回原 2n 邊形即為所求.

anan1//an10(n2)an0(n1)

a6a5//a50(4)a60(5) ,

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2.1.4 凸凸凸 n 邊邊邊形形形 (n=2k+1, kN)－－鏡鏡射射作作圖圖後後能能與與每每一一鏡鏡射射邊邊皆皆有有交

⇒ ∠1= ∠2, 又 b1a2b2c2與 b3a3b2c3皆為圓內接四邊形 ⇒ ∠3= ∠4, 同理, 在各頂點皆找到一條線使得

∠a1= ∠kn + ∠k1,∠a2 = ∠k1+ ∠k2+ · · · + ∠an = ∠kn1+ ∠kn, 如圖 12 (圖以五邊形為例):

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⇒ (∠kn + ∠k1) + (∠k1+ ∠k2) + · · · + (∠kn1+ ∠n) = (n−2)180

⇒ ∠k1+ ∠k2+ · · · + ∠kn = (n−2)90

⇒ ∠k1= (n−2)90− (∠k2+ · · · + ∠kn)

= (n−2)90− (∠a3+ ∠a5+ · · · + ∠an), 同理可得

∠ki(i=1, 2, 3,· · · , n) = (n−2)90− (∠k1+ · · · + ∠kn− ∠ki)

= (n−2)90− (∠ai+2+ ∠ai+4+ · · · + ∠ai1) (若超過 n 則由 1 重新數過, 也就是若 i+m≥n則∠ai+m 視為∠ai+mn)= (n−2)90∠ai+2開始順時針數 (n21) 個間角時, 若要使凸 n 邊形有解, 則∠k1必須使 ci, ci+1能投影到凸 n 邊形的各個邊上, 此時 0 < ∠ki <90

⇒0 < (n−2)90 由∠ai+2開始順時針數 (n21) 個間角<90

⇒ (n−2)90 >∠ai+2開始順時針數 (n21) 個間角> (n−3)90

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0(2n1)an1

0(2n2) 的 夾 角 (a5a4 與 a5

0(9)a4

0(8) 的夾角)

= [(∠a1+ ∠a3+ · · · + ∠an) − (∠a2+ ∠a4+ · · · + ∠an1)]

+ [(∠a2+ ∠a4+ · · · + ∠an1) − (∠a1+ ∠a3+ · · · + ∠an)] =0

⇒anan1//an0(2n1)an1

0(2n2) (a5a4//a50(9)a40(8))

∴ anan1 (a5a4)上的任意點到其在 an0(2n1)an1

0(2n2)a5

0(9)a4

0(8) 上的對應

anan1(←→

a5a4) 即為 L1, 而要求出凸 n 邊形內接最小 n 邊形必須 在 L1 與 L2 之間找一平行線, 使得此線也通過 anan1 (a5a4)上的任意點在 an0(n1)an10(n2) (a50(4)a4000)上的對應點.

∠San0(n1)

T = ∠an0(n1)

San

0(2n1)

(∠Sa50(4)

T = ∠a50(4)

Sa5

0(9))(內錯角)

∠anTan0(n1)

= ∠an0(n1)

San

0(2n1)

(∠a5Ta50(4)

= ∠a50(4)

Sa5

0(9))(全等)

⇒ ∠San0(n1)

T = ∠anTan0(n1)

(∠Sa50(4)

T = ∠a5Ta50(4))

⇒ anTan0(n1)

S(a5Ta50(4)

S), 為一等腰梯形, 又 U, V 為其腰上的中點

⇒anU =an0(n1)V (a5U = a50(4)V),

(a5a4)上之對應點, 且 UV // L1 // L2⇒ UV 在凸 n 邊形上的各點即為所 求.

2.2 無 無 無解 解 解情 情 情況 況 況

2.2.1 鈍鈍鈍角角角三三三角角角形形形:

4ABC 中, ∠A 為鈍角, 點 D 為 BC 上的任意點, 將此三角形對 AC, AB 鏡 射,令反射後的 D 點分別為 E, F, 如圖 15:

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⇒ A, E, F 必可形成一三角形, 且點 A 會在 AB, AC 上其中的點之連線 (即 下圖中的 HG) 與 EF 之間. 現今 G, H分別為 AC, AB 上之任意點, 分別在 AB, AC 找任意點 H, G, 則 A 為於四邊形 EFGH 內部, 連接 GH, HD, DG, 將其對 AB, AC 作反射, 得 HF, GE, 如圖 16:

4HDG的周長= FH+HG+GE,

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(等號在 AHG 重疊時會成立)4HDG 的周長 ≥2DA, 而 DA為過 A 且垂 直 BC 之垂線時會最小

⇒鈍角三角形的內接最小三角形圖形趨近於過 A 且垂直 BC 之垂線之兩倍.

2.2.2 直直直角角角三三三角角角形形形:

∵ ∠FAE =2∠BAC=180

⇒FAE三點共線

⇒直角三角形的內接最短三角形趨近於其斜邊上之高的兩倍.

2.2.3 圓圓圓內內內接接接四四四邊邊邊形形形－－－圓圓圓心心心在在在四四四邊邊邊形形形外外外的的的情情情況況況:

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AB0 與 CD 的夾角會大於 CD 與←−→

HH000 的 夾角 (←−→

HH000 是利用上述作圖法找出, 使對應的四邊形鏡射後為直線的線), 即 ←−→

HH000 同時與←→

AB0, AD 不可能同時有交點, 因此周長最小的四邊形會是 趨近於4ABH之四邊形 (CD 為較靠近圓心的那一邊).

= 1

2(

BD

AC)

= 1

2[(

CD+

BC

) − (

AB +

BC

)]

= 1

2(

CD

AB ),

HH000 的夾角為

∠I HD = 1 2

_

CD

= ∠IED =90− ∠ADB = 1

2(180

_

AB ), 當圓心落在該四邊形外時, CD

_

>

180

⇒CD與←→

AB0的夾角>CD與←−→

HH000 的夾角

⇒←−→

HH000與←→

HH000與←→

AB0有交點, 且 CD 與←→

AB0的夾角>CD與←−→

HH000 的夾角, 則 H 會比與CD 的交點更靠近 C, 使得 I 點落在AD 之外)

⇒圓內接四邊形─圓心在四邊形外的情況不可能有解.

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2.2.4 圓圓圓內內內接接接四四四邊邊邊形形形－－－圓圓圓心心心在在在四四四邊邊邊形形形上上上的的的情情情況況況:

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AB0的夾角 = 12(

_

CD

AB) = 1

2(180

_

AB

) = CD與←−→

HH000 的夾角. 而為了要使 ←−→

AB0 有交點, 必須←−→

HH000 與←→

AB0 重合, 又∠I HD

=CD與←−→

HH000 的夾角

= 12(180−AB

_

)

⇒ AEHD共圓

⇒ ∠EHD =180− ∠DAE =90, H為對角線交點對 CD 所作之垂足.

2.2.5 非非非圓圓圓內內內接接接四四四邊邊邊形形形的的的情情情況況況:

∵ ABCD為一非圓內接四邊形

∴ CD不平行 C00D000 ⇒ ∠H000HD 6= ∠HH000C00, 又∠H000HD = ∠H0(7)H000D000

⇒ ∠H0(7)H000D000 6= ∠HH000C00

⇒ ∠H0(7)H000D000與∠HH000C00 不互為對頂角

⇒ ∠HH000H0(7) 三點不可能共線.

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HH000 = I H000+H000I0(4),

I I0(4) = H000I+I H> HH000,

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∠B0(5)D000B0在四邊形內的角≥180 時, 會卡到 D 點, 又

∠DB0D000 = ∠AB0C00+ (∠C00B0D000+ ∠DB0A) = 2∠B,

∠B0(5)D00B0 = ∠C00D000A0(4)+ (∠A0(4)D00B”(5)+ ∠B0D00C00) = 2∠D, 當∠B<90時會卡到 B 點, 當∠D<90 時會卡到 D 點, 而當∠B+ ∠D>

∠A+ ∠C 時,則會卡到 A 點或 C 點⇒當非圓內接四邊形較大的兩對角有 其中一個角<90 時,它的內接最小四邊形會趨近於一三角形, 且三角形在 四邊形頂點上的角會在非圓內接四邊形較大的兩對角> 90 的那個角上.

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∠DB0D000 = ∠AB0C00+ (∠C00B0D000+ ∠DB0A) =2∠B

∠B0(5)D000B0 = ∠C00D000A0(4)+ (∠A0(4)D000B0(5)+ ∠B0D000C00) =2∠D

⇒當非圓內接四邊形較大的兩對角兩個角皆≥ 90 時,它的內接最小四邊 形會 趨近於其較短的對角線之兩倍, 也就是非圓內接四邊形較大的兩對角 之連線的兩倍.

2.2.6 凸凸凸 n 邊邊邊形形形 (n=2k, kN)－－兩兩組間角角和和相相等等, 但但卡卡點點卡卡到到的的情情況況:

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a6a1, a1a20,· · · , a50(4)a60(5) , 也就是必須找一 anan0(n1)(a6a60(5))的平行線將紅線與綠線分開, 而由於上 下卡點交錯排列, 代表此平行線不可能存在.

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2.2.7 凸凸凸 n 邊邊邊形形形 (n=2k, kN) 兩兩組間角角和和不不相相等等的的情情況況:

∵此凸 n 邊形為一非圓內接凸 n 邊形

∴ anan1(a6a5)不平行 an0(n1)an10(n2) (a50(4)a60(5)))

⇒ ∠H0(n1)Han 6= ∠HH0(n1)an1(⇒ ∠H0(5)Ha5 6= ∠HH0(5)a5).

∠H0(n1)Han = ∠H0(2n1)H0(n1)an0(n1)

(∠H0(5)Ha5 = ∠H0(11)H0(5)a60(5))

⇒∠H0(2n1)H0(n1)an0(n1)

6= ∠HH0(n1)an10(n2)

(∠H0(11)H0(5)a60(5)

6= ∠HH0(5)a50(4))

⇒ ∠H0(2n1)H0(n1)a 0(n1) 與∠HH0(n1)a 0(n2) 不互為對頂角

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= HH0(n1)(HH0(5))

= I H0(n1)+H0(n1)I0(n)(I H0(5)+H0(5)I0(6)).

= I I0(n)(I I0(6))

= I H0(n1)+H I(I H0(5)+H I)

> HH0(n1) (HH0(5)).

We refer to “total training time of method x” as the whole LIBSVM training time (where method x is used for reconstructing gradients), and “reconstruction time of method x” as

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The main purpose of this research is to compare how a traditional narrative teaching method and a GeoGebra-based computer-assisted instructional method affect

ABSTRACT Keywords : Mechanical well cleaning, Atmospheric pressure, Air-lifting Method The air-lifting method is one of the techniques used to pump water or clean wells.The method

In order to have a complete and efficient decision-making policy, it has to be done with multi-dimensional reflection and analysis; The multi-criteria decision-making analysis

The objective of this research is to conduct the theoretical and experimental studies on how to use the Empirical Mode Decomposition method to process the response of a single

The purpose of this paper is to use data mining method in semiconductor production to explore the relation of engineering data and wafer accept test.. In this paper, we use two

In this Research, the Analytic Hierarchy Process and Case Study Method are used, from which three main factors affecting the work progress were obtained: “Encountering of

This research includes the measurement of resection method, the principle of profile survey of tunnel and the application of Visual Basic programming language to

This research by the contingent valuation method（CVM）, divides into the settlement preservation overall benefit the use value with the non- use value, the use value divides

This research makes use of the Gray Relation Method to construct evaluation criteria of restaurant service quality at airports and analyzes those criteria in four categories:

Therefore, China is regarded as the biggest market of the future lighting , so that Taiwan's manufacturers use the method of Strategic alliances or joint ventures to