a geometric proof of napoleon’s theorem chrissy folsom june 8, 2000 math 495b

A Geometric Proof Of Napoleon’s Theorem

Chrissy Folsom

June 8, 2000

Math 495b

Napoleon the Mathematician?

• Theorem named after Napoleon Theorem named after Napoleon BonaparteBonaparte

• Questionable as to whether he really Questionable as to whether he really deserves credit.deserves credit.

• Some sources say he excelled in mathSome sources say he excelled in math

• Earliest definite appearance of Earliest definite appearance of theorem: 1825 by Dr. W. Rutherford in theorem: 1825 by Dr. W. Rutherford in “The Ladies Diary”“The Ladies Diary”

Napoleon’s Theorem

• Given Given anyany triangle, construct an equilateral triangle triangle, construct an equilateral triangle on each of its legs. Then the centers of the three on each of its legs. Then the centers of the three outer triangles form another equilateral triangle outer triangles form another equilateral triangle (Napoleon triangle).(Napoleon triangle).

Equilateral Triangles

(center/centroid)(center/centroid)

Defined

(center/centroid)(center/centroid)

• ABC original triangleABC original triangle

aa = BC , = BC , bb = AC , = AC , cc = AB = AB

tt

uu

ss

• G, I, H centroidsG, I, H centroids

ss = GI = GI

• We will show that all three We will show that all three sides of GHI are equal in sides of GHI are equal in length.length.

The Setup

Proof (Find Find ss in terms of in terms of a,b,ca,b,c))

30 GAB IAC

(A = both point and angle)

QuestionQuestion: Can we find : Can we find tt in terms of in terms of cc??

YES!!!YES!!!

Law of Cosines on AGI:Law of Cosines on AGI:

60AGAI

)60cos(A2222 uttus** **

Proof

• cc is the base of an is the base of an equilateral triangle, G is its equilateral triangle, G is its centroid.centroid.

tc

1

2

1)30cos(

33

2

2

1 cct So

)60cos(A2222 uttus**

Substitute for Substitute for tt and and uu in in **

GG

30

3,

3

ct

bu Likewise for Likewise for u:u:

Substitute

)60cos(A3

2

33

222

bccbs

33

ct

bu ,

)60cos(A23 222 bccbs**

)60cos(A2222 uttus**

Cosines

)60cos(A23 222 bccbs**

sin(A)2

3

2

cos(A)

)60sin(sin(A))60cos(cos(A))60cos(A Recall:Recall:

sin(A)3cos(A)3 222 bcbccbs

Plug it into Plug it into * * ::

** **

Look at ABC• Law of Cosines on ABCLaw of Cosines on ABC

• Area of ABC:Area of ABC:

sin(A)ch

hb2

1Area

hh

bb

cc

cos(A)2222 bccba (1)(1)

sin(A)2

1Area bc (2)(2)

Plug in Plug in (1)(1) and and (2)(2) to to **

Plug it in

cos(A)2222 bccba (1)(1)

sin(A)2

1Area bc (2)(2)

sin(A)3cos(A)3 222 bcbccbs **

• Plugging in:Plugging in:

ABC)of Area(32)(2

13 2222 cbas** **

Are We Done?

Hence, an equilateral triangle.Hence, an equilateral triangle.

ABC)of Area(32)(2

13 2222 cbas** **

This is symmetric in This is symmetric in a,b,ca,b,c

2

2222

)3

(32)(2

1)3

)G(H,(

ABC)of Area)H(I,(

d

cbad

So...So...

Yes, We Are Done (with the proof)!!!Yes, We Are Done (with the proof)!!!

More Neat Stuff: Tiling

(4)(4) Another equilateral triangle!! Another equilateral triangle!!

(1)(1) Rotate Rotate original original triangle 120triangle 120o o

about about centroid of centroid of each each adjacent adjacent equilateralequilateral

(2)(2) Connect exposed Connect exposed vertices to get vertices to get equilateral trianglesequilateral triangles

(3)(3) Connect vertices Connect vertices of 3 new equilateral of 3 new equilateral trianglestriangles

ConclusionsSome Generalizations:• If similar triangles of any shape are added

onto the original triangle, then any triple of corresponding points on triangles forms a triangle of same shape.

• Begin with arbitrary n-gon. Attach a regular n-gon to each side. Connect similar points and get another regular n-gon. (Napoleon when n=3).