Mean Geometry - Zachary Abel

29 Pages·2007·0.56 MB·English

Mean Geometry
Zachary R. Abel
July 9, 2007
Contents
1 Introduction 3
1.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Point Averages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Figure Averages. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.4 Finally! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Proofs of MGT 5
2.1 Proof 1: Spiral Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.2 Proof 2: Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3 Some Problems 7
3.1 Equilaterals Joined at the Hip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3.2 Napoleon’s Last Hurrah . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3.3 Extending MGT? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3.4 But Wait, There’s More! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3.5 The Asymmetric Propeller . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
4 Figure Addition 11
4.1 When Did the IMO Get So Easy? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
4.2 A Fresh Look at an Old Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
5 A Few Harder Problems 13
5.1 A Pretty(,) Busy Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
5.2 Hidden Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
5.3 Nagel Who? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
6 Additional Problems 19
7 Solutions to Additional Problems 21
1
List of Figures
1 Direct and inverse similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Point averaging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
3 Figure averaging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
4 MGT for triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
5 Similar triangles A OB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
i i
6 Similar triangles A OC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
i i
7 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
8 Napoleon’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
9 Equilateral triangle JKL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
10 More Napoleon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
11 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
12 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
13 Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
14 Point addition. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
15 Problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
16 Pompeiu’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
17 Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
18 Studying “quadrilateral” AB0C0D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
19 Quadrilateral AB00C00D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
20 Problem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
21 Circles O and O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1 2
22 Three similar triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
23 Problem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
24 Excircle I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
a
25 Problem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
26 Problem 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
27 Problem 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
28 Problem 14.c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
29 Van Aubel’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
30 Problem 16.a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
31 Problem 16.b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
32 Problem 16.c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
33 Problem 16.d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
34 Proof 2 of perpendicularization lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
35 Problem 18.b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
36 Problem 18.c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
37 Generalization of problem 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
38 Problem 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
39 Problem 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
40 Problem 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2
1 Introduction (for lack of a better section header)
There are generally two opposite approaches to Olympiad geometry. Some prefer to draw the diagram and simply
stare(labelingpointsonlycluttersthediagram!),waitingfortheinteractionsbetweentheproblem’svariouselements
to present themselves visually. Others toss the diagram onto the complex or coordinate plane and attempt to
establish the necessary connections through algebraic calculation rather than geometric insight.
Thisarticlediscussesaninterestingwaytovisualizeandapproachavarietyofgeometryproblemsbycombining
thesetwocommonmethods: syntheticandanalytic. We’llfocusonatheoremknownas“TheFundamentalTheorem
ofDirectlySimilarFigures”[4]or“TheMeanGeometryTheorem”(abbreviatedhereasMGT).Althoughtheresult
is quite simple, it nevertheless encourages a powerful new point of view.
1.1 Basic Definitions
A figure has positive orientation if its vertices are listed in counterclockwise order (like pentagons ABCDE or
LMNOP in diagram 1), and otherwise it has negative orientation (like VWXYZ or even EDCBA). Two figures
are similar if..., well, we’re all familiar with similar figures; two similar figures are directly similar if they have the
same orientation, and otherwise they’re inversely similar. For example, pentagon ABCDE is directly similar to
pentagon LMNOP, but inversely similar to VWXYZ.
Y
E Z
D
O
P
C N M X
W
A L
B
V
Figure 1: Direct and inverse similarity
1.2 Point Averages
Nowforsomenon-standardnotation. Wedefinetheaverageoftwopoints
AandB asthemidpointM ofsegmentAB,andwewrite 1A+1B =M. B
2 2
We can also compute weighted averages with weights other than 1 and
2
1, as long as the weights add to 1: the weighted average (1−k)A+(k)B 1A+1B=M
2 2 2
is the point X on line AB so that AX/AB = k. (This is consistent with
the complex-number model of the plane, though we will still treat A and N = 2A+1B
3 3
B as points rather than complex values.) For example, N = 2A+ 1B is
3 3 A
one-third of the way across from A to B (see Figure 2).
Figure 2: Point averaging
1.3 Figure Averages
We can use the concept of averaging points to define the (weighted) average of two figures, where a figure may
be a polygon, circle, or any other 2-dimensional path or region (this article will stick to polygons). To average two
figures, simply take the appropriate average of corresponding pairs of points. For polygons, we need only average
corresponding vertices:
3
A
4
C
4
1 1 B4 B3
A + B =C for i=1,2,3,4, so
2 i 2 i i A3 C3
1A A A A + 1B B B B =C C C C . A2 B2
2 1 2 3 4 2 1 2 3 4 1 2 3 4 C2
A
1 C
1 B
1
Figure 3: Figure averaging
1.4 Finally!
By now you can probably guess the punchline.
Mean Geometry Theorem. The (weighted) average of two directly similar figures is directly similar to the two
original figures.
Look again at Figure 3. Quadrilaterals A A A A and B B B B are directly similar, so the Mean Geometry
1 2 3 4 1 2 3 4
Theorem guarantees that their average, C C C C , is directly similar to both of them.
1 2 3 4
4
2 Proofs of MGT
Wementionedintheintroductionthatthisarticleincorporatessomeideas
frombothsyntheticandanalyticgeometry. Hereweoffertwoproofsofthe A3
Theorem–onebasedineachcategory–tobetterillustratetheconnection.
Although MGT holds for all types of figures, it suffices to prove result
for triangles (why?): C
3
MGT for Triangles. If 4 = A A A and 4 = B B B are di-
a 1 2 3 b 1 2 3
A
rectly similar triangles in the plane, then the weighted average 4 = 2
c
(1−k)A1A2A3 +(k)B1B2B3 = C1C2C3 is similar to both 4a and 4b C2
(Figure 4). B3 B2
2.1 Proof 1: Spiral Similarity
−→ B1 C1 A1
If 4 is simply a translation of 4 , say by vector V, then the transla-
b a
−→
tion by vector kV sends 4 to 4 , so they are directly similar (in fact Figure 4: MGT for triangles
a c
congruent).
Otherwise, there exists a unique spiral similarity1 rO that takes 4
θ a
to 4 (see [5, Theorem 4.82]). The three triangles A OB (for i=1,2,3)
b i i
have ∠A OB =θ and OB /OA =r, so they are all similar (see Figure 5).
i i i i
A A
3 3
C C
3 3
A A
2 2
C C
2 2
B3 B2 B3 B2
B1 C1 A1 B1 C1 A1
O O
Figure 5: Similar triangles A OB Figure 6: Similar triangles A OC
i i i i
Since C , C , and C are in corresponding positions in these three similar triangles (since A C /A B = k for
1 2 3 i i i i
i = 1,2,3), the three triangles A OC (i = 1,2,3) are similar to each other (Figure 6). Thus the spiral similarity
i i
r0O, whereθ0 =∠A OC andr0 =OC /OA , sends4A A A to4C C C , sothesetrianglesaredirectlysimilar,
θ0 1 1 1 1 1 2 3 1 2 3
as desired. (cid:3)QED
2.2 Proof 2: Complex Numbers
We’ll use capital letters for points and lower case for the corresponding complex numbers: point A corresponds to
1
the complex number a , and so on.
1
1i.e. arotationthroughanangleθ followedbyadilationwithratior,bothcenteredatthepointO
5
Since the triangles are similar, ∠A A A = ∠B B B and A A /A A = B B /B B . If z is the complex
2 1 3 2 1 3 1 3 1 2 1 3 1 2
number with argument ∠A2A1A3 and magnitude A1A3/A1A2, we have aa23−−aa11 =z = bb32−−bb11. Or, rearranged,
a =(1−z)a +(z)a and b =(1−z)b +(z)b . (♣)
3 1 2 3 1 2
To prove the theorem, we need to show that 4C C C behaves similarly (no pun intended), i.e. we need to show
1 2 3
that c =(1−z)c +(z)c . But since c =(1−k)a +(k)b for i=1,2,3 (by definition of 4 ), this equality follows
3 1 2 i i i c
directly from (♣):
c =(1−k)a +(k)b
3 3 3
(cid:0) (cid:1) (cid:0) (cid:1)
=(1−k) (1−z)a +(z)a +(k) (1−z)b +(z)b
1 2 1 2
(cid:0) (cid:1) (cid:0) (cid:1)
=(1−z) (1−k)a +(k)b +(z) (1−k)a +(k)b
1 1 2 2
=(1−z)c +(z)c ,
1 2
as desired. (cid:3)QED
6
3 Some Problems
The theorem has been proved, but there might be some lingering doubts about the usefulness of such a seemingly
simpleandspecializedstatement.2 Inthissection,we’llputtheTheoremtowork,andwe’lllearntorecognizewhen
and how MGT may be applied to a variety of problems.
3.1 Equilaterals Joined at the Hip
Problem 1 (Engel). OAB and OA B are positively oriented regular triangles with a common vertex O. Show
1 1
that the midpoints of OB, OA , and AB are vertices of a regular triangle. (Remember: positively oriented means
1 1
the vertices are listed in counterclockwise order.)
B
1
D
A
O
E
F
B
A
1
Figure 7: Problem 1
To show that DEF is equilateral, we’d like to express 4DEF as an average of two other directly similar
equilateral triangles. From the diagram we see that
1 1 1 1 1 1
A+ B =D, B+ O =E, and O+ A =F,
2 2 1 2 2 2 2 1
and putting these together yields
1 1
ABO+ B OA =DEF.
2 2 1 1
So we’re done by MGT, right?
Let’s make sure everything is in place. Are triangles ABO and B OA directly similar? The problem tells us
1 1
that they are both positively oriented equilateral triangles, so yes. Are we indeed taking a weighted average of the
two triangles? In other words, do the weights add to 1? Of course! 1 + 1 =1.
2 2
Now we can confidently apply the Mean Geometry Theorem and conclude that triangle DEF must be directly
similar to 4ABO and 4B1OA1, i.e. 4DEF is equilateral. (cid:3)QED
3.2 Napoleon’s Last Hurrah
Problem 2 (Napoleon’s Theorem). If equilateral triangles BCP, CAQ, ABR are erected externally on the sides of
triangle ABC, their centers X, Y, Z form an equilateral triangle (Figure 8).
2Pleasepardontheunintentionalalliteration.
7
P P
J
B X B X
L
R R
Z Z
C C
A A K
Y Y
Q Q
Figure 8: Napoleon’s Theorem Figure 9: Equilateral triangle JKL
The setup looks a lot like the configuration in the first problem, where two equilaterals share a common vertex.
We can try to mimic this earlier configuration by considering equilateral triangle 1PCB+1BAR=JKL. (Figure
2 2
9)
Now the way to reach the target points, X, Y, and Z, presents itself: X is on median JC, and it’s 1 of the way
3
across. This means that X = 2J + 1C, and likewise for Y and Z. We can write
3 3
2 1
XYZ = JKL+ CQA,
3 3
and since both 4JKL and 4CQA are negatively oriented equilateral triangles, we’re done! (cid:3)QED
3.3 Extending MGT?
Let’s look closer at that solution. First we averaged PCB and BAR: 1PCB+ 1BAR=JKL. Then, we averaged
2 2
this with CQA: 2JKL+ 1CQA=XYZ. Momentarily indulging ourselves in some questionable manipulation, we
3 3
can substitute the first equation into the second and simplify to find
1 1 1
PCB+ BAR+ CQA=XYZ. (♠)
3 3 3
This suggests that we may be able to generalize MGT to three figures, like so:
Extended MGT. Define the weighted average of 3 points ω P +ω P +ω P (where the weights ω , ω , and
1 1 2 2 3 3 1 2
ω are real numbers and add to 1) just as we would in the complex plane.3 Then if 4A A A , 4B B B , and
3 1 2 3 1 2 3
4C C C are directly similar triangles, their weighted average
1 2 3
ω A A A +ω B B B +ω C C C
a 1 2 3 b 1 2 3 c 1 2 3
is also directly similar to them. This naturally extends to include figures other than triangles (as long as they’re all
similar to each other) or more than 3 similar figures (as long as the weights add to 1).
Can you prove this? Anyway, by this generalized version of MGT, equation (♠) alone provides a succinct,
one-line proof of Napoleon’s Theorem. (Cool, huh?)
3.4 But Wait, There’s More!
Problem 3. Napoleon isn’t done with us yet: prove that triangles ABC and XYZ have the same centroid.
3Asanotablespecialcase, 13P1+ 13P2+ 31P3 isthecentroidoftriangleP1P2P3.
8
Like Napoleon’s Theorem itself, this can also be done in one line. Here’s the P
first half:
1 1 1
BXC+ CYA+ AZB =??? (♥)
3 3 3 B X
Before you read ahead, try to figure out how and why this proves the problem. R
Z
(Jeopardy song plays...) Ok, welcome back! G
Thethreeredtrianglesareisosceles30−120−30triangles,sotheyareallsimilar. C
A
IfGandH denotethecentroidsof4ABC and4XYZ respectively,thentheresult
of the expression in equation (♥) is triangle GHG, which (by the Theorem) must Y
also be a 30−120−30 triangle. But two of its vertices are at the same place! The
trianglehasthusdegeneratedintoapoint,soallthree ofitsverticesareatthesame
place, and G=H. (cid:3)QED Q
3.5 I Can’t Take Any More Equilaterals! and the Asymmet- Figure 10: More Napoleon
ric Propeller
Let’s do one more problem of a similar flavor before we move on.
Problem 4. Positively oriented equilateral triangles XAB, XCD, and XEF share a vertex X. If P, Q, and R
are the midpoints of BC, DE, and FA respectively, prove that PQR is equilateral.
Problem 5 (Crux Mathematicorum). In quadrilateral ABCD, M is the midpoint of AB, and three equilateral
triangles BCE, CDF, and DAG are constructed externally. If N is the midpoint of EF and P is the midpoint of
FG, prove that MNP is equilateral.
Problem 6. The four triangles ABC, AA A , B BB , and C C C are directly similar, and M , M , and M are
b c a c a b a b c
the midpoints of B C , C A , and A B . Show that M M M is also similar to ABC.
a a b b c c a b c
A
b
A
A c
B A
M M
P b
B
C R
M
c C
b
A
F G D C
X B
E C
B
c
E P
N
Q
C
D a
F
M
a
B
Figure 11: Problem 4 a
Figure 12: Problem 5
Figure 13: Problem 6
Hm, didn’t I just say one more problem?
Indeed, all we have to do is solve problem 6; the rest come free. The diagrams are drawn to illustrate illustrate
how the first two problems are special cases of problem 6.4 So, let’s solve problem 6, known as the Asymmetric
Propeller [2].
The problem gives us a plethora of similar triangles to work with, so our first instinct should be to try to write
triangle M M M as an average of these:
a b c
M M M =ωABC+ω AA A +ω B BB +ω C C C. (♦)
a b c a b c b a c c a b
4Inproblem4, themiddletrianglehasdegeneratedintopointX. Inproblem5, oneoftheoutsidetrianglesdegeneratesintopoint
F.
9