when did imo become easy

7/25/2019 When Did Imo Become Easy

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Mean Geometry

Zachary R. Abel

July 9, 2007


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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 Napoleons Last Hurrah . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3.3 Extending MGT? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3.4 But Wait, Theres 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

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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 AiOBi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

6 Similar triangles AiOCi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

7 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

8 Napoleons Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

9 Equilateral triangleJ KL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 Pompeius Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1217 Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

18 Studying quadrilateralAB CD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

19 Quadrilateral AB CD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

20 Problem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

21 CirclesO1 andO2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

22 Three similar triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

23 Problem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

24 ExcircleIa. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

25 Problem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

26 Problem 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

27 Problem 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2128 Problem 14.c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

29 Van Aubels 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

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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 (labeling points only clutters the diagram!), waiting for the interactions between the problems various elements

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.

This article discusses an interesting way to visualize and approach a variety of geometry problems by combining

these two common methods: synthetic and analytic. Well focus on a theorem known as The Fundamental Theorem

of Directly Similar Figures [4] or The Mean Geometry Theorem (abbreviated here as MGT). Although the result

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 hasnegative orientation (likeV W X Y Z or even EDCBA). Two figures

aresimilar if. . . , well, were all familiar with similar figures; two similar figures aredirectly similar if they have the

same orientation, and otherwise theyre inversely similar. For example, pentagon ABCDE is directly similar to

pentagonLMNOP, but inversely similar to V W X Y Z .

AB

C

D

E

L

MN

O

P

V

W

X

YZ

Figure 1: Direct and inverse similarity

1.2 Point Averages

A

N= 2

3A+

1

3B

1

2A+

1

2B = M

Figure 2: Point averaging

Now for some non-standard notation. We define theaverage of two points

Aand B as the midpointMof segmentAB, and we write 12

A + 12

B= M.

We can also compute weighted averages with weights other than 12

and1

2, as long as the weights add to 1: the weighted average (1 k)A + (k)B

is the point Xon 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

B as points rather than complex values.) For example, N = 23

A + 13

B is

one-third of the way across from A to B (see Figure 2).

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:

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2 Proofs of MGT

A1

A2

A3

B1

B2B3

C1

C2

C3

Figure 4: MGT for triangles

We mentioned in the introduction that this article incorporates some ideas

from both synthetic and analytic geometry. Here we offer two proofs of the

Theorem one based in each category to better illustrate the connection.

Although MGT holds for all types of figures, it suffices to prove result

for triangles (why?):

MGT for Triangles. Ifa = A1A2A3 andb = B1B2B3 are di-rectly similar triangles in the plane, then the weighted averagec =(1 k)A1A2A3 + (k)B1B2B3 = C1C2C3 is similar to botha andb(Figure 4).

2.1 Proof 1: Spiral Similarity

Ifb is simply a translation ofa, say by vectorV, then the transla-tion by vector k

V sendsa toc, so they are directly similar (in fact

congruent).

Otherwise, there exists a unique spiral similarity1

rO that takesatob (see [5, Theorem 4.82]). The three triangles AiOBi (fori = 1, 2, 3)have AiOBi= and OBi/OAi= r, so they are all similar (see Figure 5).

A1

A2

A3

B1

B2B3

C1

C2

C3

O

Figure 5: Similar triangles AiOBi

A1

A2

A3

B1

B2B3

C1

C2

C3

O

Figure 6: Similar triangles AiOCi

Since C1, C2, and C3 are in corresponding positions in these three similar triangles (since AiCi/AiBi = k for

i = 1, 2, 3), the three triangles AiOCi (i = 1, 2, 3) are similar to each other (Figure 6). Thus the spiral similarityr

O, where

=

A1

OC1

and r

=OC1

/OA1

, sends A1A2A3 to C1C2C3, so these triangles are directly similar,as desired.

QED

2.2 Proof 2: Complex Numbers

Well use capital letters for points and lower case for the corresponding complex numbers: point A1 corresponds to

the complex number a1, and so on.

1i.e. a rotation through an angle followed by a dilation with ratio r, both centered at the point O

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Since the triangles are similar, A2A1A3 = B2B1B3 and A1A3/A1A2 = B1B3/B1B2. Ifz is the complex

number with argument A2A1A3 and magnitude A1A3/A1A2, we have a3a1a2a1

=z = b3b1b2b1 . Or, rearranged,

a3 = (1 z)a1+ (z)a2 and b3 = (1 z)b1+ (z)b2. ()

To prove the theorem, we need to show thatC1C2C3 behaves similarly (no pun intended), i.e. we need to showthatc3 = (1

z)c1+ (z)c2. But sinceci= (1

k)ai+ (k)bi fori = 1, 2, 3 (by definition of

c), this equality follows

directly from ():

c3 = (1 k)a3+ (k)b3= (1 k)(1 z)a1+ (z)a2+ (k)(1 z)b1+ (z)b2= (1 z)(1 k)a1+ (k)b1+ (z)(1 k)a2+ (k)b2= (1 z)c1+ (z)c2,

as desired. QED

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3 Some Problems

The theorem has been proved, but there might be some lingering doubts about the usefulness of such a seemingly

simple and specialized statement.2 In this section, well put the Theorem to work, and well learn to recognize when

and how MGT may be applied to a variety of problems.

3.1 Equilaterals Joined at the HipProblem 1 (Engel). OAB andOA1B1 are positively oriented regular triangles with a common vertexO. Show

that the midpoints ofOB, OA1, andAB1 are vertices of a regular triangle. (Remember: positively oriented means

the vertices are listed in counterclockwise order.)

A

B

A1

B1

O

D

EF

Figure 7: Problem 1

To show that DEF is equilateral, wed like to expressDEF as an average of two other directly similarequilateral triangles. From the diagram we see that

1

2 A +1

2 B1 = D, 1

2 B+1

2 O= E, and 1

2 O+1

2 A1 = F,

and putting these together yields

1

2ABO+

1

2B1OA1 = DEF.

So were done by MGT, right?

Lets make sure everything is in place. Are trianglesABO and B1OA1 directly similar? The problem tells us

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! 12

+ 12

= 1.

Now we can confidently apply the Mean Geometry Theorem and conclude that triangleDEFmust be directly

similar toABO andB1OA1, i.e.DEF is equilateral. QED

3.2 Napoleons Last Hurrah

Problem 2 (Napoleons 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).

2Please pardon the unintentional alliteration.

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A

B

C

P

Q

R

X

Y

Z

Figure 8: Napoleons Theorem

A

B

C

P

Q

R

X

Y

Z

J

K

L

Figure 9: Equilateral triangle J KL

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 12

P CB + 12

BAR = J KL. (Figure

9)

Now the way to reach the target points, X, Y, andZ, presents itself: Xis on median J C, and its 13 of the wayacross. This means that X= 2

3J+ 1

3C, and likewise for Y andZ. We can write

XY Z=2

3JKL +

1

3CQA,

and since bothJKL andCQA are negatively oriented equilateral triangles, were done! QED

3.3 Extending MGT?

Lets look closer at that solution. First we averagedP CB andBAR: 12

P CB + 12

BAR = JKL. Then, we averaged

this withC QA: 23

JKL + 13

CQA = XY Z.Momentarily indulging ourselves in some questionable manipulation, we

can substitute the first equation into the second and simplify to find

1

3P CB +

1

3BAR +

1

3CQA = XY Z. ()

This suggests that we may be able to generalize MGT to three figures, like so:

Extended MGT. Define theweighted average of 3 points 1P1+ 2P2+ 3P3 (where the weights1, 2, and

3 are real numbers and add to 1) just as we would in the complex plane.3 Then ifA1A2A3,B1B2B3, and

C1C2C3 are directly similar triangles, their weighted average

aA1A2A3+ bB1B2B3+ cC1C2C3

is also directly similar to them. This naturally extends to include figures other than triangles (as long as theyre 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 Napoleons Theorem. (Cool, huh?)

3.4 But Wait, Theres More!

Problem 3. Napoleon isnt done with us yet: prove that trianglesABC andXY Zhave the same centroid.

3As a notable special case, 13P1+

1

3P2+

1

3P3 is the centroid of triangle P1P2P3.

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A

B

C

P

Q

R

X

Y

ZG

Figure 10: More Napoleon

Like Napoleons Theorem itself, this can also be done in one line. Heres the

first half:1

3BX C+

1

3CY A +

1

3AZB = ??? ()

Before you read ahead, try to figure out how and why this proves the problem.

(Jeopardy song plays. . . ) Ok, welcome back!

The three red triangles are isosceles 30

120

30 triangles, so they are all similar.

IfGandHdenote the centroids ofABCand XY Zrespectively, then the resultof the expression in equation () is triangle GHG, which (by the Theorem) mustalso be a 30 120 30 triangle. But two of its vertices are at the same place! Thetriangle has thus degenerated into a point, so all threeof its vertices are at the same

place, and G = H. QED

3.5 I Cant Take Any More Equilaterals! and the Asymmet-

ric Propeller

Lets do one more problem of a similar flavor before we move on.

Problem 4. Positively oriented equilateral trianglesXAB, XCD, andXEF share a vertexX. IfP, Q, andRare the midpoints ofB C, DE, andF A respectively, prove thatP QR is equilateral.

Problem 5 (Crux Mathematicorum). In quadrilateral ABCD, M is the midpoint of AB, and three equilateral

trianglesBC E, C DF, andDAG are constructed externally. IfNis the midpoint ofE F andP is the midpoint of

F G, prove thatM N Pis equilateral.

Problem 6. The four trianglesABC, AAbAc, BaBBc, andCaCbCare directly similar, andMa, Mb, andMc are

the midpoints ofBaCa, CbAb, andAcBc. Show thatMaMbMc is also similar to AB C.

AB

C

D

E

F

X

P

Q

R

Figure 11: Problem 4

A

B

CD

E

F

G

M

N

P


A

Ab

Ac

Ba

B

Bc

Ca

Cb

C

Ma

Mb

Mc


Hm, didnt I just say onemore 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, lets 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

triangleMaMbMc as an average of these:

MaMbMc = ABC+ aAAbAc+ bBaBBc+ cCaCbC. ()4In problem 4, the middle triangle has degenerated into point X. In problem 5, one of the outside triangles degenerates into point

F.

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The three triangles AAbAc, BaBBc, and CaCbCplay identical roles in the problem, so we have every reason to

guess thata= b= c. Now, for () to work, we need

MA=1

2(Ba+ Ca) = (+ a)A + a(Ba+ Ca).

If we set a = 1

2 and = a= 12 , then this works out perfectly (and likewise for Mb andMc). So does

MaMbMc = 12

ABC+1

2AAbAc+

1

2BaBBc+

1

2CaCbC

finish the proof?5 The triangles are all similar and the weights add to 1 (= 12

+ 12

+ 12

+ 12

), so yes. Sweet, another

one liner. QED

5Remember that negative weights are allowed, as long as the weights add to 1.

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4 Figure Addition

I keep stressing that the weights in a weighted average must add to 1. What would happen if they didnt?

A

B

M

O

P

Figure 14: Point addition

In the complex-number proof of MGT from section 2.2, theres really no reason

the weights must be k and (1 k). They could beanyreal numbers, and the proofworks the same! There is one subtle difference, though. In the diagram to the right,

use weights of1 = 2 = 1, and well try to calculate A + B. Where is it?

To add vectors or complex numbers like this, we need to know where the origin is.

If the origin is at, say, M, the midpoint between A andB, then A + B represents

the sum of the blue vectors

MA+MB =

0 = M. But if we put the origin

at a different point O, then the sum A+ B is now the sum of the red vectorsOA +

OB=

OP= P. So, withpoint addition, the sum depends on the location

of the origin, i.e. we must first specify an origin. With this minor change, MGT

extends yet again:

MGT: Figure Addition. For real numbers1 and2, define thesum of points 1P1+ 2P2 as the endpoint of

the vector1OP1+ 2

OP2, whereO is a specified origin. Then ifA1A2A3 andB1B2B3 are directly similar,

the sum of figures aA1A2A3 + bB1B2B3 (formed by adding corresponding vertices) is directly similar to the

original two triangles. As before, this naturally extends to more complicated figures and to more than two figures.

There is another way to visualize figure addition through dilation. For two similar triangles A1B1C1 and

A2B2C2, the sum A1B1C1+ A2B2C2 may be constructed by averaging the two triangles, 1

2A1B1C1+

1

2A2B2C2,

and then dilating this with center O and ratio 2.

4.1 When Did the IMO Get So Easy? (answer: 1977)

Problem 7 (IMO jury 1977 [6]). OAB andOAB are regular triangles of the same orientation, S is the centroid

ofOAB, andM andNare the midpoints ofAB andAB , respectively. Show thatSM B SN A (Figure15).

O

A B

A

B

S

M

N

W

X

Y

Z


(In the diagram, points W, X, Y , and Zare midpoints.)First of all, the red triangles both look like 30 60 90 triangles,

and if we can prove that, were done. The two equilateral triangles give

us plenty of 30 60 90s to work with; all we have to do is find theright ones!

Points O, Z, N, and W form a parallelogram, so if O is the

origin, Z + W = N. With a little experimentation, we arrive at

N SA = ZOA + W SO, and since bothZOA andW SO arepositively oriented 3060 90 triangles, so isN SA. Similarly,M SB =Y OB+ XSO (again withO as origin) proves that MSB isa negatively oriented 30

60

90 triangle. Whew, that was quick!

QED

4.2 A Fresh Look at an Old Result (1936, to be

precise)

Problem 8 (Pompeius Theorem [1]). Given an equilateral triangleABCand a pointP that does not lie on the

circumcircle of ABC, one can construct a triangle of side lengths equal to P A, P B, and P C. If P lies on the

circumcircle, then one of these three lengths is equal to the sum of the other two.

Erect equilateral triangles P CY and BP Xwith the same orientation asABC. WithPas origin, considerequilateral triangle P CY +BP X = BC A. It must be equilateral with the same orientation as ABC, which

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A

B C

P

X

Y

Figure 16: Pompeius Theorem

means A = Y +X = A, i.e. P Y A X is a parallelogram. Notice thatAP Y has AP = AP, P Y = CP, andY A= P X= BP, so if it is not degenerate,AP Y is the triangle were looking for.

When is this triangle degenerate, i.e. when are A,PandYcollinear? This happens if and only if (using directed

angles modulo 180) CP A= CP Y = 60 = CB A, i.e. quadrilateral ABCP is cyclic. And certainly, ifAP Yis degenerate, then one of its sides equals the sum of the other two.

QED

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5 A Few Harder Problems

At this point weve become proficient in utilizing the various forms of the Theorem, and weve learned to look for a

MGT approach when the problem hands us loads of similar triangles with which to play. But MGT can be useful

in many other situations as well, even if its application may be far from obvious. In this section, well look at a

few harder problems to stress that the MGT may not solve every problem immediately, but its a valuable method

to keep in mind as you explore a problem. It may be that MGT is only one of many steps in your solution. Or,

ideas related to MGT may lead you to a solution that doesnt use it at all. The point is that looking at a diagram

from an MGT viewpoint may tell you things that you formerly wouldnt have noticed.

This brings up another point: in order to recognize uses for MGT, youmust have an accurate diagram or

two, or three to look at. This article is filled with diagrams for exactly that purpose. (Whether youre solving a

geometry problem using MGT or not, its usually a good idea to have a decent diagram handy!)

On that note, lets bring on the problems.

5.1 A Pretty(,) Busy Diagram

Problem 9 (IMO Shortlist 2000, G6). LetABCD be a convex quadrilateral withAB not parallel to CD, and let

Xbe a point insideABCD such thatADX= BCX < 90 andDAX= CBX < 90. IfY is the point of

intersection of the perpendicular bisectors ofAB andC D, prove thatAY B= 2ADX.

A

B

C

D

X

Y


First thing to notice: by simply relabeling the diagram, it must also be true that DY C = 2DAX. So,

designate ADX= and DAX=. Next, notice that triangles AX D and B XCare similar. Ooh, that means

we should form another similar triangle 12

AXD + 12

BX C!Thats a decent thought, but unfortunately, even though

AXD and B XCare similar, theyre notdirectly similar.

It may not clear where to go from here, but since its necessary to start somewhere, well begin with the only

tangible fact we have: similar triangles AX Dand BXC. Its interesting that no matter how these two triangles are

hinged or scaled aroundX, pointYstill has its curious property. This suggests a possible direction for exploration:

scale the triangles and see what happens to point Y.

Lets leave triangleAXDfixed while we enlarge and shrink BX C. What size triangle would be easier to analyze?

How about zero! Consider quadrilateral ABCD, where BXC has shrunken to the degenerate triangle at point X(Figure 18). The corresponding Y is the intersection of the perpendicular bisectors ofAB =AXandDC =DX,

i.e. the circumcenterO of triangle AXD . Is it true that 2ADX = AOX? Yes, since arcAXof circle O hasmeasure 2. So when triangle B XC shrinks to zero, everything works out as expected.

What other size for triangle BX C might work well? Lets look at quadrilateralABCD, where BXC is

congruent to AXD as shown in Figure 19. To locateY, we should be looking for the perpendicular bisectors

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A

B

C

D

X = B = C

OY

Figure 18: Studying quadrilateral AB CD

ofAB and CD. But these are the same lineMN, so where is Y? We can instead locateY by finding the

point on this line so that AYB = 2, since this is another property our Ys should have. This means that

AYX= = ADX, so AX YD is cyclic, i.e. Y is the second intersection of circle O with line M N. (If the

circle happens to be tangent to line M N, second intersection simply means tangency point.)

A

B

C

D

B

C

X

M

N

O

YY

Figure 19: Quadrilateral AB CD

It seems that weve lost sight of our original problem. Weve studied quadrilaterals AXXD and ABCD,

but not the original ABCD. Luckily, a perusal of diagram 19 reveals the next step: trianglesAOX, AY B, and

AY B are similar since theyre all isosceles with vertex angle 2, so we should be able to average them. Indeed, if

XB/XB =k, then we should have (1

k)AOX+ (k)AYB =AY B. This is (almost) the last step to a complete

solution! A full, self-contained solution is given below.

Full Solution. Let XB/XA = k , and dilate B and Caround Xwith ratio 1/k to points B and C respectively,

so that ABCD is an isosceles trapezoid. M and N are the midpoints of AB and CD. Define O as the

circumcenter of triangle AX D, and let Y be the second intersection of circle ADXwith lineM N. IfADX=

and DAX= , it follows that AOX= 2, AYX= which implies AYB = 2, and likewise,DOX= 2

and DYC = 2. Therefore, the isosceles trianglesAOX and AYB are similar, as are triangles DOX and

DYC.

Define Y1 = (1 k)O+ (k)Y, and notice that triangles AY1B = (1 k)AOX+ (k)AYB and DY1C =

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(1 k)DOX+ (k)DYC are isosceles by MGT. Thus, Y1 is the intersection of the perpendicular bisectors ofABandC D, i.e. Y1= Y. Furthermore, by our application of MGT, AY B= AOX= 2, QED.

QED

5.2 Hidden Circles

Problem 10(USA TST 2005 #6). LetABCbe an acute scalene triangle withO as its circumcenter. PointP lies

inside triangleAB CwithP AB= P BCandP AC= P CB . PointQ lies on lineB CwithQA = QP. Prove

thatAQP= 2OQB.

A

BC

Q

O

M

P

X


Most of the diagram is straightforward: Ois the circumcenter,Qis the intersection ofBCwith the perpendicular

bisector ofAP, and Ive added M, the intersection ofAP with BC. Everything is simple to navigate exceptP

itself, so thats where well start investigating.

The first strange angle equality, i.e. P AB = P BC, shows thatM AB M BP. Thus, MA/MB =MB/MP, or (MA)(MP) = (MB)2. This shows, by power of a point, that the circle through A, P, and B is

tangent to line M B at B.6 Likewise, the circle throughA,P, andCis tangent to M CatC. (Call these two circles

[and their centers]O1 and O2 respectively.) Finally, (MB)2 = (MA)(MP) = (MC)2, soM is the midpoint ofBC.

A

BC

Q

OO1

O2

M

P

X

Figure 21: Circles O1 andO2

The diagram becomes much clearer from the viewpoint of the two circles. The circles intersect at A and P, and

line BC is their common tangent. PointQ, being the intersection of their axis of symmetry (line O1O2) with a

common tangent (line B C), must be the center of homothecy between circles O1 andO2.

LetXbe the midpoint ofAP. Were asked to prove that AQP= 2OQB, or equivalently, AQX= OQM.

But triangles AQX and OQM are both right triangles, so we need to prove that they are similar. Heres where

6If this circle intersected line MB at some other point B, then power of a point would show that (MB )2 = (MA)(MP) =(MB )(MB ), so MB = MB and B= B.

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Mean Geometry might come in handy: to prove that these triangles are similar, we might be able to find a third

triangle similar to, sayAQX, and then express triangle OQM as a weighted average of those two. So now wehunt for a third similar triangle. In order for this to work, this triangle should have one vertex at Q, one vertex

along line X M, and one vertex along line AO. What key points are on these two lines? JustA and P. And since

A is already being used, lets think about P. If trianglesAQX and OQMwere similar to triangle ZQP, where

would this mystery point Zhave to be? We already want it on line AO, and since AQX= XQP = ZQP, we

would also need Zto be on line QX. So define Z= AO QX, and lets see ifZQP is indeed the triangle werelooking for. First of all, is it a right triangle?

Angle QP Z is a right angle if and only ifQAZ is right, so we need QA OA, i.e. QA should be tangentto circle O at A. It turns out to be true, as follows. LetTbe the second intersection ofQA with circle O1. IfrQ (where r = QO2/QO1) is the homothecy taking circle O1 to circle O2, we have rQ(T) = A and rQ(B) = C,

so ACB = T BQ = T AB. Thus, trianglesQBA and QAC are similar, so (QB)(QC) = (QA)2, and QA is

tangent to circle O. So (by tracing backwards through a few lines of reasoning above), triangle Z QPactuallydoes

have a right angle at P. This means it is similar to triangle AQX.

A

BC

Q

OO1

O2

M

Z

T

P

X

Figure 22: Three similar triangles

The next part of our initial plan was to find a weighted average of triangles Z QP andAQXthat would produce

triangleOQM, thus completing the proof. So, all we need to show is that AO/ZO= XM/P M. Good luck.

Those segment lengths arent easy to calculate, even with plenty of paper and tons of time. For the first time

in this article, MGT fails to miraculously save the day. But weve come far enough with the MGT idea so that the

proof is moments away. Let rQ be the spiral dilation centered at Q that sendsX P toAZ. SinceMis on lineX P,rQ(M) = M

is on lineZ A. Also, since triangleMQMis similar to triangleAQX, we have QMM = 90, soM

is also on the perpendicular bisector ofB C. This means rQ(M) =M = O. Thus,OQM = AQX, as desired.

(Notice that, even though MGT was a driving force for most of the solution, not a single mention of it is necessary

in the final writeup.) QED

5.3 Nagel Who?

Problem 11 (USAMO 2001 #2). Let ABC be a triangle and let be its incircle. Denote by D1 and E1 the

points where is tangent to sides BC and AC, respectively. Denote by D2 andE2 the points on sides BC and

AC, respectively, such thatCD2 = B D1 andCE2 = AE1, and denote byP the point of intersection of segments

AD2 andB E2. Circle intersects segmentAD2 at two points, the closer of which to the vertexA is denoted byQ.

Prove thatAQ= D2P.

In the diagram Ive added F1 and F2 to preserve symmetry, and I included the midpoints of the sides of the

triangle.

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A

B CD

EF

D1

E1

F1

D2

E2

F2

Q

R

SIP


A

BC

D1

E1

F1

D2

I

Ia

Q

Figure 24: Excircle Ia

Using the standard notations a = BC, b= C A, c = AB, and s = 1

2 (a + b + c), it is relatively well known thatBD1 = s b. Indeed, ifAE1 = AF1 = x, BF1 = BD1 = y, C D1 = CE1 = z, then the system y + z = a, z + x= b,x + y= c can be solved to give B D1 = y = (c + a b)/2 = s b. Likewise, if the excircle opposite A is tangent toBCatV, a similar calculation shows that C V =s b= BD1, i.e. V =D2. SoD2, E2,F2 are the tangency pointsofB C, C A, AB with the triangles three excircles.

Another relatively well-known point in the diagram is P, commonly referred to as the Nagel point of triangle

ABC. Its simply the intersection of the three ceviansAD2,BE2, andC F2, which must concur by Cevas theorem:

BD2D2C

C E2E2A

AF2F2B

=s cs b

s as c

s bs a = 1.

A useful property of the Nagel point,7 other than its mere existence, is how its defining cevians interact with the

incircle: lineAD2intersects the incircle at point Qdiametrically opposite fromD1. To prove this, letr = AI/AIa(in

Figure 24), so that rAis the homothecy centered atA taking excircle IA to incircle I. SinceD2IA is perpendicular

to B C, its image through rA, namely QI, is also perpendicular to BC. Therefore, QD1 is a diameter of circle I.

The same goes for diameters E1R and F1S.

Now to the problem. The midpoints D , E, andF inspire us to consider

1

2D1E1F1+1

2D2E2F2 = DEF. ()

TrianglesD1E1F1and D2E2F2arent similar, so MGT doesnt tell us anything directly. But its still worth noticing.

If its true that AQ = D2P, it must also happen that B R= E2P and C S= F2P. This means that, with P as

origin, wed like to be able to show that

ABC+ D2E2F2 = QRS.

Again, these triangles are not similar, so MGT isnt applicable. But this equation has striking similarities with

(). These similarities become more pronounced if we rewrite equation () as

2DEF+ D2E2F2 = D1E1F1 ()7The Nagel point is also (and less commonly) known as the bisected perimeter point [8] or the splitting center [7], since the cevians

AD2, etc., bisect the perimeter of the triangle, i.e. AB + BD2 = AC+ CD2 = s.

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A

B CD

EF

D1

E1

F1

D2

E2

F2

Q

R

SIP


Triangle 2DEFis simply a translation ofABC, and triangle D1E1F1is a translation ofQRS. The two equationsare almost identical! This inspires the following observation:

Non-similar Figure Addition. If three triangles (figures) satisfy11+ 22 = 3, and if1 is a translatedversion of1, then11+ 22 is a translation of3, regardless of the location of the origin.

Indeed, if1 = 1+V,8 then

11+ 22 = 11+ 22+ 1V = 3+ 1V.

Now we can finish the problem. Beginning with equation (), translate2DEF to coincide with AB C. Theabove observation (with our origin still at P) proves thatQRS =ABC+ D2E2F2 must be a translation ofQRS.

But sinceA + D2 = Q lies onAD2, and likewise forR andS, trianglesQRSand QRS are also homothetic with

center P. So the two triangles must be identical, and in particular, Q = Q =A + D2. SoAQ=

PD2, proving the

desired result. QED

8i.e. a translation by vector

V

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6 Additional Problems

Problem 12. Recall problem 4: Positively oriented equilateral trianglesXAB, XC D, andXEFshare a vertex

X. IfP, Q, andR are the midpoints ofB C, DE, andF A respectively, prove thatP QR is equilateral. Prove this

problem using Napoleons Theorem.

Problem 13 (Crux Mathematicorum). A line parallel to the side AC of equilateralABC intersects B C at Mand AB at P, thus making BM P equilateral as well. D is the center ofBM P and E is the midpoint of CP.Determine the angles ofADE.

Problem 14. The following theorem appears in Geometry Revisited[5] as a special case of a theorem of Petersen

and Schoute: IfABCandABC are two directly similar triangles, whileAAA,BBB,C CC are three directly

similar triangles, thenABC is directly similar toABC.

a. Show that this theorem generalizes the triangle version of MGT.

b. What minor adjustment can be made to the statement of MGT to account for this generalization (and its

proof)?

c. Show that Napoleons theorem is a special case of this theorem.

Problem 15 (Van Aubels Theorem). Given an arbitrary planar quadrilateral, place a square outwardly on each

side, and connect the centers of opposite squares. Then these two lines perpendicular and of equal length.

Problem 16. Equilateral triangles AEB, BF C, CGD, DH A are erected outwardly on the sides of a plane

quadrilateralABCD.

a. Let M, N, O, and Pbe the midpoints of segments EG, HF, AC, and BD respectively. What is the shape

ofPMON?

b. Md andMa are the centroids ofDAH andAEB, and equilateral triangle MdT Ma is oppositely orientedwith respect to ABCD. Find the angles of triangleF T G.

c. Ma and Mc are the centroids ofAEB and CGD. Prove that segmentsMaMc and F Hare perpendicular,and, in addition,|F H| = 3 |MaMc|.

d. Equilateral triangles EW F, F XG, GY H, and HZ Eare oppositely oriented with respect to ABCD. Prove

that quadrilaterals ABCD and W XY Z have the same area.

Problem 17. Let (P,QR) denote the line through point Pperpendicular to line QR. Say thatXY Zperpen-dicularizesABC if (X,BC), (Y, CA), and (Z,AB) concur at a point. IfXY Z perpendicularizesABCandDEF, prove thatXY Zalso perpendicularizes any linear combination ofABCandDEF.

Problem 18 (Alex Zhai).

a. In triangle ABC, AD, BE, and CF are altitudes. Dc and Db are the projections ofD onto AB and AC,

respectively, and points Ea,Ec,Fa, and Fbare defined similarly. Prove that quadrilaterals BDcDbC,CEaEcA,

andAFbFaB are cyclic.

b. Let OA be the center of circle BDcDbC, and let Ta be the midpoint of altitude AD. Similarly define Ob,

Oc, Tb, and Tc. IfO is the circumcenter of triangle ABC, show that AOOaTa is a parallelogram, as well as

BOObTb andC OOcTc.

c. Prove that lines OaTa, ObTb, and OcTc are concurrent.

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Problem 19. LetP be a convex polygon in the plane. A real number is assigned to each point in the plane sothat the sum of the number assigned to the vertices of any polygon similar toP is equal to 0. Prove that all theassigned numbers are equal to 0.

Problem 20. Show that I, G, and N(the incenter, centroid, and Nagel point of a triangle) are collinear in that

order with 2 IG = GN. Hint: see problem 11.

Problem 21. Given a convex quadrilateral ABCD, construct (with ruler and compass) a square of the sameorientation with one vertex on each side ofABCD.

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7 Solutions to Additional Problems

Problem 12. Recall problem 4: Positively oriented equilateral trianglesXAB, XC D, andXEFshare a vertex

X. IfP, Q, andR are the midpoints ofB C, DE, andF A respectively, prove thatP QR is equilateral. Prove this

problem using Napoleons Theorem.

Solution. In diagram 26, trianglesBCJ,DE K,F ALare equilateral, andG,H,I,M,N,O are the centers of their

respective triangles. By napoleons theorem thrice,HGM,IH N, andGIO are equilateral, and their centersT, U, Vare also the centroids ofXBC,XDE, andXF A respectively (by problem 3). Again by napoleonstheorem (on the blue triangles),T U V is equilateral, and the dilation 3/2XcarriesT U V toP QR.

AB

C

D

E

F

X

G

H

I

J

K

L

M

N

OT

U

V

P

Q

R


AB

C

TS

X

P

E

M

D


A

B

A = C

B

C A

B

C

Figure 28: Problem 14.c

QED

Problem 13 (Crux Mathematicorum). A line parallel to the side AC of equilateralABC intersects B C at Mand AB at P, thus making BM P equilateral as well. D is the center ofBM P and E is the midpoint of CP.

Determine the angles ofADE.

Solution. S and Tare the midpoints ofCB and CA, and X is the center ofABC. Let BP/BA = k . BecauseE must lie on line ST, and sinceCS T CB A, SE/ST = BP/BA = k. Also, sinceBP D BAX,BD/BX=BP/BA= k. Thus, (1 k)ABS+ (k)AXT =ADE, soADEis a 30 60 90 triangle. QED

Problem 14. The following theorem appears in Geometry Revisited[5] as a special case of a theorem of Petersen

and Schoute: IfABCandABC are two directly similar triangles, whileAAA,BBB,C CC are three directly

similar triangles, then

ABC is directly similar to

ABC.

a. Show that this theorem generalizes the triangle version of MGT.

b. What minor adjustment can be made to the statement of MGT to account for this generalization (and its

proof)?

c. Show that Napoleons theorem is a special case of this theorem.

Solution. a. In the special case where A is on line AA, the similarity of degenerate triangles AAA, BB B,

andC CC simply means that AA/AA =BB/BB =C C/CC =k. Now, the fact that ABC is similar to

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A

B

C

D

E

F

G

H

I

J

K

L

M

N

O

P Q

RS

T

Figure 29: Van Aubels Theorem

ABC is exactly the statement of MGT, since

(1 k)ABC+ (k)ABC =ABC.

b. In the complex number proof of MGT (section 2.2), we required the two weights k and 1 k to be real numbersthat add to 1. If we allow the weights to becomplex, we obtain this generalization. The proof in section 2.2 remains

unchanged.

c. The napoleon diagram has been relabeled in figure 28 to show the correspondence. QED

Problem 15 (Van Aubels Theorem). Given an arbitrary planar quadrilateral, place a square outwardly on each

side, and connect the centers of opposite squares. Then these two lines perpendicular and of equal length.

Solution. Q, R, S, and T are the midpoints of the sides of quadrilateral MNOP. Since 14

(A+ F +G + C) =

12 (M+ N) = Q, and likewise around, averaging the four given squares proves that QRSTis itself a square:

1

4FBAE+

1

4GHCB+

1

4CIJD+

1

4ADKL= QRST.

SinceMO= 2

TS and

PN= 2

TQ, and since T S and T Q are equal in length and perpendicular, diagonals M O

andN Pmust share this property, as desired.

QED

Problem 16. Equilateral triangles AEB, BF C, CGD, DH A are erected outwardly on the sides of a plane

quadrilateralABCD.

a. Let M, N, O, and Pbe the midpoints of segments EG, HF, AC, and BD respectively. What is the shapeofPMON?

b. Md andMa are the centroids ofDAH andAEB, and equilateral triangle MdT Ma is oppositely orientedwith respect to ABCD. Find the angles of triangleF T G.

c. Ma and Mc are the centroids ofAEB and CGD. Prove that segmentsMaMc and F Hare perpendicular,and, in addition,|F H| = 3 |MaMc|.

d. Equilateral triangles EW F, F XG, GY H, and HZ Eare oppositely oriented with respect to ABCD. Prove

that quadrilaterals ABCD and W XY Z have the same area.

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A

B

CD

E

F

G

H

M

N

O

P

Figure 30: Problem 16.a

A

B

CD

E

F

G

HT

Ma

Md

Figure 31: Problem 16.b

A

B

CD

E

F

G

H

Ma

Mc

I

J

K

LMNO

P


A

B

CD

E

F

G

H

P

Q

R

S

W

X

Y

Z

Figure 33: Problem 16.d

Solution. a. Since 12

ABE+ 12

CDG = OP M and 12

BC F+ 12

DAH = P ON, trianglesOP M andP ON areequilateral, so quadrilateral MONPis a rhombus with vertex angles of 60 and 120.

b. By Napoleons theorem on triangle DAB, Tmust be the center of the equilateral triangle erected on sideDB, i.e. DT B is a 30 120 30 triangle. Now apply problem 9 to quadrilateral DBFG. Point C hasGDC = F BC = 60 and DGC = BF C = 60, so problem 9 guarantees the existence of a point Y on

the perpendicular bisectors ofDB andF Gso that DY B = GY F= 120. Tis on the perpendicular bisector of

DB and has DT B= 120, soT is Y, and thusGT F is a 30 120 30 triangle.c. This solution mimics the proof of Van Aubels Theorem (problem 15). Erect rectangles JBAI,KLCB,CMND,

andADOB with

3 : 1 side ratios as shown. The average of these four rectangles (with vertices in the listed order)

is rectangle W XY Z (not shown) which connects the midpoints of the sides of quadrilateral MaF McHand which

must be similar to the red rectangles. And as explained in problem 15, since the Varignon Parallelogram [5,

Theorem 3.11] of quadrilateral MaF McHhas sides that are perpendicular with a ratio of

3/1, the diagonalsF H

andMaMc must have this property too.

d. Recall from problem 8 that, with C as center,DGC+ BC F =XGF, proving that D+ B = X andso XDCB is a parallelogram. Likewise, ADCY is a parallelogram. Segments XB and AYare both parallel and

congruent to DC, so XAY B is a parallelogram, and AB and XY share the same midpoint P. The same goes

for Q, R, and S. Thus, quadrilaterals ABCD and W XY Z have the same Varignon Parallelogram PQRS, so

area(ABCD) = 2 area(PQRS) = area(W X Y Z ). QED

Problem 17. Let (P,QR) denote the line through point Pperpendicular to line QR. Say thatXY Zperpen-dicularizesABC if (X,BC), (Y, CA), and (Z,AB) concur at a point. IfXY Z perpendicularizesABCandDEF, prove thatXY Zalso perpendicularizes any linear combination ofABCandDEF.

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Solution. The following lemma is key: XY Z perpendicularizesABC if and only ifABC perpendicularizesXY Z. To finish problem 17 with this property, note that ifX Y Zperpendicularizes ABC andDEF, then bothABC and DEF perpendicularize XY Z, say at points P and Q respectively. Define GHI = 1ABC+ 2DEF

and R = 1P +2Q. Segment GR as a linear combination of AP and DQ is perpendicular to Y Z, and

likewise for H Rand I R. SoGH Iperpendicularizes X Y ZatR, and the above lemma guarantees that X Y Zmust

perpendicularizeGH I.

We offer two very different proofs of the lemma.

Proof 1. Well make use of the following property: lines P Qand RSare perpendicular if and only ifP R2 + QS2 =

P S2 + QR2. Indeed, using as the vector dot product,

(P R) (P R) + (Q S ) (Q S ) = (P S ) (P S ) + (Q R) (Q R)

is equivalent to (P Q) (R S ) = 0 after expanding and simplifying.

For two trianglesABCandX Y Zin the plane, let(Y,AC) and(Z,AB) meet atQ.XY ZperpendicularizesABCif and only ifX Q BC, i.e. if and only if

0 = (CQ2

QB2) + (BX2

XC2)

= (CQ2 QA2) + (BX2 XC2) + (AQ2 QB2)= (CY2 Y A2) + (BX2 XC2) + (AZ2 ZB2).

But this last expression makes no distinction between triangles ABCand X Y Z, so if one triangle perpendicularizes

another, the other must perpendicularize the first. QED

A

B C

Q

X

Y

Z

L

M

N

Figure 34: Proof 2 of perpendicularization lemma

Proof 2. AssumeXY Z perpendicularizesABC at point Q. Draw three lines: line a through X parallel toBC, lineb throughYparallel toC A, and linec throughZparallel toAB. These lines determine triangle A

BC

homothetic toABC, and so ABCperpendicularizes X Y Zif and only ifABC does.Let L, M, Nbe the projections ofA

, B

, C

onto Y Z, ZX, X Y respectively. QuadrilateralA

ZQY is cyclic,so we may calculate

CAL= 90 ZY A = 90 ZQA = QAB,

and likewise for B M andCN. Thus,

sinCAL

sinLAB sinA

BM

sinMBC sin B

CN

sinN CA =

sinQAB

sinCAQ sin QB

C

sinABQ sinQC

A

sinBCQ= 1

i.e. AL, BM, and CN do in fact concur by the trigonometric form of Cevas theorem. (They concur at the

isogonal conjugate ofQ with respect to triangle ABC.) QED

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Problem 18 (Alex Zhai).

a. In triangle ABC, AD, BE, and CF are altitudes. Dc and Db are the projections ofD onto AB and AC,

respectively, and points Ea,Ec,Fa, and Fbare defined similarly. Prove that quadrilaterals BDcDbC,CEaEcA,

andAFbFaB are cyclic.

b. Let OA be the center of circle BDcDbC, and let Ta be the midpoint of altitude AD. Similarly define Ob,

Oc, Tb, and Tc. IfO is the circumcenter of triangle ABC, show that AOOaTa is a parallelogram, as well as

BOObTb andC OOcTc.

c. Prove that lines OaTa, ObTb, and OcTc are concurrent.

Solution. a. By similar triangles ADcD and ADB, ADc/AD = AD/AB, i.e. (ADc)(AB) = (AD)2. Likewise,

(ADb)(AC) = (AD)2 = (ADc)(AB), so B DcDbC is cyclic by the converse of power-of-a-point. Similar arguments

work for the other two quadrilaterals.

A

B CD

F

Mc

Oa

H

O

Ta

Ta

Dc

Oa

Figure 35: Problem 18.b

A

B CD

E

F

Ma

MbMc

Oa

Ob

Oc

HO

Ta

Tb

Tc


b. ATaand OOaare parallel because they are both perpendicular to BC, so it is only necessary to show they have the

same length. ProjectTa, Oa, andO ontoAB at pointsTa,Oa, andMc, which must be the midpoints ofADc,AB,

and DcB respectively. We have TaA= DcA/2 and OaMc = BMc BO a = (BDc+ DcA)/2 (BDc)/2 = DcA/2.

Since the projections ofTaA and OaO ontoAB are equal in length, and sinceB A and AD are not perpendicular,

TaA= OaO, as desired.

c. It can be calculated that OAC+ AEF = (90 CB A) + (CB A) = 90, so AO EF, i.e. OaTa E F.Thus, we wish to prove that OaObOc perpendicularizes DEF (see problem 17).

Let the midpoints ofB C, CA, AB be Ma, Mb, Mc, and let Hbe the orthocenter ofABC. SinceMaMbMc is

similar toABCwith half its size, and since O is the orthocenter ofMaMbMc,OMa= 1

2AH. Since we have already

proved that OOa = 1

2AD, it follows that MaOa =

1

2HD. So with Has origin, OaObOc = MaMbMc +

1

2DEF,

i.e. DEF = 2OaObOc 2MaMbMc. And since OaObOc perpendicularizes MaMbMc (at O) and OaObOc (at itsown orthocenter), triangle OaObOc must perpendicularize the linear combination 2OaObOc 2MaMbMc = DEFby problem 17, as desired.

QED

Problem 19. LetP be a convex polygon in the plane. A real number is assigned to each point in the plane sothat the sum of the number assigned to the vertices of any polygon similar toP is equal to 0. Prove that all theassigned numbers are equal to 0.

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Solution. Letf(T) be the number associated with point Tin the plane.

The idea is to use an extended version of problem 14 (illustrated for n = 4 in diagram 37), and then to let one

of the polygons shrink to a single point X (figure 38).

A1,1

A2,1

A3,1

A4,1

A1,2

2,2

A3,2

A4,2

A1,3

A2,3

A3,3

A4,3

A1,4

A2,4

A3,4

A4,4

Figure 37: Generalization of problem 14

A1,1

A2,1

A3,1

A1,2

2,2

A3,2

A1,3

A2,3

A3,3

A1,4

A2,4

A3,4

X


An arbitrary point Xis chosen, polygonR1 = A1,1A1,2 A1,n similar toP is drawn with no vertices at X,and then polygonsBi = A1,iA2,i An1,iX (1in) are drawn all similar toP. It follows that each polygonRi = Ai,1Ai,2 Ai,n (1 i n 1) is also similar toP (see problem 14). Thus, since

bBi

f(b) = 0 andrRj

f(r) = 0 for any 1 i n and 1 j n 1, we can add over polygonsBi and over polygonsRi to cancelout most of the terms:

0 =ni=1

bBi

f(b) = n f(X) +ni=1

n1j=1

f(Aj,i) = n f(X) +n1j=1

rRj

f(r) = n f(X) + 0.

This means f(X) = 0, and since point Xwas arbitrary, this holds for all points in the plane. QED

Problem 20. Show that I, G, and N(the incenter, centroid, and Nagel point of a triangle) are collinear in that

order with 2 IG = GN. Hint: see problem 11.

Solution. Refer to figure 25 in problem 11. With originP, define DEF =2DEF and D1

E1

F1

=D1E1F1.These are the triangles used but never drawn in equation () on page 17, and they are translations ofABC

and QRS respectively. Also set X = 2D. SinceAQ =

PD2 =

D1X =

DQ, it follows that triangles DEF

and QRS are in the same relative position as AB C and QRS, i.e.AD =

QD

1. And sinceAD = 3

GP (since

ADD GDP) and QD1

= 2IP(since QD1D1 ID1P), the conclusion follows. QED

Problem 21. Given a convex quadrilateral ABCD, construct (with ruler and compass) a square of the same

orientation with one vertex on each side ofABCD.

Solution. First, we prove that it is impossible for AB CD andAD BC to occur simultaneously. Assume thisdoes happen, label AB CD = X and AD BC=Y, and assume (without loss of generality) that C is betweenXand D and between Y andB . Then

360 > B+ C+ D= 540 C >360,

contradiction. So it is safe to assume B Cis not perpendicular to DA.

Given a pointPi AB, construct square i= PiQiRiSi as follows: rotate line DA clockwise by 90 aroundPito intersect B C at Qi (this intersection exists uniquely since BC DA), and complete positively oriented square

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A

B CD

EF

D1

E1

F1

D2

E2

F2

Q

R

SIP

D

E F

D1

E1

F1

G

X


AB

C

D

P

Q

R

S

P0

Q0

R0

S0

P1

Q1

R1

S1


PiQiRiSi. This square with vertexPihas the properties thatQi BCand Si DA(by rotation), and furthermore,the construction proves that such a square is unique. Now it is only necessary to find the rightP AB so that thecorrespondingR lies onC D.

Choose two distinct points P0, P1 AB and draw squares 0 and 1 as above. Choose any other pointPt = (1 t)P0+ (t)P1 on AB, and consider t = PtQtRtSt = (1 t)0+ (t)1. Since Qt Q0Q1 BC andSt S0S1 DA, square t satisfies the defining conditions for t, i.e. t = t. In particular, Rt = Rt =(1 t)R0+ (t)R1, meaning Rt must lie on line R0R1. And sinceRt covers all of this line as t varies, the locus ofsuch points is exactly this line, i.e. R = C D

R0R1. The rest of the square can no be constructed now that the

correct ratiot = R0R/R0R1 is known.There are a few exceptional cases to consider. If lineR0R1 and line CD are identical, any point Pt AB will

produce a viable square t. If the two lines are parallel but not equal, there is no square with the desired properties.

And finally, ifR0 = R1, thenRt = R0 for all t, so all Pt work or no Pt work depending on whether or not R0 is on

lineC D. QED

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References

[1] Andreescu, T. and Gelca, R. Mathematical Olympiad Challenges.Boston, MA: Birkhauser, 2002.

[2] Bogomolny, A. Asymmetric Propeller. http://www.cut-the-knot-org/Curriculum/Geometry/

AsymmetricPropeller.shtml

[3] Bogomolny, A. A Problem in an Equilateral Triangle. http://www.cut-the-knot.org/Curriculum/

Geometry/RightInEquilateral.shtml

[4] Bogomolny, A. Three Similar Triangles. http://www.cut-the-knot.org/Curriculum/Geometry/

SimilarTriangles.shtml

[5] Coxeter, H.S.M and Greitzer, S.L. Geometry Revisited.Washington, DC: Math. Assoc. Amer., 1967.

[6] Engel, A.Problem Solving Strategies.New York: Springer, 1998.

[7] Honsberger, R. Episodes in Nineteenth and Twentieth Century Euclidean Geometry.Washington, DC: Math.

Assoc. Amer., 1995.

[8] Weisstein, Eric W. Nagel Point. FromMathworldA Wolfram Web Resource. http://mathworld.wolfram.com/NagelPoint.html

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