No title

ARBELOS

SPECIAL GEOMETRY ISSUE

VOLUME 6

Solutions to the 'Cover Problems' in Volumes 1 to 5 BY

PROFESSOR SAMUEL L. GREITZER

1906 - 1988

F

A

C

B

An arbelos Is formed by erecting semicircles on

I I. I

I

segments AC. CB. ana AB. CD J. AB.

A circle. center P. Is drawn touching the semicircle

on BC at E and touching the semicircle on AS and

tangent to CD. EP meels CD at F.

Prove that EF • AC.

Cover problem dedicated to Sam Greltzer.

,­

I

t

Copyright © 1988 Committee on the American Mathematics Competitions Mathematical Association of America.

ii -- ---------------'--­

A Tribute to Professor Samuel L Greitzer

Professor Samuel l Greitzer emigrated to the United States from Odessa, Russia in 1906. He graduated from the City College of New York in 1927 and earned his PhD degree at Yeshiva University. He had more than 25 years experience as a junior and senior high school teacher. He taught at Yeshiva University, the Polytechnic Institute of Brooklyn, Teachers College and the School of General Studies of Columbia University. His last academic teaching position was at Rutgers University. He was the author of several excellent books of which Geometry Revi!ited [co-authored with H M S Coxeter] and the International Mathematical Olympiad! 1959-1977 are best known. His untimely death, on February 22, 1988, was indeed a sad day for all who knew him. . Professor Greitzer served as a coach of the summer Mathematical Olym­ piad training program from 1974 to 1980. Professors Greitzer and Murray S Klamkin accompanied the USA team to the International Mathematical Oympiad from 1974 [the first year the USA participated] to 1980. Their suc­ .. cess in coaching the team is evidenc;ed by the fact that the team usually placed among the top three [out of 30-35 countries]. During the years Professor Greitzer worked with the highly talented math­ ematics students in the Olympiad program, he saw the need for a publication appropriate for these superior students. When he retired from these duties he initiated such a journal, the Arbelo!. This journal was designed to con­ tain material which would help talented mathematics students prepare for the prestigious USA and International Mathematical Olympiads. From 1982 through 1987 Professor Greitzer served as the editor of the Arbelo! and wrote essentially all of its artides. Consequently, most of the artides in the Arbelo8 are a reflection of his lectures in the Olympiad program for talented and gifted students. This special issue of the Arbelo!, which contains solutions to the cover problems from the Arbelo!, is indicative of Professor Greitzer's continued devo­ tion to the well-being of talented students. He wrote these solutions between the publication of the last issue of the Arbelo! in May, 1987 and his untimely death. From his hand written notes, several of his many friends assisted in editing this final version. The contributions of Professor Greitzer to the mathematical development of students and teachers from many nations will be lasting. We shall miss his humor, words of wisdom, critical comments, mathematical insight and friend­ ship.

Dept of Mathematics and Statistics Walter Mientka University of Nebraska Executive Director Lincoln, Nebraska 68588 American Mathematics Competitions

CONTENTS

The November, 1982 Arbelol Cover Problem

1

The January, 1983 Arbelol Cover Problem

2

The November, 1983 A rbelol Cover Problem The January, 1984 Arbelol Cover Problem The March, 1984 Arbelol Cover Problem A First Diversion The May, 1984 Arbelol Cover Problem A Second Diversion The The The The

4

6

8

, " .....•............... 12

14

, 17

September, 1984 Arbelol Cover Problem November, 1984 Arbelol Cover Problem January, 1986 Arbelol Cover Problem March, 1986 Arbelol Cover Problem

19

21

22

26

The May, 1985 Arbelol Cover Problem Another Diversion The September, 1986 ArbelOI Cover Problem The November, 1986 A rbelol Cover Problem The January, 1986 Arbelol Cover Problem The March, 1986 Arbelol Cover Problem The May, 1986 Arbelol Cover Problem

26

29

31

33

34

36

38

Still Another Diversion The September, 1986 Arbelo. Cover Problem Take a Break The November, 1986 Arbelo. Cover Problem

40

41

42

43

The January, 1987 Arbelo. Cover Problem The March, 1987 Arbelol Cover Problem The May, 1987 Arbelo. Cover Problem Epilogue

44

46

46

47

ii

Professor Greitler, who was the first chairman of the U.S.A. Mathe­ matical Olympiad Committee, died on February 22, 1988. These notes on the cover problems of "his journal", the Arbelo., being published posthu­ mously, are his last known mathematical writings. Among those of his friends and others who, because of their great respect for his work, transformed Sam's handwritten notes to this publica­ tion were: Dr Stanley Rabinowitz, Alliant Computer Systems Corporation [author of the cover problem for this special issue]; and Professors Robert Bumerot, Hofstra University; Darrell Horwath, John Carroll University; Murray Klamkin, University of Alberta; Walter Mientka, University of Ne­ braska; and Leo Schneider, John Carroll University.

iii

The November 1982 Arbelo" Cover Problem

1

During the brief life-span of the student journal, Arbelo", the cover problems produced the majority of correspondence. There were a few so­ lutions and many requests for solutions. Since some of th~se involv~d fine points in geometry, I have prepared these solutions in the hope that I may be able to advance the cause of an important branch of mathematics that has b~en shabbily treated. Now to the task. The first issue was actually named Aftermath. The cover had a formula for the volume ofa tetrahedron, which I will not discuss.

A

o

c

B IFigure

1.1

The November, 19S2 issue (the first Arbelo") introduced the Arbelos, which is the region bounded by the semicircles in Figure 1. Archimedes loved to play with the diagram and discovered many properties of it. Since all the curves are semi-circles, LAXC = LATB = 90°, so C X and YT are parallel. Similarly, for CY and XT. Thus, TXCY is a rectangle. Therefore, CT and XY are equal and bisect each other at O. Since 6.0XC is isosceles, LOXC = LOCX. Draw radius DX. Then 6.CDX is also isosceles, so LDXC = LDCX. Adding, LOXD = LOXC + LCXD

= LOCX + LXCD = 90°.

Hence XY is tangent to arc AXC, and, in the same way, to arc CY B. Next, let AC = a, CB will be

= b.

The area bounded by the three semicircles 1rab 4

The area of the circumcircle about TXCY equals 1r(TO)2 =

=

=

~~~ since

TC v;ib so that TO ~v;ib. Thu", the area within the Arbelo" equal" the area of the circle with TC a" diameter.

The January 1983 Arbelol Cover Problem

2

A

I\ I

I

D

T

- -__

c

p IFlaure

2.l

Now consider the cirde with radius :z: tangent to the large arc, the tangent line TC, and the small arc with radius, say ri. (See Filure 2.) Imagine the right triangle E F D formed by dropping a perpendicular EF from E, the center of the cirde with radius :z: to DC. Note that DF ri -:z: cos LEDF DE ri +:z: Let the large semicirde have radius rand P be the midpoint of AB. Apply the Law of COlinel to b.EPD to obtain

= --- = --.

:z:) . (r - :z:) 2 = (ri + :z:) 2 + (r - ri) 2 - 2( ri + :z:)( r - ri ) (ri --ri +:z: We have omitted the other semicirde. If it has radius r2, we can rewrite the above as Expanding and simplifying, we arrive at

Since this is symmetric in ri r2 we condude that the areas of the drdes inscribed in the Arbelos and tangent to CT are equal. Archimedes found these results. This says something about andent Greek geometry. The cover page of the January, 1983 Arbelol was intended to illustra.te the relative values of the arithmetic mean (A), the geometric mean (g), the

Tbe January 1983 Arbelo, Cover Problem

o IFilure

3

B

C

3.1

harmonic mean ('H) and the root-mean square ('R.). The diagram was not clear, and I have taken the liberty of presenting another. (See Filure 3.)

=

If AC a, CB Obviously A ~ 9,

= b, then OT = a+b -2- = A

and CT

= y(ib ab = g.

Now let CF be perpendicular to OT, Then

ab

2ab or TF= - - = 'H. a+b

= (TC)2 =(TF)(TO) = (TF) (a; b)

Obviously 9 that 9 2 = A'H.)

~

'H, and we now have A

~

9

~ 1(,

Finally, draw OR.L AB with length OR = OC =

(Incidentally, note

~; b.

Since AO =

a+ b we east'I y fi nd -2-'

Ja + 2

AR=

2

- 2 b- ='R.,

Now, by looking at the diagram, we see 'R. memory!

~

A

~

9

~

'H. It helps the

i \

I take the liberty of passing over the diagrams on the covers of the March, 1983 and May, 1983 Arbelo, - the first was simply the sketc.h of a multi-purpose plug which could plug a circular bole, a square hole, and a triangular hole, and the second because it is a standard diagram for Desargues' Theorem, The reader will find a proof on page 19 of the May, 1983 Arbelo"

!

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i:

4

The November 1983 Arbelo" Cover Problem

Again, I stress that Desargues' Theorem needs three dimensions for proof. Many of the "modern" proofs in two dimensions assume commu­ tativity, and there are nOll-Desarguesiall systems of geometry. The dual theorem is also true. As for the cover for the September, 1989 Arbelos, this is a way of deriving a standard theorem from spherical trigonometry, unfortunately defunct. Were we to replace some of our modern mathematics, say set theory, by spherical trigonometry, I believe it would be an improvement. The cover of the November, 1989 Arbelo" is not pure geom'etry, but I considered it original enough to use, since it ties together two concepts that appear to be completely distinct. -

We start with the Pascal Triangle (See Figure 4.) and add as indicated on a slant.

1

1 ,

1

, !

1~-2

Ii ;

1~=~

;

/~

///1_13

a

17'6 4 1

1~10

1~

15

20

10

5

1

15

6

1

etc.

Certainly, we do get the first few Fibonacci numbers. We procede by in­ duction. It appears that F.. , the nth Fibonacci number, may be

I,

I

The November 1983 Arbelo8 Cover Problem

I

5

Assume Fn+l

=

(n) + (n- 1) + (n - 2) + (n - 3) + ... 0

1

2

3

is valid for n = 0, 1, ... n. Then, Fn

(n) (n -'- 1) + (n - 1) + (n - 2) + (n - 2) + ....

+ Fn +1 = 0 +

0

1

1

2

We remember two facts: (1) For Fibonacci sequences, Fn +2 = Fnt1 + Fn , Fo = 0, F1 = 1; and (2) For the Pascal triangle (;) = (n;1) + (:=1). With these in mind, we examine this last series. First, (~) = 1 = (nt 1 ). Next, for any two successive terms, we can substitute a single term. Thus:

so that

Fn+Fn +1 =

(n+ 1) + (n) + (n - 1) + (n - 2) + (n- 3) +... = F 0

1

2

3

4

n +2.

The January 1984 Arbelo, Cover Problem

6

A

R

c p

Prove:

PQ -

PR

and:

PQ.l PRo

Jan 1984

The problem on the cover ofthe January, 198-4 Arbelo, is probably the most provocative of all the cover problems. We can solve this problem by using U A geometric figure may be moved in 'pace without changing ,ize or ,hape" in a judicious manner. Now to the problem: On the side BC = a of 6.ABC we have con­ structed equilateral triangle BCT and drawn PT. This is a crucial step, and I was led to it alter a number of unsuccessful approaches. Now we have four similar triangles: ARB, AQC, TPB, TPC. (See Figure 5.) If the sides of 6.ABC are a, 6, c, then 6.ARB

has sides

6.AQC 6.TPB

has sides

6.TPC

c, tc, uc,

b, tb, ub, has sides a,ta, ua, has sides a, ta,ua.

The January 1984 Arbelo8 Cover Problem

7

A

c

IFigure 5.1 Now we rotate 6.RBP 45° clockwise about B. Then R falls on AB, and P falls on BT. If B is held fixed and the rotated triangle is multiplied by!, the shifted and expanded RP will coincide with AT. u Similarly, we rotate 6.QCP 45° counterclockwise about C and expand by ! holding C fixed so that the shifted and similarly expanded 6.QCP u will coincide with 6.ACT. Since RP and QP coincide after these transformations, they must have been equal initially. Also, since each was rotated through 45° and were then coincident, RP and QP must have been perpendicular to start with.

8

The March 1984 A rbelos Cover Problem

Given:

Prove:

Equilateral triangles erected outwardly on the sides of triangle ABC have centers X, Y, and Z. Triangle XYZ is equilateral. Mar 1984

The cover on the March, 198i A rbelol has some interest because of its supposed history. We'll sneak up to it. (For extensive generalizations and references see Cruz Mathematicorum 6(1980) ppl85-l87.) Suppose that, on the sides of 6.ABC, we constrllc.t6.ARB, 6.BPC and 6.CQA and that LR + LP + LQ = 180°. (See Figure 6.) Then,the circumcircles about 6.ARB, 6.BPC and 6.GQA intersect at a common point OJ for, if we let 0 be the intersection of the circumcircles of 6.ARB and 6.GQA, then LAOB

= 180° -

LR

and

LAOG

= 180° -

LQ.

Hence LBOG = 360° - (180° - LR) - (180° - LQ) = LR + IQ = 180° - LP.

I

The March 1984 Arbelo8 Cover Problem

I

9

--~--------...,\Q

R

I

I

I II .

!I

p

Therefore OBPC is inscribable, and the three circumcircles have a common point, O. This is certainly true if 6ARB, 6BPC, and 6CQA are all equilateral triangles. Then their circumcenters form the Napoleon Triangle, 6XY Z, of 6ABC. In this special case, 6XY Z is an equilateral triangle. An absurdly simple proof is (See Figure 7.) : ZY is perpendicular to chord AO and ZX is perpendicular to chord BO. Hence LY ZX and LAOB are supplementary. Since LAOB and LR are supplementary, LY ZX = LR = 60°. Similarly LZXY = LXY Z = 60°. So 6XY Z is equiangular and hence equilateral. Incidentally, I recommend the reader to the treatment of this problem by I. M. Yaglom. Yaglom is a master of transformation geometry as his work in Geometric Transformations, Volume 8 of the New Mathematical Library, pp93-9-4, shows. If the equilateral triangles are constructed on the sides of 6ABC so that their interiors intersect, with the interior of 6ABC, their centers X, Y, and Z form the inner Napoleon Triangle. (See Figure 8.) Incidently, AB, AC and BC are perpendicular bisectors, respectively, of ZZ, YY, and

XX.

! .

If

The March 1984 Arbelo" Cover Problem

10

--~~-=:-------~Q

R

p

IFi,ure 1.1 For the Napoleon triangle, .6.ARB, .6.BPG, and .6.CQA are equilat­ eral. Hence, in .6.AZY, by the Law 0/ Co"inu,

(Zy)2 = (ZA)2 But ZA

+ (Ay)2

- 2(ZA)(AY) cos LZAY.

V3 2 V3 = -32 . -·-BA AY = - . -AG and 2' 3 2 '

LZAY = LBAG + LZAB + LCAY = LBAG + 30°

+ 30°,

so

For the inner Napoleon triangle, we have

(Zy)2

= (BA)2 + (AG)2...,.

2(BA)(AC) cos(LBAC _ 600).

333

The March 1984 Arbelo! Cover Problem

11

\ I

: '

l \ .

So - - , = 2(BA)(AC) [cos(LBAC - 60 0 ) - cos(LBAC + 60 0 )] (ZY) , - (ZY) -_._....3

=

~(BA)(AC)Sin

=

~(BA)(AC) sin LBAC

=

~[ABC]·.

•I

LBACsin 60 0

Similarly, ,

- -,

(X Z) - (X Z)

2(AB)(BC). = -J3-

S10 IABC

4 [ = J3 ABC]

and

(YX)' ­

Thus, since

Xy

(Xi')'

= ~.lAjiCB) sin LACB = ~[ABC].

= YZ = ZX and the inner triangle is equilateral.

v'i , V3 .. , --­ [ABC] = -.("(XZ) - -4'(XZ) = [XYZ]- [XYZ],

i

i! Also,

the difference

• 1 use the traditional notation, [RST], for the area of ~RST.

:

I

I

A First Diversion

12

in area between the outer and inner Napoleon triangles equals the area of the triangle ABC.

A First Diversion In the XXVIII International Mathematical Olympiad, held in Cuba in 1987, contestants were, as usual, given the opportunity to ask questions about the problems. One student wanted to know "What is a Euclidean plane?" Sic tran~it gloria mundi! I would like to present some ideas that may be of use to the would-be problem solver. First is the projective plane. For, us, this plane will be one where: (a) Two points are on (or detelluine )on~ line. (b) Two lines are on (or determine) one point. Note that any two lines intersect. Also note that measureml':nts (lengths, distances, area) are not part of projective geometry. Neither is parallelism or perpendicularity. Because of (a) and (b), duality can occur. From a statement about points incident with lines we automatically have a dual statement about lines incident with points. Thus, from (1) A triangle con~i~t~ of three line~ on thue point~.

point~

not on a line together with the three

we obtain (2) A triangle con~i~t~ of three point~ on thue line8.

line~

not on it point together with the three

lr · sI'19 hi f rom "A {qUadrangle} We get somet h mg t Yd'lnerent quad II'1a t eraI

of four { Pl·oints } no three on one { lin~ t } together with the mes pom on them."

con~i~t~

{lin~St } pom s .

Use the upper terms and you have a statement. Use the lower terms and you have its dual. The diagrams are in Figure 9. Notice that a conic (whatever that is) may be defined as a points or as an envelope of lines. I repeat my admonition, given on page 14 of the May, 1984 "A projective plane matriz. "

i~

a plane. It

i~

not a

~et

locu~

of

Arbelo~:

of equations nor a

Next I will discuss the affine plane. If, from a projective pillne, we remove one line, we are left with an affine plane. We can alternately just

A First Diversion

13

e

d

quadrangle 4 points, A, B, C, D 6 lines, AB, AC, AD, BC, BD, CD

quadrilateral 4 lines, a, b, e, d 6 points, ab = 1, ae = 2, ad = 3, be = 4, bd = 6, cd = 6 Note the differenee8.

IFllure

9.1

label one line without removing it. We agree to call this line the line at infinity. First, every line meets the line at infinity, so every line has a "point" at infinity. Next, if two lines are coincident at the line at infinity, we shall call these lines parallel. This permits us, for example to work with "parallelo­ grams". (See Filure 10.) Quadrilateral ABCD is a parallelogram.

--'Ilr-----------#'-line at infinity

IFigure 10.1

The May 1984 A r~lo, Cover Problem

14

Now, on with our cover problems.

-r

D~ _ _

~

C

A

Given:

ABCD

is a parallelogram.

PQIIAB.

RsIIAD. Prove:

BP, CT, DS are concurrent at O.

Ma

1984 ,"

Sometimes it is more useful to use the projective plane rather than the Euclidean plane in attacking a problem. Thus, for the cover problem of the Ma.y, 1984 Arbelo8, we might use Figure 11.

The May 1984 Arbelo. Cover Problem

15

IF'lure 11.\ A B C --......... --~--.......,~--L

"Let A, B, C be on a line L, and A', B', C' be on another line L'. Then the intersections (AB')(A' B), (BC')(B'C), and (AC')(A'C) lie on a line."

l!!i'!~ However, I have another proof. First, note P4PP'U~' Theorem which I present without prooC (See Filure 12.) Since there is nothing said about measurements in Pappus' Theorem, we can dualize it. (See Figure 13.)

r ( i

By Collowing the diagram in Filure 13 carefully. you will find that we actually do have the dual oC Pappus' Theorem. and that the dotted lines will be concurrent. Now let the points L, L' in this last diagram lie on the line at infinity, and we have the original diagram. and a proof. Just as almost nobody is aware that the Binomial expansion of (a +b)" holds for ~l ,:e~ n. few students accept the possibility that coordinate axes need not be at right angles. Once we accept this, for an alternative approach we can associate coordinates with the points in the diagram on the cover of the May, 1984 Arbelo.: A(O,O), B(b, 0), C(b, d), D(O. d), S(a, 0). R(a, d), P(O. e), Q(b, e), T(a. e). Then, for

The May 1984 Arbelo, Cover Problem

16

l

•:

AI ' -_ _.L..-

.::::::::..\

--=::::::::~

B

L'

"Let A, B, C be lines on point L, and A', B', C' be lines on point L'. Then the lines OJI (AB')(A'B), (BC/)(B'C), and (AC')(A'C) lie on & point (are concurrent). "

IFijure 13·1

line DS:

y- 0

z-a

-d a

y- 0 c- 0 -c :r:-b = O-b = b'

line BP:

line CT:

d- 0 O-a

--=--=-

y-e z_ a

= d-e b_ ~'

or

or

dz +ay= ad

or

ez

+ by = be

(d - e)z + (a - b)y

=ad -

be.

Since the equation for line CT is the difference between the equations for lines DS and BP, these three lines are concurrent. Which is a better proof is problematic, although I would prefer the projective (or affine) proof. Either way, they are "grown-up" proofs.

A Second Diversion

17

A Second Diversion Most mathematicians think that geometry ended with Euclid (ea. 300 B.C.) or perhaps Pappus (ea. 300 A.D.). We are going to look at something invented, I am told, in 1831 by L. J. Magnus - inver.!ion.

p

A

,

i

J

I

I j

Briefly, let 0 be a circle with radius r. (See Figure 14.) If tangents are drawn from a point P and if line TU intersects OP at pi, we say that points P, pI are inverse.!. Notice that op x OP' = r 2 • We say 0 is the center of inversion. It is not hard to see that a circle tangent to the cirde of inversion and passing through the center of inversion inverts into a straight line, PAt tangent to the cirde 0 at A. In fact, using similar triangles, one can show

A Second Diversion

18

that any circle passing through the center of inversion inverts into a line P A and conversely. All we need now is one more result-a eirde cutting the eirde of in­

version at rightangles inverts into itself. (See Figure IS.) Remember that if a tangent OT and a secant OPP' are drawn, then (OT)2 = (OP)(OP').

P'

IFigure 15·1 It is not hard to see some advantage of inversion. Tangent curves remain tangent after inversion, for instance. Also, the inverse of a figure may turn out to be simpler than the original.

I

[End of Second

Dive~I

The September 1984 Arbelo6 Cover Problem

19

Now back to our cover problems.

Given:

Prove:

A chain of circles is inscribed in the arbelos, each tangent to the arbelos and to the preceding circle. From the center of each circle, let a perpendicular Oi Hi be constructed to line AC. Let di be the diameter of circle 0i. The length of 0i Hi is i x ~ . (For example, 0a H2- 2 PQ.) Se 1984

I

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The cover problem on the September, 1984 Arbelo6 is due to Pappus, who wrote that it W$S old in his day. I can't help feeling that it is an Arbelo6 problem which Archimedes. might have solved. At any rate, we have an A~belos with a chain of inscribed circles 0 1 ,0" ... On' I have· stopped at circle 0, for simplicity. Let d, be the diameter of circle 0" and O,H, the perpendicular to AG. We wish to prove that O,H, = 2d, (in general, that OiHi = i: di)' Let our circle ofinversion have center A and radius equal to the tangent . from A to circle 0,. Then circle 0, inverts into itself. Since the circles with

\.

The Septemher 1984 Arbelo8 Cover Problem

20

-A

c

B

, IFilure 16·1 ,diametersAB and AC pass through the center of inversion, they invert into paralleUines. (See Fisure 16.) -

.

Circles 0 1 and O 2 will then invert into equal circles tangent to the two lines. Now just loolung shows that 02H2 is2d 2 • And in general, using center A and a radius equal to the tangent from A to 0", we can similarly show that O"H" = ndn. You can't tell if inversion will help you solve a problem if you know nothing about inversion. And you don't really have to know the fine points of inversion. You should know it exists, at least. For further applications of inversion, see Yaglom, Geometric Tran8formation8 MAA NML8, p8 or Courant and Robbins What i8 Mathematic8? Oxford Univ Press, NY, 1941, ppl.40-16.4.

21

The November 1984 Arbelo.. Cover Problem

A

1\

Given: . Prove:

The external tangents to the two circles intersect the internal tangent at P and Q. PQ -

AB. Nov 1984

The cover problem on the November, 1984 Arbelo.. is not difficult. ----~

A

J

j. I

C

Q

IFIgure

17

·1

The January 1986 Arbelo" Cover Problem

22

That AB = CD is obvious (or easily proved). (See Figure 17.) Now AP

= PX

and P B AB

Also QD

= PY, so

= P X + PY = PY + XY + PY = XY + 2PY.

= QY and QC = QX, so

= QX +QX +XY = 2QX +XY. Henee XY +2PY = AB = CD = 2QX +XY, so PY = QX. CD"= QX +QY

PQ = QX

+ XY + PY =

Finally,

AB since PY = QX.

, \

'

II

I

I

I I

II I!

y

Given:

! Prove:

On one side of line segment XY, similar triangles XYA, YXB, XCV, VOX. EYX. and are constructed. Points A, B, C, D, E and F lie on the same circle.

FXY

Jan 1985

23

The January 1985 Arbelol Cover Problem

I \. I

x

y IFigure 18·1

The cover problem on the January, 1985 Arbelol has many segments, but is not difficult. In Figure 18, all the angles we need are labeled. An angle marked 12 means that it is the sum of angles 1 and 2; an angle marked 1123 is the sum of angles 1 + 1 + 2 + 3, etc.

I

Triangle ZEF is (by symmetry) isosceles. Now, from anyone of the six similar triangles, (see 6CXY), 111223 = 180°. But LCAE+ LEFC = 1123 + 123 = 111223 = 180°. Hence C, A, E, Fare concyclie. By synune­ try, D, B, F, E are also concyclie. Therefore C, A, E, F, D, B lie on the same circle. Incidentally, the problem appears on page 289 of R.A. Johnson's Ad­ vanced Euclidean Geometry, Dover, NY, 1960. The remark is made that all six triangles have the same Brocard angle, so the six points lie on the Neuberg circle. In this connection, (a) one can get along without Brocard angles and Neuberg circles, and (b) the special case where the triangles are isosceles is worth examining. The proofin Johnson occupies 14 lines, so the esoteric geometric objects do not make the proof shorter or simpler. There is a lesson in this (for followers of Bourbaki, at any rate).

The January 1985 Arbelo$ Cover Problem

24

One of the fields of geometry I have found provocative and interesting has been that of geometric constructions. I met up with this decades ago in inscribing a square in a triangle. (See Figure 19.)

B

IFigure 19.1 What I did then was to construct a small square near B with a side on BC and a vertex on AB. Then let P be the intersection of AC with the line through B and the vertex of the square inside 6.ABC. The point R is on BC with PR.l BC, and Q is on AB with QP II BC. From similar triangles, we easily find that PQ = P R and we have three vertices of the desired square. To make it look mysterious, we can expand the square until P R equals altitude AH. Now we have a construction where we construct AP equal to and perpendicular to altitude AH and draw BP intersecting the side AC at a vertex of the desired square. (This is a simple application of homothetic figures.)

25

The March 1985 Arbelo8 Cover Problem I

!

!

i

I

E

I'

A~--------""7""-----~

c

:

B

Given:

Triangle ABC.

Construct:

line segment DE so that BD - DE - EC.

Mar 1985

For the cover problem of the March, 1985 Arbelo8, we can proceed as follows (See Filur. 20.) :

A

The May 1985 Arbelol Cover Problem (a) Marked oft' any length BX on AB. (b) Conlttructed a semi-circle with center X and radius BX. (c) Marked oft' CY = BX on AC. (d) Constructed Y Z parallel to BC meeting the semi-circle at Z. (e) Completed parallelogram WZYC. Now we have BX

= X Z = ZW with BX ZW similar to what we seek.

(f) Draw BZ meeting AC at E, (g) Construct DE parallel to XZ. Since triangles BZX and BED are simill.u, as are BZW and BEC, BD

= DE = EC and we are done.

Given:

Rectangle ABCD. inradii.

Prove:

As shown, r 1

,

r 2 , rs are

DH is an altitude.

r 1 + r2 + r s - DH . Ma

1985

The May 1986 Arbelol Cover Problem

27

r

~

a

b

IFigure

2C]

The cover problem for the May, 1986 Arbelol, which uses simple results to obtain a complicated result, is quite pretty. Recall that the area of a triangle with perimeter 2. and inradius r equals r•. * Let

(J

= LDAH = LHDC.

In

~AHD

we see that

AH· HD = rl(AH + HD+ DA), DA· (cos (J). H D = rl(DAcos (J + DAsin 9 + DA), HD'cos9 rl cos(J+sin(J+l

= ----..--.

Similarly, for

~CH D,

CH· HD = r'J(DH + HC+ CD), CD· (sin 9). HD = r'J(CDcos (J + CDsin 9 + CD), H D· sin (J r'J = . cos (J + sin 9+ 1

Next, since

~ABC:=! ~CDA,

we may write

AC· HD = ra(AD + DC+CA) AC· HD = ra(ACcos (J + ACsin 9 + AC) H D = ra(cos (} + sin 9 + 1) HD ra=----,--­ cos9+sin9+1

Adding, rl + r'J + ra = H D. There is another solution, depending on a rare datum about right triangles. (See Filure 21.)

* This can easily be proved by partitioning the triangle into three trian­ gles using the line segments from the center of the cir~e to the vertices of the given triangle. The altitude of each of the three triangles is the inradius of the original triangle.

28

The May 1986 Arbelos Cover Problem A rare datum: In any right triangle, the sum 0/ the leg, minus the hypotenu,e equal, the diameter 0/ the in,cribed circle. In Figure 21, the lines shown with equal letters are equal. The assertion is that (a + r) + (b + r) - (a + b) = 2,.. Using this, we get (See May, 19S5 cover.) 2rl =AH+HD-DA 2r, = C H + H D - DC 2ra = (DC

+ DA) -

2(rl + r, + ra) = 2HD

so that rl + r, + ra = H D.

However, you have to know the rare datum.

(AH

+ CH)

Another Diversion

29

Another Diversion I have found peace, relaxation and diversion in just gazing at a di­ agram. After a small interval of confusion, the elements of the diagram seem to so group them.elves as to produce theorems! I have not yet bad this experience while gazing at a matrix or a statement in set theory.

IFilure 22·1 One diagram I found relaxing and instructive is Filure 22: t::.ABC is a triangle inscribed in a circle with center O. Altitudes AU, BV, CW have been drawn intersecting at orthocenter H. Points U, V, Ware joined forming the orthic triangle UVW. I have extended AU, BV, CW as shown and added some dotted lines. Now we gaze and find, in order, the following results:

(1) From right triangles ABV and ACW,

= LACW. Similarly, LBAU = LBCW and LCAU = LCBV. LABV

= 90

0

-

LBAC

(2) Since we have right angles at W. U and V. quadrilaterals W B U H and VCUH are inscribable, LWBH = LWUH and LVCH =

30

Another Diversion LVUH. By (1), LWBH = LVCH, so LWUH = LVUH. Thus, the altitude', of ~ABC hued the angle! of it! orthic triangle.

(3) Triangles WHB and VHC are similar (compare angles). So WH BH

VH = CH

or

BH·HV=CH·HW =AH·HU.

(4) From (1), LBCC' = LBAA' = a. Angles BCA' and BAA' both subtend the arc BA' of the circle circumscribing ~ABC, so LBCA' = a. Since LBCC' = LBAA', ~CUH ~ ~CUA', U A' UH, and AU . U A' BU . UC AU . UH.

=

=

=

(5) Since WBCV is inscribable, AW· AB = AV· AC, ~ABC

"" ~AWV "" ~BWU "" ~CUV.

(6) Let AB be a diameter. Since LABC = LABC, triangles BAU

=

=

and BAC are simi.r. Then LCAB LBAA' a. Let P be the intersection of AB and WV. Then LAV P = LABC = LABC, so triangles APV and 'ACE are similar. Therefore, LAPV and LACE are right angles and radius AO is perpendicular to side WV. In the same way, BO.l UW and CO.l UV. ~AWV which is similar to ~ABC. Therefore corresponding lengths are propor­ tional. That is

(7) Segment AH is a diameter of the circle about

AH 2R

= WV BC

or

BC· AH

= 2R . WV,

where R is the radius of the circle circumscribing

=

~ABC.

=

Similarly AC· BH 2R· UW and AB· CH 2R· UV. Adding, we have BC· AH + AC . BH + AB . C H = 2Rp where p =

perimeter(~UVW).

BC· AU - BC . HU

+ AC . BV -

However, we may write this as

AC . HV

+ AB . CW -AB . HW =2Rp

3· 2[ABC] - 2[ABC] = 2Rp.

This means 2[ABC] = Rp, and the perimeter of the orthie triangle [ABC] equals 2 -- . R I stop here, although I did get more. Perhaps the reader might try it on other diagrams. It's better than contemplating one's navel.

I~nd

of Another

Diversi~I

31

The September 1985 Arbelol Cover Problem Now back to our cover problems.

DF-----*----~

Given:

Orthocenter H of triangle ABC is the center of a circle. P, Q, and R are the midpoints of the sides.

Prove:

AD - BE - CF. 5e

1985

The cover problem on the September, 1985 A rbelol is an example of a mess of lines hiding a solution. Look at Figure 23. We "fumble" away at a solution. Let AU, BV, and CW be altitudes of t:::.ABC, and let AU intersect RQ (extended if necessary) at J. Since DJ J.. AH, AJ = JU, and DH = r, the radius, we have

+ (JD)2 = (AJ)2 + (DH)2 - (H J)2 = (DH)2 + (AJ - H J)(AJ + H J) = r 2 + (AH)(U J + H J)

(AD)2 = (AJ)2

=r 2 +AH·HU. Similarly, (BE)2

= r 2 + BH· HV, and (CF)2 = r 2 + CH· HW.

Now we can use another "rare datum." Look at the "another diver­ sion", result (3). This tells us that AH· HU = BH· HV = CH· HW, so AD=BE=CF.

32

The September 1986 Arbelo. Cover Problem

33

The November 1985 Arbelo8 Cover Problem

A

B-"::;~""""L.-----------=-

P

Given:

c

PQR is the orthic triangle of ABC.

(Le., AP, BQ, CR are altitudes.) AX, BY, CZ are perpendicular to QR,RP,PQ respectively.

Prove:

AX, BY, CZ are concurrent.

Nov 1985 With the help of our "another diversion", ~esult (6), we can just toss oft' a solution to the cover problem on the November, 1985 Arbelo8. It shows that AX, BY and CZ are parts of radii of the circumcircle ofAABC.

i I

i'

I 34

The J...uary 1986 Ariel•• Cover ptb1em

c c

-------.JI E

&.­

A

Given:

o ABCDE is inscribed in the semicircle on AE with unit radius OA. AS-a, BC-b, CD-e, DE-d.

Prove:

a 2 + b 2 + c 2 + d 2 + abc + bcd < ...

Jan 1986

With the cover problem on the January, 1986 Arbelo", we are back to complicated mathematics.

c-.---__

A&....-------------~E 1 1 o

IFi,ure 24.'

.;

The January 1986 Arbelol Cover Problem We draw AC

= z, CE = y, (See Filure 24.) Z2

= a 2 + b2

-

2ab cos B

y2

= c2 + d 2

_

2cd cos D

Now ~ACE is a right triangle, so

Z2

35

and note that

+ y2 = 4.

Thus

We do have inscribed quadrilaterals ABCE and ACDE. Therefore

B = 180 0 cos B

=

-

LAEC

and

D = 180 0

=_!

and

cos D

-cos LAEC

2

-

LCAE

z = -cos LCAE =--. 2

Thus, 4

= a 2 + b2 + c2 + d 2 - 2(abeos B + cd cos D) = a 2 + b2 + c2 + d 2 + aby + cdz

> a 2 + b2 + c 2 + d 2 + abc + cdb. One can also show that

i

,.

!

The March 1986 A rbelo8 Cover Problem

36

'"

.'

r'

------"'"""'::-~=~-----------

Given:

Inscribed triangle ABC.

AM bisects angle

SAC.

MP is perpendicular to radius AO. Prove:

AP - AB. Mar 1986

As we go along, the cover problem gets more interesting, I think, and easier.

The'March 1986 Arbelo.. Cover Problem

37

I'

IFigure 25·1 We extend AO to B on the cirde. (See Figure 25.) Since AB is a diameter, LACB 90°. Since LPQB 90°. PQBC is inscribable. Also LABC = LB since both angles subtend the same arc..

=

=

= = =

Now LABC + LQPC 180° LAPQ + IQPC. Therefore, LAPQ = LABC LB. Also LBAM LPAM and AM AM. Hence 6.AMB:!! 6.AMP and AP = AB.

=

The above proof uses the fact that LB if LB ~ 90°?

,

i I

r

=

< 90°.

(Where¥) Is AP = AB

The May 1986 Arbelo8 Cover Problem

38

A

l'

c

r

B

Given:

Find:

Equilateral 'triangle ABC with vertices on parallel lines as shown. The distances between the lines are as shown. The area of triangle ABC. Ma

1986

In the case ofthe cover problem on the M41/, 1986 Arbelo8, there is a hint-e:onstrud the triangle on the given lines first.

A

p~~~----*::'--T

l!!lure 26·1

39

The May 1986 A rbelo6 Cover Problem

Construd LAPT = 60° and LBPT = 60°. Let APBC be the circum­ circle of 6.APB. Then since LABC and LAPC both subtend the same arc, LABC = LAPC = 60°. Similarly, LBAC = LBPC = 60°, so 6.ABC is the desired equilateral triangle. (See Figure 26.) Now a little trigonometry. !. I

i

, I

10 . 0 v'3 =SlR60 = ­ AP 2 '

AP =

v'3'

6 . ° =v'3 -=slR60 ­ BP 2 '

BP=

~

2

20

,

400 144 20 12 ° + - -2· . -.cos120 3 3 V3 v3 400 144 240 784 =3- +3- +3- = 3­ .

(AB) = -

v'3 . 784 4 3

= 196 '3

3 V". For a generali.ationto tetrahedra and simplices, see Cruz Mathemati­ corum, 13(1987), pplfO-lf,j.

Thus [ABC] =

I

,­

, I

I'

Still Another Diversion

40

Still Another Diversion' With the disappearance of geometry from our secondary schools,' we have, so to speak, occasionally "thrown out the baby with the bath water." One such instance appears to be the subject of ratio and proportion. Our copy of Plane Geometry, by Schultze and Sevenoak , in its 1926 editioll, has a short but good development of this. The first two chapters of Higher Algebra, by Hall and Knight, go even further. . "

This disappearance is unfortunate. Even Dieudonne, whose slogan, "geometry must go" was largely responsible for this, has had a change of heart. Not too long ago, he did say that perhaps some geometry had merit. I refer the reader to any old geometry text for the rules for ratio and proportion. I will proceed as if the reader knows them. A

I

,

c

B

IFigure 27 ·1

We derive Ceva's Theorem as follows (See Fisure 27.) :

BP PC .. CQ Slmtlarly, QA

=

[BOC]

= [BOA]

[BAP] [CAP]

and

AR

=

[BOP] [COP] [COAl

=

RB = [COB]'

[AOB] [AOC)"

Therefore

AR . BP , CQ _ [COAl . [AOB] . [BOC] _ 1 RB PC QA - [COB] [AOC] [BOA]­

The reader might compare this derivation with those in Altshiller-Court, Johnson, and Coxeter,

E=n=d=o=f=S=tn=l=A=n=o=t=h=e=r=D=iv=e=r=si=o::::'~ I

1:-;=1

The September 1986 Arbelo, Cover Problem

41

Now let ullook at the cover problem for the September, 1986 Arbelo,. We merely compare areas of trianglel with equal altitudes. Since AX II BY, [XBY] = [ABY]. Similarly, BY II CZ impliel [ZBY] = [CBY]. It followl that [XBY] + [ZBY] = [ABC]. Next since XA

II ZC,

[BXZ]

[ZXG] = [ZAC], so

+ [BCZ] =

[ABC]

+ [BCZ],

or [BXZ] = [ABC]. Adding, we get [XYZ] = [BXZ]

+ ([XBY] + [ZBY]) = 2· [ABC].

42

Take a Break

A

IFisure 28·1 Let's now "take a break" to contemplate Fisure 28. In it, 6ABC has sides parallel to those of 6ABC. Thus 6ABC '" 6AOC. Since the respective sides are parallel, A, B, and C bisect the sides of 6.A.OC.· The perpendicular bisectors of the sides of 6.A.OC intersect at H, the ortho­ center of 6ABC. Thus H is the center oUhe circumcircle of 6.A.BC. If 0 is the center of the circumcircle of 6ABC and OM.. is the perpendicular bisector of BC, then AH = 2 . OM... Again, the circumcircle of 6ABC is the nine-point circle of 6.A.OC, A

so the radius of the nine-point circlet of 6ABC is °2 . Anyway, when we derive the properties ofthe nine-point circle, we do find that AH = 2·0M...

• For example,

BA = 1 - CA = ....-;AC + -.._ BC - __CA- = __--"' AB- +.-.OA - -:;;;-;­ CB -:;;-;;DC

DC.AD

DA

CD

AC

DC

CA

shows that AB - CB = 0 so B is the midpoint of .A.C. t The circle whose center is midway between the circumcenter and the orthocenter and whose radius is half that of the circumcircle passes through nine notable points, namely the feet of the altitudes, the midpoints of the sides, and the midpoints of the segments from the orthocenter to the vertices.

43

The November 1986 Arbelol Cover Problem

A

I.

I

I· .

B ~--E----""'--'C

In AABC. AE, BI. CD are

alt1tudes~

If DE

~

~

13, 01

~

14. IE

15.

find the area of AABC. Nov 1986

Now we can get to the cover page for the November, 1986 Arbelol. Since AH is a diameter of the circle on ADHI, sin A =

i.

I I' I

i

~~.

(Why?)

Let the circumcircle of 6.ABC have center 0 and radius R. Then . BC 14 BC . SID A 2R' Hence AH 2R or BC . AH 28R.

=

=

=

,

Choose M 4 on BC so OM4 .L BC as in Figure 28, and choose M" and Me similarly. Now BC· 20M4 BC· AH 28R,so BC, M 4 14, Thus AC ,OM" 13R and AB, OMe 15R, so

=

=

=

=

=

I

,

1 [ABC] = 2(BC. OM4

+ AC· OM" + AB, OMe ) = 21R.

Now the area of a triangle ABC with sides 4, b, c and circumradius 4bc. 13· 14, 15 R IS [ABC] = -R' (Prove d!) Thus [DIE] = - - - = 84 where p 4 4p is the radius of the 9 point circle of t:::.ABC. Hence p = 65/8. Therefore . 21·65 1 R = 2p = 65/4. FIDally [ABC] = 21R = - . t - = 341:;, .

The January 1987 ~rbelol Cover Problem

44

Given: K is any point inside square ABCD. MN is any line through K. Circles are constructed through points AMK and CNK intersecting again at point P. frove: AC p~sses through P. Jan 1987

The cover problem for the January, 19S7Arbelo, comes from KVANT, a Russian secondary school journal in mathematics and physies which is excellent for those who can read Russian. We will sh~w that P lies on diagonal AC by showing that LAPK + LKPC 180°. Since angles AMK and APK subtend equal ares they are equal. Since angles K N C and K PC are opposite angles of an in­ scribed quadrilateral, they are supplementary. Since AM II NC, LAM K = LKNC. Thus,

=

LAPK + LKPC = LAMK + LKPC = LKNC + LKPC = 180°.

The March 1987 Arbelo6 Cover Problem

45

I. I j

I "

Given: H is the orthocenter of triangle ABC. H' is the orthocenter of triangle A'BC. Prove: AHH'A' is a parallelogram. Mar 1987

As for the cover problem for on the March, 1987 Arbelo6, we have two triangles on the same base Be, inscribed in the same circle, but with separate orthocentei•.

i· I

IFigure 29·1

46

The May 1987 A rbelol Cover Problem

Both biangles have the same eircumcenter 0, and therefore OM per­ pendicular to BC will equal half of segment AH and segment A'H'. (See Flaur• 2'.) Also; since both are perpendicular to BC, AH and A'H' are parallel. That makes AHH' A' a parallelogram.

A

B

C

x

Given: X, Y, Z are points of tangency. K is the area of triangle ABC. r is the inradius of triangle ABC. Prove: (BC) (AX) (XX')-4rK. Ma

1987

We come finally to the cover problem on the May, 1987 Arbelol. We have cel1ian AX cutting the cirde at X'. Let 0 be the center ofthe.inscribed cirde. Since XX' is involved, we draw radii OX and OX'. (See Flaur. 30.) Then OX is perpendicular to BC.· Altitude AH is also perpendicular to BC. Triangle OX X' is isosceles. Let 9 = LHAX == LAXO = LOX'X. Let OX = r and we have AH XX' AH XX'

AX = cos 9 and ~ = cos 9. Hence AX =~.

Therefore (AX)(XX') = 2r(AH). Now we multiply this by BC, and we have (BC)(AX)(XX') = 2r(AH)(BC) = 4r[ABC].

47

Epilogue A

"

B

H

c

X

I';Iure 30·1

, I

I•

Epilogue I admit that the cover problems have varied from very .imple to diffi­ cult. My excu.e for presenting them is that I found them fun, and I hoped that my readers would find them fun. We got out of step when proponents of the late "new mathematics" started fussing with the distinction between number and numeral, greater and larger, and so on. I may be wrong, of course, and this nicety in expression is valuable. Perhaps the time will come. when, instead of teaching, "the dog runs after the cat", we will become pre­ cise and teach, "cani, familia", pur,ue, feli, domelticu,." It may be that if we teach reading this way, English will eventually reach that happy state in which mathematic. finds itself today. Anyway, I hope you like the problems and solutions.

Samuel L. Greitzer