No title

/=•

••

,.•

•

•

•

•

•• •

•

•• ••

•

•

•

•

•

•

•

•

•

•

•

•

••• • ••

•• t:

'.

ARBELOS

PRODUCED FOR

PRECOLLEGE

PHILOMATHS

~.

I

1985 - 1986

EDITED BY

PROFESSOR SAMUEL L. GREITZER

RUTGERS UNIVERSITY

Copyright © 1985

Committee on the American Mathematics Competitions

Mathematical Association of America.

••• ••

•

•

••• ••

•• ••• •

•• ••• •• •• •• • •

•• ••• •• • ••

.:

,. i.i'i.. i.'.••

••••• :.••• • •• '••••.• :.•• '.••• i

i

arbelos PRODUCED FOR PRECOLLEGE PHILOMATHS

•

•• •• •• ••• •• ••

Orthocenter H of

~BC

is the center of

a circle. Lines DD', EE', FF' are on the midpoints A' ,B' ,C t of the aides. Prove: AD = BE = CF.

1985-1986:

No. 1

September, 1985

Copyright ~ The Mathematical Association of America, 1986

i PREFACE

We are surprised and grati£ied to start this £ourth year o£ the Arbelos. It £eels like a new beginning. First, acknowledgements £or solutions go to C. Kenneth Fan, Mark Kantrowitz, Ambati J. Krishna, Sherman Lau and David Morin. The solutions were to some extent usual, but there were some which showed ingenuity. Please continue to send in your solutions as soon as you can. Perhaps it is worth-while to restate the purpose o£ this publication. Part o£ it has the purpose o£ presenting material that may be new or un£amiliar to the reader. Now we do not know just what is new or un£amiliar to everyone. We can just hope that there will be enough to make the Arbelos worthy o£ readin~. Thus, Pro£essor Klamkin, writing about the cover problem £or Arbelos 4 - March, 1985, provided us with many re£erences. It was £irst apparently stated by Aleksandrov in 1882, and was not original with him. It was apparently rediscovered by Morley in 1888. There was a generalization by Rosenbaum in 192), etc. Note that Morley thought the theorem was new and prepared his own proo£. In Arbelos, we hope the reader will find a problem new to him and prepare his own - perhaps ingenious - proof. Also, Dimitrios Vathis, writing from Chalcis, Greece, pointed out that there were problems that he found in previous literature. We agree, but hope they were new to most readers. Any text on projective geometry, for example, contains the theorem that the midpoints o£ the diagonals of a complete quadrilateral are collinear. We hoped. that this would be new to enough readers to make it worth including. So = let us know what you would like to see in Arbelos, give us sug~stions, and help us make Arbelos the publication you want it to be. Samuel L. Greitzer Editor

•

•• ••• •• •• ••• •• •• •• ••• •• ••• •• •• •• •• •• •• •

•••• •• •• • ••

•• ••• •

•

•

•

•

•

•• • •

•• •• •• •• ••• •• •• ••

1

CONTINUED FRACTIONS CONTINUED We first developed some of the theory of contin­ ued fractions some time ago - in Arbelos. No.4 - of March 1983 (see p.13). There we used them to solve linear Diophantine equations. Some problems we have encountered make it useful to continue our development. Let us start with some review.

A continued fraction has the form N

b1

b

= a O + a- + ..2 1

a2 +

b

b

.J + • • • +.-ll an

a3

Each new fraction is added to the denominator of

the previous one.

We shall make some changes in this definition

for our special purposes. First. let us have each

b = 1. The resultinp; continued fractions are then

cilled Simple continued fractions. Next, for the

sake of ease in typing, we shall write the contin­ ued fraction in the form N = (a O' a 1 , a 2 , • • • , an) Finally, let us agree to have all the a

r

positive.

If we drop all fractions after a given one. we can do the arithmetic required to get a result, which we will call a partial convergent. For example, consider the fraction 39/17. We divide to get

12=2+-2=2+~=2+ 1

17

17

2 +

1

17/5

1

3 + 5/2

That is, 39/17

=2 + 1

3

+

2/5

=

1

3 + '2 + 1 2

= (2,3,2,2).

The partial converp;ents

are po/qo= 211 , P1 /q 1 = 7/3 , P2 /q 2 = 16/7. Of course, the final conver~nt will be 39/17.

2

For the moment, let PO = a O' qo

= 1.

Then

P1 = a Oa 1 + 1 and Q1 = a 1 (work it out !), and P2 = a Oa 1a 2 + a O + a 2 ' Q2 = a Oa 1 + 1. At this juncture, imagination takes over. We note that it is possible to rewrite P and Q2 as 2 Q Q2 = a 2 1 + QO· P2 = a 2P1 + Po By induction, we easily find that, in general, Pn = a nPn_1 + Pn- 2 Qn = a nQn_1 + Qn-2. These relations hold for all values of n. Notice that these last two expressions look a lot like simultaneous equations. Let us e1ininate the tems containing an from these. We get (Pnq n-1 - qnPn-1)

= -(Pn-1qn-2

- qn-1 Pn-2)

=

(-1)2(Pn_2q n_3 = qn-2Pn-3) = ••• =(-1)n(P1q O - Q1PO)· If we substitute the values for P1 ,PO,Q1,QO' we get just ~ 1. That is, (A)

If this last expression is divided by QnQn­ l' we get 1

~n-1 Frau this, we may conc1\de that successive convergents are alternately smaller and larger than the final convergent (if any) and that the set of convergents approaches a ~1mit as n increases. Moreover, we can now justify claiming that Pn , Qn are relatively prime. Any common factor of Pn and Q would have to divide 1. n

••• ••• •• •• •• •• • • •

•• :: • •

•• •• •• •• •• •• •• ••

•

'. 'I.. :.I.•

• •• • •

•

••• •• •• •• •• •• •• •• •• •• •• ••

•• ••

,.

3 Let us apply what we have to a problem. We take a problem that we have encountered so often that we can give the answer before the proposer finishes giving the problem. It is: A man cashes a check at a local lank. He spends 68¢ and then finds that he has twice the amount of the original check left. What was the amount of the original check? The original check being for 100x + y cents, the condition canes to 100y + x - 68 = 2(10Ox + y). or 98y - 199x = 68. Were we to change 199/98 to a continued fraction, we would get 199/98 = (2,32,1,2). Note: The procedure we have described for changing to a continued fraction is just the Euclidean Algorithm. Thus, 2

98)199

!2£1£ 3)98

.22)1

2 3 2 2

1)2

and we take the partial quotients. We have already remarked. that the final converg­ ent is, in this case, 199/98. From formula (A), then 199(qn_l) - 98(Pn_l) = : 1. Thus, the next-to­ last, or penultimate convergents yield a solution to 199Y - 98x = : 1. We fiddle with signs to get +1. Thus,

98(-67) + 199(33)

= 1.

Therefore

98(-67)(68) + 199(33)(68) Also

98y

- 199x

= 68

•

= 68

4 Subtracting and rearranging, we get 98 _ 113~f68~ + x 199 - Tb7 68 + y. x = 98t - 2244

Then,

y = 199t - 4556 and, for t = 23, we get x = 10 and y original check was for $10.21.

= 21.

The

Now there is nothing in the formulas we have derived. so far to indicate that they must be used. for finite continued fractions only. We therefore extend our use of continued fractions to square roots. This is a bit more difficult than before, and we will not prove all our assertions. Let us apply our results to find ~ as a continued. fraction. To do this, we repeatedly separate expressions, as they occur, into an integr~l part and a fractional part less than unity. Thus, ,/I9 = 4 + (,/19 - 4) and we let a = 4. O invert the fractional part and rationalize, thus __....1""--- =

.Jt9 -4 3

;/I9 -

2

,/19

+ 4

3

=2

..Ji9 -

+

2

3 . = -/19 + 2 = 1 + .,If9 -3

5

and a

1

= 2.

5

5~ = -J19 + 3 = 3 + .Ji9 -3

---,=:o....

:;i9 -

3

_......;;;;.2_

;/19 - 3 _-",,-5_

,/19-2 _--",,-3_ -Ji9 - 4

2

= ,fi9

+ 3

2

=1

+

5 =

.J19

.y'i9 - 4

3 + 4

-2

5

+ 2 = 2 +

= v'19

.Ji9

3

=8

+

09 -4

and a 4 = 1. and a

5

= 2.

and a6 = 8.

•• •• •• •• •• ••

•• •• •• ••• ••• •• •• •• •• •• •• •• ••• •

•• •• • •• ••• • •• •• • ••

I. I.

I.•••

•• •• ••• •• •• •• •• • ••

I.

5 We have found,J19 = (4,2,1,3,1,2,8). There are several points we must comment on at this point. First, note that we got back to at a6. Therefore, the whole process repeats from this point on. The line above the numbers on line 1 show which numbers will repeat. In general, when we reach the last number of the

-09

repeating part, an' we will fim it equal to 2a O• Next, note that, except for the last number of the repeating part, the numbers are "palindromic" ­ they read the same from left or from right. This is always true. We omit the proofs - they can be foum in texts on number theory (some of them). Assume now that we have developed a continued fraction for, say, -./N. We have reached the last repetend. Then, by virtue of our fundamental rules

involving three successive partial convergents (see

Page 2 of this article), we may write:

(If:N + a O)Pn-l

...jN = (::yrN

+ Pn-2

+ a)q + q

o n-l n-2

Now for a little algebra. We clear fractions:

q Nqn_l + (a O n-l +

qn-2)~ = (aOPn-l

Recall that, if a a = c and b = d. Nq n-l

+ ~ = c + d~ , then

For us, this means that

-ap +p

- 0 n-l n-2

&Oqn-l + qn-2

= Pp-l

+ Pn-2) +

Pn-l~

6

Again, since these look like simultaneous equations, we eliminate the terms involving a ' O 2 2 Pn-1 - N qn-2 = ~ 1. That is, to and get solve the equation x2 _ N y2 = 1 try the penultimate convergents of the continued fraction expansion of ~. We shall call a quadratic equation of this form a Fermat equation, since it was he who first used such equations. As an example, let us solve First, expand

=)

~

1

-v'f4

5

_.Ji4+)

5

so a

=

~

O =)

=1+.J14-2

5

_.J!4+2=2+.J!4-2 2 2

--=~2_ = -y'14 + ./i4 - 2 5

;Ji4'- 3

as a continued fraction.

(,/14 - ))

+

-~j14-1-4=---) -;/I4~1=4~---2

x 2 - 14 y2 = 1.

2 = 1 +

+ ) = 6 +

so a 1 = 1. so a 2 =

2

.J14 -5 )

so a)

~

so a4 = 6.

- )

= 1.

Therefore, ,jf4 = (),1,2,1,6) Before we complete the solution of this problem, suppose we try to approximate the square root of 14 from the data we have. The formulas on pa~ 2 of this article make all this mechanical. We prepare the follOWing tables

•• •• •• •• •• •• •• ••• •• ••• •• •• •• •• •

•• •• •• ••

•• •• •e • •• •• Ie

i. ,. b2 _ (c_a)2 = (b + c - a)(b - c + a), c 2 ~ c 2 _ (a_b)2 = (c + a - b)(c - a + b). Multiply, am a 2b2c 2 ~ (a + b - c)2(b + c - a)2(c + a _ b)2. Now take the square root am we are done. Finally, on Page 22, we have the following problem, where we have altered the l' notation somewhat. 0 Fim the point 0 inside the ~~r • tria~le at the right for which 0­ alp + b/q + c/r is a minimum, where p, q, rare perPendicular to sides a, b, c respectively. Here our correspoment used rectangular coordinates am partial derivatives to derive a solution. It was correct but took four pages to develop. We would like to present another solution and would welcome your comments on it.

••• •• ••• •• ••• •• •• • •• •• ••• ••• •• •• •• ••

• •

3

Note that (ap + bq + cr) = 2A (the area of the • triangle) and is therefore a constant. Then we seek

I •

I

(a b + -c)( ap + bq + cr ) = a minim\.U1l, say m. - + -

•

•

p

q

r

2 Multiply, and a + b2+ c 2 + ab(.l4S) + bc(~) + ac(!+E) • qp rq pr 222 • is to be a minimum. We discard the a + b + c • But

I •

• , • • • • •

what is left will be a minimum when each of the expres­ sions in parentheses equals 2, by a we11-nown inequa1­ ity. This occurs when p = q = r, so that the point 0 desired is the incenter of the triangle. The solutions by Ken Burton were correct, neat am a pleasure to read. We would welcome more like them.

• • • •

We have just two mild comments to make. First, we present novel theorems am proofs in the hope that the reader will remember them and use them. Thus, on Page 7, there is an example of factoring as we have , done it. Naturs11y, we hope the reader will use what • has been presented in an issue of ARBELOS to solve • problems in the manner presented. • Secom , we prefer not to use a sledgehammer to • crack a nut. We have thus far avoided problems using, • for example, calculus, am certainly not partial • derivatives. The Olympiad problems thus far have not • need calculus. However, if it becomes advisable to use mathematics • other than what we have thus far provided, am if our • readers want such materials, just let us know, and we • will present such materials.

!. • , •

the

Incidentally, Ken Burton did an excellent job on Bu1~rian problem on page 22 of the same issue.

; . Keep workinp; on the problems and sent in your • solutions. They will be duly acknowledged. •

•

•• •• • •

Dr. Samuel L. Greitzer Mathematics Department

Rutgers University

New BrunSWick, N. J. 08903

4 RELEVANCE IN MATHEMATlr-S We all know that geometry has had a hard time of it in our secondary school curricula. However, one of its strongest critics, has confessed that perhaps he has been too harsh as well as mistaken. This article is provoked by a remark in a book we were readinf.!;, stating that, " ••• the nine-point circle now has a distinct flavor of beautiful irrelevance ••• ". To add to the situation, the same book then went on to discuss the Schwarz Function. Frankly, we question which is more irrelevant - the nine-point circle or the Schwarz function. There is no question that mat~ematics changes with the times. However, nothing is actually discarded. For example, figurate numbers have not disappeared. They exist now as Ferrers Graphs. To wax historical, there was a time (until the end of the ninete~nth century, when the theorem stating that base angles of an isosceles triangle are equal was taut#lt in what we would now agree is archaic.

I'

Q

Given the triangle ABC with AB equal to AC, one first extended AB to P and AC to Q, with BP = CQ, drew lines BQ and CP, and Then proved that triangles APC and AQB were cotlgruent. One then proved triangles BPC and CQB congruent. Now, angles ABC and ACB, as supplements of equal angles, were proved equal.

Why the theorem was proved this way we do not kno""f. There were students who found this proof difficult, and who were dissuaded from further study of mathe­ matics. Also, there is a fanciful similarity of the diagram to a bridge. Therefore, this theorem with its diagram, was called the Pons Asinorum, or the Bridge of Fools.

•

•• •• •• ••• •• •• •• •• •• •• •• ••• •• •• •• •• •• ••• • Cij

i.•

•• •• •• •• • •• • •• !.• ••• 1.•• • •• •• •• •• ••• ••

5 In the early days of this century, the nine-point circle seemed to take the place of the Pons Asinorum. The would-be mathematician had to have a proof of the theorem. For the (veryfew) who do not have such a proof, we dissect the theorem and indicate a proof.

I.

A·

~.

Ha..

I.

I.

A

A

(H\ We have separated the diagram for the theorem, which will be shown later, into three diagrams. We have the same triangle, ABC. At the right, in Figure (M), we have the midpoints of the sides forming the Medial Triangle, MaMhMc • It is similar to triangle ABC and the ratio 1:2 wi~h the original. The two are homothetic. Its circumradius is 1/2 that to the triangle ABC. The diagram in the middle (E) joins the midpoints of lines AH, BH and CH. It is called the Euler tri­ angle, and points E , E , E are called Euler points. b a c Note that this triangle is also similar to triangle ABC, the ratio of corresponding lengths is also 1/2, and, in fact it is also similar to triangle ABC. Thus the medial triangle and the Euler triangle are con­ gruent. The diagram at the left (H) joins the feet of the altitudes of triangle ABC. It is called the orthic triangle and is not necessarily similar to any of the other triangles. We shall now show that, with all the triangles in one diagram, the three triangles have the same circumcircle, which explains the term nine­ point.

•

Since E

am M are midpoints of BH am BA, line b c McEbis parallel to line AH am equal to half of it. Similarly, line EcM

is parallel to line AH am equal b to half of it. Since EbE is parallel to Be which is c perPendicular to AH (or AH) , EbE MbM is a rectangle a c c and is inscriptible in a circle. This rectangle has diagonals EbM

c

am EcM , which are diameters of the b

circumcircle. In the same manner ( you may draw in the lines ), you will fim that M E E M ia a rectangle with c a c a diagonals E M am EM. Having a common diap;ona.l, a a c c it follows that we already have six points on the same circle. Finally, we should note that EaMJIaEb is also a rectanp;le, and that the three rectangles have three diagonals among them. Now the diagram looks like this:

•• •• ••• •• •• •• ••• ••• •• •• ••• •• •• •• •• •

:.• •• ,.••

7

•• •• •

•• •• •• ••• •• •• •• ••

•• •• •• •• •• •• •

We have the points E M EbMbE M all in the same a a

c c

circle. We have the three diameters EaM , EbM , EcM ' b a c Because EaHaMa is a right angle, Ha lies on the circle. Because EbHbM b is a right angle, Hb lies on the c1.rele.

Because E H M is a right angle, H lies on the circle. c c c c And thus we have arrived at the nine points which lie on the same circle. The center of the circle is at N, the intersection of the diagonals ME, MbE , EM. a a b c c We merely mention that this circle, the nine-point circle is tangent to the incircle and to the three ex-circles. A proof of this would take us too far afield. One last remark about this circle. In the diagram below, I ~i A

C

r""'--..,.-~~----...,

8 we have drawn B'C' on A parallel to BC, C'A' on B parallel to CA, and A'B' on C parallel to AB. Tr.is triangle is similar to triangle ABC with homothetic ratio 2;1. Now if we draw the circumcircle of ABC, we note that it apsses through the midpoints of the sides of triangle A'B'C', hence the circumcircle of ABC is the nine-point circle of A'B;C'. Also, AH cor­ responds to line OM , so AH = 2 'C+1 • a

a

From the two diagrams on the previous page, we can, using our imagination, derive many interesting properties of the various lines in the figures. We may, in a future article, do so. Right now, however, we have decided to concentrate on the orthic triangle.

"'-------lIl'---..;:aL'c

Remember that we have lots of right angles in this figure. Thus, LABH and LACH are equal - both are complements of LA. Next, BH HH and CH HH are cyclic. a

c

a

c

Now LH BH and LH H H are equal, as are LHbCH and c

c a

LHbHa H. Thus, angles HcHa Hand HbHa H are eq\'~a.l. That . is - each altitude bisects the angle of the orthic triangle to which it is drawn. Incidentally, if one labels the angles, one finds that MHbH "- lillH H·....· ACH H (but not &1 HbH ). This c c a a b -- a c is a sort of "trivia" to remember.

• • •• ••• •• •• •• •• •• •• •• •• ••• •• •• •• •• •• •• •

,.•••

•• •• ••• •• ••• •• •• ••• •• •• •

•• •• •• •• •• ••• •

9

Let us find the perimeter p of the orthic triangle. First, AH = b cos A and AH = c cos A, from the b c

diagram. Using the Law of Cosines,

222 222

(HbH) = b cos A + C cos A - 2bc cos Acos A 22 2 2 c = cos 2 A(b + c - 2bc cos A) = a cos A, whence HbH HaH b

=c

c

=a

cos C.

triangle is p

cos A. Similarly, H H

c a

Addin~,

=a

=b

cos Band

the perimeter of the orthic

cos A + b cos B +

C

cos C.

Using the Law of Cosines a~in, we substitute and 2+ 2 2 2+ 2 b2 2+ 2 2 ( b c - a) b(c a - ) + c(a b ~c) ~t P = a 2bc + 2ca 2ab which simDlifies to 2a2b~ + 2b~c2 + 2 c 2a 2 _ a 4 _ b4 _ c 4 p = ?abc This is a nice exercise in factoring. When we let c2

= (a

+ b)2, the numerator becomes zero, and we

have a factor. (The reader should try to substitute). Similarly, we find that, in addition to (a+b)2_ c 2, (b+c)2_ a 2 and ~c+a)2_b2 are factors. There is a problem here. The expression we wish to factor has degree four, but the product of the terms we have found to be factors has degree six. However, (a+b)2 - c 2 = (a + b + c)(A + b - c) (b+c)2 a 2 = (b + c + a)(b + c - a) (c+a)2 b2 = (c + a + b)( c + a - b) and we conclude that the actual relatively prime factors are just four in number. We test a bit, and finn p

= (a+b+c)(a+b-c)(b+c-a)(c+a-b)/(2abc.

10 This should remim us of Heron's formula for the area of a triangle. In fact, if the area of

~BC

is

A, we just multiply and divide by 16, and we get p

= &:::,2/(abc).

This formula is short and neat. We just make one more adjustment. Recall that we have shown in a previous issue that

A

= (abc)/4R, p

=

and we have, finally,

?,6./B.

If we let K be the area of the orthic triangle, and

\I

we find

the radius of the incircle of this trian9le, K = v A/R. Of course, we can keep on and

invent more formulas. Space keeps us from doing much more. However, let us do something about a value forv. Referring to our diagrams, we see that v

= HHa cos

Again labeling the parts of the diagram, we get (HHa)/a cos B : b cos C/(CH c )' which gives us HH = ab cos B cos C/(CH ) am when we substitite, a c v = (ab cos A cos B cos C)/CH • Remembering that A

= (CH c )c/2 = abc/4R, we v = 2R cos A cos

c

substitute and find that B cos C.

Now we can find the area of the orthic triangle, compare this area with that of the original triangle, and, by letting the imagination roam, invent many new and curious properties of this triangle.

A.

•• •• •• •• ••• •• •• • • •• •• ••• •• •• •• •

i.••

• •• •• •• •• •• •• ••• •• ••• ••• •• •• ••• •• •• •• ••

11

We believe we have shown enough to make it at reasonable that there is fun as well as profit from the investigation into a geometric situation. We present here some problems that the reader may like to tackle. a) What is the area of the orthic triangle ? b) What is the ratio between the area K of the orthic triangle and the area of triangle ABC ? c) Prove that, in acute triangle ABC, the orthic triangle has the least perimeter of all triangles that are inscribed in triangle ABC. (We recommend that the reader look at the proof in ''What is Mathematics ?", by Courant and Robbins, page 348. One might take a look at pages 349 and 350 as well.) d) In the diagram at the right,

perpendiculars Ha P and Ha Q

are constructed, and line

PQ drawn. Prove that the

length of PQ equals half

f3 the perimeter of the orthic triangle of ABC. e) In the diagram at the right, ABC am A' BC are inscribed

in the same circle on the

same l:ase Be (as shown).

H and H' are the ortho-

centers of the triangles.

Prove that HH' and AA'

are equal and parallel.

c.. 1-lG...

AI

"

The reader who thinks all this irrelevant can now go back to the area of mathematics considered more relevant - say catastrophe theory, normed semi-groups, etc.

•

12 PRCXttISCUOUS EXAMPLES P1)

When the polynomial

f(x)

is divided by

(x - a)(x - b), what is the remainder? P2)

Solve for

XI

eX - a)ex - b) _ ex - c)ex - d) X -

a - b

-

X -

C - d

P3)

Show that, given a,b,c > 0, then 2 2 2 2 2 a (1 + b ) + b (1 + c 2 ) + c (1 + a ) > 6abc.

P4)

Solve

(x + y)2

P5) Factor a(b4 _ c 4 ) p6)

= (x'+ + b(c4

1)(y - 1).

_ a 4) + c(a4 _ b4 )

Given p an odd prime, solve x

2

- 1

= 17p.

Prove that the product of four successive integers is never a perfect "quare. P8)

Show that three times the sum of the squares of three natural numbers is also the sum of four perfect squares.

Note I These examples can, of course, be solved by brute force, which may take some time. They can also be solved by looking at them in a "different" way. Try to find that different way.

•• •• •• •• •• •• •• •• •• •• • •• •• •• •• •• •• ••

• •• •• •• •• •• ••• •• •• ••

,e

••• • •• •• •

••• • •• •• ••• •

13 THOUGHTS ON AN OLYMPIAD PROBLEM Problem 1 on the 1985 International Mathematical Olympiad went as follows: A ~ircle has centre on the side AB of the cyclic quadrilateral ABrn. The other three sides are tangent to the circle. Prove that AD + B" = AB. Many contestants found this problem simple, and most of the solvers used trigonometry to solve it. In addition, students and adults - leaders and deputies included had their own solutions.We found the solution by Greg Patruno, USA deputy, so pretty that we wish to show it here.

D

With the dia~am as shown, let the an~le at C be 2a. Since the quadrilateral is cyclic, the an~le at A equals 180" - 2a. Since R, Sand T are tangent points angles OrR an~ oes equal a each, and the angle at "OS equals 90 - a. Also, tangentrR equals tangent rs. _Now rotate triangle oes until it takes the position arc. By addition of angles, we find that the triangle AOC is isosceles, with OA = AC = AT + TC = AT + CR. Similarly, we see that OB we finally arrive at AB = AD + Be.

= BR

+ TD. Adding these,

OA + DB = AT + CR + BR + TD, or

14

While we are at it, let us look at the fifth problem of the 1985 IMO. It goes as follows I A circle with centre 0 passes through the vertices A and C of triangle ABC, and intersects the ~eg­ ments AB and Be again at distinct points K.~~ N. respectively. The circumscribed circles of the triangles ABC and KBN intersect at exactly two distinct points B and M. Prove that angle CJt1B

is a right angle.

This appeared to be perhaps the most difficult problem of the entire IMO. This reminded us of a "theorem" we had stated in the Arbelos of March. 1985 - page 13. and we wondered i f it might be possible to use this "theorem" to solve the problem. It was !

•• •• •• •• •• ••• •• •• •• •• •• •• •• •• •• •• •• •• •• •

e e

15

e

First, line AC is the common chord of two circles. Therefore, line CAP is a radical axis. Similarly, NIl{ and BM are radical axes of their circles. These three lines must intersect at a point, P, the radical center of the circles. This gives us a large triangle PBC.

e e

Now OX is perpendicular to AC (and bisects it), and OY is perpendicular to BC ( and bisects CN.) If we can show that

e e e e e

•

•

(BM)2 _ (MP)2

+

(PX)2 _ (Xc)2

+

(Cy)2 _ (YB)2 = o.

e e e

then, from the perpend icularity of OX to CP and OY to ('!B, it will follow that OM is perpend.icular to BP. Notice first that (PM) (PB) = (PA) (PC) •

•

We now operate on each of the three pairs of tenns in the above.

e

• (BM)2 - (MP)2 = (BM + MP)(BM - MP) =

:: (BP)(BM - MP) = (BP)(BM) - (BP)(MP) = (BP)(BM) - (PA)(PC).

• (PX)2 - (Xc)2 = (PX + XC)(PX - XC) = (pc) (PA}.

:: (Cy)2 - (YA)2 • (CY + YA)(CY - YA) = -(CB)(BN}.

•

­­

e e e e

• e e

e It

e

e e e

­• e

Now add the three expressions that are underlined, and have arrived at the question - does (PB)(BM) that is, are

= (CB)(BN)

?

triangle BMN and triangle BCP similar ?

Well, since quadrilateral ACNK is cyclic, the angle Cl

at C equals angle BKN which equals angle BMN, and

angle MBN is common to both triangles. The triangles are similar, am (BM)(BP) does equal (BN)(B('!). Therefore, OM is perperrlicular to PB, am angle :sMO is a

ri~ht

angle.

This is an example of the use of simple properties of the triangle to prove a complicated result. We hope you like it.

• •• •• .

16 MAXIMA ANn MINIMA

Just as analytic p;eometry has made it possible to solve problems easily that are quite difficult to solve by synthetic geometry, calculus has made it possible to solve problems involving maxima or minima that are difficult otherwise. However, just as there .. are problems that may best be done by synthetic p;eometry. there are problems involving maxima or minima that can .. be done without calculus. •

.

In this connection, we recommend that the reader try to refer to two sources that are very interesting. The first is an article that appeared in Scripta Mathematica, Volume 18, entitled "No Calculus Please", by Butchart and Moser. We were amazed at what could be done by the two authors. The second is No. 6 in the Dolciani Series, published by the Mathematical Association, entitled "Maxima and Minima Without r.alculus", by Professor I. Niven. We ourselves have used methods of our own on occasions where they were simpler than formal calculus methods. We present here some of these methods, with a few problems. The point is that these methods offer the opportunity to readers to use their ingenuity. First, a simple observation: from the identity (x + y)2 _ (x _ y)2 = 4xy, we can see that a) if xy is a constant, (x + y) will be a minimum when (x - y)

= 0;

that is, when x

= y:

b) if (x + y) is a constant, xy will be a maximum when (x - y)

= 0;

that is, when x

= y.

Both results are easily extended to take care of the eventualities that, if the product of n terms is a constant, the sum of the terms will be smallest when all terms are equal, and if the sum of E terms is a constant, their product will be greatest when all terms are equal.

•• •• •• ••• •• •• ••• •• •• •• •• •• •• ••

e

•e •• e e •e

e

e

•e

e

e

17

Thus, consider a rectangle with terimeter 4p. Then we may consider its sides to be (p + a) and (p - a), and its area will be (p + a)(p - a) = K. Since the sum p + a + p - a = 2p is constant, the rectangle has maximum area when p + a = p - a, a = 0, and the rectangle is a square. Of course, it is even easier to note that K = p2 - a 2 which makes K a maximum when a is zero.

r - ------.-; I

,

I t

NIIl

:

e

e

e

e e

e

e e

e

e

•e

e e

-e

e

•e e e e

e e

e

e

Incidentally, this solves a popular calculus problem - that of constructing an enclosure of three sides, the f·onrth being a wall, using a given amount of fencin~. Clearly, if we reflect the enclosure as shown at the left, the resulting rectan~le must be a square. Hence y = 2x.

Next, if a triangle has constant perimeter, what shape will make its area a maximum? Here we use Heron's Formula in the form 2 K = s(s - a)(s - b)(s - c). Dropping the factor ~, which is a constant, K t,;~ill be a maximum when s - a = s - b = s - c. or a = b = c • The triangle is then equilateral. ¥ ~

yo'

Here is another popular problem: An oJ)en-top box is to be made from a square piece of cardboard whose side is ~ by cutting equal squares at the corners and foldin~ to form the box. What would be the maximum volume ?

):

We cut out the squares with side x and fold, The volume of the desired box will then equal 2 V = (a - 2x) .x •

~ a..~'

We will use a smidgeon of ingenuity here. Multiply

18 by 4 to get

4V

= (a

of the three factors

- 2x) 2 .4x. Now we have the sum

= (a

- 2x) + (a

2x) + 4x,

which is a constant. The volume will be a maximum or when x = a/6. The maximum volume will then equal 2aJ /27. Sometimes, one has to when

a - 2x

= 4x,

manipulate a bit to get a constant sum or product, and therein lies the fun and challenge. Thus, let us find when (x - J)2(x - 2) is a maximum. We multiply by

-2

to get

(x - J)2(4 - 2x). The

sum of the factors is (x - J) + (x - J) + (4 - 2x), which is a constant. Then

x - J

=4

- 2x, x

= 7/3,

and the maximum is 4/27. (a) Try your hand at maximizing where x less than a.

(a + x)2(a - x)3

Now what positive number added to its reciprocal gives the least possible sum ? Here, we write x + -1 = min. Since the product x ~ -1 = constant, x x the minimum occurs when x = 1/x, or x = 1. The minimum value is 2. To find when -a + -b + xy is a minimum, we note that x y the product of the three terms is a constant, abe a b \ We get a minimum when - =- = xy. It is easy to get x

y

y = bx/a, and, after a bit of algebra, the value of the minimum. (b) Find this minimum, from the result derived above, and then - by calculus.

(c) Find the least value of y = aekx + be-kx •

•• •• ••• •• ••• •• •• ••• •• ••• •• •• •• •• •• •• •• •••

•e e •e e

e

e e e e e e e e e

• e e

e e e e e e e e e e

•

­

e e

e e e

e

e

­• e

e

19 Sometimes, a very simple property will be enough to solve a proble~. For example, recall that the square of a real number is never negative. Now, when is

25 + 24x - 16x2

a maximum? Here, we will

"complete the square", thus: 2 25 + 24x - 16x 2 = 34 - 9 + 24x - 16x = 34 - (3-4x)2. This is obviously a maximum when () - 4x) = 0, or x = 3/4. The maximum .value is then )4. This is quite easy by calculus, but there can be more difficult ones. (d) When is 5 + 12 sin A - 9 sin2A a maximum, and what is that maximum value? Again, consider f(x) = ax 2 + bx + C (a positive). Multiply by 4a, and 4a2x 2 + 4abx + 4ac = 4f(x) .a. Completing the square we just add b2 and get

(2ax + b)2 + (4ac - b2 ) = 4af(x).

This will be· a minimum when when

(2ax + b) = 0, or

x = -b/2a.

It may be that an inequality relation can be used to solve a problem. For exami:le, we know that sin A + sin B = 2 sin A+B cos A-B 2 2 and since a cosine cannot exceed unity, we find ~

~

sin A + sin B < sin(A + B) o < A ,B < 180 • 2 2 First, this inequality can be extended to three or more an~les. That is, sin A + sin B + sin C < i (A + B + c) .3 _sn) and so on. Equality will occur when A = B = C. Similar results exist for other trigonometric functions,

20 so that

cos A + cos B

~---~-

2

< cos(

f!:

A

t'

tan A + tan B < tan(A

and

C'

-90::: A,B :: 90

o~

2

t

A,B, 90 •

These are all examples of Jensen inequalities, namely f(x)

T fey) > f(x + Y) for convex functions. The 2 2 inequalities are reversed for concave functions.

Problem: what is the minimum value for the sum of the sines of the angles of a triangle ? We may sin A + sin B + sin C = ( A + B + C) = sin 60° sin 3" 3 or sin A + sin B + sin C = 3 i/3 / 2. This will occur when A = B = C = 60' • We have already discussed the Arithmetic- mean ~eometric-

mean inequality.

ations, we have

AM

~

Usin~

the usual abbrevi­

GM , with equality

occurrin~

when all the elements are equal. We illustrate with a problem of a type communicated to us by Professor Eugene Levene, of Adelphi University. the problem is: 4 2 f(x) = 4x - ax 3 + bx - cx + 5 = 0 has four distinct roots, r , r , r , r 4 • We are given 1 2 3 I) r /2 + r /4 + r /5 + r 4 /8 = 1. 1 2 3 Find the roots and the values of a, b, c. The arithmetic mean of

terms in I is 1/4. The geometric mean of the terms in I is r r r r 1 2 3 4 320

~he

= .!.42 320 =

1 256

=

(l)~ 4;

• •• •• •• •• •• •• ••• •• •• •• •• •• •• •• •• •• ••• ••

• •• •• •• •• •• ••

•• •• •• •• ••

••• •• •• •• •• •• •• •• ••

21 Since the AM and the GM are equal, the terms must all be equal to each other (and to 1/4). Hence r 1/2 r1

= r 2/4 = r)/5 = r 4 /8 = 1/4, and, solving, = 1/2, r 2 = 1, r) = 5/4 and r4 = 2.

The rest of the problem is now trivial. We now offer some problems for the reader to do. e) For what dimensions will a box without a top and with a fixed volume have dimensions a minimum ? f) A roofer wishes to make an open gutter of maximum capacity whose bottom and sides are each 4" wide and whose sides have the same slope. What should be the width across the top? g)

Given the function

u - (x - l)(x - 6)

(x - 10)

examine for maxima and/or minima.

h)

Given the function

u

=a

sin x + b cos x,

where a and b are positive reals, examine for maxima and/or minima. i)

If

a + b +

C

= 1,

for l/x + l/y + l/z

show that the minimum value is

9.

j) Prove that the quadrilateral of maximum area inscribed in a circle of fixed radius is a square. (Hint: Use the formula for a quadrilateral with sides a, b, c, d inscribed in a circle, namely, 2 K = (s - a)(s - b)(s - c)(s - d).

22 APrERTHOUGHTS In our development of the properties of the orthic triangle, we wished to exhibit some such areas as factoring, properties of trigonometric theorems, and the like. The reader knowa that we have a special interest on reviving geometry. With this in mind, let us look once more at the orthic triangle. Examine the diagr~ at the left. Looking at it, we remember

A

that there are right angles at Ha , Hb , Hc • Therefore AHcHH b is cyclic, the circle having diameter AH. Using the Law of Sines, we get

(HbH c ) = (AH)sin A. However, 2R = a(sin A). Eliminate the sine term, we have 2R(Ha Hb )

= a(AH).

Now we have shown that, if we construct the line OM a , from the circumcenter of the circumcircle of triangle ABC to the midpoint M of side BC, then a

(AH) = 2(OM a ). Substitution gives us R(HbH c ) = aCOMa)' But, if the area of triangle (ABC) is 6, then, first R(HbH c ) = 2(OBr). Similarly, R(HcHa ) = 2(or.A) and R(HaH b ) = 2(OAB). Adding,

we get R(HbH + H H + H Hb ) = 26. Finally, the c c a a perimeter of the orthic triangle equals 26/R. We find this shorter and easier than the proof we gave previously. Moreover, it is all geometry!

••• ••• •• •• •• •• •• •• •• ••• •• •• ••• •• •• •

,.!.•• ,.••

23

••• •• ~ •• ••• •• ••• •• •• •• •• •• •• •• •• ••

· •

let H P am. H Q be a a perpendiculars from H to a the sides am. draw PQ.

A~in.

I'

Again. APHaQ is cyclic. By the Law of Sines. L..-~::"-

~

c

(PQ) = (AHa )sin A. Once more. 2R(sin A) = a. Eliminate sin A. am. we arrive at

= a(AHa ) = 26. =AIR. Comparing

2R(PQ)

from which follow

(PQ) with the first result. we see that (PQ) equals half the perimeter of the orthic triangle. Again - all geometric! Now it might be fun to fim. such additional data

as the radius of the incircle of the orthic triangle. the area of the orthic triangle. am. much more. Avoirdupois vs. Metric

Some steam seems to have gone out of the drive for metrization. but we still meet with either (or both) in problems in science and mathe~atics. Hence it is useful to have some way of changing fran Avoirdupois to Metric (am. vice verse). We present the followin~ approximationsl Kilometers 1 2 3 5 8 13 Miles 1 1 2 3 5 8 etc. As for temperatures try the followingl to change fran

Fahrenheit to Centigrade - subtract 30 and take half; Centigrade to Fahrenheit - double am. then add 30. You will fim. the results close enough for most purposes.

24 INDETERMINATE FORMS We have most probably encounteted indeterminate forms. These usually take the form % (or perhaps 00/(0) for some given value of the variable. One such form, which we will find handy to examine is f(x)

= sin

x/ x, which

is indeterminate for x = O. To accustom ourselves to some useful notation, we will be

O....c.........:;;;----~T

.5

lookin~

for

Lim sin x • Substitution does result in -00 •

x=O

x

In the early days of tri~onometry, the various functions were actually I~presented by lines in a unit circle. In the diaP,Tam above, which shows part of a unit circle, the line AT is the tan~ent of the angle x at 0, the arc TN is equal to the an~le x, and the line NS is equal to the sine of the angle x. By inspection, we have sin x < x < tan x • Dividing each of the terms by sin x, we get 1 < x < 1 sin x cos x

o

Now, if we let x Thus L1m

x=o

sin x/ x

= 1.

We conclude that we can sometimes determine the limit of an expression by usin~ a dia~m. When the diagram is a ~ph, the point of indeterminacy is called a sinRular point. If we can give the point a value so as to make the ~ph continuous, we can call the point a removable sin~larity. Our aim is to present various ways of findin~ the value of an indeterminate form, and, for awhile, we will use sin x/x for our examples.

• •• •• ••• •• •• ••• •• ••• •• ••• ••• ••• •• •• •• •• •

•• • ••• ••• •• •• •• •• •• •• •• •• •• •• •

,.

•• •• •• ••• •

•

25 Just as Gauss used to take pleasure in finding a "chiliad" or so of prime numbers, we used to take time to find expansions for various functions. We used either Maclaurin's series or Arbogast's method. for further use, we ~ive you some of our results. sin x = x - x 3/3! + x 5/5! all x cos x tan x eX In(l+x)

=1 =x =1 =x

Now, to find

4 2 - x /2! + x /4! - • • • + x 3/3 + 2x 5/15 +

all x

-~2 < x < ~2

+ x + x2/2! + x 3/3! + - x 2/2 + x3/3 _ x 2/4 + Lim sin x

x=o

x

, suppose we substitute the

series above. We may divide the denominator x into 4 2 the numerator, getting 1 - x /3! + x /5! ­ and substitute x = 0, to get Lim sin x = 1. Somehow x=O x we believe this is easier than using the diagram.

"

Most of us have seen I'Hopital's rule, which we just sketch for our readers. (There is a story that I'Hopital paid Leibnitz for the formula, but we do not listen to libelous stories - as a rule.) Suppose we have functions f(x) and ~(x) such that f(a) = 0 and g(a) = O. Then f(a)/g(a) = 0/0 which is indeterminate. Let us rewrite the form above

~s f(a+h)h- f(a)/ g(a+h)h- g(a) • But in the limit, each expression becomes a derivative, and we have

.

Lim fL!l = Lim £:~ There is a hint here - if x=a SIX) x=a g (XJ the ratio of the d~rived expressions is indeterminate, we can continue to use second derivatives, etc.

•

26 A Lim sin x , we find that Using l'Hopita1's rule on x=O x Lim cos x when we differentiate, we get = 1.

x=O

1

We now have three ways of operating on an indeter­ minate form. The skill lies in selecting the one that is easiest. C id th Lim -Ja + x - -.,fa - x (= -0°) • ons er e x=O x We wouldn't think of a diagram for this. We might expand each by using the binomial theorem, but we think this might be too complicated. Of course, we could use l'H~pita1's rule, but this looks quite complicated, too. So - we will multiply numerator and denominator by ~a + x + -Ja - x. This gives 2x

xC,fa + x + ~a _ x)

• Cancelling and letting

x = 0, we find the value to be

1/"';;:.

It is a fact that many students, once they are aware of a general method for doing anything, stick to that method through thick and thin. I~enuity consists of selecting the shortest, fastest and simplest method. For example, examine Lim x - sin x

x=o

x

3

Using l'Hipita1's rule reqUires three differentia­ tions. Substitution is shortest and easiest. Question: What is the value of this form ? Suppose we try Lim x(1/1-x). Think first of x=1 what method or methods we might use. Then, take logarithms. This gives us log x

1 - x •

Now we may use 1 'H~pi tal's rule, which.

will give us the value -1, whence the limit is e- 1 •

•• ••• •• •• •• ••• ••• •• •• ••• •• •• •• •• •• ••• •

I.i.I. •• ,.•• ••

•• •• •• •• • ••

•• •• •• •• •• •• •• •• •• ••

I.

27 Here is another problem, with some hints for its solution: Find -2 x Lim(tan x) x=O x Here we took logarithms first and then substituted f or the tangent term. Question: what is the value of this indeterminate ? Since 00 is not a number, an expression that takes the form 00/00 is indeterminate, and we can use our methods on such forms •Anyway , we can always change such forms jo the 2form 0/0, if need be. Thus for Li m 2x - 23x + 43 x=o 5x - x - 7x we merely divide allover by x3 and we easily find

the desired limit.

Question: what is the limiting value for this?

We conclude with some problems for which you should

try to find the best way to proceed. Find Lim (ex _ e- x _ 2x)/(e x _ 1)3

x=o

Find Lim (sin x)sin x •

x=O

Find

Lim x + tan x - tan 2x x=O 2x + tan x - tan 3x

Find

Lim x loge1 + x) x=o 1 - cos x

28

OLYMPIAD PROBLEMS 6}

A polyhedron has 12 faces and is such that: (i) all faces are isosceles triangles, (ii) all edges have length either x or y, (iii) at each vertex e. ither 3 or 6 edges meet, and (iv) all dihedral angles are equal. Find the ratio x/yo

(BraZil)

7) Let A be a set of positive integers such that for any two elements x,y of A, x - Y ~ xy/25. Prove that A contains at most nine elements. Give an example of such a set of nine elements. (Czechoslovakia)

8) Prove that a triangle with angles a, ~, y, circumradius R and area A satisfies tan ,g + tan ~ + tan.x < 9R 2/4A. 2 2 2 ­ (German Democratic Rep) The weight w(p) of a polynomial P, where p(x) = Eaix i , with integer coefficients a i is defined as the number of its odd coefficients. For i

= 0,1,2, ••• ,

let qi(x)

= (1

+ x)i. Prove

that for any finite sequence 0 ~ i 1 < i ••• < in 2

The inequality

w(q + q + • • • + q ) > w(q ) i i i i n 1 2 1 (Netherlands) 10)

Decompose the number 51985 into a product of three integers, each of which is larger than

100

5 • (USSR)

Please send your solutions to your editor,

Dr. Samuel L. Greitzer Mathematics Department Rutgers University New Brunswick, NJ 08903

• •• •• •• •• •• •• ••• •• •• •• •• •• •• •• •• •• •• ••

,. •

•• •• ••• •

•• •• •• •• •• •• ••• •• ••

,.

••• •• •• •• ••

29

.. , KURSCHAK KORNER This column features those problems of the famous Hungarian Ktirschak Contest which have not yet appeared in English trans­ lation. For the problems of 1894-1928, the reader should consult Hungarian Problem Books I & II in the MAA's New Mathematical Library series (available from the MAA hdqrs.); the problems (with brief summaries of their solutions) covering the years 1966-i981 were translated into English by Prof. Csirma~ of Hungary in 1982. (The author of this column will gladly send a copy of the latter to interested readers.) Student readers are also invited to set aside an uninterrupted 4-hoUf period to compose complete, well-written solutiom to the problems below, and to submit their work to the address given below for critical evaluation.

1/1962. Let n be a natural number. Prove that the number of ordered pairs of natural numbers, (u, v) , for which the least common multiple of u and v is n, is equal to the number of positive divisors of n 2 • 2/1962. Prove that it is impossible to choose more than n diagonals in a convex polygon of n sides se that each pair of diagonals should have a common point. 3/1962. Let ABCD be a tetrahedron and assume that P is a point, distinct from D, in its interior or on its boundary. Prove that among the segments PA, PB and PC there is one which is shorter than one of the segments DA, DB and DC. Dr. George Berzsenyi 2040 Chevy Chase Beaumont, TX 77706

CONTENTS COVER: Geometry Problem i

Preface

1 Thoughts on Problem Solving 4 Relevance in Mathematics 12 Promiscuous Examples 1) Thoughts on an Olympiad Problem

16 ~ Maxima and Minima 22 Afterthoughts 2) Avoirdupois vs. Metric 24 Indeterminate Forms 28

Olympiad Problems

29 Ktirschak Korner

•• •• •• •• •• •• •• • •• •• •• •• •• •• •• •• •• •

e e e e e e

•

e •

e e •

e

arbelos

PRODUCED F'OR PRECOLLE~E

P}TIIJOMA THS

e

e e e e e

ABeD is inscribed in semicircle AE with unit radius GA. (

e

e e e e e e e e e e

-------4l-------4

1Io-=....

f=­

cn = c, DE = d. 2 2 Prove: a + b + c 2 + d 2 - a'be - 'bed is less than 4.

AB

= a, Be = b,

e

e

e e tit tit

e e

­

e

e

e

1985-1986: Copyright~

No. 3

January, 1986

The Mathematical Association of America, 1986

1

COMMENTS ·We are happy to report that our mail has oocome much heavier and interesting during the Fall. We will do our oost to reply to some of it. But - keep your letters coming. We enjoy hearing from you all. Ms. Regina Grieco, of New Windsor, N.Y., was quite complementary. We do include solutions to problems occasionally, but prefer that readers supply them. Professor p,. Levine, of Adelphi University, has sent us some ooautiful problems, requiring ingenuity to solve. 'ile have included one in this issue, ann hope to present others in the future. Thomas Chung, of Apple Valley, Minnesota, sent in a fine solution to the "Lewis Carroll" Diophantine equation problem, and equally good solutions to our trigonometry exercises. Congratulations are due. Ambati Jaya Krishna has sent us several problems thus far. 'I'l e fear that they are not qui te "right "for Arbelos. Remember that it is "Produced for Precollege Polymaths". We have simplified one problem and it will 00 found in this issue. We hope to receive more problems which may be appropriate. Credit is here given to David Morin, of South Paris, f1aine, for his solutions to our "term-enders". S. Nam, of Dayton, Ohio, has sent in an excellent solution to our Cover problem of September, 1985. Incidentally, I used Stewart's Theorem to get a short proof. Stephen Smith, of Thornhill, Ontario, Canada wrote in suggesting that there be more references to some of the mathematical material. We do try to make what we write as self-contained as possible, but we have occasionally provided references. We are dubious about his suggstion that some articles on computers be included in future issues.

.•

••

•• •• •• •• •• •• ••• •• •• •• •• •• •• •• ••• •••

e

e

e e

e e e e

e e e e e

e

e

e

•

e e

e e e e e

e

e

e e e

e

e e e e e

e e e e

.e

e

e

2

E.L. Wilmer, of Stuyvesant High School, in New York City, solved one of our cover pag-e problems. He asks for permission to publish some Arbelos problems in the school mathematics publication. We are not sure this can be legally done. We suggest that he write Professor Mientka about this. The address will be found on the back cover. Through the y8ars, there have been repeated requests fer problems involving calculus. We may have, to accede to these requests, not to solve problems usually found in texts, but, as is our purpose, to involve calculus in special problems, provided such solutions are better than solutions not involving calculus.

We have a special interest in Stuyvesant High

School, since we were a student there at the beginning of the century. In fact, that is where we encountered the first problem of our "Many ~heerful Facts", way back in 1920. There are some schools that are quite ~ood in New York City, like the Bronx High School of Science and Hunter ~ollege Hi~h School. When we tau~ht at Bronx Science, that school was per-eminent in mathematics. It is now starting to re~in some of its excellence, and we eX])ect this improvement to continu8. '?ight now, we are trying to proo.uce an index of subject matter covered in our Arbc:os issues to date. This is surprisingly difficult to do, since we had no idea of how much we had presented. We hope to have something available before -the summe~ of 1986. At any rate, many thanks for the letters and for the complementary r~marks contained in them. Send in solutions, ideas for subjects you would like us to cover, suggestions for improvement, etc. We read them with pleasure and they make the task of preparing futur~ issues leas of a chore. Write! Dr. Samuel L. Greitzer Mathematics Department Rutgers Universi~y New Brunswick, NJ

08903

3 THE ELLIPSE Several problems and articles have come to our attention in recent weeks ~volvin~ the ellipse, so we thought it might be useful to consider this curve. We hope that the reader will find something useful and new in this article. First, "ellipsis" is defined in our dictionary as " omission of a word or words in a sentence". :;:xample, "He is taller than his brother (is tall)". Peaders should look up "parabole" and Hyperbole". I early r.reece, ronic sections were considered to be sections of a ri~ht circular cone (with one nappe) by a plane perpendicular to one element. acute right obtuse

(a)

(b)

• •• ••• •• •• •• •• •• ••

/L ••• (c)

When the angle at the vertex was acute, the section was called an ellip~~; when the angle at the vertex was a ri~ht an~le, the section was called a parabola; when the angle at the vertex was obtuse, the section was called an hyperbola. Note two peculiarities of this definition: no way a section be a circle, and only half of any hyperbola was obtained. In later Greek mathematics, Appolonius and Archimedes, as well as Euclid, used a definition that corrected these omissions • co~li

There are numerous ways of defining the ellipse.

•• •• •• •• •• •• ••• •

••• •• • •• •• ••• •• •• •• •• •• •• •• •• •• ••• •• ••

,. '.

'. ,.

4

The most popular is to think of one as the locus of points such that the sum of the distances any point to two distinct points, called foci (sin~lar;focus) is a constant.

Although Dandelin's method of describing a conic is well-known, we will sketch it for the ellipse. The diagram above shows a cone and a section. We insert two balloons, one above the inteTse~ting plane and one below, so that each is tangent to the intersecting plane and to the surface of the cone. The points of tangency are F1 and F • For any point P on the curve 2 of intersectioi'!., PF 1 will equal PM, because tangents to a sphere from an external point are equal, and PF 2 will equal PN. Since PM + PN = MN is constant, we have P? 1 + PF2 = MN = constant. This, then, is a second way of-definin~ an ellipse, and it is the most common. Precisely the same procedure shows that the section of a circular cylinder by a plane not perpen­ dicular to an element is also an ellipse. This gives us a third definition of an u~lipse. What is more, we can now ~et the circle as the locus of points on a section whose plane is perpendic­ ular to an element.

5

Y

•

If we use algebra, let us place F

at (-c,a) and 1 at (c,a) on a rectangular coordinate system as

F

2 shown above. The definition ,tells us that FP

equals a constant - say 2a. That is, + c)2 + ~2 + c)2 + y2

V«X

.,i'cx _

1

+ PF

2

= 2a.

When we rationalize tbe above ( a pedestrian bit of alge bra), we end with 2 _ = 1.

y_ L 2 + _.oL. 222

a

Let 2 L 2 a

a

2

_ c2

+

x:b2

a

- c

= b2 , and we end with the familiar

=

•• •• •• •• •• •• •

1, or b2x 2 + a 2y2

= a 2 b2

We will call either one of these the "canonical" form for the equation of an ellipse. From it, we see that the curve is s~~etric with respect to both axes. Where it cuts the X-axis, we have x = a and x = -a. Where it cuts the Y-axis, we have y = band y = -b. The curve can be inscribed in a rectangle whose base is 2a and whose altitude is 2b. (Not) incidentally, the ratio cia = e is called the " excentric ity" of the ellipse, and it is less than one - another case of ellipsis. Bemember c = ae. Next, consider the diagram at the top of the next page:

• •• •• ••• •• •••

,.

•• • • •• •• •• ••• •• •• •• •• •• •• •• •• •• •• •• ••

•• • •

6 \/

We have two concentric circles. The smaller one has radius = OM = b. The larger one has radius = ON = a. Then, from the diagram, OV = a.cos e and PV = MU = b.sin e. Eliminating the trigonometric functions, we have once more, x 2/a 2 + y2/b2 = 1, the canonical form for the ellipse - still another definition

= ~(x + c)2 + y2. If we solve in 1 2 . the =a b 2 for 2 y and substitute 1n first expression (letting b2 = a 2 _ c 2 ), or b2 = a 2 (1 _ e 2 ), we find (more destrian algebra) Now, let 2 2 b x + a 2 y2

PF

1

=

(x +

PF

~e)2

2

+ (1 - e )(a

2

- x

2

=

a + ex.

e(x + ~). Now (x +~) is the e e equation of a line perpendicular to the X-axis, and

Write this last as

we have found that the distance from a point on the ellipse from the focus (-c,a) is equal to

e

times

its distance from the line x = -a/e. We call this line a directrix. Hence we have the last definition for an ellipse - it is the locus of points whose distance from a given focus F equals e times its 1

distance from a fixed line - the directrix. There are two directrices - one involving F 1 and one inv~l"'ing F2. Also, we have at five ways of defining - and constructing an ellipse.

7

v

Suppose (see diagram above) that we have tangent

UV a tangent to the ellipse b2x 2 + a 2 y2 -_ a.2~2 ,_. Th e easiest way to find the slope of the tangent is by calculus. We ~et b2x.dx + a 2y.dy = 0, from which the

*

2

2

slope is = - b x/a y. Draw PN perpendicular to the

tan~nt

at P and

the X=ax-s at N. Such a line is called a normaL Because of perpendicularity, the slope of 2 2 PN = + a Y/b x. The equation of the line on P and N would then be intersectin~

(y - Yt)/(x - Xt) = a2Yt/b2Xt· To find the coordinates of N, we let y = 0, and get (after some algebraic manipulation)

x = xte 2 •

We now remind the reader that a line bisecting an angle of a triangle divides the side it cuts into segments proportional to the sides ad.:jacent to those segments. Were PN and angle bisector, therefore, we would have (F P)/(F N) = (F P)/(F N). Let us see. 2 2 t 1 The length of F N is (c + e 2x) and the length of 1 F2N is (c - e 2 x). Remember that c = ae, and we can

•• •• •• •• •• •

•• •• •• •• •• •• •• •• ••

•• •• •• • •• •• •• •• ••

•• ••

•• •• ••

•• •• •• •• •• •• ••

'.

8

substitute, thus: for FiN, c + e 2x

= ae

2

+ e x

=

e(a + ex). Similarly, F N equals e(a - ex). However, 2 we have already shown that FiP = (a + ex). Also, F P = (a - ex). Therefore, we really have 2 (F P)/(F N) = (F P)/(F N). That is, normal PN is i i 2 2 really the bisector of the angle F PF • Finally, the i 2 angles marked a. are equal, so that the tangent makes equal angles with the radii vectores at point P. If there were such a thing as an elliptical pool table, a ball placed at F1 and struck would bounce off the cushion and pass through F2' regardless of

the direction in which it was struck. A noise made at one focus of an ellipse would travel outward and finally pass through F • This is the principle of

2 whispering galleries. One speaks in a normal voice at one focus, and the only place one can hear this voice would be at the other focus. Books have been written examining only the ellipse,

and we will not take the space to provide more of

its properties. We add only one further property:

2 2 2 2 2 2 the area of the ellipse b x + a y = a b is equal to nab. We can justify this by resorting to an orthogonal projection of a circle onto a plane not parallel to the plane of the circle. Or, one could use calculus. We will provide the reader with some problems based on ideas that have already been covered in previous issues of Arbelos. Concepts of projective geometry will be found very useful. Try them, for fun.

9

1)

Determine the envelope of all ellipses whose axes lie on the coordinate axes, whose centers are at the origin, and which have equal areas.

2)

An ellipse is inscribed in triangle ABC. From a focus F. lines are drawn to vertices A, B, C and to tangent point D on side AB. Prove that LAFD + LBFC = 1800 •

3)

Two pairs of paralle I tangents are drawn to an ellipse. Parallels to these are drawn through a focus and intersect these tangents in four points. Show that the four points lie on a circle.

4)

All escribed parallelograms of an ellipse have the same area.

5)

F P and F P are constructed perpendicular to 1 1 2 2 a line tangent to an ellipse at point T. Prove

6)

7) 8)

that (F P )(F P ) is a constant. 2 2 1 1 PQRS is a trapezoid inscribed in an ellipse, and PR and QS intersect at X. Tangents at Q and R intersect at T. Show that XT and PQ are parallel. The major axis of an ellipse is greater than any other line passing through the center of the ellipse and terminating on the ellipse. If F P and F P form a right angle, what is a 1 2

special relation between a and b ?

•

•• •• •• ••• ••• • •

•• •• •• •• •• ••• •• ••

R from the axis about which rotation will take place, this axis lying in the plane of the c irc Ie. The center of gravity 2f the circle is at its center and its area is nR • Hence the volume of the

2

2

torus is (nR )(2nr) = 2n2R r. Incidentally, what is the formula for the surface area of this torus ? Now assume we have an equilateral triangle with side s. If this triangle be revolved about one edge as axis, what is the volume of the solid thus formed ? And what if the triangle were to be revolved about an axis passing through one vertex and parallel to the side opposite? Physicists can use Pappus' theorems to locate centers of gravity. Where would the center of gravity of a semt~trcular arc be located? Well, i f the arc were to be revolved about the semidiameter, we would end up with a spherical surface with area 4nR 2 • The length of the arc is nR. Hence 4rrR 2= (nR) (2ny), or y = 2R/n • Where would the center of gravity of a hemisphere be located ? Here we know the volume of the hemi­ sphere, ~3. It is formed when the semicircle with (

radius R is revolved 180 • Therefore, we have 2 3 ~

= (nR 2 /2)(21TY)

We end with one problem as physicists might find quadrilateral, how would gravity of such a figure

2R

, or y ~ • that mathematicians as ~ell interesting; Given a one locate the center of ?

• •• •• •• •• •• ••• •• •• • •• •• •• ••• ••• •• •• •

•

•e •• e

22

Pythagorean Diversions

We are all aware that, if a = m2 - n2 , b = 2mn, and c = m2 + n2 , then a, b, c can be sides of a right triangle. If m and n are relatively prime,

It

e

e e e

e

e

with one odd and the other even, then a, b, c may form a primitive

pytha~orean

to present some curiosities

e

Diophantos has the following problem: Find. a pythagorean triangle in which the bisector of one of the acute angles is rational.

­

e

A

fit

e

•e

f t

\3

CL

e e

­­ -­e e e

e

e e

e

e

e e

--e

involvin~ p~~hagorean

triples.

e

e

triple. We would like

We know that an angle bisector divides the side to which it is drawn in the same ratio as the adjacent sides. Thus AT = xc and ~T = xa. Since xa + xc = b, we finally find that ab CT = a + c • Now we use the

theorem on trian~le BeT to find BT. We leave the details of the computations to the reader

pytha~orean

and give our result only- t 2

= a 2c/m2 •

If we let

c be a perfect square, we can find a rational t. For example, let m a

=4

2

- ]2

= 7,

b

=4

= 24,

and

and c

n

=4

= 3. 2

+

Then

32

= 25.

The bisector of the acute angle that cuts the side whose measure is an even number (namely b), should then equal 8 ]/4. We note that the bisector must end at the side that has even number measure. Can you prove this? Next, what are the dimensions of another right triangle, not congruent to the above, which has a rational bisector ?

•

23

We remember that the Pythagoreans held numbers in reverence. They had lucky and unlucky numbers, abundant numbers and deficient numbers, perfect numbers, and so on. They noticed that the numbers 222 1 ,5, 7 or 1, 25, 49 formed an arithmetic progression. The common difference, 24

was called

a congruum (not to be confused with congruence). Naturally, they wondered whether there were any other similar triples. a congruum is difficult, but we can do in a special case. Consider the following: 2 2 a 2 + b = c, and c 2+ k = x2 ,c 2 - k = y2 Findin~

somethin~

If we allow the congruum k to equal 2ab, we ~t (a? + b2 ) + 2ab = (a + b)2, (a2 + b2 ) - 2ab = (a - b)~ and we have three squares in arithmetic progression. 2 2 When we let m = 4, n = 3, we get (4 + 3 ) = 25, The congruum is 2x4x3 example.

=

24, which explains our

For a triangle with sides 5 - 12 - 13, we have 52 + 122 = 13 2 • We take as congruum k = 2x5x12 = 120. And we have

169 + 120

= 289 = 49.

120 169 Obviously, there are an infinity of such relations available. We have noticed that every congruum, for the special case we have accepted, is a multiple of 24. Can any reader prove this? And, finally, are there any cases where four perfect squares are in arithmetic progression? We have not (yet) found a solution to this problem.

••• •• •• •• •• •• •• •• •• ••• •• • ••• •• •• •

•

e e •e •e It

24

We read, with interest, a note from Lewis Carroll in which he said he had worked on a problem for hours without solving it. The problem - find three pythagorean triangles with e~(-lal areas.

e

We can get two triangles easily. The triangles with sides 20 - 21 - 29 and 12 - 35 - 37 both have areas of 210. But a third ?

•

Thanks to Henry Dudeney's "Canterbury Puzzles", we have a formula that will p;ive us three such trian~les. It is the following: Let m = u 2 + v 2 + uv

e e e e e e

e e

It

e

­

•

e

e e e e e e e

e e e e e e e e e

­

e e

n = u P

2

- v

= 2,.uv

2

+ v

2

If you form three pythago~an triangles with generators m and n, m and p, m and n+p, says Lewis Carroll, you will get three triangles with equal areas •

Starting with u m = 37,

n

= 7,

= 4, p

v

= J,

he got, first,

= 33.

The generators of the three triangles were then

37 and 7 , 37 and 33 , 37 and 40. The right triangles had dimensions (a) (b) (c)

1418

280

1320 2442

2)1

2960

2969

518

2458

\-le feel that there are smaller triangles than the ones above. A few minutes on a computer should ~ive us better results.

Finally, Montucla says that, if one accepts fractional results, one can get an infinite number of such triangles. He missed the possibility of multiplying by the common denominator of these fractional sides. Hence it is possible to find, say, four right triangles with equal areas.

•

25 Olympiad Problems 11)

Find the maximum value of 2 2 2 sin e + sin e + • • • + sin en 1 2 subject to the restrictions

e1 12)

T

e2

+ ••• +

0
4.

•• •• •• •• •• ••• •• •• •• •• • •• •• •• •• •• •• ••• ••

e

e e

•e •e fit

e e e e e e e

• e

•e

­ e

e e e e e e e «It e e e e e e e

•

e e e

e

e

5

Applying the substitutions in (7) and (8) show that this problem is equivalent to

(t - 3)(5t - 1) 2 < o. l+t The quantity on the left is 0 when t == 3 or when t == 1/5. Since 1 + t 2 is always positive, the sign of the left-hand side depends only on the sign of (t - 3)(t - 1/5). Between 1/5 and 3 this expression is negative; elsewhere it is positive. Thus

t lies in the interval (1/5,3) and hence x lies in the interval (2 arctan k, 2 arctan 3). Example 3 (Integration). For those of you who know some calculus, the substitution t == tan

~

will let you integrate any

rational function of sin x and cos x by changing it to a rational function of t. This latter expression can then be integrated by expanding into partial fractions. For details, consult a calculus textbook such as [3]. As we have seen, the substitution t == tan

~

is a very effec­

tive method of solving certain types of problems. However, it should only be used as a last resort because its use can involve very complicated expressions. Its effectiveness makes it ideal for computer applications. For hand calculation, it is advis­ able to first look for other tricks or

substitutions~

Ingenuity

will normally lead to a simpler solution than brute force. The following example illustrates this point.

Example 4. Murray Klamkin asked in [2] to find all x in

(0, 27r) satisfying ·

81s1n

10

81 x+cos 10 x==-. 256

••

6

In this case, the simpler substitution, t

= sin 2 x,

1- t

= cos 2 x

can be used. This substitution may be used any time that all the exponents of sin x and cos x are even. In this case we get

8It 5

+ (1 -

t)5 = ­

81 256

which expands out to

4096t 5

+ 256t 4

-

512t 3 + 512t - 256t

+ 35 =

O.

The powers of 2 appearing as coefficients suggests that we make the further substitution t = T / 4 which yields

4T 5

+T4

-

8T 3 + 32T 2 - 64T

+ 35

•• •• •• •• •• ••

= O.

In this form, T = 1 is seen to be a solution (twice) and the polynomial factors as

Since T = 4 sin 2 x is always non-negative, 4T3 can never be 0 and so T

t =

i- and sin x = ± t.

+ 9T2 + 6T + 35

= 1 is the only solution. This yields

Thus, x

= 1f /6,

51f /6, 71f /6, or ll1f /6.

References [1] J. T. Groenman, Problem 1020, CRUX Mathematicorum. 11(1985)51.

[21 Murray S. Klamkin, Problem D-3, AMATYC Review. 4(1983)63. [3] George B. Thomas, Jr., Calculus and Analytic Geome­

try (third edition). Addison-Wesley Publishing Company, Inc. Reading: 1960.

•• ••• •• •• ••• ••• •• •

•• •• ••• •• •• ••• •• • ••• •• ••• •• •• •• ••• •• •• •• •

j.

7

THE PENTAGON Recently, we have come upon several articles about the regular polygon of seventeen sides, and how to construct it. This was certainly one of the great achievements of Gauss, being the construction of a new polygon unknown to the Greek geometers. We couldn't help noting that, of late, there has been nothing in our geometry relating to the construction of, say, a re~lar

pentap;on or decap;on. We had found. such con­

structions fun. To find out about such constructions, we went lack to our old geometry text, "Plane Geometry", by Schultze and Sevenoak, printed in 1925. First, we came again on the golden mean, or golden section, or golden ratio.

I' b _-', A__a.._--.ol--_--..:.._

13

If the line AB is divided at point P into pieces a, b a b so that b = a + b ' we say that AB has been divided in "extreme and mean ratio'" The rectangle below with

h

b

a.~lr

dimensions according to the golden mean was considered by the ancient Greeks to have the most pleasing shape.

•

8 This may well be so - our flag has about the proper dimensions.

ab = a+b b = ba + 1,

Let us write Then

-r

=1

+ 1, or

-r 2 - -r - 1

= O.

quadratic ;or -r, we easily find that

and let

ab = -r.

Solving this

-r

= ~2+

1.

This, then, is the golden mean. The number

-r

is well-known. We have referred to it

in previous articles. As a continued fraction, examp I e, ~• -- 1 + I1 + I1+1 I + • • ., so tha t ~• and we are back to the equation for the golden

for 1 -- 1 + :r­ mean.

Note also that since we can construct a segment equal to

~ (in several ways), we can construct a length

equal to -r. We can use the golden mean to construct a regular pentagon or a regular decagon as did the Greeks, but we prefer to do a little extra work because it may be helpful for the future. We will work with a circle with unit radius. The extension to a circle of any radius is obvious. A

• •• •• •• •• •• •• •• • •• • •• •• ••• •• •

•

•e •e e •e e

e

e e e

e e

• •e e e e

e

e

e

e

e

e e e e

e e e

e e e

e e tit e e

--

9

Let AB be a side of a regular pentagon inscribed in the unit circle. Call this side

~.

c Then LADE = 72 ,

c am, in right triangle MOB, LMOB = 36 ,and we have

~ = sin 36°. We must fim sin 36°. (We have done some of this before, but it doesn't hurt to repeat it.) First, note that sin 36° = cos 54". Let am we may write

x = 1ff ,

sin 2x = cos 3x. From De Moivre 's

Theorem, (cos x + i sin x)3

= cos

3x + i sin 3x.

We expand on the left and separate real fran complex tenns. We then equate am find that 2 cos 3x = cos3x - 3 sin x.cos x 2 sin 3x = 3 cos x.sin x - sin3x •

Using the first of these equalities, (and line 7 above)

we have

sin 2x = cosJx - 3 sin2x.cos x 2 sin x.cos x = cos3x -

3 sin2x.cos x

and dividing by cos x (which is not zero) we get 2 sin x

= cos 2x

2 - 3 sin x

= 1 - 4

2

sin x. This is a

quadratic in sin x, namely 4 sin2x + 2 sin x - 1

= O.

Solving, we have fourd that

o

sin x = sin 18

= -J5-1 4

o Now it is easy to construct ~, then ~ -1, 18 ,

o

and eventually, sin )6. then

se~ent S,

Now we construct segment s/2,

and we have constrmted a side of our

regular pentagon (in a unit circle). Admittedly, this is complicated and messy, but it will

work~

10

The reader has certainly noticed that if, in the diagram above, PQ is a side of a regular decagon, then LPOQ is

36' ,

then PQ/2

and, if M is the midpoint of side PQ,

= t = sin

c

1S ,

•

Perhaps it is easier, if one

wants a regular pentagon, to first construct a regular decagon and then connect alternate vertices. This 1s certainly easier but still messy. ,€

Let us find cos 36 • We start with cos 2x

=1 -

2 sin2x.

and we have found 2 0 cos 36

=1 -

2 sin 36

=

= 1 - 2(~15 -

cos 36

2 ., sin 36

()

=

1)2/16

= ,}54+

1

-J5+1 4 • Now

= 6 +16G"'-/5

,so that

16 '15 . and a side of the regular pentagon

10 - 2

in a unit circle is

...J.10

-22..}5. Constructing this

would indeed be a canp11cated procedure, but we can

•• •• •• •• •• •• •• •• •• •• ••

construct our regular pentagon, at least. With a 11t tIe labor (and a lot of trigonanetry) , we have managed to derive ways to construct our regular pentagon and our regular decagon. Where does the golden mean cane in ? Let us tie things together.

•• ••• •• •• ••

e e

11

•e

e e e e e e e ee e

e e e e e e e e e e e e e e e

e e e e e e e

e e

e

tit

e

e

The Greeks did their construction as follows. Let OA be a radius of a circle. Divide OA in extreme and mean ratio at D. Mark off AB = a. Then the angle at center 0 equals J6e , and AB is a side of a regular decagon. We leave it to the reader to prove that triangles ABD and ABO are similar isosceles triangles, that the angle at A is 72v and the angle at 0 is J 6tJ • This should raise the roader's estimation of the Greek geometers.

D

.14.' ~-~-----f------------I

A

12

We now present our own way of constructing a regular pentagon. In the diagram, we have a unit circle with perpemicular diameters. We work as follows. (a) Bisect radius OA at M; then OM

(b) with radius MB, draw arc

= 1/2.

Be meeting radius OA'

at C. (c) since OM

oc =

= 1/2

(-/5 -

and OB

= 1,

MB

= -/5/2.

Hence

1) /2 •

(d) With radius Be, mark off chord BD. Since Be

2

= B02

Be = BD =

+ OC

2

=1

+ (6 - 2...[5)/4

= (10

-

2-J"5)/4.

--aha - 20 2

Notice that, in line with all our

~vious

computations,

OC equals a side of the regular inscribed dec.agon, and BD equals a side of the regular inscribed pentagon. We happen to think this

t~

As an added curiosity"

easiest method. since OB would be a side

of a regular inscribed hexagon, we have proved that "When the square of a side of a regular hexagon is added to the square of a side of a regular decagon in the same c ire Ie, the sum equals the square of a side of the regular inscribed pentagon in the same circle. " We invite the reader to get out the proper tools am perform the construction - just for fun.

•• •• •• ••• •• •• • ••• •• •• •• •• •• •• •• •• •• •

•• •• •••

••e •

e e • e •

e

13 Constructions afford a way of seeing some of the

relations one derives by other methods. Once one has solved a construction problem, it remains in the

memory. Moreover, constructions provide a way of using sane of the things one has learned elsewhere. For example, how would one construct the canmon

tangents to two circles1 We add some problems (with

hints). Given H , Hb , H the feet of the altitudes of a a c triangle in position, reconstruct the triangle.

e e e

•e •e

e e e e e e e e e e e e

e

e e e e

e

(Hint: Draw a triangle with Ha , Hb , Hc the feet of the altitudes, then join them to make the orthic triangle. What relation is there between an angle of this triangle and the altitude at the vertex ?) If P, Q, R are midpoints of three sides of a parallelogram, construct the parallelogram. (Hint: you should be able to draw the diagonals.

Construct an eqUilateral triangle with one vertex on

each of three parallel lines.

(Hint: See the problem on the cover page.)

From a point outside a e1rcle, draw a line cutting

the circle so that the chord inside the circle is

half the lellgth of the whole line.

14

MANY CHEERFUL FACTS

We have already presented a method. of fiming the sum of a series by means of finite differences (see our issue of September, 1982). Sometimes this method. is too complicated am may even not be applicable in some given case. We would like to present another method. that may be easier. Given a series S = u

+ u + • • • + un. If we can 2 1 write u l = sl - s2' u 2 = s2 - s), ~ = s) - s4' am so on, the series takes the form. S = (sl- s 2) + (s2- s )) + (s)~s4) + ••• + (sn_l- s n), then the terms other than the first am last cancel each other, am we have

S = s1 - sn. Such a series

is called a "telescoping" series.

•• •• •• •• •• ••

The simplest example we can think of is the series + _1_ + _1_ + + 1 W ite it S = _1_ lx2 2x) )x4 ••• n(n+l)· r as _ ( 1) (1 1) (1 S - 1 -"2 + "2 - J + J am the sum is obviously 1 _ n 1 - n+l - n + 1 •

-"41)

+ ••• +

n-

(1

1) n+1

Usually, a bit of work is needed to put the series in the desired form. For example, let us consider S =_1_+_1_+_1_+ lx4 4x7 7xl0 •

. .

We cannot write this simply in the form

_ ( 1 -"41)

S -

+

(1"4 -"71)

+. •• +

( )n-2 1

1)

- )n+l •

because the expression in any parenthesis does not equal the correspoming term of the original series.

••• •• •• •• •• •• •• •

••• ••• •• • •• •• •• •• •• •• •• •• •• •• •• ••

,.

•• •• •• ••

15 In this case, we make use of partial fractions. We 1 write the general term, (3t-2)(3t+1) in the form

=

1

-rO::-:t~-~2)r'i('"='3~t+~1~)

A B + Ot+l) Ot-2) , am remelll be r t hat

the numerator 3(A + B)t + (2A - B) must equal 1. Then (A + B) A = 1/3

=0 am

Solving these, we have

B = -1/3. Now we may write the series

= (ill _ ill) 1 4

S

= 1.

(2A ­ B)

and

+

(ill _ ill) 4 ?

+

...

+

(..1L1 _ !L.L) 3n-2 3n+1'

telescope, simplify the result, am fim that S -

n

- 3n + 1·

Exercise. Fim an expression for the sum of the seriest T

= 1X~X3

+ 2x5x4 + • • • + n(n+ltCn+2) •

The philomath should. add partial fractions am the method of telescoping series to his/her armamentarium. As another instance of telescoping series am to present another rule that students will fim useful, let us examine the seriest S = sin(a) + sin(a+d) + sin(a+2d) + ••• + sin(a+ n-l d). Here the angles are in arithmetic progression, but the sines are not. Here we will need sane trigonaaetry. First,

cos(x - y) cos(x + y)

so

= cos = cos

x.cos y + sin x.sin y x.cos y - sin x.sin y.

cos(x - y) - cos(x + y)

=2

sin x.sin y.

Next, we multiply each term in our sine series by 2 sin

d '2.

The

~neral

2 sin ~ sin(a + td)

=

term then becomes 0 ~ t ~ n-1.

=

Now let x a + td am y d/2 in our formula for the difference of two cosines (see above) , and we have

16 the general term in the form cos(a + t - 1!2 d) - cos(a + t + 1!2 d). Substituting successive values for t, from 0 to n-1, we fUrl cos(a - 1/2.d) - cos(a + 1/2.d) cos(a + 1/2.d) - cos(a + 3/2.d) cos(a + 3/2.d) - cos(a + 5/2.d)

. . . . . .

cos(a + n-3!2.~) - cos(a + n-l!2.d) which telescopes to 2. sin % • S

= cos(a

- 1/2.d) - cos(a + n-l!2 .d).

•• •• •• •• •• •

Using the formula for the difference of cosines again, we let

x - y

=a

and get 2.sin %.S

- 1/2.d

x + y

= 2.sin(a

=a

+ n-l!2.d, and

+ n-l!2.d).sin(nd/2),

We divide by 2.sin d/2, am finally arrive at _ sin nd 2( ( / ~ S - sin d 2 sin a + n-1 2.d//. This formula is worth adding to your list of cheerful facts. We can make it easier to remember by noting the multiplier

s;~n~ ~ and that the term inside

the other sine term is the average of the first and last terms of the angles in the original series. nd 2 ( i first angle + last angle) Th a t i s, S = sin sin d 2 s n 2 The reader who is willing to go through the same process for the sum of a series of cosines of angles in arithmetic progression will find that theformula is

••

•• •• •• •• •• •• •• •• •

•• •• ••• •• •• •• ••• •• •• ••• •• •• •• ••• •• •• •• •• •

'.

17

c = sin

m. 2 ( cos first ahgle + last angle) sin d 2 2 •

Exercises & Show that

2

sin x + sin 3x + sin 5x + ••• + sin(2n-1)x sin 2x + sin 4x + sin ox + ••• + sin 2nx

nx = sin sin x·

=

sin nx. sin (n+1)x sin x cos 2x + cos 4x + cos 6x +

... + cos 2nx =

sin nx. cos(n+1)x sin x F ind

1'1' + J.!! + cos 2!! + 171'1' cos 19 cos 19 19 ••• + cos 19 •

If we are faced with the problem of summing a set of squares of sines or a square of. cosines, we can always proceed by first using the formulas 2 2 2 sin a = 1 - cos 2a or 2 cos a = 1 + cos 2a am. then using our formulas already derived. Note I We have tried without success to derive the formulas for the sum of sines am. the sum of cosines by other methods, but have not had much success. We tried using De Moivre's Theorem, Euler's formulas and other devices. This is because we consider the multiplication by 2 sin

'2d

a "trick". We welcome

other methods that are more "natural" from our readers. Please write if you have any other method. S. Greitzer

18

LIGHT FARE

1)

Piranhas and perch are fighting. A perch can kill one or two piranhas, is he ld at bay by three piranhas, am can be killed by four piranhas in three minutes. How long will a fight between four perch am thirteen piranhas last ?

2)

Three arabs have 17 camels, which they are to divide in the ratios 1/2, 1/3 am 1/9. The usual solution has another arab donate one camel, whereupon the three take 1/2 of 18 = 9, 1/3 of 18 = 6, and 1/9 of 18 = 2. This would leave one camel, which the fourth arab takes back. This solution is wrong. How can the camels be shared correctly ?

3)

On one side of a bad scale, a cheese weighed 16 poums. On the other side, it weighed 9 poums. What is the true weight of the cheese?

4)

It is said that the mathe~tician De Morgan was x years old in the year x • However, Jenkins was a

2 + b2 years old in the year a 4 + b 4 ,

2 2m years old in the year 2m , and. 4 3n years old in the year 3n •

How old was Jenkins each time ?

5)

We have two three-digit numbers of which the first is three times the sec om. If we reverse the digits of the secom number, that number will be four times the number obtained by reversing the first number. What are the numbers?

6)

A triangular field has a square field on each of its sides. One field has an area of 74 acres, another has an area of 116 acres, am the third field has an area of 370 acres. Fim the area of the triangular field.

You may send in your solutions, preferably by the end. of July. Label your answers "LF" followed by the number.

••• •• •• •• ••• •

•• •• ••• •• •• •• •

••• •

•e •• e

19

MORE ON DIOPHANTINE EQUATIONS

In the Arbelos for March, 198) (pp 10-17), we first investigated the linear Diophantine Equation. Among the methods we presented for solving these, we included (a) guessing, (b) Euler's method, (c) using the Farey e sequence, (d) using continued fractions, and (e) using tit con~uences. Our own favorate is guessinp;.

• •

e

e

To refresh the memory, suppose we are given the linear Diophantine Equation ax + by = c. We restrict ourselves to positive values for a, b, c, x, y and assume that (a,b,c) = 1.

e e

e • :

We first guess x o' YO such that ax O + byO = 1. Then acx O + bcyO = c.

Subtract ax + by = c , and we have

• •

a(cx O - x) + b(cyO - y) = O. We transform this into a proportion, as follows I

tt

ee

cy - y _0_ _

b and we may write

x - cx o cyO - y

a

•

e e •

-=

= at

and

Finally, we solve these for x, y,

e :

x = cx

O + bt

e

e

e e

e

= bt.

and arrive at

t

will produce

to positive values for

cyo -

at >

o.

Let us apply all this to a problem. We wish to

: find. positive x, y to satisfy

e

O

Y = cyO - at.

Substituting any integer value for

e solutions x and y. e If we restrict ourselves e x and y, we must have e cx + bt > 0 a.nd o e

x - cx

7x + 9Y = 400.

20

First, guess

7(4) + 9(-3)

=1

7(1600) + 9(-1200)

So Since

+

7x

9Y

7(1600 - x) + 9(-1200 - y)

1 = 1200

Therefore,

+ y 1600 - x'

9 am. finally,

= 1600

x

•

= 400 • = 400, = o.

7t

= 1200

+ Y

9t

= 1600

- x

- 9t

J = 7t - 1200

Substitute any integer value for

t , am. we get

solutions of the original equation.

There are some interesting problems that may arise. First, given

= c,

ax + by

with a, b, c

= 1,

positive, and with (a,b,c)

given, all

how many solutions

• ••• •• ••• •• • • ..

are there where x and y are positive? Second, given

..

ax + by, what values of

• •

c

will produce

n

solutions?

In the problem we have just solved, we would like to have

1600 - 9t > 0

and

7t - 1200 > O.

The first inequality gtves us t >

sec ond gives us

172

171~. ~

t

~

t < 17?§ • The

In integer values for t, 177 •

There will be positive solutions for x, y t = 172, 173, 174, 175, 176, 177. For example, let that

x

= 25

am.

t

= 175, y = 25.

when

•

..

•

• • •

..

am. we fim., on substituting, .. •

•• •.

•

••• • •• • •

21

Going back to the inequafities

cx

.cyO - at > 0, and solving these for

O

t

+ bt > 0

and

, we find

cyo

< -a­

.Since we know a, b, c, x

. • (integer) limits for

o am Yo' we can detennine the

t. The number of solutions in

.positive integers for x, y

will equal the number of

•

.integers between these limits. •

When we do not know the value of

c , the problem

.becanes more canpl1cated. At this point, we decide to \ : use the properties of the Greatest Integer function • • This can be defined in three ways, as followsl

.(a)

[ x ]

~ x < [ x]

.(b)

x - 1

< [ x ]

:(c)

0

~

~

+ 1

x

x - [ x ] < 1

.We agree to use whichever definition is handiest.

•

Assuming that

.let us replace :(d)

x

is not an integer, and using (b), by

-x. This gives us

-x - 1 < [-x] < -x

~~

x-1 x, so that

is the least integer greater than

x.

22

Let us return to the inequality at the top of the previous page. Using the properties of the Greatest Integer function, we findl the smallest value for

t

_ [cx OJ b eyo -J. a

is

r

the greatest value for t is CX cyO O The difference is [ aJ + [ T J. This should tell us the number of positive values for a given It is easy to show that, where integers,

Lu

u

and.

v

c•

are not

J + [ v J < [u + v J < [ u J + [ v J + 1­

Substitution of the expressions above in this, we cyO cx O cyO CX cyO CX O O have [ a J + [b J < [ a + b J < [ a J + [b J + 1. However, the middle term in this inequality equals c ( ) _ c ab axo + byo - ab Therefore, if

•

N is the number of

are N positive solutions, then

t's for which there

N~· [c/abJ

or

N .~ [c/ab J + 1. In the illustrative example, c/ab

= 400/63,

and

(400/63J is 6, so we expect to have six solutions. Can you find a value of

c

for which

7x + 11y = c

will have four solutions in positive integers ? Because of our restrictions on a, b, c, x, y, the solutions for N usually boil down to N ~ [c/abJ.

••• •• ••• •• •• •• • ••

•

•• •• •• •• •• •• •••

• • • •

23

• The same issue of Arbelos had the following problem • to which nobody subnitted an answers • • • • • • •

On a certain planet, there are two inimical foms

of life. The Septicapita have seven heads but only

two legs each, while the Pentapoda have only two heads but do have five legs each. One day an odd lot of Septicapita encountered an odd lot of Pentapoda am a wild me lee ensued. Heads and legs were flying allover - one observer counted 180 of both. How many of each type were involved in the fracas ?

• Solutions The Septicapita contributed 7 heads am 2 . • legs each. The Pentapoda contributed 2 heads am 5 , legs each. Let x = the number of heads and y = the • number of legs. Then • 9x + 7y = 180 .Now 9(4) + 7(-5) = 1 , .so 9(720) + 7(-900) = 180 • • Subtracting, .or •

•

=0

.2_y+900 7 - 720 - x

• am

•

9(x - 720) + 7(y + 900)

= 720

x If

x

and

- 7t y

Y

= 9t

- 900.

are to be positive, we must have

• 900/9 < t < 720/7

.There are two possible solutions - when t = 101 or

.102. For t = 101, we find 13 Septicapita am 9 Penta­

.poda. For t = 102, there are 6 of one and 18 of the

• other.

• However, we must discard the second solution, since • we did say that there were an odd lot at each animal. • Our apologies for not presenting a complete set of

.rules for solving this type of problem, but we do

• believe there is enough here to enable a student to

.solve similar problems more easily.

•• •• ••

24 OLYMPIAD PROBLEMS 21)

Let p be a prime. For which k can the set (1,2,3, ••• ,k) be partitioned into p subsets with equal sum of elements?

22)

(Poland)

A collection of 2n letters contains two each of n different letters. The collection is partitioned into n pairs, each pair

containin~

two letters

which may be the same or different. Denote the number of distinct partitions by u • (Partitions n

differing in the order of the pairs in the partition or in the order of the two letters in the pairs are not considered distinct.) Prove that u = (n + l)un _ n(n2- 1) un_2 •

n+l (Great Britain)

23)

For each P inside the triangle ABC, let A(P), B(P) and C(P) be the points of intersection of the lines AP, BP and CP with the sides opposite to A, Band C respectively. Determine P in such a way that the area of the triangle A(P)B(P)C(P) is as large as possible. (Monp;ol1a)

24)

A

non-ne~tive inte~r

fen) is assip;ned to each

•

•• ••• •• ••• •• • • •• •• •• •• ••• •

positive integer n in such a way that the followi~ • conditions are satisfied: (a) f(mn) = f(m) + fen) for all positive integers m, n;

(b)

(c)

= 0 whenever f(10) = O. fen)

the units digit of n is 3;

Find the value of f(1985).

(Colanbia)

•• •• •• • •

••

25



•

.. ' , !

I

­ .

-

• e ••e •

e

i,..

f

'"' ~, :i

'!<

•

• •

•

,

•

e•

•• e e

e e

;

.'e e

,

,. , . t.: I

• •• .' •