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Mat he m

at ico ru m

C R U X M A T H E M A T I C O R U M

Vol. 13. No. 3 March 1987

Published by the Canadian Mathematical Society/ Publie pax la Societe Mathematique du Canada

The support of the University of Calgary Department of Mathematics and Statistics is gratefully acknowledged.

X X X

CRUX MATHEMATICORUM! is a problem-solving journal at the senior secondary and university undergraduate levels for those who practise or teach mathe­matics. Its purpose is primarily educational, but it serves also those who read it for professional, cultural, or recreational reasons.

It is published monthly (except July and August). The yearly subscrip­tion rate for ten issues is $22.50 for members of the Canadian Mathematical Society and $25 for nonmembers. Back issues: $2.75 each. Bound volumes with index: Vols. 1 & 2 (combined) and each of Vols. 3-10: $20. All prices quoted are in Canadian dollars. Cheques and money orders, payable to CRUX MATHEMATICORUM, should be sent to the Managing Editor.

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Founding Editors: Leo Sauve, Frederick G.B. Maskell.

Editor: G.W. Sands, Department of Mathematics and Statistics, University of Calgary, 2500 University Drive N.W., Calgary, Alberta, Canada, T2N 1N4.

Managing Editor: Dr. Kenneth S. Williams, Canadian Mathematical Society, 577 King Edward Avenue, Ottawa, Ontario, Canada, KIN 6N5.

ISSN 0705 - 0348.

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x * *

CONTENTS

The Olympiad Corner: 8 3 . . . - R.E. Woodrow 70

Problems: 1221-1230 . 85

Solutions: 1032, 1033, 1077, 1081-1086 . . 87

- 69 -

- 70 -

THE OLYMPIAD CORNER: 83

R.E. WOODROW

All communications about this column should be sent directly to Professor

R.E. Woodroio, Department of Mathematics and Statistics, The University of

Calgary, Calgary, Alberta, Canada, T2N INk.

We begin this column with two sets of problems. The first is from the

6 Brazilian Mathematical Olympiad, September 15, 1984, for which we thank

Professor Dr. Angelo Barone Netto, Sao Paulo, Brazil.

The first three problems of the list were proposed to all participants.

The last three were chosen by regional committees from a list of 15 problems.

Those presented here were chosen for Rio de Janeiro and Sao Paulo.

i. Find all natural numbers n and k such that (n + 1) - 1 = n!

2. The 289 students enrolled in a course are to be distributed into 17

classes each having 17 students, for each of several units of

instruction. No two students may be assigned to the same class for different

units. What is the largest number of units for which an assignment is

possible following this rule?

3. Let F and F', E and E' respectively be two pairs of opposite faces

of a regular dodecahedron. From the centres of these four faces,

perpendicular segments of length m are raised outward from the solid. Let A,

C, B and D, respectively, be the outer extremes of these perpendicular

segments. Show that ABCD is a rectangle and find the ratio of its sides.

4. Let D be a point on the hypotenuse BC of a right triangle ABC.

Suppose points E and F9 on the legs of the triangle, are such that

DE and DF are perpendicular to the corresponding leg. Find the position of D

that minimizes the length EF.

5. Let ABCD be a convex quadrilateral, and let E, F, G, H be the

centres of the external squares erected upon sides AB, BC9 CD, DA

respectively. Show that segments EG and FE have the same length and are

perpendicular. [Note: This appeared as Crux 1179(a) [1986: 206].]

- 71

6. Figure 1 shows a board used in a game called "One Left". The game

starts with one marker on each square of the board except for the

central square which has no marker. Let A 9 B 9 C ( o r C 9 B 9 A ) b e three

adjacent squares in a horizontal row or a vertical column. If A and B are

occupied and C is not, then the marker at A may be jumped to C, and the marker

at C removed (see Figure 2). Is it possible to end the game with one marker

in the position shown in Figure 3?

-L*-U' m

Figure 1 Figure 2 Figure 3

The second set of four problems we pose are from the Second Balkan

Mathematical Olympiad, May 69 1985. We thank Ivan Tonov for forwarding them.

i. Proposed by Bulgaria.

Let 0 be the centre of the circle through the points As B9 C9 and

let II be the midpoint of AB. Let E be the centroid of the triangle ACD.

Prove that the line CD is perpendicular to the line OE if and only if AB = AC.

and

2. Proposed by Romania.

Let a9 fo9 c9 d be real numbers in the interval

sin a + sin b + sin c 4- sin d = 1

10 cos 2a -t- cos 2b + cos 2c + cos 2d > ~^

-IT ir

2 ' 2

Prove that as b9 c9 d actually must belong to the interval

such that

o £ ' 6

- 72 -

3. Proposed by Greece.

Let E denote the set of real numbers, and let S be the set of

elements of E of the form 19x + 85y where x and y are positive Integers.

Colour the elements of S red and the remaining Integer elements of E green.

Is there a real number a (not necessarily integral) such that for every two

integers b and c if b and c have the same colour then b and c are not

symmetrically placed about a (i.e. [a - b| / ja - c|)?

[Note: As we received it this problem is pretty trivial. Any a not of the

'form (m + n)/2 will do. With the restriction a = (m + n)/2, m, n Integers,

what is the answer?]

4. Proposed by Romania.

1985 people take part in an international meeting. In any group of

three there are at least two individuals who speak the same language. If each

person speaks at most five languages, then prove that there are at least 200

people who speak the same language.

As promised last month, we give the official answers to the First Round

of the 1986 Dutch Mathematical Olympiad.

Al. 37.8

A4. 35

Bl. 523152 and 523656 B2. 60'

B3. 7 B4. 1932 + 932 or 2132 4- 232

CI. n = 11, in = 3 C2. 8 C3. 0

*€ X K

We next present solutions to some problems posed in the April 1985

Olympiad Corner.

51. [1985: 102] Proposed by Australia.

Let P be a regular convex 2n-gon. Show that there is a 2n-gon Q

with the same vertices as P (but in a different order) such that Q has exactly

one pair of parallel sides.

A2.

A5.

A: 26,

150

B: 14, C: 8

o

A3.

A6.

4

4

- 73 -

Solution by Aage Bondesen, Royal Danish School of Educational Studies.

The cases n = 1 and n = 2 are trivial. Assume n > 2. Enumerate the

vertices of P In order p 0pr °-

p2rit~i °

W h e n i s P0 Pi P a r a l l e l t o pfe

p^ w i t h

k < i? The only possibilities are that 0 < k < £ < i and fc = i - 5, or that

i < h < £ < 2n and 2n - « = fe - I.

Now Imagine we enumerate the vertices PpJP* Pa * » »P9 In such a way

that the (clockwise) distance ip . - L on P Is either n or n - 1 except for

one pair whose distance is -1. Also suppose that every vertex, except for the

endpoints of the pair at distance -1, has both distances n, n - 1 for its two

neighbors, and that the neighbors of each end of the side of length one are at

distance n. Then It Is easy to see that the only parallel sides possible

Involve as one side the pair at distance -1. This is because two pairs at

distance n determine diameters, and were two pairs P. P0 and Pa P. at

distance n - 1 parallel, we would have i, + n = ifi (mod 2n) and i* + n = i,

(mod 2n), thus P. = P. and P. = P. , thus k + 2 = £ + 1 (mod 2n) lfe+2 l«+l l£+2 lfe+l

and £ + 2 = fe + 1 (mod 2n), a contradiction.

For n even the sequence

n - l9n9n - l9n9 . ..fnf-l,ntn ~r l,...,n

st of distances, with a "-1" In the n + 1 position, makes the first side

st parallel to the n + 1 side since the common distance between endpoints Is

2 n - ( n - l + l ) _ n 2 " 2"

For n odd the sequence

n9n - 1,...,n,-l,n,n - 1,..., n9n - 1

st of distances also produces the desired result, with the first and n + 1

sides parallel, since the common distance between endpoints is

2n - (n + 1) _ n - 1 2 ~ 2 •

The situations with n = 4 and n = 5 are Illustrated below.

- 74

52. [1985: 102] Proposed by Belgium.

If n > 2 is an integer and [x] denotes the greatest integer < x,

show that n(n + 1) 4n - 2

\n + l] ~ L 4 J

Solutions independently by Beno Arbei, Department of Mathematics, Tel

Aviv University; Aage Bondesen, Royal Danish School of Educational Studies;

Bob Prielipp, University of Wisconsin, Oshkosh; and Edward T.H. Wang, Wilfrid

Laitrier University, Waterloo, Ontario.

Since

it suffices to show that

n + 1 * n(n + 1) 4 4n - 2

n(n + 1) . 4n - 2

n + 1 + 1. (*)

Straightforward calculation gives

and for a > 2 we have

Thus

n(n + 1) _ n + 1 n + 1 4n - 2 ~ 4 4(2n - 1)

n + 1 < 1 4(2n - 1) 4

Hl!ULii< n+ 1 I 4n - 2 4 4 '

There are two cases:

- 75 -

(i) n = 4k + r, r = 0,1,2. Then

[=•£!] • i - k • i

while n 4- 1 1 — 4 — + 4 < fe * 1

so (*) follows.

(li) n = 4fe + 3. Then

n + 1 + 1 = fe + 2.

But then

n + 1 . 1 u . i . 1 4 4 4

and the Inequality (*) holds.

54. [1985: 102] Proposed by Bulgaria.

The Incircle of triangle AJA2A3 touches the sides A2A39 A3At, AtA2

at the points T±9 T2, T3, respectively. If M±9 M2, M3 are the midpoints of

the sides A2A39 A3Ai, AtA2s respectively, prove that the perpendiculars

through the points M±, I2, M3 to the lines T2T39 T3T1$ T±T2, respectively, are

concurrent.

Solution by J.T. Groerwum, Arnhem, The Netherlands.

Notice first that A3T2 = A3Tt and that the line d3 through A3 and the

Incentre I bisects ZA3. Let ds bisect the exterior angle at A3 as Indicated,

and let D H e on the same side of AtA3 as does A2. Notice that

ZlA3T± + ZT ±A3B = 90°. Thus ZT ±A3B is the complement of ZlA3T±. In the

Isosceles triangle TtT2A3 we easily see that ZA3T±T2 Is the complement of

ZlA3T±. Thus TiT2 Is parallel to d3. The perpendicular from Jf3 to T±T2 Is

- 76

thus perpendicular to da and thus is parallel to d3. Similarly for 1 = 1,2 we

see that the perpendicular from M. to the corresponding side of T±T2T3 is

parallel to the angle bisector d. of ZA..

Consider now triangle M±M2M3. Now AMiS^Ma is similar to AA*A2A3 and

moreover corresponding sides are parallel. Thus the bisectors clf e2, e3 of

angles H±9 M2, M3 are parallel to the bisectors di9 d2s d3 of Al9 A2, A3

respectively. These coincide at the centre of M±M2M3. But elf e2, and e3 are

just the perpendiculars referred to in the last paragraph, completing the

proof.

56. [1985: 102] Proposed by Canada.

Given that a1sa~,...,a9 are distinct integers such that the

equation

(x - ax){% - a2) ... (x - a^) + {-l)U~ (n!) = 0

has an integer solution r, show that r = (a1 + a~ + ... + a~ )/2n.

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,

Ontario.

We first establish two simple lemmas:

Lemma 1. If d. < d~ < ... < d are distinct positive integers such that

d^dy...d = n! then d, = k for k = 1,2,...,n.

Proof. Since d1 > 1 we have d, > fe for all fe. Then d^d~.. .d > n! 1 " I * ~ 1 2 n ~

with equality if and only if d- = fe for all fe.

Lemma 2. If b^,b^,...,b^ are positive integers with b1 < b~ < ... < b~

where no three consecutive terms are equal and b-.bp.. .b~ = (n!)2 then

(b1.b2,b3''*",b2n^ = C1'1'2'2 n,n).

Proof. Consider the sequences bltb~,...,bp - and b~,b4,...,bp .

Since b1 > 1 we have b~ > 2, b~ .. > j and b^^.-.b^ . > n!. Similarly

b? > j and b2b4'•-b2n. - n!" B u t blb2" " b2n = ^n!^2 a n d h e n c e blb3,,ab2n-l

= b^b....b^ = n! and the conclusion follows from Lemma 1. 2 4 2n

Now we prove the main result.

By assumption,

- 77 -

(r - ai)(r - a2).0.(r - a^) = (-lf(n!)2

and r - at / r - a for i / j. Let b-.bg, . . . 8 b ^ denote the numbers |r - a. |

arranged in nondecreasing order of magnitude. Then 1 < b- < b0 < . . . < b0

where no three consecutive terms can be equal. Since b.|bp...b«> = (n!)2 we

conclude from Lemma 2 that

Let a° a° ..„, a' denote a rearrangement of a1}a«}...,a« such that

|r - aj| = bj. Then from b 2 % = bg = j we obtain l -a' j | = |r-a' | = j

for j = 1,2, .. . ,n. Hence for each j one of r - a' 1 and r - a' . must be j zj-i zj

and the other -j. Therefore we get

0 = Ji {(P " ^ - l 3 + (P ~ a2J)}

= " j 2 [a2J-l + a2j]

2n = 2nr - 2 a, .

From this r = (a- + cu + . .. + a^ )/2n is immediate.

Finally, the above argument shows that if c- < c^ < ... < c~ are

o integers such that jCjCp.-.Cpl = ^n^ then necessarily

(c ,...,*:,*) = (-n9-ri 4- 1,..., -1,1,2,...,n)

and hence c-Cp. . .Cp = (-1) (n!) . In other words, although the sign (-1)

is not necessary for the entire argument to go through, it is nonetheless the

correct sign as it is a consequence of the other conditions.

63. [1985: 104] Proposed by Great Britain.

Prove that the product of five consecutive integers cannot be a

perfect square.

Solution by Aaye Bondesen, Royal Danish School of Educational Studies,

Copenhagen.

- 78 -

The assertion Is trivially true if all five integers are negative. If

one of them is zero it is false. Thus we should assume that the integers are

all positive.

So suppose there are successive positive integers al9 a2, a3, a4, a5 with aia2a3a4a5 a perfect square. If for some i € {1,2,3.4,5} a. has a prime

factor p > 5 then p divides none of the other a.'s. It follows that the

exponent of p in the prime factorization of a. is even. Consequently, each of t

a±, a2, a3, a4, as is of one of the four forms:

(I) a perfect square;

(II) twice a perfect square;

(iii) three times a perfect square;

(iv) six times a perfect square.

By the pigeon-hole principle two of the a.'s have the same form. Now if two

positive integers are of form (ii), (iii) or (iv), their difference Is at

least 6, which Is impossible. If both are of the form (i) their difference is

3 or at least 5. Clearly we can accept only difference 3. But then two of

the numbers must be 1 and 4. Then alt a2, a3, a4, as are 1, 2, 3, 4, 5 with

product 120, which is not a perfect square.

[Bob Prielipp, University of Wisconsin, Oshfeosh also pointed out the

error, itfiile John Morvay, Dallas, Texas submitted a different solution.]

67. [1985: 104] Proposed by Poland.

Let a, b, c be positive numbers such that

Prove that the system

V5~~~a + Vz - a = 1

< Vz - b + Vx - b = 1

Vx - c + -Jy - c = 1

has exactly one real solution (x,y9z).

Comment by George Euagelopoulos, Law student, University of Athens,

Greece.

Bjorn M. Poonen, who was a USAMO winner in 1985 and a member of the USA

team at the 26th HO, gave a very nice solution to the problem. His solution

- 79 -

was published in The College Mathematics Journal (Volume 16, Number 5,

November 1985, p.343).

71. [1985: 105] Proposed by Spain.

Construct a nonisosceles triangle ABC such that

a(tan B - tan C) = b(tan A - tan C),

where a and b are the side lengths opposite angles A and B, respectively.

Solution by J.T. Groenman, Arnhem, The Netherlands.

We look for a solution with an obtuse angle at A and with a > b. Refer

to the figure, where D denotes the foot of the altitude from A, and E that

from B.

Now a(tan B - tan C) = b(tan A - tan C) is equivalent to

(AD _ AD] a|DB DCJ "

BE _ BE AE CE (1)

i_ _ i_ BB DC

1_ _ L. AE CD

since A is obtuse. Next a(AD) = b(BE) = 2 Area(ABC) so (1) is equivalent to

(2)

Further, BB = c cos B, DC = b cos C9 AE = -c cos A9 and CE = a cos C, again

using the fact that A is obtuse. Hence we have (2) equivalent to

1 1 1 1 c cos B b cos C c cos A a cos C (3)

Using

cos A = b2 -f c2 - a2

2hc cos B = a2 + c2 - b2

2ac cos C =

a2 + b2 - cs

2ab

we have (3) equivalent to

a a a2 + c2 _ b2 a2 + b 2 _ Q2 b2 + c2 - a2 a2 + b2 - c2

- 80 -

which in turn is equivalent to

abz - etc2 a2b - he'

a2 + cz - b2 b2 + c2 - a2 (4)

Cross multiplying and collecting terms in powers of c gives that (4) is

equivalent to

(b - a)c4 - c2(b3 - a3) -f ab(b3 - a3) + a2b2(b - a) = 0. (5)

Since we seek a nonisosceles triangle we impose b / a to obtain (5) equivalent

to

c4 - c2 (a2 + ab -f b2) + ab(a + b} 2 = 0. (6)

To find an example we may take a = 5b to get

c4 - 31c2b2 + 180b4 = 0,

f31 + VJ2411 b2 as one solution. Checking, we see that A is obtuse giving c

since

As

2 _

b2 + c2 - a2 = ! + ?1 +JSE _ b2 < 0.

5, 1, 31 + v§5T

are constructible numbers, a nonisosceles example can be constructed.

M M *€

We have not received solutions to the remaining problems posed in the

April 1985 Olympiad Corner. This is also true for the May 1985 issue. Send

in your elegant solutions!

We next give some of the solutions to problems posed in the June 1985

number of the Corner. The remaining solutions submitted will be discussed

next month.

3. [1985: 169] The 198*f Batch Olympiad.

For n = 1,2,3,..., let

1*4*7* ... *(3rt - 2) a = h ~ 2-5*8* ... *(3R - 1) '

Prove that, for all n,

- 81 -

(3n + 1) 1/2 < an < (3n + 1) 1/3.

Solution by Beno Arbel, School of Mathematics, Tel Aulu University,

Israel.

We shall prove a more general result: for a and b positive real numbers

with a > b,

<r n a + 3(fe - l)b I £ ^ , " a + (3k - 2)b ^ 34a + :

a ' " ' " - CD Ja + 3n,b - . . a + (3fe - 2)b 3nb

with strict inequalities for n > 1, or if a > b. The problem follows by

setting a = h = 1.

We first prove that for each k = 1,2,...

a + 3(fe - l)b ^T+3(k - l)b a 4- (3fe - 2)b a + 3kb

This Inequality is equivalent to

[a -i- 3(fe - l)b]2(a + 3kb) < [a + (3k - 2)b]3.

But, from the AM-GM inequality,

[a + 3(k - l)b]2(a + 3kb) < [2a + 6(k - l)b + a + 3kb

fea + 9kb - 6bl3

= [a + (3k - 2)b]3

with strict inequality because a + 3(k - l)b j* a + 3kb. Now

S a + 3(fe - l )b y JJ la + 3(k - T)b [ a " a + (3k - 2)b ^ A 3>i a + 3kb ~ 3>la + : fcl a + (3fe - 2)b k=1 3nb

the last equality being obtained by cancellation.

For the left inequality of (1) we prove that for each k = 1,2,3,...

a + 3(k - l)b . la + 3(k - T)b a + (3k - 2)b ~ A a + 3kb (3k - 2)b

which is equivalent to

[a + 3(k - l)b][a + 3kb] > [a + (3k - 2)b]2.

Expanding we obtain

a2 + 6okb + 9k2b2 - 3ab - 9kb2 I a2 + 9k2b2 + 4b2 + 6akb - 4ab - 12kb2

which simplifies to give the equivalent inequality

b(4 - 3k) < a.

- 82 -

This is true for all k since b < a. Note that the inequality is strict if

b < a or k > 2. Now

U a * 3( f e - 1 ) b s n fa^^k^l^b _ f a k^x a + (3fc - 2)b ~ ^ J a + 3feb " 4a 4- 3rib

with strict inequality if b < a or n > 2. This completes the proof.

Remark. One can prove slightly more. For a and b positive real numbers

and for all m and n positive integers the following inequality holds".

n a + m(fe - l)b . K„I

a + (^ ~ m + l)b m a a + mnb

This follows from

[a + m(k - l)b]m X(a -*- mfeb) < [a + (mfe - m + l)b]m

for k = l,2,...,n9 which in turn follows by application of the AM-GM

inequality.

[Comment: This problem was also solved by George Euagelopouios, Laii)

student, Athens, Greece by a different method.]

4. [1985: 169] The 198k Dutch Olympiad.

By inserting parentheses in the expression 1: 2: 3, we get two

different numerical values, (1: 2): 3 = 1/6 and 1: (2: 3) = 3/2. Now

parentheses are inserted in the expression

1: 2: 3: 4: 5: 6: 7: 8.

(a) What are the maximum and minimum numerical values that can be

obtained in this way?

(b) How many different numerical values can be obtained in this way?

Solution by Curtis Cooper, Central Missouri State Uniuersiiy.

Consider the expression

Ai: Ao: ... : K 1 2 n

By fully parenthesizing this expression, we can obtain (possibly)

different values. A fully parenthesized expression can be transformed into

its post-fix expression by taking the fully parenthesized expression, removing

the left parentheses and colons (:), and replacing the right parentheses by

slashes (/). For example, if n = 5 the fully parenthesized expression

((Xt: X 2): ((Xa: X 4): X5))

- 83 -

In post-fix form is

xix2/x3x4/x5//. We can evaluate a fully parenthesized expression by examining its post­

fix expression and using a stack. As we scan the post-fix expression from

left to right, we push the elements X. onto the stack. If a slash is

encountered, we take the top two elements of the stack, perform the division

of the second stack element by the first, and push the result back on the top

of the stack. After the expression is scanned, the item on top of the stack

is the value of the fully parenthesized expression. Notice that the final

algebraic expression has both a numerator and denominator, since there is

always a last slash.

Also note that X± is necessarily in the numerator. Also, for

t = l,2,...,n-l, X„ and X. , i are both in the numerator or both in the i l+l

denominator of the fraction if the number of slashes between X. and X.Bi is i i+l

odd; otherwise X. and X.Bi are in different parts of the fraction. To see i i+l *^

this, notice that when X. Is scanned It is put on the stack. An even number

of slashes between X. and X.ai puts X. In the numerator of the result which is i i+l r i

placed on top of the stack; while an odd number puts X. in the denominator of

the result sitting on top of the stack. Now when X. - Is scanned and put on top of the stack, It is in the numerator of the result on top of the stack.

This element X. . will remain in the numerator of a stack element until a i+l

slash combines it with the result on the stack containing X.. If there are an

even number of slashes between X. and X.si, X. and X.f1 end up in different i i+l i i+l

parts of the fraction; but an odd number of slashes puts X. and X . in the

same part of the fraction. Finally, this relationship Is preserved throughout

the course of the remaining stack operations.

From the above discussion, X± is always in the numerator and X2 is always

In the denominator of the result. Also the value of XiX2X3/X4/. . .X // Is

A1A3A4...X

and the value of X1X0/X3/X4/...X / Is ri

- 84 -

81

Thus for n = 8 with X* = 1,....X8 = 8, the maximum value, from (1), is TT" and

the minimum value, from (2), is ~y%

We next construct inductively, for n > 2, a set S of post-fix

expressions s so that evaluating s gives a 1-1 correspondence with the

collection of algebraic quotients which arise from fully parenthesizing

Xt: X2: ... : X . It will be evident that JS I = 2 . With n = 2 we set

Xi S>2 = {XiX2/}. Note that the only possible quotient Is - establishing the

claim for n = 2. Now suppose S has been constructed. Note that every

element s of S ends with at least one slash. For s € S let s denote the n n

string that results when the last slash of s is removed. Form S . by

collecting expressions of the form sX V and s X +1//- Clearly

|S - I = 2 . For i = 1,2,... ,n - 1 the number of slashes between X, and

X. - has not changed. Thus different values of s give distinct algebraic

expressions. Comparing sX -/ and s X ~// we note that the parity of the

number of slashes between X and X ... is different so that X and X ... are in n n+1 n n+1

the same part of the evaluated expression for one case and in different parts

for the other. Therefore different post-fix expressions In S . give

distinct algebraic quotients. To see that all allowable algebraic expressions

E arise, notice that each expression is determined by a choice of a subset

from {X3,...,X A to act (with X±) as numerator. This is possible In 2

ways so all are accounted for by evaluating post-fix expressions in S -.

When we set Xt = 1,X2 = 2, . . . ,X8 = 8 and n = 8 we must ask when different

post-fix expressions give rise to the same rational. This can only occur

because of a product of some elements of the numerator being equal to the

product of some elements of the denominator. This does not give rise to

duplication if one of the numbers involved is 1 or 2 since X± = 1 must be in

the numerator and X2 in the denominator. Also 5 and 7 obviously cannot be

Involved. This leaves 3*8 = 4*6. There are 2 2 possibilities with 1, 3, 8 In

the numerator and 2, 4, 6 in the denominator corresponding to the choice of

- 85 -

subset of {5,7} for the numerator. These are also realized by post-fix

expressions giving X49 X6 in the numerator and X3„ XQ in the denominator.

Thus the correct accounting for (b) is 2 6 - 2^ = 60 different numerical

values. *€ K X

P R O B L E M S

S^rohlem proposals and solutions should he sent to the editor, whose ocWieaa appears on the float pag,e of this Issue. Proposals should, whenever posslhle, he accompanied hy a solution, references, aad other Insights which ate likely, to he of help to the editor, sin asterisk (*) after a numher Indicates a prohlem submitted without a solution.

Original prohlems wie particularly' bought, Ait other Interesting, prohlems may, also he acceptahle provided they, are not too well known, mid references one plwea as to their provenance. Ordinarily,, If the originator of a prohlem can he located, It should not he submitted hy, somehody, else without his or hen, permission.

So facilitate theln, consideration, you/i solutions, typewritten on, neatly handwritten on signed, separate sheets, should preferably, he mailed to the editor hefore Ocloher 1, 1987, although solutions received after that date will also he considered until the time when a solution lb pahllshed.

1221. Proposed by D.S. Mitrinovic and J.E. Pecaric, University of

Belgrade, Belgrade, Yugoslavia.

Let u8 u9 ID be nonnegative numbers and let 0 < t < 2. If a9 bs

c are the sides of a triangle and if F is its area, prove that

It —7—(be) + — T ~ ( c a ) + —^—fob) > =•

[See Solution II of Crux 1051 [1986: 252].]

4F

v5

1222. Proposed by George Szekeres, University of New South Wales,

Kens ington, Aus trai ia.

Evaluate the symmetric n x n determinant D in which

d. . rt = d. rt . = -1 for I = lt...,n - 2t d. . = 1 otherwise. Also evaluate D I 9I+2 i+2,i tj n

in which d. . = - d. . for i j* f, d.. = d... [See the solution to Crux 1033

[1987: 90].]

- 86 -

1223. Proposed by E.S.M. Coxeter, University of Toronto, Toronto,

Ontario.

(a) Show that, if all four faces of a tetrahedron in

Euclidean space are right-angled triangles, there must be two vertices at each

of which two right angles occur. In other words, the tetrahedron must be an

orthoscheme (Coxeter, Introduction to Geometry, Wiley, New York, 1969, p. 156).

(b) Show that this also holds in hyperbolic space but not in elliptic

space.

1224. Proposed by George Tsintsifas, Thessalonihi, Greece.

AtA2A3 is a triangle with circumcircle 0. Let x± < X± be the

radii of the two circles tangent to AiA2, AiA3, and arc A2A3 of Q. Let x2,

X2, x3, X3 be defined analogously. Prove that:

3 x. (a) 2 y- = 1 ;

1=1 Ai

3 3 (b) 2 X > 3 2 x. > 12r,

1=1 t 1=1 l

where r is the lnradius of AAtA2A3.

1225. Proposed by David Singmaster, The Polytechnic of the South

Bank, London, England.

What convex subset S of a unit cube gives the maximum value for

V/A, where V is the volume of S and A is its surface area? (For the two-

dimensional case, see Crux 870 [1986: 180].)

1226. Proposed by Hidetosi Fukagawa, Yokosuha High School, Aichi,

Japan.

Let ABCD be a quadrilateral inscribed in a circle, and let 0±,

02, 03, 04 be the inscribed circles of triangles BCD, CDA, DAB, ABC

respectively.

(a) Show that the centers of these four circles are the vertices of a

rectangle.

(b) Show that rt •*- r3 = r2 + r4, where r. is the radius of 0..

1227. Proposed by Stanley Rabtnoioltz, Alilant Computer Systems Corp.,

Littleton, Massachuse11s.

- 87 -

Find all angles 0 in [0f2ir) for which

sin 0 + cos 0 + tan 0 + cot 0 + sec 0 + esc 0 = 6.4,

1228. Proposed by J. Garfunkel, Flushing, New York and C. Gardner,

Austin, Texas.

If QRS is the equilateral triangle of minimum perimeter that

can be inscribed in a triangle ABCS show that the perimeter of QRS is at most

half the perimeter of ABCB with equality when ABC is equilateral.

1229. Proposed by Edward T.fl. ¥angs Wilfrid Laurier University,

Waterloo, Ontario.

Characterize all positive integers a and b such that

(a.b)Ca'b] i [a,b]<a'b>

and determine when equality holds. (As usual, (a,b) and [a,b] denote

respectively the g.c.d. and i.c.m. of a and b.)

1230. Proposed by Jordi Dou, Barcelona, Spain.

Let ABCD be a quadrilateral inscribed in a circle Q. Let

P = AC fl BB and let s be a line through P cutting AD at E and BC at F. Prove

that there exists a conic tangent to AD at E, to AC at F, and twice tangent to

0.

tt K X

S O L U T I O N S

Mo pads-lem la e^ea permanently closed. Jhe edlto% mill alway>& 4e pleaded to consider $o% publication new &olution& on, nem ln&lght& on pa&t t>iob>lem&.

1032. [1985: 121] Proposed by J.T. Groenman, Arnhem, The

Netherlands.

The following formulas are given in R.A. Johnson's Advanced

Euclidean Geometry (Dover, New York, 1960, p.205):

JH2 = 2p2 - 2Rr and OH2 = R2 - 4Rr,

where, in Johnson's notation, 0, I, fl, R, p, r are the circumcenter, incenter,

orthocenter, circumradius, inradius, and inradius of the orthic triangle,

respectively, of a given triangle. Johnson claims (at least tacitly) that

- 88 -

these formulas both hold for all triangles. Prove that neither formula holds

for obtuse triangles.

I. Editor's Comment.

There were two submissions (other than the proposer's) concerning

this problem, one agreeing with the proposal and one against. After sitting

on this problem for some time, wondering what to do about it, I've decided

simply to print the two opposing views, both of which I suspect have some

validity, and let the readership decide. Perhaps someone out there would care

to contribute the definitive treatise on the subject.

II. Comment by Waither Janous, Ursulinengymnasium, Innsbruck, Austria.

It should be noted that the same error also appears in [1], page 41.

It basically stems from an uncareful application of the formula

r = 2R cos A cos B cos C

which is only valid for acute triangles. In [1], page 27, for obtuse

triangles the following formulae are derived. If A > TT/2, then

s' = 2R sin A cos B cos C

R2

F* = - ~- sin 2A sin 2B sin 2C

where s' and F' denote respectively the semiperimeter and area of the orthic

triangle. Therefore, F'

r = —r = -2R cos A sin B sin C s

for obtuse triangles with A > ir/2.

Reference-

[1] E. Donath, Die merkwirdigen Punkte und Linien des ebenen Dreiecks,

Berlin, 1969.

III. Comment by Leon Bankoff t Los Angeles, California.

The proposer claims that Johnson claims (at least tacitly) that the

two formulas involving the orthocenter H hold for all triangles. Johnson does

nothing of the sort. He claims explicitly that these formulas were used by

Feuerbach in the mainly algebraic proof of his celebrated theorem. Feuerbach

claims rightfully that

r = 2R cos A cos B cos C

(Johnson, 299g, page 191) so that the formulas for IH2 and OH2 require a

negative value for r when one of the cosines is negative, as would be the case

- 89 -

In obtuse triangles.

M M K

1033? [1985: 121; 1986: 207] Proposed by f.R. Utx, University of

Missouri-Columbia.

Let D be any symmetric determinant of order n in which the

elements In the principal diagonal are all l's and all other elements are

either l9s or -l's, and let D be the determinant obtained from D by n n

replacing the non-principal-diagonal elements by their negatives. It is easy to show that B2B2 = 0 for all D2 and B3B3 = 0 for all D3» For which n > 3 is

It true that D W = 0 for all D ? n n n

Solution by G. Szefeeres, Uniuersity of Neio South Wales, Kensington,

Australia.

As suggested by the Editor [1986: 211], the problem Is Indeed do-able.

The answer Is of course that for every n > 4 there exists a B with D B / 0. J n n n

What we need Is a sequence of determinants B which can be evaluated directly.

The most obvious examples are symmetric "clrculant" determinants B = (d. .) In n tj

which d. .t . = d., . , = -1 for i = 1,...,n and for j In some given subset J 1fi+J i+J,t n

of I = {1,...,[n/2]}9 all other entries of D being +1. (The convention used

here Is that if i + j > n then d, ., , = d, . . and d.. . . = d.. . _ ..) ifi+j lfl+j-n i*J*i i+j-n,l

We

want to determine J so that D D / 0. Note that D is also circulant with J n n ri n n

being the complement of J in I . It is well known (see e.g. Muir9 A treatise

on the theory of determinants, Dover, New York, I960, p.444) that the value of

D (for a given J ) Is obtained as follows: Let ui- = L H U . ....H> be the nth

roots of unity, and let

where

n-1 u . = 1 + 2

J=l

£ . = 6 . = - 1

. e .10"?

J i

i f

i =

J € J

9n

e - e , = +1 If j € J = I ~ J . n-j j J n n ^n

Then, apart from a ± sign, D is equal to IUIU. . .u , and so D # 0 If and only

- 90 -

if each u. # 0. In particular we must have u± £ 0. But u± = 0 can only

happen if n = 2m is even and

2|j I = m + 1 if m € J , 2\j I = m if m € J .

Similarly for D ^ 0 we must have u± j£ 0, where

n-1 . u. = 1 - 2 £fW

J=l J

and e. is defined as above. u± = 0 can only happen if n = 2m is even and

2lj I = m if m € J , 2JJ I = m - 1 if in C J . 1 n1 n ! n1 n

Thus the condition for u±u± £ 0 is that either n is odd, or R = 2m and

if m € J then 2\j \ £ m or m + 1, n n'

if m € J then 2|j I # m or m - 1 , n ' n1

(1)

that

Henceforth ID will always denote an nth root of unity different from 1, so

O r»—1 1 + 1 0 + 10 + . . . + 1 D = 0 .

Then u. # 0 for i = 2,...,n is equivalent to

2 (w3 + u) •*) # 1

J€J ri

n/2 10

0

if n is even and n/2 € J n

otherwise,

for w = w , t = 2,... ,n. Similarly u # 0 for i = 2 n is equivalent to

1 + 2 (IDJ + ID •*) * •]

J€J, n

n/2 w

0

if n is even and n/2 € J n

otherwise.

So a necessary and sufficient condition for D D ^ 0 is

m (WJ + ID J ) i*

n

u> or 10 - 1 if n is even and n/2 € J n

0 or -1 otherwise, (2)

n for every w £ 1 such that 10 = 1, together with condition (1) if n is even. p

Consider first the case when n = p is a prime power.

Case (i):

2 < n/2 for r > 1. We want to show that

p = 3, r > 1. Take J = {1,2}, which is admissible since

- 91 -

w + w2 + ID"2 + m"1 j* 0 or -1.

Supposing

ID -f ID2 + ID"2 + l)"1 = 0 B

then

(1 + 1D)(1 + ID3) = l4-|D4-|D 3-flD 4=0

and hence either ID = -1 or ID3 = -1, both of which conflict with ID = 1.

Also9 If

ID + ID2 4- l)"2 4- ID""1 = -1

3r

then w = 1 which with ID = 1 gives ID = 1, contrary to assumption.

Case (ii): p > 3 9 r > 0. Take J = {1}. Then ID + ID""1 = 0 gives

ID2 = -1 and so ID4 = 1, while 1 + ID + ID"1 = 0 gives ID3 = 1. Both Imply (with r

ur = 1) that ID = 1, a contradiction.

Case (iii): p = 2, r > 4. Take

n r-1

Then m = 2 € J „ and condition (1) requires that

2|jJ = 2r / 2r_1

which is so if r > 4. Therefore we want to show (2); equivalently, that

2 4 2r~"1 -2 r~ 2 -1 i D - f i D + i D " f . . e 4 - i D + ^ + . . . + in = 0 or -1 (3)

does not; hold. Now w must satisfy exactly one of the Irreducible equations

r-1 ID + 1 = 0, ID2 4- 1 = 09 ID4 4- 1 = 0 10 4-1=0.

If ID = -1 then (3) obviously does not hold unless r = 2. If ID2 = -1, then

clearly (3) does not hold unless r = 2 or 3. Suppose then that ID 4 - 1 = 0

for some 1 < j < r - 1. Then If (3) holds,

2 4 2 -2^ -1 ID4-ID 4- ID 4- ... 4- ID 4-|D + ... + H)

Is equal to an integer, giving for ID an equation of degree 2J with integer

2j coefficients and distinct from ID 4-1=0, which Is Impossible.

Case (iv): n = 16. Take J = {1,3,4}, and consider

S = ID 4- ID3 4- D 4 4- ID"4 4- M>"3 4- ID"1 .

- 92 -

I f ID = - 1 t h e n S = - 2 , i f w2 = - 1 t h e n

S = (ID + H)" 3 ) ( l + ID 2 ) + ID4 + ID"4 = 2 ,

i f ID4 = - 1 t h e n

S = ( I D - 1 + I D " 3 ) ( I D 4 + 1) + ID4 + ID"4 = - 2 ,

and i f ID8 = - 1 t h e n

S = ID + ID3 - ID5 - ID7

which is certainly not an integer. Hence condition (2) is satisfied in all

four cases. Condition (1) is satisfied too since \j 1 = 3 . n

Thus we have a cyclic D with D D / 0 for all prime powers except 2, 3,

4 and 8.

Now suppose we have already found a suitable J, for some proper divisor d

of n = dq, q > 1. Let m = 1; then wq is a dth root of unity, and we may

take

Jn = qJd = <qxlx € Jd>

to satisfy condition (2) for every ID such that ID = 1 . Furthermore n/2 € J

if and only if d/2 € J, , and since in Case (iii) we had 2|j [ < n/2 and in

Case (iv) we had 21J | < n/2 - 1, condition (1) will also be satisfied by the

above construction. It follows that there exists a cyclic D with D D & 0 n n n p s

for every n which has a prime divisor p > 3, or for n = 2 3 with s > 1, or r* p

for n = 2 or n = 2 *3 with r > 3. So we have settled all cases except n = 2,

3, 4, 6, 8, 12, 24. For 2, 3, 4 and 6 the problem has already been settled by

previous solvers, so the only cases left are 8, 12 and 24.

To produce a suitable example for these values, take a D with d. . ~ =

d..o . = -1. t = l,...,n - 2, all other entries being +1. These D and the i+z,i n

corresponding D can be evaluated explicitly and give non-zero values whenever

n = 0 or 8 (mod 12) [see Problem 1222, this issue]. In the particular cases

of 8, 12 and 24 the values are easily calculated to be

D8 = 5J = -256,

D12 = -3-212, 577= -2 1 2,

D24 = -5-224, DaT = -3-224.

Neither of these methods gives an example for n = 6.

M * *

- 93 -

1077. [1985: 249] Proposed by Jack Garfunkel, Flushing, ff.Y.

For i = 19293 let C be the center and r the radius of the

Halfatti circle nearest A in triangle AiA2A3o Prove that

AtC.-AaCa-AaCa > (rl + ra + rg)° ~ 3rxr8r9

When does equality occur?

Editor's comment-

There have been no solutions to this problem submitted as yet.

1081. [1985: 288] Proposed by Loren C. Larson, St. Olaf College,

florthfield, Minnesota.

For a given integer b > 1, evaluate

( K M - • (The floor of x9 denoted by [xj, is the largest integer < x; and the

ceiling of x9 denoted by [x], is the smallest integer > x.)

Solution by Zui Margaiiot, Grade 11, A.B. Lucas Secondary School, London,

Ontario.

Let9s denote

C(x) = M

D(x) = logb M and

E(x) = log. M x

Since C(x) = 1 for x > 1, we observe that D(x) = 0 for all x > 1.

Therefore all we have to consider is 0 i x < 1.

For x = b""1^""2, . . . ,b"~r\ E(x) = l,2,...,n respectively, since

= bk and llog^I = h. Thus

for b~* < x < 1, E(x) = 0,

for b"2 < x < b"1, E(x) = 1.

for b~3 < x < b"29 E(x) = 2, etc,

which enables us to sketch E(x):

fb - * i . i.

- 94 -

5 -

4 -

^ « o •

9 -Z "

1 -1 "

i s

- - j 8 1 i i

R4

i i i i i

R3

1 I l b<

R2

"1 I S

1

1 I 1 b 3

i §

i

1 ! 1 b 2

I s S

I

I 1 b

The area to be calculated will be divided into rectangles Rl9R2>... as

shown. Hence 00

00

I E(x)dbc = Z A 0 1=1 t

where A. is the area of R.. Since A. = b and b > 1, i i t

£K M x b b 2 b 3

b

*-h 1 "

Also solued by FRANK P. BATTLES, Massachusetts Maritime Academy, Buzzards

Bay, Massachusetts; RICHARD A. GIBBS, Fort Lewis College, Durango, Colorado;

RICHARD I. HESS, Rancho Paios Verdes, Calif ornia; WALTHER JANOUS,

Ursuiinengymrtasium, Innsbruck, Austria; FRIEND H. KIERSTEAD JR. , Cuyahoga

Falls, Ohio; LEROY F. MEYERS, The Ohio State University, Columbus, Ohio; PAT

SURRY, Grade 11, A.B. Lucas Secondary School, London, Ontario; and the

proposer.

Gibbs (and probably others) observed that b need not be an integer but

only a real number greater than 1.

* M *

1082. [1985: 288] Proposed by 0. Bottema, Delft, The Netherlands.

The midpoints of the edges A3A4, A4Ai, AiA2 of a tetrahedron

A1A2A3A4 are B2, B3, B4, respectively; £± is the line through k± parallel to

A2A3; and £2, i39 £4 are the lines A2B2, A3B3, A4B4, respectively.

- 95 -

(a) Show that the four lines £. have two conjugate Imaginary

transversals t and i'.

(b) If S. Is the intersection of t and £., and SI that of t1 and £.

(t = 1, 2, 3S 4), show that S. and S! are equianharmonic quadruples of points.

Solution by the proposer.

We introduce homogeneous barycentric point coordinates x, y, z9 u> with

respect to the tetrahedron. Then A± = (1,0,0,0), etc. The plane at infinity

has the equation x + y + z + u) = 0; hence the point at infinity on A2A3 is

(0,1,-1,0). Furthermore B2 = (0,0,1,1), B3 = (1,0,0,1), B4 = (1,1,0,0). For

the lines £. we obtain the following (Pliicker) line-coordinates p. .:

' l

2

3

P i 2

1

0

0

0

P l 3

-1

0

- 1

0

P i 4

0

0

0

-1

P 3 4

0

0

1

0

P 4 2

0

- 1

0

1

P 2 3

0

1

0

0

Two lines p. . and q. . intersect if

V±2Q04 + Pi3<342 + Pi4<323 + P34<2i2 + P42<l i3 + P23<2i4 = 0 .

Hence q. .intersects the four lines £. if tj t

Q34 "" <242 = 0» -Qi3 + Qi4 = 0, -q42 + q12 = 0, -q23 + q13 = 0.

Hence

<?34 = ^42 = <2l2 • *Ji3 = Qi4 = ^23 •

or

qi2 = u, q13 = us q14 = u, q34 = u, q42 = u, q23 = u.

The fundlamental equation 2 qi2*3ta4 = 0 gives us

u2 -i- uu + u2 = 0,

and therefore

u/v = |(-1 ± i ^ ) .

These numbers are the Imaginary cubic roots of 1. Hence the resulting lines t

and t8 aire conjugate imaginary transversals.

If we introduce the number w = ={1 + i^J), with CJ3 = -1, our roots are -w

and -co"1. For one of the transversals, t say, we obtain

q12 = -d), q13 = 18 qi4 = lf q34 = -6), q42 = -GJ, q23 = 1,

- 96 -

with or2 - G) + 1 = 0 . The intersections of t and 2. are those of t and the

planes w = 0, x = 0, y = 0, z = 0. Hence

S± = (l.ai.-G),0). S 2 = (0,-G),l,l), S 3 = (w, O.l.co), S 4 = (1,1,0,o).

A parameter representation of i is given by

x = X, y = 6)(X - 1), z = 1 - GJX, ID = 1,

where Slt S2, S3i S 4 correspond to X = w, X = 0, X = l , X = &T1. Hence the

cross ratio (Si ,S2,S3,S4) is equal to (^.O.I.GT1) = aT1, which shows that the

quadruple of points is equianharmonic; the same holds for the set S±, 82. S3,

Si.

1083? [1985: 288] Proposed by Jack Garfunfeel, Flushing, N.Y.

Consider the double inequality

-rr 2 sin A i Z cos -—<z— < -77 2 cos - ,

where the sums are cyclic over the angles A, B, C of a triangle. The left

inequality has already been established in this journal (Problem 613 [1982:

55, 67, 138]). Prove or disprove the right inequality.

Solution by Vedixla N. Marty, Pennsyluania State Uniuerslty, Mlddletoiun,

Pcnnsyluania.

Let

a = —-—• , p = — - — , <y = 2 • K " 2 ' 2

Then a, /?, nr represent the angles of a triangle. Moreover

2 cos — = - — = E cos(j3 - T ) ,

2 cos <j = 2 sin a.

Hence the inequality in question is established if we prove

2 cos(j3 - nr) < -TT- 2 sin a. (1)

Using the relations given in Crux [1982: 67] we have

v rn \ x 2 + y2 + 2x - 2 2 cos(p - nr) = *—^ ,

2 sin a = y,

where x = r/R and y = s/R, r, R, and s being the inradius, circumradius, and

semiperimeter of a triangle with angles a, j3, T. Thus (1) is established if

- 97 -

we show that

x2 4- y2 + 2% - 2

or

h§]R)4^-¥h<>. (2)

It Is well known that 0 < x £ -^ and 0 < y < -^-, and hence (2) holds, with

equality If and only If a = 0 = nr. Thus the required Inequality holds with

equality If and only if A = B = C.

Also solved by LEON BANKOFF, Los Angeles, California; and WALTEFR JANOUS,

Ursulinengymnasium, Innsbruckf Austrta.

1084. [1985: 288] Proposed by Allan Wm. Johnson Jr., Washington,

D.C.

Prove that every symmetrical fourth-order magic square can be

written as

F + ID + y

| F - x - y

1 F + x - y

F - ID + y

F - to + z

F + x - z

p _ x _ z

F + D -§" Z

— — — " •

F - ID - z

F -f- x + z

F - x + z

F + ID - z

1 F + w - y

F - x + y

F -4" x + y

F - ID - y

Solution by the proposer.

The algebraic form

F + A

F + J

I

F " M

F - B

F + B

F + K

F - L

F - C

F + C

F + L

F - K

F - B

1 F + D

F + X

F - J

F - A

( * )

- 98 -

is symmetrical, and becomes a magic square with magic sum 4F only if

A + B = -B - C (1)

K + L = -J - M (2)

A - D = -J + Jf (3)

K - L = -B + C. (4)

By (1) and (3)

A = h-B - C - J + M) (5) 2V

D = i{-B - C + J - M) (6)

while by (2) and (4)

K = |{-B + c - J - M) (7)

L = |(B - C - J - M). (8)

Comparing (*) to the square in the problem proposal yields the equations

B = -ID + zf C = -ID - zf J = -x - y, M = -x + y, (9)

and so from (5)-(8)

A = ID + y, D = ii)-y, K = x - z , L = x + z. (10)

Substitution of (9) and (10) into (*) produces the square in the problem

proposal. Also, (9) gives

u> = -|{B + C), z = |<B - C), x = -|{J + M), y = |(-J + M)

whose substitution into the square in the problem proposal, using (5)-(S),

reproduces (*). Hence the square in the problem proposal is universal, that

is, it is a form in which any symmetrical fourth-order magic square can be

written.

M M *

1085. [1985: 289] Proposed by George Tsintsifas, Thessalonihi,

Greece.

Let o = A~A....A be a regular n-simplex in R , and let ir. be n 0 1 n & ^ i

the hyperplane containing the face o - = A^A.... A. .A. ... .A . If B. € TT. J* * & n-1 0 1 i-1 t+1 n i l

for i = 0,1 n, show that -*i v n + 1

2 IB B | I ^ ^ e .

where e is the edge length of o .

- 99 -

Solution by the proposer.

Let 0 be the origin and let X = OX for a point X„ Using barycentric

coordinates we have

n 1 , ~ ik R

k=0

n where p.. = 0 and 2 p., = 1. Thus

11 k=0 L

A.A,2 - B4B. • A .A. = (A7 - AT + S7 - B T ) ( A 7 - IT) J t i j it x i J i 3 i J

= ( t - t + t - B?)(A^ ~ A?) v i j i j J X x j /

n n

,*,».!. V - V *t - . i * * ^ (A. - A.)

n n. (A. - A.)

n n = 2 p . , A .A, • A.A. + 2 p . , A.A, • A.A..

k=0 l h J k J l fcdO Jfe l fc l J

But JA A I = e and Zk A AA =60 for distinct r, s, t; thus (1) becomes 1 r s' r s t v '

(1)

n n e 2 - B.B. • A.A. = e 2 cos 6 0 - 2 p . , + e 2 cos 60 * 2 p . .

1 J J L k=0 I f e k=0 J k

k/ i f j fert, j

so t h a t

Using

we have

and hence

B.B. • A.A. = s K p . . + p . . ) .

o ; . y ^ < IO;I . irri = eio;i

I O ; I > icptj + p^).

2 | 0 ? | > | 2 (p + p ) 0<i<j£n J 0<i<j<n l J J

= §<» + 1).

- 100 -

1086. [1985: 289] Proposed by M.S. Klamkin, University of Alberta.

The medians of an n-dimensional simplex A0At...A in K

intersect at the centroid G and are extended to meet the circumsphere again in

the points B09B±,...,B , respectively.

(a) Prove that

A0G 4- AtG -f ... + A G £ BQG + B±G + ... 4- B G. u 1 n ° 1 n

(b) Determine all other points P such that

AQP * AtP + ... + A P £ BQP + B±P + ... 4- B P.

I. Solution and generalization of (a) by George Tslritslfas,

Thessalonlfel, Greece.

Let P be a point which is inside the simplex and is also in the

(closed) ball with diameter 0G, where 0 is the circumcenter of the simplex.

Let B. denote the intersection of A.P with the circumsphere. We prove that

AQP + A±P 4- ... 4- A P £ B0P 4- BiP 4- ... 4- B P. u x n u 2 n

In particular, putting P == G we get (a).

From the triangle OPG we get

(0G)2 > (OP)2 4- (PG)2,

so

(n 4- 1)2(0G)2 > (n 4- 1)2((0P)2 + (PG)Z). (1)

Leibniz's formula for the polar moment of inertia asserts that for any point

X, n.4-1 l,n4-l 2 X.(A.X)2 = (QX)2 4- 2 X.X.(A.A.)2

1=1 L Z j<i L J L J

where Q is a point with barycentric coordinates X-,Xp,...,X -. We put Q = G

and X = 0. Leibniz's formula is transformed into

l,n4-l R2 = (0G)2 + 2 (A A.)2/(n + I) 2, (2)

j<l J

where R is the circumradius. Therefore from (1) and (2) we have

l,n+l (n + 1)2R2 - 2 (A.A,)2 > (n + 1)2((0P)2 + (PG)2).

J<1

Hence

- 101 -

l,n+l (n + 1)2(R2 - (OP)2) Z (A A,) 2 + (n + 1)2(PG):

j<t l J

or l,n+l

(n + 1)2J(P) > (n + 1)(PG)2 + Z (A.A .)2/(n+l), j<i l J

where 2)(P) is the p>ower of the point P with respect to the circumsphere.

Putting X = P and Q = G in Leibniz's formula, we find

n+l l,n+l Z a2 = (n + 1)(PG)2 + Z (A.A.)2/(n + 1). 1=1 l j<t l J

where a. = A.P. From (3) and (4) we get

n+l (n + 1)2!(P) > Z a2 ,

1=1 l

(3)

(4)

or n+l

(n + 1)22J(P) > (n + 1) Z a? . 1=1 l

The well known arithmetic mean and root-mean-square inequality of the numbers

al , a2 s " * " 9Ctn+l S*ves

n+l Z a? 1=1 l

n + l

n+l Z a. 1=1 l

n + l

Hence

(n + 1)2®(P) 1 n+l Z a.

i=i l

But from the A.M.-G.M. inequality follows

n+l Z a. 1=1 l

n+l Z 1/a. 1=1 l

1 (n + 1)!

or n+l Z a. 1=1 l

n+l Z 1/a. 1=1 l

2>(P) > n+l Z a. 1=1 l

or n+l Z 1/a 1=1 l

n+l 2>(P) > Z a..

1=1 l (5)

- 102 -

Now, letting b. = B.P, and bearing in mind that a.b. = 2J(P) for all i, from

(5) we immediately conclude

n+1 n+1 Z b. > S a. . • 1 l ~ 4 1 l

i=l 1=1

II. Comments by the proposer.

There is equality in part (a) if and only if all the JA. | are equal,

i.e., if and only if the centroid G coincides with the circumcenter. For

n = 2 this requires the triangle to be equilateral. For n > 3, the simplex

need not be regular, although for n = 3 the tetrahedron will have to be

isosceles.

It is to be noted that the inequality for the special case n = 2

corresponds to Problem E2959, American Mathematical Monthly 89 (1982) 498,

proposed by Jack Garfunkel.

It is also to be noted that more generally if one is given n points

AltA2.....A lying on an m-dimensional sphere and with centroid G, then

A±G + A2G + ... + A G < B±G + B2G + ... + B G

where B. is the point where A G extended intersects the sphere.

Also solued. (part (a)) by the proposer.

A solution to part (a), by P. Boente, appears in the Monthly, May 1985,

page 360, as a generalization of problem E2959 mentioned aboue.

Part (b) remains open. Tsintsifas' generalization of (a) presumably is

not what loas meant in (b).

* * *