Solutions for May. 3 x + 7 = 4 x x +

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Mercy Hamilton
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1 Solutios for May 493. Prove that there is a atural umber with the followig characteristics: a) it is a multiple of 007; b) the first four digits i its decimal represetatio are 009; c) the last four digits i its decimal represetatio are 009. Solutio. First, we show that there is a positive iteger for which is a multiple of 007. For otherwise, if oe of the umbers of the form with = 0, 1,,, 006 were a multiple of 007, the, by the Pigeohole Priciple, there must be two of them cogruet modulo 007. Hece, their differece, a umber of the form k m) 10 4 with 0 k, m 006 would be a multiple of 007, a impossibility. Now use this umber to compose the umber M = Sice M = k = k k ) for 5 k 8, M is a multiple of 007. Sice it has all three of the desired characteristics, the problem is solved a) Fid all real umbers x that satisfy the equatio 8x 56) 3 x = 30x x 97. b) Fid all real umbers x that satisfy the equatio x + 3 x + 7 = 4 x Solutio. a) We must have x 3. The equatio ca be rewritte Let y be positive with y = 3 x. The Hece y =, so that x = 1. This solutio is valid. 0 = 8x 7) 3 x + x 30x = 8y + 4)y + y 3) 303 y ) + 97 = y 4 8y 3 + 4y 3y + 16 = y ) 4. b) The domai of the equatio is give by x 0. Oe solutio is x = 1; we prove that it is the oly solutio. The equatio is equivalet to Let x + 4x 3/ x + 7) 1/3 + 6xx + 7) /3 + 4x 1/ x + 7) + x + 7) 4/3 = x fx) = 4x 3/ x + 7) 1/3 + 6xx + 7) /3 + 4x 1/ x + 7) + x + 7) 4/3. This fuctio fx) is icreasig ad f1) = 80. If x > 1, the also x > x ad fx) > 80, so that x + fx) > x Similarly, whe 0 x < 1, the x < x ad fx) < 80, so that x + fx) < x Hece, there is o solutio to the equatio save Let 3. A regular go has area S. Squares are costructed exterally o its sides, ad the vertices of adjacet squares that are ot vertices of the polygo are coected to form a sided polygo, whose area is T. Prove that T )S. For what values of does equality hold? 1

2 Solutio. Wolog, let the sidelegth of the give polygo be 1. The sided polygo is composed of the regular go, squares with sidelegth 1 ad isosceles triagles with equal sides of legth 1 ad agle betwee these sides equal to Therefore π π + π ) )π + = π. T = S si π. O the other had, S is the sum of the areas of isosceles triagles, each with base 1, apex agle π/ ad height 1/) cotπ/). Hece S = /4) cotπ/), so that = 4S taπ/). Therefore T = S si π = S ta π + ta π si π ). Apply si θ = ta θ/1 + ta θ) to siπ/) to obtai that T = S ta π + 4 ta π ) 1 + ta π S ta π + 4 ta π ) 1 + ta π [ = S ta π )] ta π. Sice 3 ad the taget fuctio is icreasig, 0 < taπ/) taπ/3) = 3. so that ta π ta π/) ) = Therefore. T 4S 3+1), as desired. Equality holds whe = 3 ad the polygo is a equilateral triagle Is the hudreds digit of N = eve or odd? Justify your aswer. Solutio. Observe that However, modulo 100, ad for each positive iteger. Hece N = ) = = ) = = N = ) mod 100). Deote the hudreds digit of N by h. Sice N is a multiple of 8, the three digit umber h48 must be a multiple of 8 as well. This is possible oly if h is eve. Thus, the hudreds digit of the umber N is eve Give 4 poits i the plae with o three colliear, costruct all segmets coectig two of these poits. It is kow that the legth of each of these segmets is a positive iteger. Prove that the legths of at least 1/6 of the segmets are multiples of 3.

3 Solutio. First, we prove a lemma: Four poits are give i the plae with o three colliear. The legth of each of the segmets joiig two of these poits is a iteger. Therefore, at least oe of the segmets has a legth divisible by 3. Deote the four poits by A, B, C, D; wolog, assume that BAD = BAC + CAD. Let BAC = α, CAD = β ad BAD = γ, so that γ = α + β. Applyig the Law of Cosies to triagles ABC, ACD ad ABD, we fid that ad BC = AB + AC AB AC cos α CD = AD + AC AD AC cos β BD = AB + AD AB AD cos γ. Assume, if possible, that the legths of all six segmets AB, AC, AD, BC, BD, CD are ot multiples of 3. The AB AC AD BC BD CD 1 modulo 3, from which it follows that modulo 3. Therefore modulo 3. AB AC cos α AD AC cos β AB AD cos γ 1 AC AB AD cos α cos β AB AC cos α)ad AC cos β) 1 From the foregoig equatios, each of cos α, cos β ad cos γ are ratioal. Let cos α = p/q ad cos β = r/s, i lowest terms, where p, q, r, s are itegers. Noe of these four itegers ca be multiples of 3. The deomiators q ad s caot be multiples of 3 for they must cacel ito side legths, ad the umerators p ad r caot be multiples of 3 sice the terms cotaiig the cosies are ot divisible by 3. Hece p q r s 1 mod 3). Sice cos γ = cos α cos β si α si β, we have, from AB AD cos γ 1 ad AC 1, modulo 3, that AC AB AD cos γ 1 AC AB AD cos α cos β AC AB AD The secod product o the left side is a multiple of 3, so that q p s r q s 1. AC AB AD cos α cos β 1. This cotradicts a earlier statemet ad establishes the lemma. Let 4. There are 4) sets of four poits, so by the lemma, there are at least this may segmets whose legths are multiples of 3, coutig multiplicity some couted more tha oce). Sice each of the segmets is couted at most ) times for the sets of four poits cotaiig the edpoits of the segmet), it follows that there are at least ) 4 / ) distict segmets whose legths are multiples of 3. Sice 4) ) = 1) ) 3) 4! ) 3) = 1 6 1) = 1 6 ), there are at least as may segmets with legths divisible by 3 as oe-sixths of the umber of pairs of segmets, ad the result follows. 3

4 498. Let a be a real parameter. Cosider the simultaeous sytem of two equatios: 1 x + y + x = a 1 ; 1) x x + y = a. ) a) For what value of the parameter a does the system have exactly oe solutio? b) Let < a < 3. Suppose that x, y) satisfies the sytem. For which value of a i the stated rage does x/y) + y/x) reach its maximum value? Solutio. From the idetificatio of the coefficiets of a quadratic i terms of the sum ad product of the roots, we see that 1/x + y) ad x are the solutios of the quadratic equatio There are two optios. Optio 1. 1/x + y) = a, x = 1, so that with a. 0 = t a 1)t + a ) = t a )t 1). x, y) = Optio. 1/x + y) = 1, x = a, so that 1, 3 a ) a x, y) = a, 3 a). a) For the system to have exactly oe solutio, either the two optios produce the same pair or oly oe of the optios is possible. I the first istace, we have a = 3 ad the uique solutio x, y) = 1, 0) ad i the secod, we have a = ad the uique solutio is x, y) = 0, 1). b) Whe < a < 3, both x/y ad y/x are positive for either solutio of the system. By the Arithmetic- Geometric Meas Iequality, x y + y x with equality if ad oly if x/y = y/x. The coditio for equality is equivalet to a )/3 a) = 3 a)/a ), or a = 5/. Thus, x/y) + y/x) attais its miimum value of whe a = 5/ The triagle ABC has all acute agles. The bisector of agle ACB itersects AB at L. Segmets LM ad LN with M AC ad N BC are costructed, perpedicular to the sides AC ad BC respectively. Suppose that AN ad BM itersect at P. Prove that CP is perpedicular to AB. Solutio 1. Let m be a lie through C parallel to AB ad let AN ad BM itersect m at F ad E, respectively. Let CP ad AB itersect at D. Triagles ADP ad F CP are similar, as are triagles DBP ad CEP. Hece AD : CF = P D : P C = DB : CE. Therefor AD : BD = CF : CE. O the other had, triagles ABM ad CEM are similar, ad triagles ABN ad F CN are similar. Therefore, AM : MC = AB : CE ad BN : CN = AB : CF. However, right triagles CLM ad CLN are cogruet ad CM = CN, so that AM : BN = CF : CE. Together with AD : BD = CF : CE, this yields AD : BD = AM : BN ad AM : AD = BN : BD. 4

5 Let the altitude from C to AB itersects AB at a poit H. Sice triagles ALM ad ACH are similar, AL : AC = AM : AH. Similarly, from the similarity of triagles BLN ad BCH, BL : BC = BN : BH. By the agle-bisector theorem, AL : AC = BL : BC. It follows that AM : AH = BN : BH. Sice AM : AD = BN : BD ad AM : AH = BN : BH, it follows that D ad H divide AB iterally i the same ratio, ad so D = H. Thus, CP AB ad the statemet is established. Solutio. [J. Kileel] Use the same otatio as i Solutio 1. Let H be the foot of the perpedicular from C to AB. It suffices to show that AN, BM ad CH are cocurret. By Ceva s Theorem, this is equivalet to showig that AM MC CN NB BH HA = 1. Triagles MCL ad NCL are cogruet ASA), so that CM = CN. Triagles ALM ad ACH are similar, so that AM : HA = LM : HC. Likewise, triagles BLN ad BCH are similar; therefore, BH : NB = HC : LN = HC : LM. It follows that AM MC CN NB BH HA = AM HA BH NB CN MC = LM HC HC LM CN MC = 1. 5

Coffee Hour Problems of the Week (solutions)

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