Hong Kong Maths Competition questions.

16  Can't see a neat geometric solution to this, but here's a co-ordinate geometry one.

Set the triangle up so that B is at the origin, BA lies along the x-axis and BC along the y-axis.

Then, AX^2 + BX^2 + CX^2 = {(x-3)^2 + (y-0)^2} + {(x-0)^2 + (y-0)^2} + {(x-0)^2 + (y-4)^2}

                                             = 3x^2 + 3y^2 - 6x - 8y + 25

and, (completing the square on both x and y),

                                             = 3(x - 1)^2 + 3(y - 4/3)^2 + 50/3,

so the minimum will be 50/3 when x = 1 and y = 4/3.

Tiggsy

7.   The smallest  integer  Pythagorean triple is when AD  = 3 , AB  = 4 and DB  = 5

Perimeter of ABD  =  12

Area of ABCD  = AB * AD  =  [ 3 * 4 ]    = 12

Scaling ABCD  up by  a factor of 2  produces   AD  = 6, AB  = 8  and DB  = 10

Perimeter of ABD  =  24

Area  of ABCD  = [ 6 * 8 ]  = 48  = minimum value of S  for integer sides

11)

M=1! x 2! x 3!........10! =6,658,606,584,104,736,522,240,000,000.

[6,658,606,584,104,736,522,240,000,000]^1/3

=1,881,313,269 Perfect cubes!!.

Cphill: Please check this! Is there a shortcut to this? Thanks.

8.

There are 2016 different proper fractions. The sum of  any 2015 of them is a proper fraction with denominator 2017 when expressed in its simplest form. Find the sum of these 2016 proper fractions.

CPhill: Does this form an arithmetic series with the first term=1/2017..........and the last term=2016/2017, the common difference =1/2017, number of terms =2016???. If so, then it sums up to =1,008???.

19. Find the highest(greatest) common factor of 3^2046 - 1 and 3^2016 - 1.

Wolfram/Alpha gives =728. I have NO IDEA how they got it short of factoring each number individually, which too much to expect for such large numbers!!!???. Can anybody explain it? Thanks.

15. In ∆ ABC, D is a point on AB such that CD bisects ∠ACB, ∠A = 2∠B, AC = 11 and AD = 2

      Find the length of BC.

By Euclid......BC/DB  = 11/2  = 5.5   so  BC  = 5.5DB

And  < A  = 2 <B 

So

sinA/ BC  = sin B /11

sin (2B) /BC  = sin B / 11

[2sinBcosB] / BC  = sinB/ 11

2cosB/BC  = 1/11 

22cosB = BC

cosB  = BC/22 = 5.5DB/22   =  DB/4  

Using the Law  of Cosines....we have

11^2  = (BC)^2 + (DB+2)^2  – 2(BC)(DB+2)cosB

11^2  =  (5.5DB)^2 + (DB+2)^2 – 2(5.5DB)(DB+2)(DB/4)

let x  = DB

121  = 30.25x^2 + (x + 2)^2  – 2.75x^2 * (x + 2)

121 = 30.25x^2 + x^2 + 4x + 4 – 2.75x^3 – 5.5x^2

2.75x^3 - 25.75x^2  – 4x + 117  = 0

275x^3 – 2575x^2 –400x + 11700  =  0

Solving  this for x  produces two  possible positive  solutions   x = 26/11  or x = 9

So  DB  = 26/11   or DB  = 9     →   BC  = 5.5(DB)  = 5.5(26/11) = 13  

Or

BC  = 5.5(9)  = 49.5

But  BC  cannot  = 49.5  because cosB  = BC/22  = 49.5/22  which is impossible

So.......BC  = 13

19)  Solution for Greatest Common factor of ((3^2046) - 1) and ((3^2016) - 1)

General formula

\(\gcd(b^m-1,b^n-1)= (b^{\gcd(m,n)}-1). \\ \gcd(3^{2046} - 1, 3^{2016}-1)= (3^{\gcd(2046,\;2016)}-1)\\\)

\(3^{2046} = 2046*log(3)\ \& \ 3^{2016} = 2016*log(3) \text { |Note that exponents become factors: } \\ \gcd (2046), (2016) = 6 \text {| Find the GCF of the exponents. }\\\)

\((3^{6}) -1 \text {| Use this factor as an exponent of the base and subtract the one. }\\ (3^{6}) -1 = 728 \text { |Solve in base ten. }\\\) \(\text {Descriptive mathematics for the example. }\\ (3^{2046}- 1) = ((3^{6})-1)((3^{2040})+(3^{2034})+(3^{2028})+(3^{2022})+ (3^{2016}) . . . +(3^{6}) + 1)\\ (3^{2016} - 1) = ((3^{6})-1)((3^{2010})+ (3^{2004})+(3^{1998})+(3^{1992})+ (3^{1988}) . . . +(3^{6}) + 1)\\ \)

(My mentor – Lancelot Link, taught this method to me.)

18. Figure: http://imgur.com/a/euSF9

In the figure, ABCD is a rhombus. E is a point on BC such that AE⊥BC, cos B = 4/5 and EC = 2. P is a point on AB. Find the least value of PE + PC.

I worked my  way through this....[ well, almost ].....I believe that P minimizes PE + PC  when

P = (256/45 , 64/15  )

To start, we need to find the coordinates of E  and C.....let B  = (0, 0)

And since cosB = 4/5, then BE / AB  = 4/5  → BE  = (4/5)AB  and EC = 2

And BC  = AB....so

(4/5)AB + 2  = AB

2  = (1/5)AB  → AB  =10 = BC......so BE  = 8

So  E  = (8,0)   and  C  = (10,0)....and ABE  is a 6, 8, 10 right triangle, with A = (8,6)....therefore, the tanB  = 6/8  = 3/4....and for any point x on P, y =(3/4)x

So......we have this

D  = PE + PC

D = √[ (x - 8)^2  + [(3/4)x]^2 ]  + √[ (x - 10)^2  + [(3/4)x]^2 ]

D = √[ (x^2 - 16x  + 64  + (9/16)x^2 ]  + √[ (x^2 - 20x  + 100  + (9/16)x^2 ] 

D = √[ (25/16)x^2 - 16x  + 64  ] + √[ (25/16)x^2 - 20x  + 100  ]

So.....taking the derivative, we have

D'  =  [ (25/8)x - 16] /  √[ (25/16)x^2 - 16x  + 64  ]  + [ (25/8)x - 20] /  √[ (25/16)x^2 - 20x  + 100  ] 

Set this = 0  and solve for x...[ I let WolframAlpha do the heavy lifting here.....if I have time  tomorrow, I might  see if I can go through the messy Algebra associated with this....but....I'm not promising anything.........I know you're not supposed to use CAS help....but I'm not in the competition....!!! ]

Anyway....setting to 0 and solving for x  should result  in 

P = [x, (3/4)x ]  = (256/45 , 64/15  )

MaxWong

I think we should use Euclidean Algorithm for that.

Wrong, Max......

GA  applied the [ lesser known] "Lancelot Link Falling Off A LOG Algorithm"  to solve this.....

Those chimps of his don't monkey around.......

P.S -  Rumor has it that  Sir LaTex  also provided some technical assistance.........

MaxWong

Wow. Any other methods? My classmates do not know calculus!!

I'm sure there is a geometric solution....but....I don't know how to go about it.....

18      Here's a geometrical method.

Begin by finding the length of BE.

Let BE = 4L, then BA = 5L, but the figure is a rhombus, so 4L + 2 = 5L, so L = 2 in which case, BE = 8.

Now, take a line through E perpendicular to AB and extend it to the other side of AB to a point E* so that E and E* are equidistant from the line AB.

CE* then cuts AB at the required point P, (for minimum CP + PE).

To see that this is the case, see that PE = PE* so that CP + PE = CP + PE* = CE*, and that if P moves either way along AB, then CP + PE will become bigger.

(Alternatively the line perpendicular to AB could be taken through C, extended to C*. C*E (equal in length to CE*) then gets you the same point P as earlier).

The length of CE*  can be found by some simple trig, tanB in the smaller triangle at B followed by the cosine rule in the triangle CEE*.

I got the answer as 2sqrt(745)/5, approx 10.9179.

Tiggsy

17

Suppose that the 'heights' of the rectangles in the top row are A, those of the second row B and those of the third row C, and suppose that the widths of the rectangles in the  first column are D those in the second column E and those in the third column F, then we can write down the set of nine equations,

(first column)

(1) A + D = 5 , (2) B + D = s/2 ,  (3) C + D = 8,

(second column)

(4) A + E = t/2 , (5) B + E = 5 ,  (6) C + E = q/2,

(third column)

(7) A + F = 8 ,  (8) B + F = r/2 ,  (9) C + F = p/2.

Eq(3) - Eq(1) : C - A = 3,

Eq( 9) - Eq(7): C - A = p/2 - 8, 

so p/2 - 8 = 3 , p = 22.

Eq(6) - Eq(4): C - A = q/2 - t/2 = 10 - t/2 (since q = 20) = 3, so t = 14 etc, etc.

Tiggsy

BTW, I can't read the whole of questions 8, 9 and 11, the @@ Interest questions box gets in the way.

Anyone else have this problem ?

Tiggsy

Here are 8, 9 and 11 questions:

8.

There are 2016 different proper fractions. The sum of  any 2015 of them is a proper fraction with denominator 2017 when expressed in its simplest form. Find the sum of these 2016 proper fractions.

9.

It is given that the sum of any 2 digits of a 3-digit number is equal to a multiple of the remaining digit. How many such 3-digit numbers are there?

11.

For positive integer n, denote 1 x 2 x 3 x ... x n = n!. If M = 1! x 2! x ... x 10!, find the number of positive factors of M whose are perfect cubes.

I hope you can see them.

Thanks, Tiggsy  for those excellent answers.....17  wasn't  clear to me, but I now see that it  isn't that difficult to solve......the geometric proof of 18 was very  nice.....!!!!!!