Olympiad Practice Problems Part 3

After Part 1 and Part 2, here's Part 3, with some problems from previous Maths Olympiads.

1 Solve the following system for real triplets (x,y,z)(x,y,z) x+yz=4;x2y2+z2=4;xyz=6x+y-z=4; x^2-y^2+z^2=-4; xyz=6

2 Let A1,A2,..............,AnA_1, A_2,..............,A_n be the vertices of an nn sided regular polygon in that order. Determine nn, given that 1A1A2=1A1A3+1A1A4\dfrac{1}{A_1A_2}=\dfrac{1}{A_1A_3}+\dfrac{1}{A_1A_4}

3 Let ABCDABCD be a quadrilateral inscribed in a circle. Suppose AB=2+2\overline{AB}=\sqrt{2+\sqrt{2}} and AB\overline{AB} subtends an angle of 135135^{\circ} at the centre of the circle. Find the maximum possible area of the quadrilateral ABCDABCD.

4 Let a,b,ca,b,c be positive real numbers such that a3+b3=c3a^3+b^3=c^3. Prove that a2+b2c2>6(ca)(cb)a^2+b^2-c^2>6(c-a)(c-b)

5 Find all real numbers xx such that [x2+2x]=[x]2+2[x][x^2+2x]=[x]^2+2[x], where [n][n] denotes the greatest integer less than or equal to nn.

6 In a book of page numbers from 11 to 100100, some pages are torn off. The sum of the numbers on the remaining pages is 49494949. How many pages are torn off?

Try 'em all, and as always, please post your approaches as comments and don't forget to like, reshare, and enjoy!

Note by Satvik Golechha
6 years, 1 month ago

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3.Is the answer for 6) 33+542\frac{3\sqrt{3}+5}{4\sqrt{2}}

Souryajit Roy - 6 years ago

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  1. Rearranging first equation
xz=4yx - z = 4 - y


x2+z22xz=16+y28yx^{2} + z^{2} - 2xz = 16 + y^{2} - 8y

x2y2+z2=16+2xz8yx^{2} - y^{2} + z^{2} = 16 + 2xz -8y

With the help of second equation

16+2xz8y=416 + 2xz - 8y = -4

With the help of third equation

4+3y2y=14 + \dfrac{3}{y} - 2y = -1

From here we get

y=3,12y = 3, \dfrac{-1}{2}

Now x,zx, z can easily be found

When y=3y = 3 we get 2 pairs of (x,z)(x,z)

(2,1)(2,1), (1,2)(-1,-2)

When y=12y = \frac{-1}{2} x,zx,z comes out imaginary

Finally we get 2 solutions of (x,y,z)(x,y,z)



2) Answer is 7 can be proved with complex numbers

5) Answer is \infty put x any integer

Krishna Sharma - 6 years, 1 month ago

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Please post solution of answer 22 also. I want to see your complex numbers method. Thanks!

Satvik Golechha - 6 years, 1 month ago

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The length of side of regular polygon can be represented in terms of angle made at centre & circumradius as (proved with complex numbers)

a=2Rsin(θ2)a = 2R sin(\dfrac{\theta}{2})

Where θ=2πn\theta = \dfrac{2\pi}{n} is angle made at centre

1a1a2=1a1a3+1a1a4\dfrac{1}{a_1{a}_2} = \dfrac{1}{{a}_1{a}_3} + \dfrac{1}{{a}_1{a}_4}

12Rsin(πn)=12Rsin(2πn)+12Rsin(3πn)\dfrac{1}{2R sin(\frac{\pi}{n})} = \dfrac{1}{2R sin(\frac{2\pi}{n})} + \dfrac{1}{2Rsin(\frac{3\pi}{n})}

Now let x=πn\displaystyle x = \dfrac{\pi}{n}

1sinx=1sin2x+1sin3x\displaystyle \dfrac{1}{sinx} = \dfrac{1}{sin2x} + \dfrac{1}{sin3x}

sin2x.sin3xsinx=sin2x+sin3x\displaystyle \dfrac{sin2x.sin3x}{sinx} = sin2x + sin3x

2cosx.sin3x=sin2x+sin3x\displaystyle 2cosx.sin3x = sin2x + sin3x

sin4x+sin2x=sin2x+sin3x\displaystyle sin4x + sin2x = sin2x + sin3x

sin4x=sin3xsin4x = sin3x

4x+3x=π4x + 3x = \pi

x=π7=πnx = \dfrac{\pi}{7} = \dfrac{\pi}{n}

n=7n = \boxed{\boxed{7}}

Krishna Sharma - 6 years, 1 month ago

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@Krishna Sharma Did it the same way!

Racchit Jain - 5 years ago

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For 5, you are asked to "find all real numbers xx ...". You are not asked to find the total number of real numbers.

Calvin Lin Staff - 6 years, 1 month ago

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This one would be messy, ok so when x is integer it satifies the condition we have to check for

x = I + f

Substituting to get final result as

f2+2f(I+1)=0\lfloor{f^{2} + 2f(I +1)}\rfloor = 0

We have to make cases

When I<1,f0I < -1, f ≠ 0

No value of x satify the condition

When I=1I = -1 we will get 0<f<10 < f < 1

When I=0I = 0 we will get 0<f<21 0 < f < \sqrt{2} - 1

As we go on increasing value of 'I' the range value of 'f' will go on decreasing.

This doesn't look good I think there is a better solution

Krishna Sharma - 6 years, 1 month ago

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So would putting x{x} is any integer be a suitable answer?

Curtis Clement - 6 years ago

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Ans.6 Only 2 pages are torn off The sum of integers from 1 to 100 is 5050.

If two pages are torn off the sum of the remaining pages is 4949.

101 is the sum of the torn pages which can not be achieved in the tearing of 3 pages.

Mehul Arora - 6 years, 1 month ago

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Actually wouldn't you tear out 1 page because if the pages have a number on the front and back (like any book) then you would just tear out the one with a 50 on one side and a 51 on the other.

Curtis Clement - 6 years ago

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Yeah. That could be a possible solution too!!

Mehul Arora - 6 years ago

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I don't understand the problem How the pages are torn off?

How about these 3 pages torn

83+17+1=10183 + 17 + 1 = 101

Krishna Sharma - 6 years, 1 month ago

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I think that since it's a book, you must tear pairs of pages. That is, you can't tear page 8383 without tearing 8484 with it.

Satvik Golechha - 6 years, 1 month ago

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@Satvik Golechha What is the answer of 3?

I m getting 1+121 + \dfrac{1}{\sqrt{2}}

Krishna Sharma - 6 years, 1 month ago

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@Krishna Sharma How?

Satvik Golechha - 6 years, 1 month ago

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@Satvik Golechha Oops! did a mistake sorry,

I think 1 variable is not adequate you have to take 2 variables

Krishna Sharma - 6 years, 1 month ago

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@Satvik Golechha Yes. I believe one page is torn and the page numbers it contain are 50 & 51

Aditya Chauhan - 5 years, 5 months ago

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Along with page 83, Either of page 82 or 84 will be torn. With 17 either 18 or 16 will be torn. Hence this is not a possible solution.

Mehul Arora - 6 years ago

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Thanks for the questions. Also, do you have the solutions to these questions?

Curtis Clement - 6 years ago

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solution to problem2:- sum of no on 100 pages =5050 sum of torn pages =5050-4949=101 sum on tearing 1 page with no r=r +r+1=2r+1 if n pages are torn 2<r1+r2+r3...............+rn>+n=101 observe that sum of no on each side of page is of form 4k +3 i.e. (2k+1)+(2k+2) =>4<k1 +................................+kn>+3n=101 =>3n is1 (mod4)=>n is3(mod4) therefore n=4r+3 for some r an integer >or=0 if r>0,n>or =7 => 4 <k1+.....................kn>+3n> 4<0+1+2+3+4+5+6.>+21=4<21>+21=105>101, contradict therefore n=3 there fore 3 pages are torn

Sauditya YO YO - 6 years ago

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2)The question number 2 is easy to solve with complex numbers.But also found a geometric solution.

Let the polygon be nn- sided.

Suppose the vertex after A4A_{4} be A5A_{5}.Hence the 55 consecutive vertices are A1,A2,A3,A4,A5A_{1},A_{2},A_{3},A_{4},A_{5} ( arranged anticlockwise from A1A_{1} to A5A_{5},say)

Firstly,1A1A2=1A1A3+1A1A4\frac{1}{A_{1}A_{2}}=\frac{1}{A_{1}A_{3}}+\frac{1}{A_{1}A_{4}} gives


One can easily show that quadrilateral A1A3A4A5A_{1}A_{3}A_{4}A_{5} is cyclic.

So,by Ptolemy's theorem, A1A5.A3A4+A1A3.A5A4=A1A4.A5A3A_{1}A_{5}.A_{3}A_{4}+A_{1}A_{3}.A_{5}A_{4}=A_{1}A_{4}.A_{5}A_{3}...(2)(2)

Also,since the polygon is regular,we have A1A2=A3A4=A4A5A_{1}A_{2}=A_{3}A_{4}=A_{4}A_{5} and A5A3=A1A3A_{5}A_{3}=A_{1}A_{3}.

So,from (1)(1), A1A2.A1A4+A1A2.A1A3=A1A3.A1A4=A3A5.A1A4=A1A5.A3A4+A1A3.A5A4=A1A5.A1A2+A1A3.A1A2A_{1}A_{2}.A_{1}A_{4}+A_{1}A_{2}.A_{1}A_{3}=A_{1}A_{3}.A_{1}A_{4}=A_{3}A_{5}.A_{1}A_{4}=A_{1}A_{5}.A_{3}A_{4}+A_{1}A_{3}.A_{5}A_{4}=A_{1}A_{5}.A_{1}A_{2}+A_{1}A_{3}.A_{1}A_{2}


Since,these two diagonals are equal,number of vertices between A1A_{1} and A5A_{5}(moving clockwise) equals the number of vertices between A1A_{1} and A4A_{4}(moving anti-clockwise).

Hence,n5=2n-5=2 or n=7n=7.

Souryajit Roy - 6 years ago

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5)Adding 11 to both sides,[(x+1)]2]=([x]+1)2=([x+1])2[(x+1)]^{2}]=([x]+1)^{2}=([x+1])^{2}

Case 1: x+10x+1≤0


Hence,there must be equality in every step.So,[x+1]=x+1[x+1]=x+1 i.e, x is an integer.So,x=1,2,3,....x=-1,-2,-3,....

Case 2: x+1>0x+1>0

Now,(x+1)2[(x+1)2]=([x+1])2=(x+1)^2≥[(x+1)^{2}]=([x+1])^{2}= and also,(x+1)21+[(x+1)2]=1+([x+1])2(x+1)^2≤1+[(x+1)^{2}]=1+([x+1])^{2}


Hence,xx belongs to [n,1+(n+1)21][n,\sqrt{1+(n+1)^{2}}-1] for any integer n1n≥-1.

Souryajit Roy - 6 years ago

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6) Suppose xx pages are torn off.The page numbers on both sides of a page are of the form 2a12a-1 and 2a2a and their sum is 4a14a-1.This is the crux move!

So,the sum of numbers of pages torn=(4a11)+..(4ax1)=4(a1+...+ax)x(4a_{1}-1)+..(4a_{x}-1)=4(a_{1}+...+a_{x})-x

Sum total of all pages=1+2+...+101=50501+2+...+101=5050

So,4(a1+...+ax)x=1014(a_{1}+...+a_{x})-x=101.So,x3x≡3 (mod 44).

Now,show that x<7x<7 (proceed by contradiction)

So,x=3x=3 is the only option.Also, note that a1+a2+a3=26a_{1}+a_{2}+a_{3}=26.One can choose distinct positive integers a1,a2,a3a_{1},a_{2},a_{3} in several ways.

Souryajit Roy - 6 years ago

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Q2. Ans is n=7. I did it using trigo, but later found out that this can be done using Ptolemy's theorem too

Racchit Jain - 5 years ago

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