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 \((x,y,z)\) \[x+y-z=4; x^2-y^2+z^2=-4; xyz=6\]

**2** Let \(A_1, A_2,..............,A_n\) be the vertices of an \(n\) sided regular polygon in that order. Determine \(n\), given that
\[\dfrac{1}{A_1A_2}=\dfrac{1}{A_1A_3}+\dfrac{1}{A_1A_4}\]

**3** Let \(ABCD\) be a quadrilateral inscribed in a circle. Suppose \(\overline{AB}=\sqrt{2+\sqrt{2}}\) and \(\overline{AB}\) subtends an angle of \(135^{\circ}\) at the centre of the circle. Find the maximum possible area of the quadrilateral \(ABCD\).

**4** Let \(a,b,c\) be positive real numbers such that \(a^3+b^3=c^3\). Prove that \(a^2+b^2-c^2>6(c-a)(c-b)\)

**5** Find all real numbers \(x\) such that \([x^2+2x]=[x]^2+2[x]\), where \([n]\) denotes the greatest integer less than or equal to \(n\).

**6** In a book of page numbers from \(1\) to \(100\), some pages are torn off. The sum of the numbers on the remaining pages is \(4949\). 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!

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## Comments

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TopNewest3.Is the answer for 6) \(\frac{3\sqrt{3}+5}{4\sqrt{2}}\)

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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

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

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

Sum total of all pages=\(1+2+...+101=5050\)

So,\(4(a_{1}+...+a_{x})-x=101\).So,\(x≡3\) (mod \(4\)).

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

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

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

Case 1:\(x+1≤0\)Now,\([(x+1)]^{2}]≥(x+1)^2≥([x+1])^{2}=[(x+1)]^{2}]\).

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

Case 2:\(x+1>0\)Now,\((x+1)^2≥[(x+1)^{2}]=([x+1])^{2}=\) and also,\((x+1)^2≤1+[(x+1)^{2}]=1+([x+1])^{2}\)

So,\([x]+1=[x+1]≤x+1≤\sqrt{1+([x+1])^{2}}=\sqrt{1+([x]+1)^{2}}\)

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

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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 \(n\)- sided.

Suppose the vertex after \(A_{4}\) be \(A_{5}\).Hence the \(5\) consecutive vertices are \(A_{1},A_{2},A_{3},A_{4},A_{5}\) ( arranged anticlockwise from \(A_{1}\) to \(A_{5}\),say)

Firstly,\(\frac{1}{A_{1}A_{2}}=\frac{1}{A_{1}A_{3}}+\frac{1}{A_{1}A_{4}}\) gives

\(A_{1}A_{3}.A_{1}A_{4}=A_{1}A_{2}.A_{1}A_{4}+A_{1}A_{2}.A_{1}A_{3}\)...\((1)\)

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

So,by Ptolemy's theorem, \(A_{1}A_{5}.A_{3}A_{4}+A_{1}A_{3}.A_{5}A_{4}=A_{1}A_{4}.A_{5}A_{3}\)...\((2)\)

Also,since the polygon is regular,we have \(A_{1}A_{2}=A_{3}A_{4}=A_{4}A_{5}\) and \(A_{5}A_{3}=A_{1}A_{3}\).

So,from \((1)\), \(A_{1}A_{2}.A_{1}A_{4}+A_{1}A_{2}.A_{1}A_{3}=A_{1}A_{3}.A_{1}A_{4}=A_{3}A_{5}.A_{1}A_{4}=A_{1}A_{5}.A_{3}A_{4}+A_{1}A_{3}.A_{5}A_{4}=A_{1}A_{5}.A_{1}A_{2}+A_{1}A_{3}.A_{1}A_{2}\)

Hence,\(A_{1}A_{4}=A_{1}A_{5}\).

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

Hence,\(n-5=2\) or \(n=7\).

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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

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

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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.

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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.

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Yeah. That could be a possible solution too!!

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

How about these 3 pages torn

\(83 + 17 + 1 = 101\)

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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 \(83\) without tearing \(84\) with it.

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I m getting \(1 + \dfrac{1}{\sqrt{2}}\)

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

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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.

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Squaring

\(x^{2} + z^{2} - 2xz = 16 + y^{2} - 8y\)

\(x^{2} - y^{2} + z^{2} = 16 + 2xz -8y\)

With the help of second equation

\(16 + 2xz - 8y = -4\)

With the help of third equation

\(4 + \dfrac{3}{y} - 2y = -1\)

From here we get

\(y = 3, \dfrac{-1}{2}\)

Now \(x, z\) can easily be found

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

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

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

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

\((2,3,1)\)

\((-1,3,-2)\)

2) Answer is

7can be proved with complex numbers5) Answer is \(\infty \) put x any integer

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Please post solution of answer \(2\) also. I want to see your complex numbers method. Thanks!

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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)

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

Where \(\theta = \dfrac{2\pi}{n}\) is angle made at centre

\[\dfrac{1}{a_1{a}_2} = \dfrac{1}{{a}_1{a}_3} + \dfrac{1}{{a}_1{a}_4} \]

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

Now let \(\displaystyle x = \dfrac{\pi}{n}\)

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

\(\displaystyle \dfrac{sin2x.sin3x}{sinx} = sin2x + sin3x\)

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

\(\displaystyle sin4x + sin2x = sin2x + sin3x\)

\(sin4x = sin3x\)

\(4x + 3x = \pi\)

\(x = \dfrac{\pi}{7} = \dfrac{\pi}{n}\)

\(n = \boxed{\boxed{7}}\)

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For 5, you are asked to "find all real numbers \(x\) ...". You are not asked to find the total number of real numbers.

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

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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

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

We have to make cases

When \(I < -1, f ≠ 0 \)

No value of x satify the condition

When \(I = -1 \) we will get \(0 < f < 1\)

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

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

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