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IMO 1983 – Problem 3


Let N and k be positive integers and  let S be a set of   n points in the plane such that

(i) no three points of S  are collinear, and
(ii)  for any point P of  S, there are at least k points of S equidistant from P

Prove that  k  <  \frac{1}{2}  +  \sqrt{2n}

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Solution will be updated soon

IMO 2012 problems


This year IMO problems !!!!

Problem 1 :

Given triangle ABC the point J is the centre of the excircle opposite the vertex A. This excircle is tangent to the side BC at M, and to the lines AB and AC at K and L, respectively. The lines LM and BJ meet at F, and the lines KM and CJ meet at G. Let S be the point of intersection of the lines AF and BC, and let T be the point of intersection of the lines AG and BC. Prove that M is the midpoint of ST.

Problem 2 :

Let {n\ge 3} be an integer, and let {a_2,a_3,\ldots ,a_n} be positive real numbers such that  {a_{2}a_{3}\cdots a_{n}=1}  Prove that

\displaystyle \left(a_{2}+1\right)^{2}\left(a_{3}+1\right)^{3}\dots\left(a_{n}+1\right)^{n}>n^{n}.

Problem 3 :

The liar’s guessing game is a game played between two players {A} and {B}. The rules of the game depend on two positive integers {k} and {n} which are known to both players.

At the start of the game {A} chooses integers {x} and {N} with {1 \le x \le N.} Player {A} keeps {x}secret, and truthfully tells {N} to player {B}. Player {B} now tries to obtain information about {x} by asking player {A} questions as follows: each question consists of {B} specifying an arbitrary set {S} of positive integers (possibly one specified in some previous question), and asking {A}whether {x} belongs to {S}. Player {B} may ask as many questions as he wishes. After each question, player {A} must immediately answer it with [i]yes[/i] or [i]no[/i], but is allowed to lie as many times as she wants; the only restriction is that, among any {k+1} consecutive answers, at least one answer must be truthful.

After {B} has asked as many questions as he wants, he must specify a set {X} of at most {n}positive integers. If {x} belongs to {X}, then {B} wins; otherwise, he loses. Prove that:

1. If {n \ge 2^k,} then {B} can guarantee a win. 2. For all sufficiently large {k}, there exists an integer {n \ge (1.99)^k} such that {B} cannot guarantee a win.

Problem 4 :

Find all functions {f:\mathbb Z\rightarrow \mathbb Z} such that, for all integers {a,b,c} that satisfy {a+b+c=0}, the following equality holds:

\displaystyle f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).

 Problem 5 :

Let {ABC} be a triangle with {\angle BCA=90^{\circ}}, and let {D} be the foot of the altitude from {C}. Let {X} be a point in the interior of the segment {CD}. Let {K} be the point on the segment {AX}such that {BK=BC}. Similarly, let {L} be the point on the segment {BX} such that {AL=AC}. Let {M} be the point of intersection of {AL} and {BK}.

Show that {MK=ML}.

Problem 6 :

Find all positive integers {n} for which there exist non-negative integers {a_1, a_2, \ldots, a_n} such that

\displaystyle \frac{1}{2^{a_1}} + \frac{1}{2^{a_2}} + \cdots + \frac{1}{2^{a_n}} = \frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \cdots + \frac{n}{3^{a_n}} = 1.

You may download the PDF version :
IMO 2012


Thank you and good luck for these delicious problems

Bumper Problems


These are the set of 5 problems  each of  5 marks


The person who get more than 20 points will get a mathematics book on any topic he want…
So try these beautiful problems to test your mathematics abilities and getting new things…

Ways by which you can answer :
I am adding a form in the end of this post  , you can answer there ….

Note : You have give solution of your answer too..

Problem 1)

The bisectors of the angles A and B of the \bigtriangleup ABC  meet the sides
BC and CA at the points D and E , respectively.
Assuming that AE+BD =AB, determine the angle C

Problem 2)

Given a  \bigtriangleup ABC and  D be point on side  AC such that  AB = DC,
\angle BAC= 60-2X ,   \angle DBC= 5X and  \angle BCA= 3X
Find the value of  X

Problem 3)

If p and q are natural numbers so that

Prove that p is divisible by 1979 .


Problem 4)

Find highest degree n of 1991 for which 1991ⁿ  divides the number :

Problem 5)

Let ƒ(n) denote the sum of the digits of n. Let N = 4444⁴⁴⁴⁴
Find ƒ(ƒ(ƒ(n))))

You can use these symbols to write solutions more conveniently

Mathematical Operators
Exponents  :   ⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁼ ⁽ ⁾ ₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ₊ ₋ ₌ ₍ ₎ 

Important theorems of Geometry(Triangles) – 1


Today we will be discussing about  2 important Triangle theorems which have very High applications in field of Geometry

1) Angle Bisector theorem
2) Stewart’s Theorem

Angle bisector theorem :

Introduction :

The Angle Bisector Theorem states that given triangle \triangle ABC and angle bisector AD, where D is on side BC, then c/m= b/n . Likewise, the converse of this theorem holds as well.

size(200);defaultpen(fontsize(10));real a,b,c,d;pair A=(1,4), B=(-5,0), C=(3,0), D;b = abs(C-A);c = abs(B-A);D = (b*B+c*C)/(b...

Proof :

Because of the ratios and equal angles in the theorem, we think of similar triangles. There are not any similar triangles in the figure as it now stands, however. So, we think to draw in a carefully chosen line or two. Extending AD until it hits the line through C parallel to AB does just the trick:

size(200);defaultpen(fontsize(10));real a,b,c,d,m,n;pair A=(1,4), B=(-5,0), C=(3,0), D, E;b = abs(C-A);c = abs(B-A);D = (b*B+...

Since AB and CE are parallel, we know that \angle BAE=\angle CEA and \angle BCE=\angle ABC. Triangle ACE is isosceles, with AC = CE.

By AA similarity\triangle DAB \cong \triangle DEC. By the properties of similar triangles, we arrive at our desired result:

c/m = b/n

Stewarts Theorem :

Introduction :

Given a triangle \triangle ABC with sides of length a, b, c opposite vertices ABC, respectively. If cevian AD is drawn so that BD = mDC = n and AD = d, we have that b^2m + c^2n = amn + d^2a. (This is also often written man + dad = bmb + cnc, a form which invites mnemonic memorization, e.g. “A man and his dad put a bomb in the sink.”)

Stewart's theorem.png

Proof :

Applying the Law of Cosines in triangle \triangle ABD at angle \angle ADB and in triangle \triangle ACD at angle \angle CDA, we get the equations

  • n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2}
  • m^{2} + d^{2} - 2md\cos{\angle ADB} = c^{2}

Because angles \angle ADB and \angle CDA are supplementarym\angle ADB = 180^\circ - m\angle CDA. We can therefore solve both equations for the cosine term. Using the trigonometric identity  \cos{\theta} = -\cos{(180^\circ - \theta)} gives us

  • \frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}
  • \frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}

Setting the two left-hand sides equal and clearing denominators, we arrive at the equation: c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n. However, m+n = a som^2n + n^2m = (m + n)mn and we can rewrite this as man + dad= bmb + cnc (A man and his dad put a bomb in the sink).
Thank you

Source :
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