Sunday, September 28, 2014

Geometry Inequality



Prove that in any quadrilateral, the sum of the diagonals is greater than the half of its perimeter.

Consider the quad. In the above diagram. Let E be the point of the intersection.

Now, AE+EB > AB
EB+EC > BC
AE+ED > AD
EC+ED > DC (Using Triangle Inequality)
Adding the above four inequalities we get
2(AE+EC+EB+ED) > AB+BC+AD+DC
=> AC + BD > ½(AB+BC+AD+DC)
Thus sum of the diagonals is greater than the half of its perimeter Q.E.D



In any triangle four times the sum of its medians is greater than 3 times its perimeter.

We know that difference of any two sides of a triangle is less than the third side (prove it)
In triangle ABE,
AE > AB-BE
In triangle ACE,
AE > AC-CE
Adding above two inequalities we get,
2AE > AB + AC -(BE+CE)
=> AE > ½(AB+AC-BC)
=> 4AE > 2(AB+AC-BC).........(1)
Similarly,
4BD > 2(AB+BC-AC).............(2) and 4CF > 2(AC+BC-AB)........(3)
Adding (1),(2) and (3) we have,
4(AE+BD+CF) > 2(AB+AC-BC+AB+BC-AC+AC+BC-AB)
=> 4(AE+BD+CF) > 2(AC+AB+BC)
=> sum of the lengths of the medians is greater than half the perimeter
We can strengthen the inequality by using the fact that the point 'O' divides the medians AE,BD,CF internally in the ration 2:1
Therefore, OD:OB = 1:2
=> (OB+OD):OB = (1+2):2
=> BD:OB=3:2
=> OB = 2/3 BD........(a)
Similarly, OC = 2/3 CF.........(b) and OA = 2/3 AE.......(c)
Now in triangle OBC, OB+OC> BC
=>2/3(BD+CF)>BC [using (a) and (b)]
=> 2(BD+CF)>3BC
Similarly, 2(CF+AE)>3AC and 2(BD+AE)>3AB
Adding the last three inequalities we get 4(AE+BD+CF) > 3(AB+BC+CA)

In the triangle ABC, AE,BD and CF are the medians where O is the point of there intersection

Sunday, September 21, 2014

Monday, May 26, 2014

Solved Problems on Circles (Tangent Properties)

In these material we explore the properties of circles, tangents and common tangents. Using congruency and similarity of triangles many of the desired property has been deduced.


Wednesday, May 7, 2014

Olympiad Problems













Find the common roots of the equation x2000+x2002+1 = 0 and x3+2x2+2x+1 = 0.
#IIT #CMI #ISI

x3+2x2+2x+1 = 0
=> x3+1+2x(x+1) = 0
=> (x+1)(x2-x+1)+2x(x+1) = 0
=> (x+1)(x2-x+1+2x) = 0
=> (x+1)(x2+x+1) =0
=> x = -1,w,w2 where 'w' is the cube root of unity
Now putting these values in the expression x2000+x2002+1 we see that w and w2 only reduce it to zero. So the common roots are w,w2


Find real x for which 1/[x] + 1/[2x] = {x} + 1/3, where [x] = greatest integer less than equal to x and {x} = x – {x}.

Let x = z + f, where z is the integer part and 0 <= f < 1
Now we have two cases (a) 0 <= f < 1/2 (b) 1/2 <= f < 1
Also note that the R.H.S of the given equation i.e., {x} + 1/3 > 0 so x has to be positive.

Case (a) 0 <= f < 1/2, [x] = z and 2x = 2z + 2f , [2x] = 2z since 0 <= 2f < 1
and {x} = x – [x] = z + f – z = f.
This reduces the equation to
1/z + 1/2z = f + 1/3 ........... (1)
=> 1/z + 1/2z >= 1/3 (Since f >= 0 )
=> 3/2z >= 1/3
=> 9 >= 2z
=> 2z – 9 <= 0
=> z = 1,2,3,4
for z = 1, {x} = f = 1 which is invalid in this case
for z = 2, using equation (1) {x} = f = 5/12 < 1/2
for z = 3, using equation (1) {x} = f = 1/6 < 1/2
for z = 4, using equation (1) {x} = f = 1/24 < 1/2
so possible values of x in this case is 2+5/12, 3+1/6 and 4+1/24
i.e., x = 29/12, 19/6 and 97/24
Case(b) Following the steps of case (a) show that in this case no solution exists.
Show that abc(a³-b³)(b³-c³)(c³-a³) is always divisible by 7 where a, b and c are non-negative integers.

Result: Cube of any integer leaves remainder 0, 1 or 6 on dividing by 7.
if any one of a,b or c is divisible by 7, then abc(a³-b³)(b³-c³)(c³-a³) is divisible by 7. So we assume that none of the numbers a,b or c is divisible by 7.
Now using the result stated above a³, b³,c³ leaves remainder 1 or 6 (due to the assumption) otherwise if 7 | or 7 | or 7 | => 7 | a or 7 | b or 7 | c ( since 7 is prime) A contradiction to the assumption.
Now by Pigeon Hole Principle two of the numbers a³, b³,c³ have the same remainder. Hence their difference is divisible by 7, thus
7 | abc(a³-b³)(b³-c³)(c³-a³) in any case.

Find the remainder when 2 is divided by 1990

Result: If p is prime & p do not divide a then ap is congruent a (modulo p) (Fermat's Theorem)
Note that 1990 = 199 X 10 and 199 is a prime.
Using Fermats theorem we have 2199 is congruent 2 (modulo 199)
=> (2199 )10 is congruent 210 (modulo 199)
=> 2 is congruent 1024 (modulo 199)
i.e 199 | 2 - 1024
Now 2 and 1024 has the same last digit, which is 4 ( try to find it for 2 ). Therefore
10 | 2 - 1024
Now g.c.d (199, 10) = 1 therefore 1990 | 2 - 1024
So the remainder is 1024

Tuesday, May 6, 2014

Coordinate Geometry

Solved Problems on Coordinate Geometry for 10th grade. Problems based on distance formula, section ratio formula   
 








Sunday, May 4, 2014

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Friday, May 2, 2014

GEOMETRY Thales Theorem or Basic Proportionality Theorem Solved Problems



Solved Problems on Thales Theorem useful for students of 10th grade of CBSE and ICSE and other state board. This material is also useful for the preparation of Regional Mathematics Olympiad. Basic Geometry is a pre-requisite  for any Olympiad.


Monday, April 28, 2014

Probability Solved problems for CBSE, ICSE 10th grade


Post by Maths.

Wednesday, April 2, 2014

Probability

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