Geometry Inequality

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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

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M.C.A Entrance Question Paper

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 | a³ or 7 | b³ or 7 | c³ => 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 a^p 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

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