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prahalada reddy

# Find the no. of isosceles triangle with integer sides, if no sides exceeds 1994.

Avinash Bhat

raja tomar

How can you say that all of the triangles would exist

Divya Alok Gupta

dude i think u wud also need help of the inequality that sum of 2 sides is always greater than the thirs side here in many cases this does not match

Arvind Bhusnurmath

well consider it this way

let the equal sides be of length 1 .. then we have only one choice for the non-equal side, that would be 1. Since if we take 2 then triangle inequality will mess things up

if equal sides are of length 2 .. third side can either be 1 or 2 or 3... basically until we hit 2*2.

if equal sides are of length 3 .. third side can be 1 or 2 or 3 or 4 or 5.

now this logic continues until we hit 998 ..since 2*998=1996 we can basically pick anything for the third side from 1 to 1994

So it looks like the sum should be 1+3+5+...1993+ 997*1994

Avinash Bhat

OH !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

I was really mad over there. SORRY !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Avinash Bhat

NOW LOOK AT THIS ONE :

Given that the isosceles triangles should have integral sides without exceeding 1994.

Let 'a' be one of the sides while 'b', the measurement of the other two sides. Now let us consider the no. of isosceles triangles that may be formed.

NOW : a<b

When a = 1; b { 2, 3, 4 .......................... 1993, 1994 }

SO : A total no. of 1993 triangles may be formed.

When a = 2 ; b { 3, 4, 5, 6 .............................. 1993, 1994 }

SO : A total no. of 1992 triangles may be formed.

.

When a = 1992 ; b { 1993, 1994 }

SO : A total no. of 2 triangles may be formed.

When a = 1993 ; b { 1994 }

SO : 1 triangle may be formed.

BUT AS ALL THE EQUILATERAL TRIANGLES ARE ALSO ISOSCLELES, NOW LET US CONSIDER THE TOTAL NO. OF EQUILATERAL TRIANGLES THAT MAY BE FORMED :

Sides are integers without exceeding 1994.

SO : 1994 equilateral triangles may be formed.

Now : a>b

When b = 1 ; a = { }

When b = 2 ; a = {3}

When b = 3 ; a = {4, 5}

.

When b = 997 ; a = { 996 valued }

When b = 997 ; a = { 996 values }

.

When b = 1992 ; a = {1993, 1994}

When b = 1993 ; a = { 1994 }

When b = 1994 ; a = { }

HENCE, TOTAL NO. OF TRIANGLES :

(1 + 2 + 3 + ................. + 1992 + 1993) + 1994 + 2(1 + 2 + 3 + .................... + 996)

Arvind Bhusnurmath

the problem with this solution seems to be situations like

when a=2 b=2 c=3 is also a valid isoceles triangle. this is not accounted for anywhere. i tried doing it this way initially and things got messy which is why the other approach was taken.

also equilateral triangles are isoceles. this is a minor fix however.

shitij agarwal

Let two equal sides of an isosceles

be p units each and let remaining sides be q units.
Case1: p>q
q can take values 1,2,3,.....,p-1(if p-1>0), condition for p,p,q be sides of a

is automatically satisfied here. For each positive integer p>1,
we have p-1 isosceles

s i.e.

_{[p=2 ]}

^{[1994 ]} (p-1)

=1+2+3....+1993
=1998721

Case2:p<q
In order that p,p,q may be sides of

we must have 2p>q i.e.
p<q<2p
If p is even say 2m ,then q can take value 1,2,...m-1
If p is odd say 2m-1,then q can take values 1,2,...,m-1=(p-1)/2
No. of possible isosceles triangles is
(1-1)/2+(3-1)/2+....+(1993-1)/2+1+2+3+....
for q=1994,p+q>q is true.
Also ,we must have q/2<p<q
There are in all isosceles

_{[q even ]}

(q-2)/2 + _{[q odd ]}

(q-1)/2

s
1

q

1994 1

q

1994
i.e.(1+2+3.....+996)+(1+2+3+...+996) triangles
=993012 triangles
Total no of isosceles triangles=1998721+993012
=2991733
Hope it will help u ,I gave my best to solve this
TC.......
SHITIJ

prahalada reddy

Thanks Mr. shitij 4 ur answer ,its really helped me

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