A right relation of triangular numbers

Algebra Level 3

The perimeter of a triangle is 84 84 and its sides form an arithmetic progression (AP) with integral terms a a , b b and c c .

By setting T n = 1 + 2 + 3 + . . . + n T_{n}=1+2+3+...+n , the following relation holds true:

T 2 a − 2 T a + 8 T b 2 − 1 + 2 b − 8 T c − 1 2 = 1 T_{2a}-2T_{a}+8T_{\frac{b}{2}-1}+2b-8T_{\frac{c-1}{2}}=1

Find the common difference of the AP.


The answer is 7.

Thanos Petropoulos by Thanos Petropoulos , 61, Greece

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1 solution

It is well known that T n = 1 + 2 + . . . + n = n ( n + 1 ) 2 T_{n}=1+2+...+n=\frac{n(n+1)}{2} . Using this formula, we get:

  • T 2 a − 2 T a = 2 a ( 2 a + 1 ) 2 − 2 a ( a + 1 ) 2 = 2 a 2 + a − a 2 − a = a 2 T_{2a}-2T_{a}=\frac{2a(2a+1)}{2}-2\frac{a(a+1)}{2}=2a^2+a-a^2-a=a^2

  • 8 T b 2 − 1 + 2 b = 8 ( b 2 − 1 ) b 2 2 + 2 b = 4 ( b 2 4 − b 2 ) + 2 b = b 2 8T_{\frac{b}{2}-1}+2b=8\frac{(\frac{b}{2}-1)\frac{b}{2}}{2}+2b=4(\frac{b^2}{4}-\frac{b}{2})+2b=b^2

  • 8 T c − 1 2 = 8 c − 1 2 ( c − 1 2 + 1 ) 2 = 4 c − 1 2 c + 1 2 = c 2 − 1 8T_{\frac{c-1}{2}}=8\frac{\frac{c-1}{2}(\frac{c-1}{2}+1)}{2}=4\frac{c-1}{2}\frac{c+1}{2}=c^2-1

Hence, the given relation yields

a 2 + b 2 − ( c 2 − 1 ) = 1 a^2+b^2-(c^2-1)=1 , that is a 2 + b 2 = c 2 a^2+b^2=c^2 . This means that we have a right angled triangle and ( a , b , c ) (a, b, c) is a Pythagorean triple (Pt).

Now, there are only two Pt which add up to 84 84 : ( 12 , 35 , 37 ) (12, 35, 37) and ( 21 , 28 , 35 ) (21, 28, 35) .

Since a , b , c a, b, c are in AP, we get a = 21 , b = 28 , c = 35 a=21, b=28, c=35 , thus, the common difference is 7 \boxed{ 7 } □ _\square .

3 Helpful 0 Interesting 1 Brilliant 0 Confused

Finding triplets is really difficult for large numbers. I took the Arithmetic Progression a , b , c a,b,c as x − y , x , x + y x-y,x,x+y where y is the common difference. So x − y + x + x + y = 84 x-y + x + x+y = 84 which means x = 28 x = 28 . After substituting these values in the equation a 2 + b 2 = c 2 a^2+b^2=c^2 y y can be found.

Abha Vishwakarma - 2 years, 7 months ago

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