Question 4

Geometry Level 3

If a a , b b , c c are the sides of a triangle and 2 a + 3 b + 4 c = 4 2 a − 2 + 6 3 b − 3 + 8 4 c − 4 − 20 2a +3b+4c = 4\sqrt{2a-2}+ 6\sqrt{3b-3}+8\sqrt{4c-4}-20 then the area of triangle is:

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Samarpit Swain by Samarpit Swain , 23, India

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1 solution

Jubayer Nirjhor
Mar 26, 2015

Set 2 a − 2 = x , 3 b − 3 = y , 4 c − 4 = z \sqrt{2a-2}=x,~\sqrt{3b-3}=y,~\sqrt{4c-4}=z so that 2 a = x 2 + 2 , 3 b = y 2 + 3 , 4 c = z 2 + 4. 2a=x^2+2,~3b=y^2+3,~4c=z^2+4. And the equation becomes x 2 + y 2 + z 2 + 9 = 4 x + 6 y + 8 z − 20 ⟹ ( x − 2 ) 2 + ( y − 3 ) 2 + ( z − 4 ) 2 = 0 ⟹ x = 2 , y = 3 , z = 4. \begin{aligned}&~ & x^2+y^2+z^2+9=4x+6y+8z-20 \\ \\ &\implies & (x-2)^2+(y-3)^2+(z-4)^2=0 \\ \\ &\implies & x=2,~y=3,~z=4.\end{aligned} From here we can find ( a , b , c ) = ( 3 , 4 , 5 ) (a,b,c)=(3,4,5) so the triangle is right with area 6 \fbox{6} .

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