2.7182818284590452353602874

Geometry Level 5

2 graphs of functions y = x 2 y={ x }^{ 2 } and y = x x y=\sqrt [ x ]{ x } intersect at points A and B. At both A and B, x is not equal to 0. At max y = x x ( y ) \max _{ y=\sqrt [ x ]{ x } }{ (y) } on the second graph, abscissa is p. Find the perimeter of triangle ABC if the point (p,0) is considered as point C. (Till 2 decimal points)


The answer is 5.16.

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1 solution

Mas Mus
May 12, 2015

Taking logarithm and equating both of functions:

ln x 2 = ln x 1 x ( 2 1 x ) ln x = 0 x = 1 and x = 1 2 \ln{x^2}=\ln{x^{\frac{1}{x}}}\\\left(2-\dfrac{1}{x}\right)\ln{x}=0\implies{x=1~~\text{and}~~x=\dfrac{1}{2}}

So, we have A ( 1 , 1 ) and B ( 1 2 , 1 4 ) A\left(1, 1\right)~~\text{and}~~B\left(\dfrac{1}{2}, \dfrac{1}{4}\right) .

Now, taking the derivative of the second function,

ln y = 1 x ln x y y = 1 x 2 ln x + 1 x × 1 x = 1 x 2 ( 1 ln x ) x x x 2 ( 1 ln x ) = 0 x = e = p \ln{y}=\dfrac{1}{x}\ln{x}\\\dfrac{y'}{y}=-\dfrac{1}{x^2}\ln{x}+\dfrac{1}{x}\times{\dfrac{1}{x}}=\dfrac{1}{x^2}\left(1-\ln{x}\right)\\\dfrac{\sqrt[x]{x}}{x^2}\left(1-\ln{x}\right)=0\implies{x=\large{e}=p}

and we have C ( e , 0 ) C\left(\large{e}, 0\right) . Now, the distance of point A A to B B , A A to C C , and B B to C C are 0.901 0.901 , 1.988 1.988 , and 2.232 2.232 respectively, and, finally the perimeter of triangle ABC is 5.12 5.12

I agree with this solution so perhaps the Brilliant Staff or the Original Poster could change the answer from 5.16 to 5.12

Bob Kadylo - 4 years, 4 months ago

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